-----Заголовок-----
Источник: http://picxxx.info
Ссылка на PDF: http://picxxx.info/pml.php?action=GETCONTENT&md5=d695ca7bd45d52aeb3ad90aaace37a38
-----Конец заголовка-----
This page intentionally left blank
Introduction to Classical Mechanics
This textbook covers all the standard introductory topics in classical mechanics,
including Newton’s laws, oscillations, energy, momentum, angular momentum,
planetary motion, and special relativity. It also explores more advanced topics,
such as normal modes, the Lagrangian method, gyroscopic motion, fictitious
forces, 4-vectors, and general relativity.
It contains more than 250 problems with detailed solutions so students can
easily check their understanding of the topic. There are also over 350 unworked
exercises, which are ideal for homework assignments. Password-protected
solutions are available to instructors at www.cambridge.org/9780521876223.
The vast number of problems alone makes it an ideal supplementary book for
all levels of undergraduate physics courses in classical mechanics. The text also
includes many additional remarks which discuss issues that are often glossed
over in other textbooks, and it is thoroughly illustrated with more than 600
figures to help demonstrate key concepts.
David Morin is a Lecturer in the Physics Department at Harvard University. He
received his Ph.D. in theoretical particle physics from Harvard in 1996. When not
writing physics limericks or thinking of new problems whose answers involve e
or the golden ratio, he can be found running along the Charles River or hiking in
the White Mountains of New Hampshire.
Introduction to Classical
Mechanics
With Problems and Solutions
David Morin
Harvard University
CAMBRIDGE UNIVERSITY PRESS
Cambridge, New York, Melbourne, Madrid, Cape Town, Singapore, São Paulo
Cambridge University Press
The Edinburgh Building, Cambridge CB2 8RU, UK
Published in the United States of America by Cambridge University Press, New York
www.cambridge.org
Information on this title: www.cambridge.org/9780521876223
© D. Morin 2007
This publication is in copyright. Subject to statutory exception and to the provision of
relevant collective licensing agreements, no reproduction of any part may take place
without the written permission of Cambridge University Press.
First published in print format 2008
ISBN-13 978-0-511-37723-5
eBook (EBL)
ISBN-13
hardback
978-0-521-87622-3
Cambridge University Press has no responsibility for the persistence or accuracy of urls
for external or third-party internet websites referred to in this publication, and does not
guarantee that any content on such websites is, or will remain, accurate or appropriate.
To Allen Gerry and Neil Tame,
who took the time
to give a group of kids
some really cool problems
There once was a classical theory,
Of which quantum disciples were leery.
They said, “Why spend so long
On a theory that’s wrong?”
Well, it works for your everyday query!
Contents
Preface
page xiii
1 Strategies for solving problems
1.1 General strategies
1.2 Units, dimensional analysis
1.3 Approximations, limiting cases
1.4 Solving differential equations numerically
1.5 Problems
1.6 Exercises
1.7 Solutions
1
1
4
7
11
14
15
18
2 Statics
2.1 Balancing forces
2.2 Balancing torques
2.3 Problems
2.4 Exercises
2.5 Solutions
22
22
27
30
35
39
3 Using F = ma
3.1 Newton’s laws
3.2 Free-body diagrams
3.3 Solving differential equations
3.4 Projectile motion
3.5 Motion in a plane, polar coordinates
3.6 Problems
3.7 Exercises
3.8 Solutions
51
51
55
60
65
68
70
75
84
4 Oscillations
4.1 Linear differential equations
4.2 Simple harmonic motion
101
101
105
vii
viii
Contents
4.3
4.4
4.5
4.6
4.7
4.8
Damped harmonic motion
Driven (and damped) harmonic motion
Coupled oscillators
Problems
Exercises
Solutions
107
109
115
120
122
127
5 Conservation of energy and momentum
5.1 Conservation of energy in one dimension
5.2 Small oscillations
5.3 Conservation of energy in three dimensions
5.4 Gravity
5.5 Momentum
5.6 The center of mass frame
5.7 Collisions
5.8 Inherently inelastic processes
5.9 Problems
5.10 Exercises
5.11 Solutions
138
138
147
148
152
156
161
164
167
173
180
194
6 The Lagrangian method
6.1 The Euler–Lagrange equations
6.2 The principle of stationary action
6.3 Forces of constraint
6.4 Change of coordinates
6.5 Conservation laws
6.6 Noether’s theorem
6.7 Small oscillations
6.8 Other applications
6.9 Problems
6.10 Exercises
6.11 Solutions
218
218
221
227
229
232
236
239
242
246
251
255
7 Central forces
7.1 Conservation of angular momentum
7.2 The effective potential
7.3 Solving the equations of motion
7.4 Gravity, Kepler’s laws
7.5 Problems
7.6 Exercises
7.7 Solutions
281
281
283
285
287
296
298
300
Contents
ˆ
8 Angular momentum, Part I (Constant L)
8.1 Pancake object in x-y plane
8.2 Nonplanar objects
8.3 Calculating moments of inertia
8.4 Torque
8.5 Collisions
8.6 Angular impulse
8.7 Problems
8.8 Exercises
8.9 Solutions
309
310
316
318
322
328
331
333
339
349
ˆ
9 Angular momentum, Part II (General L)
9.1 Preliminaries concerning rotations
9.2 The inertia tensor
9.3 Principal axes
9.4 Two basic types of problems
9.5 Euler’s equations
9.6 Free symmetric top
9.7 Heavy symmetric top
9.8 Problems
9.9 Exercises
9.10 Solutions
371
371
376
383
388
393
396
399
415
421
428
10 Accelerating frames of reference
10.1 Relating the coordinates
10.2 The fictitious forces
10.3 Tides
10.4 Problems
10.5 Exercises
10.6 Solutions
457
458
460
471
477
482
486
11 Relativity (Kinematics)
11.1 Motivation
11.2 The postulates
11.3 The fundamental effects
11.4 The Lorentz transformations
11.5 Velocity addition
11.6 The invariant interval
11.7 Minkowski diagrams
11.8 The Doppler effect
11.9 Rapidity
11.10 Relativity without c
501
502
509
511
523
529
533
536
539
543
546
ix
x
Contents
11.11 Problems
11.12 Exercises
11.13 Solutions
549
556
565
12 Relativity (Dynamics)
12.1 Energy and momentum
12.2 Transformations of E and p
12.3 Collisions and decays
12.4 Particle-physics units
12.5 Force
12.6 Rocket motion
12.7 Relativistic strings
12.8 Problems
12.9 Exercises
12.10 Solutions
584
584
594
596
600
601
606
609
611
615
619
13 4-vectors
13.1 Definition of 4-vectors
13.2 Examples of 4-vectors
13.3 Properties of 4-vectors
13.4 Energy, momentum
13.5 Force and acceleration
13.6 The form of physical laws
13.7 Problems
13.8 Exercises
13.9 Solutions
634
634
635
637
639
640
643
645
645
646
14 General Relativity
14.1 The Equivalence Principle
14.2 Time dilation
14.3 Uniformly accelerating frame
14.4 Maximal-proper-time principle
14.5 Twin paradox revisited
14.6 Problems
14.7 Exercises
14.8 Solutions
649
649
650
653
656
658
660
663
666
Appendix A
Appendix B
Appendix C
Appendix D
Appendix E
Appendix F
675
679
690
693
696
698
Useful formulas
Multivariable, vector calculus
F = ma vs. F = dp/dt
Existence of principal axes
Diagonalizing matrices
Qualitative relativity questions
Contents
Appendix G
Appendix H
Appendix I
Appendix J
References
Index
Derivations of the Lv/c2 result
Resolutions to the twin paradox
Lorentz transformations
Physical constants and data
704
706
708
711
713
716
xi
Preface
This book grew out of Harvard University’s honors freshman mechanics course.
It is essentially two books in one. Roughly half of each chapter follows the form of
a normal textbook, consisting of text, along with exercises suitable for homework
assignments. The other half takes the form of a “problem book,” with all sorts
of problems (and solutions) of varying degrees of difficulty. I’ve always thought
that doing problems is the best way to learn, so if you’ve been searching for a
supply to puzzle over, I think this will keep you busy for a while.
This book is somewhat of a quirky one, so let me say right at the start how I
imagine it being used:
• As the primary text for honors freshman mechanics courses. My original motivation
for writing it was the fact that there didn’t exist a suitable book for Harvard’s freshman
course. So after nine years of using updated versions in the class, here is the finished
product.
• As a supplementary text for standard freshman courses for physics majors. Although
this book starts at the beginning of mechanics and is self contained, it doesn’t spend
as much time on the introductory material as other freshman books do. I therefore
don’t recommend using this as the only text for a standard freshman mechanics course.
However, it will make an extremely useful supplement, both as a problem book for all
students, and as a more advanced textbook for students who want to dive further into
certain topics.
• As a supplementary text for upper-level mechanics courses, or as the primary text which
is supplemented with another book for additional topics often covered in upper-level
courses, such as Hamilton’s equations, fluids, chaos, Fourier analysis, electricity and
magnetism applications, etc. With all of the worked examples and in-depth discussions,
you really can’t go wrong in pairing up this book with another one.
• As a problem book for anyone who likes solving physics problems. This audience
ranges from advanced high-school students, who I think will have a ball with it, to
undergraduate and graduate students who want some amusing problems to ponder, to
professors who are looking for a new supply of problems to use in their classes, and
finally to anyone with a desire to learn about physics by doing problems. If you want,
you can consider this to be a problem book that also happens to have comprehensive
xiii
xiv
Preface
introductions to each topic’s set of problems. With about 250 problems (with included
solutions) and 350 exercises (without included solutions), in addition to all the examples
in the text, I think you’ll get your money’s worth! But just in case, I threw in 600 figures,
50 limericks, nine appearances of the golden ratio, and one cameo of e−π .
The prerequisites for the book are solid high-school foundations in mechanics
(no electricity and magnetism required) and single-variable calculus. There are
two minor exceptions to this. First, a few sections rely on multivariable calculus, so I have given a review of this in Appendix B. The bulk of it comes in
Section 5.3 (which involves the curl), but this section can easily be skipped on
a first reading. Other than that, there are just some partial derivatives, dot products, and cross products (all of which are reviewed in Appendix B) sprinkled
throughout the book. Second, a few sections (4.5, 9.2–9.3, and Appendices D
and E) rely on matrices and other elementary topics from linear algebra. But a
basic understanding of matrices should suffice here.
A brief outline of the book is as follows. Chapter 1 discusses various problemsolving strategies. This material is extremely important, so if you read only one
chapter in the book, make it this one. You should keep these strategies on the
tip of your brain as you march through the rest of the book. Chapter 2 covers
statics. Most of this will likely be familiar, but you’ll find some fun problems.
In Chapter 3, we learn about forces and how to apply F = ma. There’s a bit of
math here needed for solving some simple differential equations. Chapter 4 deals
with oscillations and coupled oscillators. Again, there’s a fair bit of math needed
for solving linear differential equations, but there’s no way to avoid it. Chapter 5
deals with conservation of energy and momentum. You’ve probably seen much
of this before, but it has lots of neat problems.
In Chapter 6, we introduce the Lagrangian method, which will most likely be
new to you. It looks rather formidable at first, but it’s really not all that rough.
There are difficult concepts at the heart of the subject, but the nice thing is that the
technique is easy to apply. The situation here is analogous to taking a derivative
in calculus; there are substantive concepts on which the theory rests, but the act
of taking a derivative is fairly straightforward.
Chapter 7 deals with central forces and planetary motion. Chapter 8 covers
the easier type of angular momentum situations, where the direction of the
angular momentum vector is fixed. Chapter 9 covers the more difficult type,
where the direction changes. Spinning tops and other perplexing objects fall into
this category. Chapter 10 deals with accelerating reference frames and fictitious
forces.
Chapters 11 through 14 cover relativity. Chapter 11 deals with relativistic
kinematics – abstract particles flying through space and time. Chapter 12 covers
relativistic dynamics – energy, momentum, force, etc. Chapter 13 introduces the
important concept of “4-vectors.” The material in this chapter could alternatively
be put in the previous two, but for various reasons I thought it best to create a
Preface
separate chapter for it. Chapter 14 covers a few topics from General Relativity.
It’s impossible for one chapter to do this subject justice, of course, so we’ll just
look at some basic (but still very interesting) examples. Finally, the appendices
cover various useful, but slightly tangential, topics.
Throughout the book, I have included many “Remarks.” These are written
in a slightly smaller font than the surrounding text. They begin with a smallcapital “Remark” and end with a shamrock (♣). The purpose of these remarks is
to say something that needs to be said, without disrupting the overall flow of the
argument. In some sense these are “extra” thoughts, although they are invariably
useful in understanding what is going on. They are usually more informal than
the rest of the text, and I reserve the right to use them to occasionally babble
about things that I find interesting, but that you may find tangential. For the most
part, however, the remarks address issues that arise naturally in the course of the
discussion. I often make use of “Remarks” at the ends of the solutions to problems,
where the obvious thing to do is to check limiting cases (this topic is discussed in
Chapter 1). However, in this case, the remarks are not “extra” thoughts, because
checking limiting cases of your answer is something you should always do.
For your reading pleasure (I hope!), I have included limericks throughout the
text. I suppose that these might be viewed as educational, but they certainly don’t
represent any deep insight I have into the teaching of physics. I have written them
for the sole purpose of lightening things up. Some are funny. Some are stupid.
But at least they’re all physically accurate (give or take).
As mentioned above, this book contains a huge number of problems. The ones
with included solutions are called “Problems,” and the ones without included
solutions, which are intended to be used for homework assignments, are called
“Exercises.” There is no fundamental difference between these two types, except
for the existence of written-up solutions. I have chosen to include the solutions
to the problems for two reasons. First, students invariably want extra practice
problems, with solutions, to work on. And second, I had a thoroughly enjoyable
time writing them up. But a warning on these problems and exercises: Some are
easy, but many are very difficult. I think you’ll find them quite interesting, but
don’t get discouraged if you have trouble solving them. Some are designed to be
brooded over for hours. Or days, or weeks, or months (as I can attest to!).
The problems (and exercises) are marked with a number of stars (actually
asterisks). Harder problems earn more stars, on a scale from zero to four. Of
course, you may disagree with my judgment of difficulty, but I think that an
arbitrary weighting scheme is better than none at all. As a rough idea of what I
mean by the number of stars, one-star problems are solid problems that require
some thought, and four-star problems are really, really, really hard. Try a few
and you’ll see what I mean. Even if you understand the material in the text
backwards and forwards, the four-star (and many of the three-star) problems will
still be extremely challenging. But that’s how it should be. My goal was to create
an unreachable upper bound on the number (and difficulty) of problems, because
xv
xvi
Preface
it would be an unfortunate circumstance if you were left twiddling your thumbs,
having run out of problems to solve. I hope I have succeeded.
For the problems you choose to work on, be careful not to look at the solution
too soon. There’s nothing wrong with putting a problem aside for a while and
coming back to it later. Indeed, this is probably the best way to learn things. If
you head to the solution at the first sign of not being able to solve a problem,
then you have wasted the problem.
Remark: This gives me an opportunity for my first remark (and first limerick, too). A fact that
often gets overlooked is that you need to know more than the correct way(s) to do a problem; you
also need to be familiar with many incorrect ways of doing it. Otherwise, when you come upon
a new problem, there may be a number of decent-looking approaches to take, and you won’t be
able to immediately weed out the poor ones. Struggling a bit with a problem invariably leads
you down some wrong paths, and this is an essential part of learning. To understand something,
you not only have to know what’s right about the right things; you also have to know what’s
wrong about the wrong things. Learning takes a serious amount of effort, many wrong turns,
and a lot of sweat. Alas, there are no shortcuts to understanding physics.
The ad said, For one little fee,
You can skip all that course-work ennui.
So send your tuition,
For boundless fruition!
Get your mail-order physics degree! ♣
Any book that takes ten years to write is bound to contain the (greatly appreciated) input of many people. I am particularly thankful for Howard Georgi’s help
over the years, with his numerous suggestions, ideas for many problems, and
physics sanity checks. I would also like to thank Don Page for his entertaining
and meticulous comments and suggestions, and an eye for catching errors in earlier versions. Other friends and colleagues who have helped make this book what
it is (and who have made it all the more fun to write) are John Bechhoefer, Wes
Campbell, Michelle Cyrier, Alex Dahlen, Gary Feldman, Lukasz Fidkowski,
Jason Gallicchio, Doug Goodale, Bertrand Halperin, Matt Headrick, Jenny
Hoffman, Paul Horowitz, Alex Johnson, Yevgeny Kats, Can Kilic, Ben Krefetz,
Daniel Larson, Jaime Lush, Rakhi Mahbubani, Chris Montanaro, Theresa Morin,
Megha Padi, Dave Patterson, Konstantin Penanen, Courtney Peterson, Mala
Radhakrishnan, Esteban Real, Daniel Rosenberg, Wolfgang Rueckner, Aqil
Sajjad, Alexia Schulz, Daniel Sherman, Oleg Shpyrko, David Simmons-Duffin,
Steve Simon, Joe Swingle, Edwin Taylor, Sam Williams, Alex Wissner-Gross,
and Eric Zaslow. I’m sure that I have forgotten others, especially from the earlier
years where my memory fades, so please accept my apologies.
I am also grateful for the highly professional work done by the editorial and
production group at Cambridge University Press in transforming this into an
actual book. It has been a pleasure working with Lindsay Barnes, Simon Capelin,
Margaret Patterson, and Dawn Preston.
Finally, and perhaps most importantly, I would like to thank all the students
(both at Harvard and elsewhere) who provided input during the past decade.
Preface
The names here are literally too numerous to write down, so let me simply say a
big thank you, and that I hope other students will enjoy what you helped create.
Despite the painstaking proofreading and all the eyes that have passed over
earlier versions, there is at most an exponentially small probability that the
book is error free. So if something looks amiss, please check the webpage
(www.cambridge.org/9780521876223) for a list of typos, updates, etc. And
please let me know if you discover something that isn’t already posted. I’m
sure that eventually I will post some new problems and supplementary material,
so be sure to check the webpage for additions. Information for instructors will
also be available on this site.
Happy problem solving – I hope you enjoy the book!
xvii
Chapter 1
Strategies for solving problems
Physics involves a great deal of problem solving. Whether you are doing
cutting-edge research or reading a book on a well-known subject, you are going
to need to solve some problems. In the latter case (the presently relevant one,
given what is in your hand right now), it is fairly safe to say that the true test
of understanding something is the ability to solve problems on it. Reading about
a topic is often a necessary step in the learning process, but it is by no means
a sufficient one. The more important step is spending as much time as possible
solving problems (which is inevitably an active task) beyond the time you spend
reading (which is generally a more passive task). I have therefore included a very
large number of problems/exercises in this book.
However, if I’m going to throw all these problems at you, I should at least give
you some general strategies for solving them. These strategies are the subject of
the present chapter. They are things you should always keep in the back of your
mind when tackling a problem. Of course, they are generally not sufficient by
themselves; you won’t get too far without understanding the physical concepts
behind the subject at hand. But when you add these strategies to your physical
understanding, they can make your life a lot easier.
1.1
General strategies
There are a number of general strategies you should invoke without hesitation
when solving a problem. They are:
1. Draw a diagram, if appropriate.
In the diagram, be sure to label clearly all the relevant quantities (forces, lengths,
masses, etc.). Diagrams are absolutely critical in certain types of problems. For
example, in problems involving “free-body” diagrams (discussed in Chapter 3) or
relativistic kinematics (discussed in Chapter 11), drawing a diagram can change a
hopelessly complicated problem into a near-trivial one. And even in cases where
diagrams aren’t this crucial, they’re invariably very helpful. A picture is definitely
worth a thousand words (and even a few more, if you label things!).
1
2
Strategies for solving problems
2. Write down what you know, and what you are trying to find.
In a simple problem, you may just do this in your head without realizing it. But in
more difficult problems, it is very useful to explicitly write things out. For example,
if there are three unknowns that you’re trying to find, but you’ve written down
only two facts, then you know there must be another fact you’re missing (assuming
that the problem is in fact solvable), so you can go searching for it. It might be a
conservation law, or an F = ma equation, etc.
3. Solve things symbolically.
If you are solving a problem where the given quantities are specified numerically,
you should immediately change the numbers to letters and solve the problem in terms
of the letters. After you obtain an answer in terms of the letters, you can plug in the
actual numerical values to obtain a numerical answer. There are many advantages
to using letters:
• It’s quicker. It’s much easier to multiply a g by an by writing them down on a
piece of paper next to each other, than it is to multiply them together on a calculator.
And with the latter strategy, you’d undoubtedly have to pick up your calculator at
least a few times during the course of a problem.
• You’re less likely to make a mistake. It’s very easy to mistype an 8 for a 9 in
a calculator, but you’re probably not going to miswrite a q for a g on a piece of
paper. But if you do, you’ll quickly realize that it should be a g. You certainly
won’t just give up on the problem and deem it unsolvable because no one gave
you the value of q!
• You can do the problem once and for all. If someone comes along and says,
oops, the value of is actually 2.4 m instead of 2.3 m, then you won’t have to do
the whole problem again. You can simply plug the new value of into your final
symbolic answer.
• You can see the general dependence of your answer on the various given quantities. For example, you can see that it grows with quantities a and b, decreases with
c, and doesn’t depend on d. There is much, much more information contained in a
symbolic answer than in a numerical one. And besides, symbolic answers nearly
always look nice and pretty.
• You can check units and special cases. These checks go hand-in-hand with the
previous “general dependence” advantage. But since they’re so important, we’ll
postpone their discussion and devote Sections 1.2 and 1.3 to them.
Having said all this, it should be noted that there are occasionally times when things
get a bit messy when working with letters. For example, solving a system of three
equations in three unknowns might be rather cumbersome unless you plug in the
actual numbers. But in the vast majority of problems, it is highly advantageous to
work entirely with letters.
4. Consider units/dimensions.
This is extremely important. See Section 1.2 for a detailed discussion.
1.1 General strategies
5. Check limiting/special cases.
This is also extremely important. See Section 1.3 for a detailed discussion.
6. Check order of magnitude if you end up getting a numerical answer.
If you end up with an actual numerical answer to a problem, be sure to do a sanity check to see if the number is reasonable. If you’ve calculated the distance
along the ground that a car skids before it comes to rest, and if you’ve gotten
an answer of a kilometer or a millimeter, then you know you’ve probably done
something wrong. Errors of this sort often come from forgetting some powers of
10 (say, when converting kilometers to meters) or from multiplying something
instead of dividing (although you should be able to catch this by checking your
units, too).
You will inevitably encounter problems, physics ones or otherwise, where
you don’t end up obtaining a rigorous answer, either because the calculation is
intractable, or because you just don’t feel like doing it. But in these cases it’s
usually still possible to make an educated guess, to the nearest power of 10. For
example, if you walk past a building and happen to wonder how many bricks
are in it, or what the labor cost was in constructing it, then you can probably
give a reasonable answer without doing any severe computations. The physicist
Enrico Fermi was known for his ability to estimate things quickly and produce
order-of-magnitude guesses with only minimal calculation. Hence, a problem
where the goal is to simply obtain the nearest power-of-10 estimate is known as a
“Fermi problem.” Of course, sometimes in life you need to know things to better
accuracy than the nearest power of 10 . . .
How Fermi could estimate things!
Like the well-known Olympic ten rings,
And the one hundred states,
And weeks with ten dates,
And birds that all fly with one . . . wings.
In the following two sections, we’ll discuss the very important strategies of
checking units and special cases. Then in Section 1.4 we’ll discuss the technique
of solving problems numerically, which is what you need to do when you end up
with a set of equations you can’t figure out how to solve. Section 1.4 isn’t quite
analogous to Sections 1.2 and 1.3, in that these first two are relevant to basically
any problem you’ll ever do, whereas solving equations numerically is something
you’ll do only for occasional problems. But it’s nevertheless something that every
physics student should know.
In all three of these sections, we’ll invoke various results derived later in the
book. For the present purposes, the derivations of these results are completely
irrelevant, so don’t worry at all about the physics behind them – there will be
3
4
Strategies for solving problems
plenty of opportunity for that later on! The main point here is to learn what to do
with the result of a problem once you’ve obtained it.
1.2
Units, dimensional analysis
The units, or dimensions, of a quantity are the powers of mass, length, and time
associated with it. For example, the units of a speed are length per time. The
consideration of units offers two main benefits. First, looking at units before
you start a problem can tell you roughly what the answer has to look like, up
to numerical factors. Second, checking units at the end of a calculation (which
is something you should always do) can tell you if your answer has a chance at
being correct. It won’t tell you that your answer is definitely correct, but it might
tell you that your answer is definitely incorrect. For example, if your goal in a
problem is to find a length, and if you end up with a mass, then you know it’s
time to look back over your work.
“Your units are wrong!” cried the teacher.
“Your church weighs six joules – what a feature!
And the people inside
Are four hours wide,
And eight gauss away from the preacher!”
In practice, the second of the above two benefits is what you will generally
make use of. But let’s do a few examples relating to the first benefit, because
these can be a little more exciting. To solve the three examples below exactly, we
would need to invoke results derived in later chapters. But let’s just see how far we
can get by using only dimensional analysis. We’ll use the “[ ]” notation for units,
and we’ll let M stand for mass, L for length, and T for time. For example, we’ll
write a speed as [v] = L/T and the gravitational constant as [G] = L3 /(MT 2 )
(you can figure this out by noting that Gm1 m2 /r 2 has the dimensions of force,
which in turn has dimensions ML/T 2 , from F = ma). Alternatively, you can just
use the mks units, kg, m, s, instead of M , L, T , respectively.1
θ
Example (Pendulum): A mass m hangs from a massless string of length
(see Fig. 1.1) and swings back and forth in the plane of the paper. The acceleration
due to gravity is g. What can we say about the frequency of oscillations?
g
l
Solution: The only dimensionful quantities given in the problem are [m] = M ,
[ ] = L, and [g] = L/T 2 . But there is one more quantity, the maximum angle θ0 ,
which is dimensionless (and easy to forget). Our goal is to find the frequency, which
m
Fig. 1.1
1
When you check units at the end of a calculation, you will invariably be working with the kg,m,s
notation. So that notation will inevitably get used more. But I’ll use the M , L, T notation here,
because I think it’s a little more instructive. At any rate, just remember that the letter m (or M )
stands for “meter” in one case, and “mass” in the other.
1.2 Units, dimensional analysis
has units of 1/T . The only combination of our given dimensionful quantities that has
√
units of 1/T is g/ . But we can’t rule out any θ0 dependence, so the most general
possible form of the frequency is2
ω = f (θ0 )
g
,
(1.1)
where f is a dimensionless function of the dimensionless variable θ0 .
Remarks:
1. It just so happens that for small oscillations,
f (θ0 ) is essentially equal to 1, so the
√
frequency is essentially equal to g/ . But there is no way to show this by using
only dimensional analysis; you actually have to solve the problem for real. For
larger values of θ0 , the higher-order terms in the expansion of f become important.
Exercise 4.23 deals with the leading correction, and the answer turns out to be f (θ0 ) =
1 − θ02 /16 + · · · .
2. Since there is only one mass in the problem, there is no way that the frequency (with
units of 1/T ) can depend on [m] = M . If it did, there would be nothing to cancel the
units of mass and produce a pure inverse-time.
3. We claimed above that the
√ only combination of our given dimensionful quantities
that has units of 1/T is g/ . This is easy to see here, but in more complicated
problems where the correct combination isn’t so obvious, the following method will
always work. Write down a general product of the given dimensionful quantities
raised to arbitrary powers (ma b g c in this problem), and then write out the units of
this product in terms of a, b, and c. If we want to obtain units of 1/T here, then
we need
M a Lb
L
T2
c
=
1
.
T
(1.2)
Matching up the powers of the three kinds of units on each side of this equation gives
M : a = 0,
L : b + c = 0,
T : −2c = −1.
(1.3)
The solution to this system
of equations is a = 0, b = −1/2, and c = 1/2, so we
√
have reproduced the g/ result. ♣
What can we say about the total energy of the pendulum (with the potential energy
measured relative to the lowest point)? We’ll talk about energy in Chapter 5, but
the only thing we need to know here is that energy has units of ML2 /T 2 . The only
combination of the given dimensionful constants of this form is mg . But again, we
can’t rule out any θ0 dependence, so the energy must take the form f (θ0 )mg , where
f is some function. That’s as far as we can go with dimensional analysis. However,
if we actually invoke a little physics, we can say that the total energy equals the
potential energy at the highest point, which is mg (1 − cos θ0 ). Using the Taylor
expansion for cos θ (see Appendix A for a discussion of Taylor series), we see that
f (θ0 ) = θ02 /2 − θ04 /24 + · · · . So in contrast with the frequency result above, the
maximum angle θ0 plays a critical role in the energy.
2
We’ll measure frequency here in radians per second, denoted by ω. So we’re actually talking about
the “angular frequency.” Just divide by 2π (which doesn’t affect the units) to obtain the “regular”
frequency in cycles per second (hertz), usually denoted by ν. We’ll talk at great length about
oscillations in Chapter 4.
5
6
Strategies for solving problems
Example (Spring): A spring with spring constant k has a mass m on its end
(see Fig. 1.2). The spring force is F(x) = −kx, where x is the displacement from the
equilibrium position. What can we say about the frequency of oscillations?
k
m
Solution: The only dimensionful quantities in this problem are [m] = M , [k] =
M /T 2 (obtained by noting that kx has the dimensions of force), and the maximum
displacement from the equilibrium, [x0 ] = L. (There is also the equilibrium length,
but the force doesn’t depend on this, so there is no way it can come into the answer.)
Our goal is to find the frequency, which has units of 1/T . The only combination of
our given dimensionful quantities with these units is
Fig. 1.2
ω=C
k
,
m
(1.4)
where C is a dimensionless number. It just so happens that C is equal to 1 (assuming
that we’re measuring ω in radians per second), but there is no way to show this by
using only dimensional analysis. Note that, in contrast with the pendulum above, the
frequency cannot have any dependence on the maximum displacement.
What can we say about the total energy of the spring? Energy has units of ML2 /T 2 ,
and the only combination of the given dimensionful constants of this form is Bkx02 ,
where B is a dimensionless number. It turns out that B = 1/2, so the total energy
equals kx02 /2.
Remark: A real spring doesn’t have a perfectly parabolic potential (that is, a perfectly
linear force), so the force actually looks something like F(x) = −kx + bx2 + · · · . If we
truncate the series at the second term, then we have one more dimensionful quantity to
work with, [b] = M /LT 2 . To form a quantity with the dimensions of frequency, 1/T , we
need x0 and b to appear in the combination x0 b, because this is the only way to get rid
of the L. You can then see (by using the strategy of writing out a general product of the
variables, discussed in the third remark in the pendulum example above) that the frequency
must be of the form f (x0 b/k) k/m, where f is some function. We can therefore have x0
dependence in this case. This answer must reduce to C k/m for b = 0. Hence, f must be
of the form f (y) = C + c1 y + c2 y2 + · · · . ♣
Example (Low-orbit satellite): A satellite of mass m travels in a circular orbit
just above the earth’s surface. What can we say about its speed?
Solution: The only dimensionful quantities in the problem are [m] = M , [g] =
L/T 2 , and the radius of the earth [R] = L. 3 Our goal is to find the speed, which has
units of L/T . The only combination of our dimensionful quantities with these units is
v = C gR.
(1.5)
It turns out that C = 1.
3
You might argue that the mass of the earth, ME , and Newton’s gravitational constant, G, should
be also included here, because Newton’s gravitational force law for a particle on the surface of the
earth is F = GME m/R2 . But since this force can be written as m(GME /R2 ) ≡ mg, we can absorb
the effects of ME and G into g.
1.3 Approximations, limiting cases
1.3
Approximations, limiting cases
As with units, the consideration of limiting cases (or perhaps we should say
special cases) offers two main benefits. First, it can help you get started on a
problem. If you’re having trouble figuring out how a given system behaves,
then you can imagine making, for example, a certain length become very large or
very small, and then you can see what happens to the behavior. Having convinced
yourself that the length actually affects the system in extreme cases (or perhaps
you will discover that the length doesn’t affect things at all), it will then be
easier to understand how it affects the system in general, which will then make
it easier to write down the relevant quantitative equations (conservation laws,
F = ma equations, etc.), which will allow you to fully solve the problem. In
short, modifying the various parameters and observing the effects on the system
can lead to an enormous amount of information.
Second, as with checking units, checking limiting cases (or special cases)
is something you should always do at the end of a calculation. But as with
checking units, it won’t tell you that your answer is definitely correct, but
it might tell you that your answer is definitely incorrect. It is generally true
that your intuition about limiting cases is much better than your intuition
about generic values of the parameters. You should use this fact to your
advantage.
Let’s do a few examples relating to the second benefit. The initial expressions
given in each example below are taken from various examples throughout the
book, so just accept them for now. For the most part, I’ll repeat here what I’ll
say later on when we work through the problems for real. A tool that comes up
often in checking limiting cases is the Taylor series approximations; the series
for many functions are given in Appendix A.
Example (Dropped ball): A beach ball is dropped from rest at height h. Assume
that the drag force from the air takes the form Fd = −mαv. We’ll find in Section 3.3
that the ball’s velocity and position are given by
v(t) = −
g
1 − e−αt ,
α
and
y(t) = h −
g
α
t−
1
1 − e−αt
α
.
(1.6)
These expressions are a bit complicated, so for all you know, I could have made a
typo in writing them down. Or worse, I could have completely botched the solution. So let’s look at some limiting cases. If these limiting cases yield expected
results, then we can feel a little more confident that the answers are actually
correct.
If t is very small (more precisely, if αt
1; see the discussion following this example), then we can use the Taylor series, e−x ≈ 1 − x + x2 /2, to make approximations
7
8
Strategies for solving problems
to leading order in αt. The v(t) in Eq. (1.6) becomes
v(t) = −
g
α
1 − 1 − αt +
(αt)2
− ···
2
≈ −gt,
(1.7)
plus terms of higher order in αt. This answer is expected, because the drag force is
negligible at the start, so we essentially have a freely falling body with acceleration
g downward. For small t, Eq. (1.6) also gives
y(t) = h −
≈h−
g
1
t−
α
α
1 − 1 − αt +
gt 2
,
2
(αt)2
− ···
2
(1.8)
plus terms of higher order in αt. Again, this answer is expected, because we essentially
have a freely falling body at the start, so the distance fallen is the standard gt 2 /2.
We can also look at large t (or rather, large αt). In this case, e−αt is essentially
zero, so the v(t) in Eq. (1.6) becomes (there’s no need for a Taylor series in this case)
g
v(t) ≈ − .
α
(1.9)
This is the “terminal velocity.” Its value makes sense, because it is the velocity for
which the total force, −mg − mαv, vanishes. For large t, Eq. (1.6) also gives
y(t) ≈ h −
gt
g
+ 2.
α
α
(1.10)
Apparently for large t, g/α 2 is the distance (and this does indeed have units of
length, because α has units of T −1 , because mαv has units of force) that our ball lags
behind another ball that started out already at the terminal velocity, −g/α.
Whenever you derive approximate answers as we just did, you gain something
and you lose something. You lose some truth, of course, because your new answer
is technically not correct. But you gain some aesthetics. Your new answer is
invariably much cleaner (sometimes involving only one term), and this makes it
a lot easier to see what’s going on.
In the above example, it actually makes no sense to look at the limit where
t is small or large, because t has dimensions. Is a year a large or small time?
How about a hundredth of a second? There is no way to answer this without
knowing what problem you’re dealing with. A year is short on the time scale
of galactic evolution, but a hundredth of a second is long on the time scale of
a nuclear process. It makes sense only to look at the limit of a small (or large)
dimensionless quantity. In the above example, this quantity is αt. The given
constant α has units of T −1 , so 1/α sets a typical time scale for the system. It
1.3 Approximations, limiting cases
therefore makes sense to look at the limit where t
1/α (that is, αt
1), or
where t
1/α (that is, αt
1). In the limit of a small dimensionless quantity,
a Taylor series can be used to expand an answer in powers of the small quantity,
as we did above. We sometimes get sloppy and say things like, “In the limit
of small t.” But you know that we really mean, “In the limit of some small
dimensionless quantity that has a t in the numerator,” or, “In the limit where t is
much smaller that a certain quantity that has the dimensions of time.”
Remark: As mentioned above, checking special cases tells you that either (1) your answer is
consistent with your intuition, or (2) it’s wrong. It never tells you that it’s definitely correct.
This is the same as what happens with the scientific method. In the real world, everything comes
down to experiment. If you have a theory that you think is correct, then you need to check that
its predictions are consistent with experiments. The specific experiments you do are the analog
of the special cases you check after solving a problem; these two things represent what you
know is true. If the results of the experiments are inconsistent with your theory, then you need
to go back and fix your theory, just as you would need to go back and fix your answer. If, on
the other hand, the results are consistent, then although this is good, the only thing it really
tells you is that your theory might be correct. And considering the way things usually turn out,
the odds are that it’s not actually correct, but rather the limiting case of a more correct theory
(just as Newtonian physics is a limiting case of relativistic physics, which is a limiting case of
quantum field theory, etc.). That’s how physics works. You can’t prove anything, so you learn
to settle for the things you can’t disprove.
Consider, when seeking gestalts,
The theories that physics exalts.
It’s not that they’re known
To be written in stone.
It’s just that we can’t say they’re false. ♣
When making approximations, how do you know how many terms in the
Taylor series to keep? In the example above, we used e−x ≈ 1 − x + x2 /2.
But why did we stop at the x2 term? The honest (but slightly facetious) answer
is, “Because I had already done this problem before writing it up, so I knew
how many terms to keep.” But the more informative (although perhaps no more
helpful) answer is that before you do the calculation, there’s really no way of
knowing how many terms to keep. So you should just keep a few and see what
happens. If everything ends up canceling out, then this tells you that you need to
repeat the calculation with another term in the series. For example, in Eq. (1.8),
if we had stopped the Taylor series at e−x ≈ 1 − x, then we would have obtained
y(t) = h − 0, which isn’t very useful, since the general goal is to get the leadingorder behavior in the parameter we’re looking at (which is t here). So in this
case we’d know we’d have to go back and include the x2 /2 term in the series.
If we were doing a problem in which there was still no t (or whatever variable)
dependence at that order, then we’d have to go back and include the −x3 /6 term
in the series. Of course, you could just play it safe and keep terms up to, say,
fifth order. But that’s invariably a poor strategy, because you’ll probably never
in your life have to go out that far in a series. So just start with one or two terms
and see what it gives you. Note that in Eq. (1.7), we actually didn’t need the
9
10
Strategies for solving problems
second-order term, so we in fact could have gotten by with only e−x ≈ 1 − x.
But having the extra term here didn’t end up causing much heartache.
After you make an approximation, how do you know if it’s a “good” one? Well,
just as it makes no sense to ask if a dimensionful quantity is large or small without
comparing it to another quantity, it makes no sense to ask if an approximation is
“good” or “bad” without stating the accuracy you want. In the above example, if
you’re looking at a t value for which αt ≈ 1/100, then the term we ignored in
Eq. (1.7) is smaller than gt by a factor αt/2 ≈ 1/200. So the error is on the order
of 1%. If this is enough accuracy for whatever purpose you have in mind, then
the approximation is a good one. If not, it’s a bad one, and you should add more
terms in the series until you get your desired accuracy.
The results of checking limits generally fall into two categories. Most of the
time you know what the result should be, so this provides a double-check on your
answer. But sometimes an interesting limit pops up that you might not expect.
Such is the case in the following examples.
m
M
v
Fig. 1.3
Example (Two masses in 1-D): A mass m with speed v approaches a stationary
mass M (see Fig. 1.3). The masses bounce off each other elastically. Assume that
all motion takes place in one dimension. We’ll find in Section 5.6.1 that the final
velocities of the particles are
vm =
(m − M )v
,
m+M
and
vM =
2mv
.
m+M
(1.11)
There are three special cases that beg to be checked:
• If m = M , then Eq. (1.11) tells us that m stops, and M picks up a speed v. This is
fairly believable (and even more so for pool players). And it becomes quite clear
once you realize that these final speeds certainly satisfy conservation of energy
and momentum with the initial conditions.
• If M
m, then m bounces backward with speed ≈ v, and M hardly moves.
This makes sense, because M is basically a brick wall.
• If m
M , then m keeps plowing along at speed ≈ v, and M picks up a speed
of ≈ 2v. This 2v is an unexpected and interesting result (it’s easier to see if you
consider what’s happening in the reference frame of the heavy mass m), and it
leads to some neat effects, as in Problem 5.23.
g
θ
l
m
Example (Circular pendulum): A mass hangs from a massless string of length .
Conditions have been set up so that the mass swings around in a horizontal circle,
with the string making a constant angle θ with the vertical (see Fig. 1.4). We’ll find
in Section 3.5 that the angular frequency, ω, of this motion is
ω=
Fig. 1.4
g
.
cos θ
(1.12)
1.4 Solving differential equations numerically
As far as θ is concerned, there are two limits we should definitely check:
• If θ → 90◦ , then ω → ∞. This makes sense; the mass has to spin very quickly
to avoid flopping down.
√
• If θ → 0, then ω → g/ , which is the same as the frequency of a standard
“plane” pendulum of length (for small oscillations). This is a cool result and
not at all obvious. (But once we get to F = ma in Chapter 3, you can convince
yourself why this is true by looking at the projection of the force on a given
horizontal line.)
In the above examples, we checked limiting and special cases of answers
that were correct (I hope!). This whole process is more useful (and a bit more
fun, actually) when you check the limits of an answer that is incorrect. In this
case, you gain the unequivocal information that your answer is wrong. But
rather than leading you into despair, this information is in fact something you
should be quite happy about, considering that the alternative is to carry on in
a state of blissful ignorance. Once you know that your answer is wrong, you
can go back through your work and figure out where the error is (perhaps by
checking limits at various stages to narrow down where the error could be).
Personally, if there’s any way I’d like to discover that my answer is garbage,
this is it. At any rate, checking limiting cases can often save you a lot of trouble
in the long run…
The lemmings get set for their race.
With one step and two steps they pace.
They take three and four,
And then head on for more,
Without checking the limiting case.
1.4
Solving differential equations numerically
Solving a physics problem often involves solving a differential equation.
A differential equation is one that involves derivatives (usually with respect to
time, in our physics problems) of the variable you’re trying to solve for. The
differential equation invariably comes about from using F = ma, and/or τ = I α,
or the Lagrangian technique we’ll discuss in Chapter 6. For example, consider a
falling body. F = ma gives −mg = ma, which can be written as −g = y¨ , where
a dot denotes a time derivative. This is a rather simple differential equation, and
you can quickly guess that y(t) = −gt 2 /2 is a solution. Or, more generally with
the constants of integration thrown in, y(t) = y0 + v0 t − gt 2 /2.
However, the differential equations produced in some problems can get rather
complicated, so sooner or later you will encounter one that you can’t solve exactly
(either because it’s in fact impossible to solve, or because you can’t think of the
11
12
Strategies for solving problems
appropriate clever trick). Having resigned yourself to not getting the exact answer,
you should ponder how to obtain a decent approximation to it. Fortunately, it’s
easy to write a short program that will give you a very good numerical answer to
your problem. Given enough computer time, you can obtain any desired accuracy
(assuming that the system isn’t chaotic, but we won’t have to worry about this
for the systems we’ll be dealing with).
We’ll demonstrate the procedure by considering a standard problem, one
that we’ll solve exactly and in great depth in Chapter 4. Consider the
equation,
x¨ = −ω2 x.
(1.13)
This is the equation for a mass on a spring, with ω = k/m. We’ll find in
Chapter 4 that the solution can be written, among other ways, as
x(t) = A cos(ωt + φ).
(1.14)
But let’s pretend we don’t know this. If someone comes along and gives us the
values of x(0) and x˙ (0), then it seems that somehow we should be able to find
x(t) and x˙ (t) for any later t, just by using Eq. (1.13). Basically, if we’re told how
the system starts, and if we know how it evolves, via Eq. (1.13), then we should
know everything about it. So here’s how we find x(t) and x˙ (t).
The plan is to discretize time into intervals of some small unit (call it ), and
to then determine what happens at each successive point in time. If we know x(t)
and x˙ (t), then we can easily find (approximately) the value of x at a slightly later
time, by using the definition of x˙ . Similarly, if we know x˙ (t) and x¨ (t), then we
can easily find (approximately) the value of x˙ at a slightly later time, by using the
definition of x¨ . Using the definitions of the derivatives, the relations are simply
x(t + ) ≈ x(t) + x˙ (t),
x˙ (t + ) ≈ x˙ (t) + x¨ (t).
(1.15)
These two equations, combined with (1.13), which gives us x¨ in terms of x, allow
us to march along in time, obtaining successive values for x, x˙ , and x¨ .4
Here’s what a typical program might look like.5 (This is a Maple program, but
even if you aren’t familiar with this, the general idea should be clear.) Let’s say
4
5
Of course, another expression for x¨ is the definitional one, analogous to Eqs. (1.15), involving
the third derivative. But this would then require knowledge of the third derivative, and so on with
higher derivatives, and we would end up with an infinite chain of relations. An equation of motion
such as Eq. (1.13) (which in general could be an F = ma, τ = I α, or Euler–Lagrange equation)
relates x¨ back to x (and possibly x˙ ), thereby creating an intertwined relation among x, x˙ , and x¨ , and
eliminating the need for an infinite and useless chain.
We’ve written the program in the most straightforward way, without any concern for efficiency,
because computing time isn’t an issue in this simple system. But in more complex systems that
require programs for which computing time is an issue, a major part of the problem-solving process
is developing a program that is as efficient as possible.
1.4 Solving differential equations numerically
that the particle starts from rest at position x = 2, and let’s pick ω2 = 5. We’ll
use the notation where x1 stands for x˙ , and x2 stands for x¨ . And e stands for .
Let’s calculate x at, say, t = 3.
x:=2:
x1:=0:
e:=.01:
for i to 300 do
x2:=-5*x:
x:=x+e*x1:
x1:=x1+e*x2:
end do:
x;
#
#
#
#
#
#
#
#
#
initial position
initial velocity
small time interval
do 300 steps (ie, up to 3 seconds)
the given equation
how x changes, by definition of x1
how x1 changes, by definition of x2
the Maple command to stop the do loop
print the value of x
This procedure won’t give the exact value for x, because x and x˙ don’t really
change according to Eqs. (1.15). These equations are just first-order approximations to the full Taylor series with higher-order terms. Said differently, there
is no way the above procedure can be exactly correct, because there are ambiguities in how the program can be written. Should line 5 come before or after
line 7? That is, in determining x˙ at time t + , should we use the x¨ at time t or
t + ? And should line 7 come before or after line 6? The point is that for very
small , the order doesn’t matter much. And in the limit → 0, the order doesn’t
matter at all.
If we want to obtain a better approximation, we can just shorten down to
0.001 and increase the number of steps to 3000. If the result looks basically the
same as with = 0.01, then we know we pretty much have the right answer. In the
present example, = 0.01 yields x ≈ 1.965 after 3 seconds. If we set = 0.001,
then we obtain x ≈ 1.836. And if we set = 0.0001, then we get x ≈ 1.823.
The correct answer must therefore be somewhere around
√x = 1.82. And indeed,
if we solve the problem exactly, we obtain x(t) = 2 cos( 5 t). Plugging in t = 3
gives x ≈ 1.822.
This is a wonderful procedure, but it shouldn’t be abused. It’s nice to know that
we can always obtain a decent numerical approximation if all else fails. But we
should set our initial goal on obtaining the correct algebraic expression, because
this allows us to see the overall behavior of the system. And besides, nothing
beats the truth. People tend to rely a bit too much on computers and calculators
nowadays, without pausing to think about what is actually going on in a problem.
The skill to do math on a page
Has declined to the point of outrage.
Equations quadratica
Are solved on Math’matica,
And on birthdays we don’t know our age.
13
14
Strategies for solving problems
1.5
Problems
Section 1.2: Units, dimensional analysis
1.1. Escape velocity *
As given below in Exercise 1.9, show that the escape velocity from the
√
earth is v = 2GME /R, up to numerical factors. You can use the fact
that the form of Newton’s gravitation force law implies that the acceleration (and hence overall motion) of the particle doesn’t depend on
its mass.
m
M
l
Fig. 1.5
1.2. Mass in a tube *
A tube of mass M and length is free to swing around a pivot at one end.
A mass m is positioned inside the (frictionless) tube at this end. The tube
is held horizontal and then released (see Fig. 1.5). Let η be the fraction of
the tube that the mass has traversed by the time the tube becomes vertical.
Does η depend on ?
1.3. Waves in a fluid *
How does the speed of waves in a fluid depend on its density, ρ, and
“bulk modulus,” B (which has units of pressure, which is force per area)?
1.4. Vibrating star *
Consider a vibrating star, whose frequency ν depends (at most) on its
radius R, mass density ρ, and Newton’s gravitational constant G. How
does ν depend on R, ρ, and G?
1.5. Damping **
A particle with mass m and initial speed V is subject to a velocitydependent damping force of the form bv n .
(a) For n = 0, 1, 2, . . . , determine how the stopping time depends on
m, V , and b.
(b) For n = 0, 1, 2, . . . , determine how the stopping distance depends
on m, V , and b.
Be careful! See if your answers make sense. Dimensional analysis gives
the answer only up to a numerical factor. This is a tricky problem, so don’t
let it discourage you from using dimensional analysis. Most applications
of dimensional analysis are quite straightforward.
Section 1.3: Approximations, limiting cases
1.6. Projectile distance *
A person throws a ball (at an angle of her choosing, to achieve the maximum distance) with speed v from the edge of a cliff of height h. Assuming
1.6 Exercises
15
that one of the following quantities is the maximum horizontal distance
the ball can travel, which one is it? (Don’t solve the problem from scratch,
just check special cases.)
gh2
,
v2
v2
,
g
v2 h
,
g
2gh
v2
1+ 2 ,
g
v
v2
g
1+
2gh
,
v2
v 2 /g
1 − 2gh
v2
.
Section 1.4: Solving differential equations numerically
1.7. Two masses, one swinging **
Two equal masses are connected by a string that hangs over two pulleys
(of negligible size), as shown in Fig. 1.6. The left mass moves in a vertical
line, but the right mass is free to swing back and forth in the plane of the
masses and pulleys. It can be shown (see Problem 6.4) that the equations
of motion for r and θ (labeled in the figure) are
r
m
2¨r = r θ˙ 2 − g(1 − cos θ),
θ¨ = −
2˙r θ˙
g sin θ
−
.
r
r
(1.16)
Fig. 1.6
Assume that both masses start out at rest, with the right mass making an
initial angle of 10◦ = π/18 with the vertical. If the initial value of r is
1 m, how much time does it take for it to reach a length of 2 m? Write a
program to solve this numerically. Use g = 9.8 m/s2 .
1.6
Exercises
Section 1.2: Units, dimensional analysis
1.8. Pendulum on the moon
If a pendulum has a period of 3 s on the earth, what would its period be
if it were placed on the moon? Use gM /gE ≈ 1/6.
1.9. Escape velocity *
The escape velocity on the surface of a planet is given by
v=
2GM
,
R
m
(1.17)
where M and R are the mass and radius of the planet, respectively, and G
is Newton’s gravitational constant. (The escape velocity is the velocity
16
Strategies for solving problems
needed to refute the “What goes up must come down” maxim, neglecting
air resistance.)
(a) Write v in terms of the average mass density ρ, instead of M .
(b) Assuming that the average density of the earth is four times that of
Jupiter, and that the radius of Jupiter is 11 times that of the earth,
what is vJ /vE ?
1.10. Downhill projectile *
A hill is sloped downward at an angle θ with respect to the horizontal.
A projectile of mass m is fired with speed v0 perpendicular to the hill.
When it eventually lands on the hill, let its velocity make an angle β with
respect to the horizontal. Which of the quantities θ, m, v0 , and g does
the angle β depend on?
1.11. Waves on a string *
How does the speed of waves on a string depend on its mass M , length
L, and tension (that is, force) T ?
1.12. Vibrating water drop *
Consider a vibrating water drop, whose frequency ν depends on its radius
R, mass density ρ, and surface tension S. The units of surface tension
are (force)/(length). How does ν depend on R, ρ, and S?
Section 1.3: Approximations, limiting cases
m3
m1
1.13. Atwood’s machine *
Consider the “Atwood’s” machine shown in Fig. 1.7, consisting of three
masses and three frictionless pulleys. It can be shown that the acceleration
of m1 is given by (just accept this):
m2
a1 = g
3m2 m3 − m1 (4m3 + m2 )
,
m2 m3 + m1 (4m3 + m2 )
(1.18)
Fig. 1.7
with upward taken to be positive. Find a1 in the following special
cases:
(a)
(b)
(c)
(d)
(e)
m2
m1
m1
m2
m1
= 2m1 = 2m3 .
much larger than both m2 and m3 .
much smaller than both m2 and m3 .
m1 = m3 .
= m2 = m3 .
1.14. Cone frustum *
A cone frustum has base radius b, top radius a, and height h, as shown
in Fig. 1.8. Assuming that one of the following quantities is the volume
1.6 Exercises
17
of the frustum, which one is it? (Don’t solve the problem from scratch,
just check special cases.)
πh 2
πh 2
(a + b2 ),
(a + b2 ),
3
2
πh a4 + b4
·
, π hab.
3 a2 + b 2
πh 2
(a + ab + b2 ),
3
h
b
1.15. Landing at the corner *
A ball is thrown at an angle θ up to the top of a cliff of height L, from
a point a distance L from the base, as shown in Fig. 1.9. Assuming that
one of the following quantities is the initial speed required to make the
ball hit right at the edge of the cliff, which one is it? (Don’t solve the
problem from scratch, just check special cases.)
gL
1
,
2(tan θ − 1) cos θ
gL
1
,
2(tan θ − 1) cos θ
gL
,
2(tan θ + 1)
gL tan θ
.
2(tan θ + 1)
Fig. 1.8
L
v0
θ
Fig. 1.9
1.16. Projectile with drag **
Consider a projectile subject to a drag force F = −mαv. If it is fired
with speed v0 at an angle θ , it can be shown that the height as a function of time is given by (just accept this here; it’s one of the tasks of
Exercise 3.53)
y(t) =
a
1
g
v0 sin θ +
α
α
1 − e−αt −
gt
.
α
(1.19)
Show that this reduces to the usual projectile expression, y(t) =
(v0 sin θ )t − gt 2 /2, in the limit of small α. What exactly is meant by
“small α”?
Section 1.4: Solving differential equations numerically
1.17. Pendulum **
A pendulum of length is released from the horizontal position. It can
be shown that the tangential F = ma equation is (where θ is measured
with respect to the vertical)
θ¨ = −
g sin θ
.
(1.20)
If = 1 m, and g = 9.8 m/s2 , write a program to show that the time
it takes the pendulum to swing down through the vertical position is
L
18
Strategies for solving problems
√
t ≈ 0.592 s. This happens to be about 1.18 times the (π/2) /g ≈
0.502 s it would take the pendulum to swing down if it were released
from very close to the vertical (this is 1/4 of the standard period of
√
2π /g for a pendulum). It also happens to be about 1.31 times the
√
2 /g ≈ 0.452 s it would take a mass to simply freefall a height .
1.18. Distance with damping **
A mass is subject to a damping force proportional to its velocity, which
means that the equation of motion takes the form x¨ = −A˙x, where A is
some constant. If the initial speed is 2 m/s, and if A = 1 s−1 , how far has
the mass traveled at 1 s? 10 s? 100 s? You should find that the distance
approaches a limiting value.
Now assume that the mass is subject to a damping force proportional
to the square of its velocity, which means that the equation of motion
now takes the form x¨ = −A˙x2 , where A is some constant. If the initial
speed is 2 m/s, and if A = 1 m−1 , how far has the mass traveled at 1 s?
10 s? 100 s? How about some larger powers of 10? You should find that
the distance keeps growing, but slowly like the log of t. (The results
for these two forms of the damping are consistent with the results of
Problem 1.5.)
1.7
Solutions
1.1. Escape velocity
It is tempting to use the same reasoning as in the√
low-orbit √
satellite example in Section
1.2. This reasoning gives the same
√ result, v = C gR = C GME /R, where C is some
number (it turns out that C = 2). Although this solution yields the correct answer, it
isn’t quite rigorous, in view of the footnote in the low-orbit satellite example. Because
the particle isn’t always at the same radius, the force changes, so it isn’t obvious that we
can absorb the ME and G dependence into one quantity, g, as we did with the orbiting
satellite. Let us therefore be more rigorous with the following reasoning.
The dimensionful quantities in the problem are [m] = M , the radius of the earth
[R] = L, the mass of the earth [ME ] = M , and Newton’s gravitational constant [G] = L3 /MT 2 . These units for G follow from the gravitational force law,
other than these given quantities, then there
F = Gm1 m2 /r 2 . If we use no information
√
is no way to arrive at the speed of C GME /R, because for all we know, there could be a
factor of (m/ME )7 in the answer. This number is dimensionless, so it wouldn’t mess up
the units.
If we want to make any progress in this problem, we have to use the fact that the
gravitational force takes the form of GME m/r 2 . This then implies (as was stated in the
problem) that the acceleration is independent of m. And since the path of the particle is
determined by its acceleration, we see that the answer can’t depend on m. We are therefore left with the quantities G, R, and ME , and you can show
√ that the only combination
of these quantities that gives the units of speed is v = C GME /R.
1.2. Mass in a tube
The dimensionful quantities are [g] = L/T 2 , [ ] = L, [m] = M , and [M ] = M . We
want to produce a dimensionless number η. Since g is the only constant involving time,
η cannot depend on g. This then implies that η cannot depend on , which is the only
1.7 Solutions
length remaining. Therefore, η depends only on m and M (and furthermore only on the
ratio m/M , since we want a dimensionless number). So the answer to the stated problem
is, “No.”
It turns out that you have to solve the problem numerically if you actually want to
find η (see Problem 8.5). Some results are: If m
M , then η ≈ 0.349. If m = M , then
η ≈ 0.378. And if m = 2M , then η ≈ 0.410.
1.3. Waves in a fluid
We want to make a speed, [v ] = L/T , out of the quantities [ρ] = M /L3 , and [B] =
[F/A] = (ML/T 2 )/(L2 ) = M /(LT 2 ). We can play around with these quantities to find
the combination that has the correct units, but let’s do it the no-fail way. If v ∝ ρ a Bb ,
then we have
M a M b
L
=
.
(1.21)
T
L3
LT 2
Matching up the powers of the three kinds of units on each side of this equation gives
M : 0 = a + b, L : 1 = −3a − b, T : −1 = −2b.
(1.22)
The solution to√this system of equations is a = −1/2 and b = 1/2. Therefore, our
answer is v ∝ B/ρ. Fortunately, there was a solution to this system of three equations
in two unknowns.
1.4. Vibrating star
We want to make a frequency, [ν] = 1/T , out of the quantities [R] = L, [ρ] = M /L3 ,
and [G] = L3 /(MT 2 ). These units for G follow from the gravitational force law, F =
Gm1 m2 /r 2 . As in the previous problem, we can play around with these quantities to find
the combination that has the correct units, but let’s do it the no-fail way. If ν ∝ Ra ρ b G c ,
then we have
c
M b
L3
1
= La
.
(1.23)
T
L3
MT 2
Matching up the powers of the three kinds of units on each side of this equation gives
M : 0 = b − c, L : 0 = a − 3b + 3c, T : −1 = −2c.
(1.24)
The solution to√this system of equations is a = 0, and b = c = 1/2. Therefore, our
answer is ν ∝ ρG. So it turns out that there is no R dependence.
Remark: Note the difference in the given quantities in this problem (R, ρ, and G)
and the ones in Exercise 1.12 (R, ρ, and S). In this problem with the star, the mass
is large enough so that we can ignore the surface tension, S. And in Exercise 1.12
with the drop, the mass is small enough so that we can ignore the gravitational force,
and hence G. ♣
1.5. Damping
(a) The constant b has units [b] = [Force][v −n ] = (ML/T 2 )(T n /Ln ). The other
quantities are [m] = M and [V ] = L/T . There is also n, which is dimensionless. You can show that the only combination of these quantities that has
units of T is
m
(1.25)
t = f (n) n−1 ,
bV
where f (n) is a dimensionless function of n.
For n = 0, we have t = f (0) mV /b. This increases with m and V , and decreases
with b, as it should.
For n = 1, we have t = f (1) m/b. So we seem to have t ∼ m/b. This, however,
cannot be correct, because t should definitely grow with V . A large initial speed
V1 requires some nonzero time to slow down to a smaller speed V2 , after which
time we simply have the same scenario with initial speed V2 . So where did we go
wrong? After all, dimensional analysis tells us that the answer does have to look
like t = f (1) m/b, where f (1) is a numerical factor. The resolution to this puzzle
19
20
Strategies for solving problems
is that f (1) is infinite. If we worked out the problem using F = ma, we would
encounter an integral that diverges. So for any V , we would find an infinite t. 6
Similarly, for n ≥ 2, there is at least one power of V in the denominator of t.
This certainly cannot be correct, because t should not decrease with V . So f (n)
must likewise be infinite for all of these cases.
The moral of this exercise is that sometimes you have to be careful when
using dimensional analysis. The numerical factor in front of your answer
nearly always turns out to be of order 1, but in some strange cases it turns
out to be 0 or ∞.
Remark: For n ≥ 1, the expression in Eq. (1.25) still has relevance. For example,
for n = 2, the m/(Vb) expression is relevant if you want to know how long it
takes to go from V to some final speed Vf . The answer involves m/(Vf b), which
diverges as Vf → 0. ♣
(b) You can show that the only combination of the quantities that has units of L is
m
= g(n) n−2 ,
(1.26)
bV
where g(n) is a dimensionless function of n.
For n = 0, we have = g(0) mV 2 /b. This increases with V , as it should.
For n = 1, we have = g(1) mV /b. This increases with V , as it should.
For n = 2 we have = g(2) m/b. So we seem to have ∼ m/b. But as in
part (a), this cannot be correct, because should definitely depend on V . A large
initial speed V1 requires some nonzero distance to slow down to a smaller speed
V2 , after which point we simply have the same scenario with initial speed V2 . So,
from the reasoning in part (a), the total distance is infinite for n ≥ 2, because the
function g is infinite.
Remark: Note that for integral n = 1, t and are either both finite or both
infinite. For n = 1, however, the total time is infinite, whereas the total distance
is finite. This situation actually holds for 1 ≤ n < 2, if we want to consider
fractional n. ♣
1.6. Projectile distance
All of the possible answers have the correct units, so we’ll have to figure things out by
looking at special cases. Let’s look at each choice in turn:
gh2
: Incorrect, because the answer shouldn’t be zero for h = 0. Also, it shouldn’t
2
v
grow with g. And even worse, it shouldn’t be infinite for v → 0.
v2
g
: Incorrect, because the answer should depend on h.
v2 h
g
v2
g
v2
: Incorrect, because the answer shouldn’t be zero for h = 0.
1+
2gh
v2
: Can’t rule this out, and it happens to be the correct answer.
2gh
1 + 2 : Incorrect, because the answer should be zero for v → 0. But this
g
v
expression goes to 2h for v → 0.
v 2 /g
: Incorrect, because the answer shouldn’t be infinite for v 2 = 2gh.
1 − 2gh
2
v
6
The total time t is actually undefined, because the particle never comes to rest. But t does grow
with V , in the sense that if t is defined to be the time to slow down to some certain small speed,
then t grows with V .
1.7 Solutions
1.7. Two masses, one swinging
As in Section 1.4, we’ll write a Maple program. We’ll let q stand for θ , and we’ll use
˙ and q2 stands for θ.
¨ Likewise for r. We’ll run the
the notation where q1 stands for θ,
program for as long as r < 2. As soon as r exceeds 2, the program will stop and print
the value of the time.
r:=1:
r1:=0:
q:=3.14/18:
q1:=0:
e:=.001:
i:=0:
while r<2 do
i:=i+1:
r2:=(r*q1ˆ2-9.8*(1-cos(q)))/2:
r:=r+e*r1:
r1:=r1+e*r2:
q2:=-2*r1*q1/r-9.8*sin(q)/r:
q:=q+e*q1:
q1:=q1+e*q2:
end do:
i*e;
#
#
#
#
#
#
#
#
#
#
#
#
#
#
#
#
initial r value
initial r velocity
initial angle
initial angular velocity
small time interval
i counts the number of time steps
run the program until r=2
increase the counter by 1
the first of the given eqs
how r changes, by definition of r1
how r1 changes, by definition of r2
the second of the given eqs
how q changes, by definition of q1
how q1 changes, by definition of q2
the Maple command to stop the do loop
print the value of the time
This yields a time of t = 8.057 s. If we instead use a time interval of 0.0001 s, we obtain
t = 8.1377 s. And a time interval of 0.00001 s gives t = 8.14591 s. So the correct time
must be somewhere around 8.15 s.
21
Chapter 2
Statics
The subject of statics often appears in later chapters in other books, after force
and torque have been discussed. However, the way that force and torque are
used in statics problems is fairly minimal, at least compared with what we’ll
be doing later in this book. Therefore, since we won’t be needing much of the
machinery that we’ll be developing later on, I’ll introduce here the bare minimum of force and torque concepts necessary for statics problems. This will
open up a whole class of problems for us. But even though the underlying
principles of statics are quick to state, statics problems can be unexpectedly tricky. So be sure to tackle a lot of them to make sure you understand
things.
2.1
Balancing forces
A “static” setup is one where all the objects are motionless. If an object remains
motionless, then Newton’s second law, F = ma (which we’ll discuss in great
detail in the next chapter), tells us that the total external force acting on the object
must be zero. The converse is not true, of course. The total external force on an
object is also zero if it moves with constant nonzero velocity. But we’ll deal only
with statics problems here. The whole goal in a statics problem is to find out what
the various forces have to be so that there is zero net force acting on each object
(and zero net torque, too, but that’s the topic of Section 2.2). Because a force is
a vector, this goal involves breaking the force up into its components. You can
pick Cartesian coordinates, polar coordinates, or another set. It is usually clear
from the problem which system will make your calculations easiest. Once you
pick a system, you simply have to demand that the total external force in each
direction is zero.
There are many different types of forces in the world, most of which are largescale effects of complicated things going on at smaller scales. For example, the
tension in a rope comes from the chemical bonds that hold the molecules in
the rope together, and these chemical forces are electrical forces. In doing a
mechanics problem involving a rope, there is certainly no need to analyze all the
details of the forces taking place at the molecular scale. You just call the force in
22
2.1 Balancing forces
the rope a “tension” and get on with the problem. Four types of forces come up
repeatedly:
Tension
Tension is the general name for a force that a rope, stick, etc., exerts when it
is pulled on. Every piece of the rope feels a tension force in both directions,
except the end points, which feel a tension on one side and a force on the other
side from whatever object is attached to the end. In some cases, the tension may
vary along the rope. The “Rope wrapped around a pole” example at the end of
this section is a good illustration of this. In other cases, the tension must be the
same everywhere. For example, in a hanging massless rope, or in a massless
rope hanging over a frictionless pulley, the tension must be the same at all points,
because otherwise there would be a net force on at least some part of the rope,
and then F = ma would yield an infinite acceleration for this (massless) piece.
Normal force
This is the force perpendicular to a surface that the surface applies to an object.
The total force applied by a surface is usually a combination of the normal force
and the friction force (see below). But for frictionless surfaces such as greasy
ones or ice, only the normal force exists. The normal force comes about because
the surface actually compresses a tiny bit and acts like a very rigid spring. The
surface gets squashed until the restoring force equals the force necessary to keep
the object from squashing in any more.
For the most part, the only difference between a “tension” and a “normal
force” is the direction of the force. Both situations can be modeled by a spring.
In the case of a tension, the spring (a rope, a stick, or whatever) is stretched,
and the force on the given object is directed toward the spring. In the case of
a normal force, the spring is compressed, and the force on the given object is
directed away from the spring. Things like sticks can provide both normal forces
and tensions. But a rope, for example, has a hard time providing a normal force.
In practice, in the case of elongated objects such as sticks, a compressive force
is usually called a “compressive tension,” or a “negative tension,” instead of a
normal force. So by these definitions, a tension can point either way. At any rate,
it’s just semantics. If you use any of these descriptions for a compressed stick,
people will know what you mean.
Friction
Friction is the force parallel to a surface that a surface applies to an object.
Some surfaces, such as sandpaper, have a great deal of friction. Some, such as
greasy ones, have essentially no friction. There are two types of friction, called
“kinetic” friction and “static” friction. Kinetic friction (which we won’t cover in
this chapter) deals with two objects moving relative to each other. It is usually
a good approximation to say that the kinetic friction between two objects is
23
24
Statics
proportional to the normal force between them. The constant of proportionality
is called µk (the “coefficient of kinetic friction”), where µk depends on the two
surfaces involved. Thus, F = µk N , where N is the normal force. The direction
of the force is opposite to the motion.
Static friction deals with two objects at rest relative to each other. In the static
case, we have F ≤ µs N (where µs is the “coefficient of static friction”). Note
the inequality sign. All we can say prior to solving a problem is that the static
friction force has a maximum value equal to Fmax = µs N . In a given problem,
it is most likely less than this. For example, if a block of large mass M sits on a
surface with coefficient of friction µs , and you give the block a tiny push to the
right (tiny enough so that it doesn’t move), then the friction force is of course not
equal to µs N = µs Mg to the left. Such a force would send the block sailing off
to the left. The true friction force is simply equal and opposite to the tiny force
you apply. What the coefficient µs tells us is that if you apply a force larger than
µs Mg (the maximum friction force on a horizontal table), then the block will end
up moving to the right.
Gravity
Consider two point objects, with masses M and m, separated by a distance R.
Newton’s gravitational force law says that the force between these objects is
attractive and has magnitude F = GMm/R2 , where G = 6.67·10−11 m3 /(kg s2 ).
As we’ll show in Chapter 5, the same law also applies to spheres of nonzero size.
That is, a sphere may be treated like a point mass located at its center. Therefore,
an object on the surface of the earth feels a gravitational force equal to
F =m
GM
R2
≡ mg,
(2.1)
where M is the mass of the earth, and R is its radius. This equation defines g.
Plugging in the numerical values, we obtain g ≈ 9.8 m/s2 , as you can check.
Every object on the surface of the earth feels a force of mg downward (g varies
slightly over the surface of the earth, but let’s ignore this). If the object is not
accelerating, then there must be other forces present (normal forces, etc.) to make
the total force be equal to zero.
Another common force is the Hooke’s-law spring force, F = −kx. But we’ll
postpone the discussion of springs until Chapter 4, where we’ll spend a whole
chapter on them in depth.
Mg
θ
Fig. 2.1
M
Example (Block on a plane): A block of mass M rests on a fixed plane inclined
at an angle θ . You apply a horizontal force of Mg on the block, as shown in Fig. 2.1.
Assume that the friction force between the block and the plane is large enough to keep
the block at rest. What are the normal and friction forces (call them N and Ff ) that
the plane exerts on the block? If the coefficient of static friction is µ, for what range
of angles θ will the block in fact remain at rest?
2.1 Balancing forces
25
Solution: Let’s break the forces up into components parallel and perpendicular
to the plane. (The horizontal and vertical components would also work, but the
calculation would be a little longer.) The forces are N , Ff , the applied Mg, and
the weight Mg, as shown in Fig. 2.2. Balancing the forces parallel and perpendicular to the plane gives, respectively (with upward along the plane taken to be
positive),
N
Ff = Mg sin θ − Mg cos θ,
Ff
(2.2)
N = Mg cos θ + Mg sin θ.
θ
Mg
Intermediate remarks:
1. If tan θ > 1, then Ff is positive (that is, it points up the plane). And if tan θ < 1, then
Ff is negative (that is, it points down the plane). There is no need to worry about
which way it points when drawing the diagram. Just pick a direction to be positive,
and if Ff comes out to be negative (as it does in the figure above, because θ < 45◦ ),
then it actually points in the other direction.
2. Ff ranges from −Mg to Mg as θ ranges from 0 to π/2 (convince yourself that these
limiting values make sense). As√an exercise, you can show that N is maximum when
tan θ = 1, in which case N = 2Mg and Ff = 0.
3. The sin θ and cos θ factors in Eq. (2.2) follow from the angles θ drawn in Fig. 2.2.
However, when solving problems like this one, it’s easy to make a mistake in the
geometry and then label an angle as θ when it really should be 90◦ − θ . So two
pieces of advice: (1) Never draw an angle close to 45◦ in a figure, because if you do,
you won’t be able to tell the θ angles from the 90◦ − θ ones. (2) Always check your
results by letting θ go to 0 or 90◦ (in other words, does virtually all of a force, or
virtually none of it, act in a certain direction when the plane is, say, horizontal). Once
you do this a few times, you’ll realize that you probably don’t even need to work out
the geometry in the first place. Since you know that any given component is going
to involve either sin θ or cos θ, you can just pick the one that works correctly in a
certain limit. ♣
The coefficient µ tells us that |Ff | ≤ µN . Using Eq. (2.2), this inequality becomes
Mg| sin θ − cos θ | ≤ µMg(cos θ + sin θ).
(2.3)
The absolute value here signifies that we must consider two cases:
• If tan θ ≥ 1, then Eq. (2.3) becomes
sin θ − cos θ ≤ µ(cos θ + sin θ )
=⇒
tan θ ≤
1+µ
.
1−µ
(2.4)
We divided by 1 − µ, so this inequality is valid only if µ < 1. But if µ ≥ 1, we
see from the first inequality here that any value of θ (subject to our assumption,
tan θ ≥ 1) works.
• If tan θ ≤ 1, then Eq. (2.3) becomes
− sin θ + cos θ ≤ µ(cos θ + sin θ )
=⇒
tan θ ≥
1−µ
.
1+µ
(2.5)
θ Mg
Fig. 2.2
26
Statics
Putting these two ranges for θ together, we have
1+µ
1−µ
≤ tan θ ≤
.
1+µ
1−µ
(2.6)
Remarks: For very small µ, these bounds both approach 1, which means that θ must be
very close to 45◦ . This makes sense. If there is very little friction, then the components
along the plane of the horizontal and vertical Mg forces must nearly cancel; hence, θ ≈ 45◦ .
A special value for µ is 1, because from Eq. (2.6), we see that µ = 1 is the cutoff value
that allows θ to reach both 0 and π/2. If µ ≥ 1, then any tilt of the plane is allowed. We’ve
been assuming throughout this example that 0 ≤ θ ≤ π/2. The task of Exercise 2.20 is to
deal with the case where θ > π/2, where the block is under an overhang. ♣
Let’s now do an example involving a rope in which the tension varies with
position. We’ll need to consider differential pieces of the rope to solve this
problem.
Example (Rope wrapped around a pole): A rope wraps an angle θ around a
pole. You grab one end and pull with a tension T0 . The other end is attached to a large
object, say, a boat. If the coefficient of static friction between the rope and the pole
is µ, what is the largest force the rope can exert on the boat, if the rope is not to slip
around the pole?
Solution: Consider a small piece of the rope that subtends an angle dθ. Let the
tension in this piece be T (which varies slightly over the small length). As shown in
Fig. 2.3, the pole exerts a small outward normal force, Ndθ , on the piece. This normal
force exists to balance the “inward” components of the tensions at the ends. These
inward components have magnitude T sin(dθ/2).1 Therefore, Ndθ = 2T sin(dθ/2).
The small-angle approximation, sin x ≈ x, allows us to write this as Ndθ = T dθ.
The friction force on the little piece of rope satisfies Fdθ ≤ µNdθ = µT dθ. This
friction force is what gives rise to the difference in tension between the two ends of
the piece. In other words, the tension, as a function of θ, satisfies
dθ
T
Ndθ
T sin dθ/2
Fig. 2.3
T (θ + dθ) ≤ T (θ) + µT dθ
=⇒
dT
≤
T
=⇒
=⇒
=⇒
1
dT ≤ µT dθ
µ dθ
ln T ≤ µθ + C
T ≤ T0 eµθ ,
(2.7)
One of them actually has magnitude (T + dT ) sin(dθ/2), where dT is the increase in tension along
the small piece. But the extra term this produces, (dT ) sin(dθ/2), is a second-order small quantity,
so it can be ignored.
2.2 Balancing torques
27
where we have used the fact that T = T0 when θ = 0. The exponential behavior
here is quite strong (as exponential behaviors tend to be). If we let µ = 1, then just a
quarter turn around the pole produces a factor of eπ/2 ≈ 5. One full revolution yields
a factor of e2π ≈ 530, and two full revolutions yield a factor of e4π ≈ 300 000.
Needless to say, the limiting factor in such a case is not your strength, but rather the
structural integrity of the pole around which the rope winds.
2.2
Balancing torques
In addition to balancing forces in a statics problem, we must also balance torques.
We’ll have much more to say about torque in Chapters 8 and 9, but we’ll need
one important fact here. Consider the situation in Fig. 2.4, where three forces
are applied perpendicular to a stick, which is assumed to remain motionless. F1
and F2 are the forces at the ends, and F3 is the force in the interior. We have, of
course, F3 = F1 + F2 , because the stick is at rest. But we also have the following
relation:
Claim 2.1 If the system is motionless, then F3 a = F2 (a + b). In other words,
the torques (force times distance) around the left end cancel.2 And you can show
that they cancel around any other point, too.
We’ll prove this claim in Chapter 8 by using angular momentum, but let’s give a
short proof here.
Proof: We’ll make one reasonable assumption, namely, that the correct
relationship between the forces and distances is of the form,
F3 f (a) = F2 f (a + b),
(2.8)
where f (x) is a function to be determined.3 Applying this assumption with the
roles of “left” and “right” reversed in Fig. 2.4 gives
F3 f (b) = F1 f (a + b).
(2.9)
Adding Eqs. (2.8) and (2.9), and using F3 = F1 + F2 , yields
f (a) + f (b) = f (a + b).
(2.10)
This equation implies that f (rx) = rf (x) for any x and for any rational number
r, as you can show (see Exercise 2.28). Therefore, assuming f (x) is continuous,
2
3
Another proof of this claim is given in Problem 2.11.
What we’re doing here is simply assuming linearity in F. That is, two forces of F applied at a point
should be the same as a force of 2F applied at that point. You can’t really argue with that.
F1
F2
a
b
F3
Fig. 2.4
28
Statics
it must be a linear function, f (x) = Ax, as we wanted to show. The constant A is
irrelevant, because it cancels in Eq. (2.8).
Note that dividing Eq. (2.8) by Eq. (2.9) gives F1 f (a) = F2 f (b), and hence
F1 a = F2 b, which says that the torques cancel around the point where F3 is
applied. You can show that the torques cancel around any arbitrary pivot point.
When adding up all the torques in a given physical setup, it is of course required
that you use the same pivot point when calculating each torque.
In the case where the forces aren’t perpendicular to the stick, the above claim
applies to the components of the forces perpendicular to the stick. This makes
sense, because the components parallel to the stick have no effect on the rotation
of the stick around the pivot point. Therefore, referring to Figs. 2.5 and 2.6, the
equality of the torques can be written as
Fa a sin θa = Fb b sin θb .
Fa sin θa
Fa
θa
a
Fb sinθb
Fb
θb
b
Fig. 2.5
Fa
θa
a
a sin θa
Fig. 2.6
Fb
θb
b
b sinθb
(2.11)
This equation can be viewed in two ways:
• (Fa sin θa )a = (Fb sin θb )b. In other words, we effectively have smaller forces acting
on the given “lever arms,” as shown in Fig. 2.5.
• Fa (a sin θa ) = Fb (b sin θb ). In other words, we effectively have the given forces acting
on smaller “lever arms,” as shown in Fig. 2.6.
Claim 2.1 shows that even if you apply only a tiny force, you can balance
the torque due to a very large force, provided that you make your lever arm
sufficiently long. This fact led a well-known mathematician of long ago to claim
that he could move the earth if given a long enough lever arm.
One morning while eating my Wheaties,
I felt the earth move ‘neath my feeties.
The cause for alarm
Was a long lever arm,
At the end of which grinned Archimedes!
One handy fact that comes up often is that the gravitational torque on a
stick of mass M is the same as the gravitational torque due to a point-mass
M located at the center of the stick. The truth of this statement relies on the fact
that torque is a linear function of the distance to the pivot point (see Exercise
2.27). More generally, the gravitational torque on an object of mass M may be
treated simply as the gravitational torque due to a force Mg located at the center
of mass.
We’ll talk more about torque in Chapters 8 and 9, but for now we’ll just
use the fact that in a statics problem the torques around any given point must
balance.
2.2 Balancing torques
Example (Leaning ladder): A ladder leans against a frictionless wall. If the
coefficient of friction with the ground is µ, what is the smallest angle the ladder can
make with the ground and not slip?
Solution: Let the ladder have mass m and length . As shown in Fig. 2.7, we have
three unknown forces: the friction force F, and the normal forces N1 and N2 . And
to solve for these three forces we fortunately have three equations: Fvert = 0,
Fhoriz = 0, and τ = 0 (τ is the standard symbol for torque). Looking at the
vertical forces, we see that N1 = mg. And then looking at the horizontal forces, we
see that N2 = F. So we have quickly reduced the unknowns from three to one.
We will now use τ = 0 to find N2 (or F). But first we must pick the “pivot” point
around which we will calculate the torques. Any stationary point will work fine, but
certain choices make the calculation easier than others. The best choice for the pivot
is generally the point at which the most forces act, because then the τ = 0 equation
will have the smallest number of terms in it (because a force provides no torque around
the point where it acts, since the lever arm is zero). In this problem, there are two
forces acting at the bottom end of the ladder, so this is the point we’ll choose for the
pivot (but you should verify that other choices for the pivot, for example, the middle
or top of the ladder, give the same result). Balancing the torques due to gravity and
N2 , we have
mg
.
(2.12)
N2 sin θ = mg( /2) cos θ =⇒ N2 =
2 tan θ
29
N2
l
mg
θ
N1
F
Fig. 2.7
This is also the value of the friction force F. The condition F ≤ µN1 = µmg
therefore becomes
mg
≤ µmg
2 tan θ
=⇒
tan θ ≥
1
.
2µ
(2.13)
Remarks: Note that the total force exerted on the ladder by the floor points up at an angle
given by tan β = N1 /F = (mg)/(mg/2 tan θ) = 2 tan θ . We see that this force does not
point along the ladder. There is simply no reason why it should. But there is a nice reason
why it should point upward with twice the slope of the ladder. This is the direction that
causes the lines of the three forces on the ladder to be concurrent (that is, pass through a
common point), as shown in Fig. 2.8. This concurrency is a neat little theorem for statics
problems involving three forces. The proof is simple. If the three lines weren’t concurrent,
then one force would produce a nonzero torque around the intersection point of the other
two lines of force.4
This theorem provides a quick way to solve the ladder problem in the more general
case where the center of mass is a fraction f of the way up. In this case, the concurrency
theorem tells us that the slope of the total force from the floor is (1/f ) tan θ, consistent
with the f = 1/2 result from above. The vertical component is still mg, so the horizontal
(friction) component is now fmg/ tan θ. Demanding that this be less than or equal to
µmg gives tan θ ≥ f /µ, consistent with the f = 1/2 result. Since this result depends
4
The one exception to this reasoning is where no two of the lines intersect, that is, where all three
lines are parallel. Equilibrium is certainly possible in such a scenario, as we saw in Claim 2.1. But
you can hang on to the concurrency theorem in this case if you consider the parallel lines to meet
at infinity.
N2
mg
θ
Fig. 2.8
Ffloor
30
Statics
only on the location of the center of mass, and not on the exact distribution of mass, a
corollary is that if you climb up a ladder (resting on a frictionless wall), your presence
makes the ladder more likely to slip if you are above the center of mass (because you have
raised the center of mass of the entire system and thus increased f ), and less likely if you
are below. ♣
The examples we’ve done in this chapter have consisted of only one object.
But many problems involve more than one object (as you’ll find in the problems
and exercises for this chapter), and there’s one additional fact you’ll often need to
invoke for these, namely Newton’s third law. This states that the force that object
A exerts on object B is equal and opposite to the force that B exerts on A (we’ll
talk more about Newton’s laws in Chapter 3). So if you want to find, say, the
normal force between two objects, you might be able to figure it out by looking
at forces and torques on either object, depending on how much you already know
about the other forces acting on each. Once you’ve found the force by dealing
with, say, object A, you can then use the equal and opposite force to help figure
out things about B. Depending on the problem, one object is often more useful
than the other to use first.
Note, however, that if you pick your subsystem (on which you’re going to
consider forces and torques) to include both A and B, then this won’t tell you
anything at all about the normal force (or friction) between them. This is true
because the normal force is an internal force between the objects (when considered together as a system), whereas only external forces are relevant in calculating
the total force and torque on the system (because all the internal forces cancel in
pairs, by Newton’s third law). The only way to determine a given force is to deal
with it as an external force on some subsystem(s).
Statics problems often involve a number of decisions. If there are various parts
to the system, then you must decide which subsystems you want to balance the
external forces and torques on. And furthermore, you must decide which point to
use as the origin for calculating the torques. There are invariably many choices
that will give you the information you need, but some will make your calculations
much cleaner than others (Exercise 2.35 is a good example of this). The only way
to know how to choose wisely is to start solving problems, so you may as well
tackle some . . .
2.3
Problems
Section 2.1: Balancing forces
2.1. Hanging rope
A rope with length L and mass density per unit length ρ is suspended
vertically from one end. Find the tension as a function of height along
the rope.
2.3 Problems
31
2.2. Block on a plane
A block sits on a plane that is inclined at an angle θ . Assume that
the friction force is large enough to keep the block at rest. What are the
horizontal components of the friction and normal forces acting on the
block? For what θ are these horizontal components maximum?
2.3. Motionless chain *
A frictionless tube lies in the vertical plane and is in the shape of a function that has its endpoints at the same height but is otherwise arbitrary.
A chain with uniform mass per unit length lies in the tube from end to
end, as shown in Fig. 2.9. Show, by considering the net force of gravity
along the curve, that the chain doesn’t move.
2.4. Keeping a book up *
A book of mass M is positioned against a vertical wall. The coefficient of friction between the book and the wall is µ. You wish to keep
the book from falling by pushing on it with a force F applied at an
angle θ with respect to the horizontal (−π/2 < θ < π/2), as shown
in Fig. 2.10.
(a) For a given θ, what is the minimum F required?
(b) For what θ is this minimum F the smallest? What is the
corresponding minimum F?
(c) What is the limiting value of θ, below which there does not exist
an F that keeps the book up?
Fig. 2.9
M
F
θ
Fig. 2.10
2.5. Rope on a plane *
A rope with length L and mass density per unit length ρ lies on a plane
inclined at an angle θ (see Fig. 2.11). The top end is nailed to the plane,
and the coefficient of friction between the rope and the plane is µ. What
are the possible values for the tension at the top of the rope?
2.6. Supporting a disk **
L
µ
θ
Fig. 2.11
(a) A disk of mass M and radius R is held up by a massless string, as
shown in Fig. 2.12. The surface of the disk is frictionless. What is
the tension in the string? What is the normal force per unit length
that the string applies to the disk?
(b) Let there now be friction between the disk and the string, with
coefficient µ. What is the smallest possible tension in the string
at its lowest point?
2.7. Objects between circles **
Each of the following planar objects is placed, as shown in Fig. 2.13,
between two frictionless circles of radius R. The mass density per unit
µ
M
R
Fig. 2.12
32
Statics
area of each object is σ , and the radii to the points of contact make an
angle θ with the horizontal. For each case, find the horizontal force that
must be applied to the circles to keep them together. For what θ is this
force maximum or minimum?
L
F
F
θ
R
(a) An isosceles triangle with common side length L.
(b) A rectangle with height L.
(c) A circle.
L
θ
2.8. Hanging chain ****
θ
Fig. 2.13
d
λ
(a) A chain with uniform mass density per unit length hangs between
two given points on two walls. Find the general shape of the chain.
Aside from an arbitrary additive constant, the function describing
the shape should contain one unknown constant. (The shape of a
hanging chain is known as a catenary.)
(b) The unknown constant in your answer depends on the horizontal
distance d between the walls, the vertical distance λ between the
support points, and the length of the chain (see Fig. 2.14). Find
an equation involving these given quantities that determines the
unknown constant.
l
Fig. 2.14
2d
l=?
Fig. 2.15
α
Fig. 2.16
2.9. Hanging gently **
A chain with uniform mass density per unit length hangs between two
supports located at the same height, a distance 2d apart (see Fig. 2.15).
What should the length of the chain be so that the magnitude of the
force at the supports is minimized? You may use the fact that a hanging
chain takes the form, y(x) = (1/α) cosh(αx). You will eventually need
to solve an equation numerically.
2.10. Mountain climber ****
A mountain climber wishes to climb up a frictionless conical mountain. He wants to do this by throwing a lasso (a rope with a loop) over
the top and climbing up along the rope. Assume that the climber is of
negligible height, so that the rope lies along the mountain, as shown
in Fig. 2.16. At the bottom of the mountain are two stores. One sells
“cheap” lassos (made of a segment of rope tied to a loop of fixed
length); see Fig. 2.17. The other sells “deluxe” lassos (made of one
piece of rope with a loop of variable length; the loop’s length may
change without any friction of the rope with itself). When viewed from
the side, the conical mountain has an angle α at its peak. For what
angles α can the climber climb up along the mountain if he uses a
“cheap” lasso? A “deluxe” lasso? (Hint: The answer in the “cheap” case
isn’t α < 90◦ .)
2.3 Problems
33
2.11. Equality of torques **
This problem gives another way of demonstrating Claim 2.1, using an
inductive argument. We’ll get you started, and then you can do the
general case.
Consider the situation where forces F are applied upward at the ends
of a stick of length , and a force 2F is applied downward at the midpoint
(see Fig. 2.18). The stick doesn’t rotate (by symmetry), and it doesn’t
translate (because the net force is zero). If we wish, we may consider
the stick to have a pivot at the left end. If we then erase the force F on
the right end and replace it with a force 2F at the middle, then the two
2F forces in the middle cancel, so the stick remains at rest.5 Therefore,
we see that a force F applied at a distance from a pivot is equivalent
to a force 2F applied at a distance /2 from the pivot, in the sense that
they both have the same effect in canceling out the rotational effect of
the downwards 2F force.
Now consider the situation where forces F are applied upward at the
ends, and forces F are applied downward at the /3 and 2 /3 marks
(see Fig. 2.19). The stick doesn’t rotate (by symmetry), and it doesn’t
translate (because the net force is zero). Consider the stick to have a
pivot at the left end. From the above paragraph, the force F at 2 /3
is equivalent to a force 2F at /3. Making this replacement, we now
have a total force of 3F at the /3 mark. Therefore, we see that a
force F applied at a distance is equivalent to a force 3F applied at a
distance /3.
Your task is to now use induction to show that a force F applied at a
distance is equivalent to a force nF applied at a distance /n, and to
then argue why this demonstrates Claim 2.1.
cheap
Section 2.2: Balancing torques
2.12. Direction of the tension *
Show that the tension in a completely flexible rope, massive or massless,
points along the rope everywhere in the rope.
2.13. Find the force *
A stick of mass M is held up by supports at each end, with each support
providing a force of Mg/2. Now put another support somewhere in the
middle, say, at a distance a from one support and b from the other;
see Fig. 2.20. What forces do the three supports now provide? Is this
solvable?
5
There is now a different force applied at the pivot, namely zero, but the purpose of the pivot is to
simply apply whatever force is necessary to keep the left end motionless.
deluxe
Fig. 2.17
F
F
2F
2F
2F
Fig. 2.18
F
F
F
F
F
F
2F
Fig. 2.19
M
a
Fig. 2.20
b
34
Statics
2.14. Leaning sticks *
One stick leans on another as shown in Fig. 2.21. A right angle is formed
where they meet, and the right stick makes an angle θ with the horizontal.
The left stick extends infinitesimally beyond the end of the right stick.
The coefficient of friction between the two sticks is µ. The sticks have
the same mass density per unit length and are both hinged at the ground.
What is the minimum angle θ for which the sticks don’t fall?
θ
Fig. 2.21
2.15. Supporting a ladder *
A ladder of length L and mass M has its bottom end attached to the
ground by a pivot. It makes an angle θ with the horizontal and is held
up by a massless stick of length that is also attached to the ground by a
pivot (see Fig. 2.22). The ladder and the stick are perpendicular to each
other. Find the force that the stick exerts on the ladder.
M
L
l
θ
2.16. Balancing the stick **
Given a semi-infinite stick (that is, one that goes off to infinity in one
direction), determine how its density should depend on position so that it
has the following property: If the stick is cut at an arbitrary location, the
remaining semi-infinite piece will balance on a support that is located a
distance from the end (see Fig. 2.23).
Fig. 2.22
l
Fig. 2.23
R
T
r
θ
Fig. 2.24
R
Fig. 2.25
2.17. The spool **
A spool consists of an axle of radius r and an outside circle of radius R
which rolls on the ground. A thread is wrapped around the axle and is
pulled with tension T at an angle θ with the horizontal (see Fig. 2.24).
(a) Given R and r, what should θ be so that the spool doesn’t move?
Assume that the friction between the spool and the ground is large
enough so that the spool doesn’t slip.
(b) Given R, r, and the coefficient of friction µ between the spool
and the ground, what is the largest value of T for which the spool
remains at rest?
(c) Given R and µ, what should r be so that you can make the spool
slip from the static position with as small a T as possible? That
is, what should r be so that the upper bound on T in part (b) is as
small as possible? What is the resulting value of T ?
2.18. Stick on a circle **
A stick of mass density per unit length ρ rests on a circle of radius R (see
Fig. 2.25). The stick makes an angle θ with the horizontal and is tangent
to the circle at its upper end. Friction exists at all points of contact, and
assume that it is large enough to keep the system at rest. Find the friction
force between the ground and the circle.
2.4 Exercises
2.19. Leaning sticks and circles ***
A large number of sticks (with mass density per unit length ρ) and
circles (with radius R) lean on each other, as shown in Fig. 2.26. Each
stick makes an angle θ with the horizontal and is tangent to the next
circle at its upper end. The sticks are hinged to the ground, and every
other surface is frictionless (unlike in the previous problem). In the limit
of a very large number of sticks and circles, what is the normal force
between a stick and the circle it rests on, very far to the right? Assume
that the last circle leans against a wall, to keep it from moving.
2.4
35
θ
...
R
Fig. 2.26
Exercises
Section 2.1: Balancing forces
2.20. Block under an overhang *
A block of mass M is positioned underneath an overhang that makes an
angle β with the horizontal. You apply a horizontal force of Mg on the
block, as shown in Fig. 2.27. Assume that the friction force between the
block and the overhang is large enough to keep the block at rest. What
are the normal and friction forces (call them N and Ff ) that the overhang
exerts on the block? If the coefficient of static friction is µ, for what
range of angles β does the block in fact remain at rest?
Mg
M
β
Fig. 2.27
2.21. Pulling a block *
A person pulls on a block with a force F, at an angle θ with respect to the
horizontal. The coefficient of friction between the block and the ground
is µ. For what θ is the F required to make the block slip a minimum?
What is the corresponding F?
2.22. Holding a cone *
With two fingers, you hold an ice cream cone motionless upside down,
as shown in Fig. 2.28. The mass of the cone is m, and the coefficient
of static friction between your fingers and the cone is µ. When viewed
from the side, the angle of the tip is 2θ . What is the minimum normal
force you must apply with each finger in order to hold up the cone? In
terms of θ , what is the minimum value of µ that allows you to hold up
the cone? Assume that you can supply as large a normal force as needed.
2.23. Keeping a book up **
The task of Problem 2.4 is to find the minimum force required to keep a
book up. What is the maximum allowable force, as a function of θ and
µ? Is there a special angle that arises? Given µ, make a rough plot of
the allowed values of F for −π/2 < θ < π/2.
finger
finger
θ θ
Fig. 2.28
36
Statics
2.24. Bridges **
(a) Consider the first bridge in Fig. 2.29, made of three equilateral
triangles of beams. Assume that the seven beams are massless and
that the connection between any two of them is a hinge. If a car
of mass m is located at the middle of the bridge, find the forces
(and specify tension or compression) in the beams. Assume that
the supports provide no horizontal forces on the bridge.
(b) Same question, but now with the second bridge in Fig. 2.29, made
of seven equilateral triangles.
(c) Same question, but now with the general case of 4n−1 equilateral
triangles.
m
m
Fig. 2.29
θ
2.25. Rope between inclines **
A rope rests on two platforms that are both inclined at an angle θ (which
you are free to pick), as shown in Fig. 2.30. The rope has uniform mass
density, and the coefficient of friction between it and the platforms is 1.
The system has left-right symmetry. What is the largest possible fraction
of the rope that does not touch the platforms? What angle θ allows this
maximum fraction?
θ
Fig. 2.30
2.26. Hanging chain **
A chain with mass M hangs between two walls, with its ends at the
same height. The chain makes an angle θ with each wall, as shown in
Fig. 2.31. Find the tension in the chain at the lowest point. Solve this in
two different ways:
θ
θ
M
Fig. 2.31
(a) Consider the forces on half of the chain. (This is the quick way.)
(b) Use the fact (see Problem 2.8) that the height of a hanging chain is
given by y(x) = (1/α) cosh(αx), and consider the vertical forces
on an infinitesimal piece at the bottom. This will give you the
tension in terms of α. Then find an expression for α in terms of
the given angle θ . (This is the long way.)
Section 2.2: Balancing torques
2.27. Gravitational torque
A horizontal stick of mass M and length L is pivoted at one end. Integrate
the gravitational torque along the stick (relative to the pivot), and show
that the result is the same as the torque due to a mass M located at the
center of the stick.
2.28. Linear function *
Show that if a function satisfies f (a) + f (b) = f (a + b), then f (rx) =
rf (x) for any x and for any rational number r.
2.4 Exercises
2.29. Direction of the force *
A stick is connected to other parts of a static system by hinges at its
ends. Show that (1) if the stick is massless, then the forces it feels at the
hinges are directed along the stick, but (2) if the stick is massive, then
the forces need not point along the stick.
2.30. Ball on a wall *
A ball is held up by a string, as shown in Fig. 2.32, with the string tangent
to the ball. If the angle between the string and the wall is θ , what is the
minimum coefficient of static friction between the ball and the wall that
keeps the ball from falling?
2.31. Cylinder and hanging mass *
A uniform cylinder of mass M sits on a fixed plane inclined at an
angle θ . A string is tied to the cylinder’s rightmost point, and a mass m
hangs from the string, as shown in Fig. 2.33. Assume that the coefficient of friction between the cylinder and the plane is sufficiently large
to prevent slipping. What is m, in terms of M and θ , if the setup is
static?
2.32. Ladder on a corner **
A ladder of mass M and length L leans against a frictionless wall, with
a quarter of its length hanging over a corner, as shown in Fig. 2.34. It
makes an angle θ with the horizontal. What angle θ requires the smallest
coefficient of friction at the corner to keep the ladder at rest? (Different
values of θ require different ladder lengths, but assume that the mass is
M for any length.)
2.33. Stick on a corner **
You support one end of a stick of mass M and length L with the tip of
your finger. A quarter of the way up the stick, it rests on a frictionless
corner of a table, as shown in Fig. 2.35. The stick makes an angle θ
with the horizontal. What is the magnitude of the force your finger must
apply to keep the stick in this position? For what angle θ does your force
point horizontally?
2.34. Stick and a cylinder **
A horizontal stick of mass m has its left end attached to a pivot on a plane
inclined at an angle θ , while its right end rests on the top of a cylinder
also of mass m which in turn rests on the plane, as shown in Fig. 2.36.
The coefficient of friction between the cylinder and both the stick and
the plane is µ.
37
θ
µ
Fig. 2.32
M
m
θ
Fig. 2.33
1/4 of the
L
length
θ
M
Fig. 2.34
M
L
θ
Fig. 2.35
µ
m
m
µ
(a) Assuming that the system is at rest, what is the normal force from
the plane on the cylinder?
Fig. 2.36
θ
38
Statics
m
θ θ
l
(b) What is the smallest value of µ (in terms of θ) for which the
system doesn’t slip anywhere?
m
2.35. Two sticks and a string **
Two sticks, each of mass m and length , are connected by a hinge at their
top ends. They each make an angle θ with the vertical. A massless string
connects the bottom of the left stick to the right stick, perpendicularly
as shown in Fig. 2.37. The whole setup stands on a frictionless table.
l
ng
stri
(a) What is the tension in the string?
(b) What force does the left stick exert on the right stick at the hinge?
Hint: No messy calculations required!
Fig. 2.37
2.36. Two sticks and a wall **
Two sticks are connected, with hinges, to each other and to a wall. The
bottom stick is horizontal and has length L, and the sticks make an angle
of θ with each other, as shown in Fig. 2.38. If both sticks have the same
mass per unit length, ρ, find the horizontal and vertical components
of the force that the wall exerts on the top hinge, and show that the
magnitude goes to infinity for both θ → 0 and θ → π/2. 6
θ
L
2.37. Stick on a circle **
Using the results from Problem 2.18 for the setup shown in Fig. 2.39,
show that if the system is to remain at rest, then the coefficient of friction:
Fig. 2.38
(a) between the stick and the circle must satisfy
µ≥
R
θ
sin θ
.
1 + cos θ
(2.14)
(b) between the stick and the ground must satisfy7
Fig. 2.39
µ≥
l
sin θ cos θ
.
(1 + cos θ)(2 − cos θ)
(2.15)
2.38. Stacking blocks **
N blocks of length are stacked on top of each other at the edge of a
table, as shown in Fig. 2.40 for N = 4. What is the largest horizontal
Fig. 2.40
6
7
The force must therefore achieve a minimum at some intermediate angle. If you want to go through
√
the algebra, you can show that this minimum occurs when cos θ = 3 − 1, which gives θ ≈ 43◦ .
If you want to go through the algebra, you can show that the right-hand side achieves a maximum
√
when cos θ = 3 − 1, which gives θ ≈ 43◦ . (Yes, I did just cut and paste this from the previous
footnote. But it’s still correct!) This is the angle for which the stick is most likely to slip on the
ground.
2.5 Solutions
39
distance the rightmost point on the top block can hang out beyond the
table? How does your answer behave for N → ∞?8
2.5
Solutions
2.1. Hanging rope
Let T (y) be the tension as a function of height. Consider a small piece of the rope
between y and y + dy (0 ≤ y ≤ L). The forces on this piece are T (y + dy) upward,
T (y) downward, and the weight ρg dy downward. Since the rope is at rest, we have
T (y + dy) = T (y) + ρg dy. Expanding this to first order in dy gives T (y) = ρg. The
tension in the bottom of the rope is zero, so integrating from y = 0 up to a position
y gives
T (y) = ρgy.
(2.16)
As a double-check, at the top end we have T (L) = ρgL, which is the weight of the
entire rope, as it should be.
Alternatively, you can simply write down the answer, T (y) = ρgy, by noting that
the tension at a given point in the rope is what supports the weight of all the rope
below it.
2.2. Block on a plane
Balancing the forces shown in Fig. 2.41, parallel and perpendicular to the plane, we
see that F = mg sin θ and N = mg cos θ. The horizontal components of these are
F cos θ = mg sin θ cos θ (to the right), and N sin θ = mg cos θ sin θ (to the left).
These are equal, as they must be, because the net horizontal force on the block is zero.
To maximize the value of mg sin θ cos θ , we can either take the derivative, or we can
write it as (mg/2) sin 2θ , from which it is clear that the maximum occurs at θ = π/4.
The maximum value is mg/2.
N
F
θ mg cosθ
mg
Fig. 2.41
2.3. Motionless chain
Let the curve be described by the function f (x), and let it run from x = a to x = b.
Consider a little piece of the chain between x and x + dx (see Fig. 2.42). The length
of this piece is 1 + f 2 dx, so its mass is ρ 1 + f 2 dx, where ρ is the mass per unit
length. The component of the gravitational acceleration along the curve is −g sin θ =
−gf / 1 + f 2 (using tan θ = f ), with positive corresponding to moving along the
curve from a to b. The total force along the curve is therefore
b
F=
a
b
(−g sin θ) dm =
a
−gf
1+f
2
·ρ 1+f
f 'dx
θ
2 dx
b
= −gρ
x
f dx
a
= −gρ f (a) − f (b)
= 0.
Fig. 2.42
(2.17)
2.4. Keeping a book up
(a) The normal force from the wall is F cos θ, so the friction force Ff holding
the book up is at most µF cos θ. The other vertical forces on the book are the
8
mg sinθ
It turns out that the method of stacking shown in Fig. 2.40 (with the blocks simply stacked on top
of each other) doesn’t yield the optimal overhang. See Hall (2005) for an interesting discussion of
other methods.
x+dx
40
Statics
gravitational force, which is −Mg, and the vertical component of F, which is
F sin θ . If the book is to stay at rest, we must have F sin θ + Ff − Mg = 0.
Combining this with the condition Ff ≤ µF cos θ gives
F(sin θ + µ cos θ ) ≥ Mg.
(2.18)
Therefore, F must satisfy
Mg
,
(2.19)
sin θ + µ cos θ
assuming that sin θ +µ cos θ is positive. If it is negative, then there is no solution
for F.
(b) To minimize this lower bound, we must maximize the denominator. Taking the
derivative gives cos θ − µ sin θ = 0, so tan θ = 1/µ. Plugging this value of θ
back into Eq. (2.19) gives
mg
(with tan θ = 1/µ).
(2.20)
F≥
1 + µ2
This is the smallest possible F that keeps the book up, and the angle must be
θ = tan−1 (1/µ) for it to work. We see that if µ is very small, then to minimize
your F, you should push essentially vertically with a force mg. But if µ is very
large, you should push essentially horizontally with a force mg/µ.
(c) There is no possible F that satisfies the condition in Eq. (2.19) if the right-hand
side is infinite (more precisely, there is no F that satisfies Eq. (2.18) if the
coefficient of F is zero or negative). This occurs when
F≥
tan θ = −µ.
(2.21)
If θ is more negative than this, then it is impossible to keep the book up, no
matter how hard you push.
2.5. Rope on a plane
The component of the gravitational force along the plane is (ρL)g sin θ, and the maximum value of the friction force is µN = µ(ρL)g cos θ . Therefore, you might think
that the tension at the top of the rope is ρLg sin θ − µρLg cos θ. However, this is
not necessarily the case. The tension at the top depends on how the rope is placed on
the plane. If, for example, the rope is placed on the plane without being stretched,
then the friction force points upwards, and the tension at the top does indeed equal
ρLg sin θ − µρLg cos θ . Or it equals zero if µρLg cos θ > ρLg sin θ, in which case
the friction force need not achieve its maximum value.
If, on the other hand, the rope is placed on the plane after being stretched (or
equivalently, it is dragged up along the plane and then nailed down at its top end),
then the friction force points downwards, and the tension at the top equals ρLg sin θ +
µρLg cos θ.
Another special case occurs when the rope is placed on a frictionless plane, and then
the coefficient of friction is “turned on” to µ. The friction force is still zero. Changing
the plane from ice to sandpaper (somehow without moving the rope) doesn’t suddenly
cause there to be a friction force. Therefore, the tension at the top equals ρLg sin θ.
In general, depending on how the rope is placed on the plane, the tension at the top
can take any value from a maximum of ρLg sin θ + µρLg cos θ , down to a minimum
of ρLg sin θ − µρLg cos θ (or zero, whichever is larger). If the rope is replaced by a
stick (which can support a compressive force), then the tension can achieve negative
values down to ρLg sin θ − µρLg cos θ, if this happens to be negative.
2.6. Supporting a disk
(a) The gravitational force downward on the disk is Mg, and the force upward is
2T . These forces must balance, so
Mg
.
(2.22)
T =
2
2.5 Solutions
41
We can find the normal force per unit length that the string applies to the disk
in two ways.
First method: Let N dθ be the normal force on an arc of the disk that subtends
an angle dθ . Such an arc has length R dθ, so N /R is the desired normal force
per unit arclength. The tension in the string is the same throughout it, because
the string is massless. So all points are equivalent, and hence N is constant,
independent of θ . The upward component of the normal force is N dθ cos θ,
where θ is measured from the vertical (that is, −π/2 ≤ θ ≤ π/2 here). Since
the total upward force is Mg, we must have
π/2
−π/2
N cos θ dθ = Mg.
(2.23)
The integral equals 2N , so we have N = Mg/2. The normal force per unit
length, N /R, therefore equals Mg/2R.
Second method: Consider the normal force, N dθ , on a small arc of the disk
that subtends an angle dθ. The tension forces on each end of the corresponding
small piece of string almost cancel, but they don’t exactly, because they point
in slightly different directions. Their nonzero sum is what produces the normal
force on the disk. From Fig. 2.43, we see that the two forces have a sum of
2T sin(dθ/2), directed “inward”. Since dθ is small, we can use sin x ≈ x to
approximate this as T dθ. Therefore, N dθ = T dθ, and so N = T . The normal
force per unit arclength, N /R, therefore equals T /R. Using T = Mg/2 from Eq.
(2.22), we arrive at N /R = Mg/2R.
(b) Let T (θ ) be the tension, as a function of θ, for −π/2 ≤ θ ≤ π/2. T now
depends on θ , because there is a tangential friction force. Most of the work for
this problem was already done in the “Rope wrapped around a pole” example in
Section 2.1. We’ll simply invoke Eq. (2.7), which in the present language says9
T (θ ) ≤ T (0)eµθ .
Letting θ = π/2, and using T (π/2) = Mg/2, gives Mg/2 ≤
therefore see that the tension at the bottom point must satisfy
T (0) ≥
Mg −µπ/2
e
.
2
dθ
T
T sin dθ/2
Fig. 2.43
(2.24)
T (0)eµπ/2 .
We
(2.25)
Remark: This minimum value of T (0) goes to Mg/2 as µ → 0, as it should.
And it goes to zero as µ → ∞, as it should (imagine a very rough surface, so
that the friction force from the rope near θ = π/2 accounts for essentially all the
weight). But interestingly, the tension at the bottom doesn’t exactly equal zero,
no matter now large µ is. Basically, the smaller T is, the smaller N is. But the
smaller N is, the smaller the change in T is (because N determines the friction
force). So T doesn’t decrease much when it’s small, and this results in it never
being able to reach zero. ♣
N
2θ
θ
2.7. Objects between circles
(a) Let N be the normal force between the circles and the triangle. The goal in
this problem is to find the horizontal component of N , that is, N cos θ. From
Fig. 2.44, we see that the upward force on the triangle from the normal forces is
2N sin θ . This must equal the weight of the triangle, which is gσ times the area.
Since the bottom angle of the isosceles triangle is 2θ, the top side has length
9
This holds for θ > 0. There would be a minus sign on the right-hand side if θ < 0. But since the
tension is symmetric around θ = 0 in the case we’re concerned with, we’ll just deal with θ > 0.
Fig. 2.44
42
Statics
2L sin θ, and the altitude to this side is L cos θ. So the area of the triangle is
L2 sin θ cos θ . The mass is therefore σ L2 sin θ cos θ . Equating the weight with
the upward component of the normal forces gives N = (gσ L2 /2) cos θ . The
horizontal component of N is therefore
gσ L2 cos2 θ
.
(2.26)
2
This equals zero when θ = π/2, and it increases as θ decreases, even though
the triangle is getting smaller. It has the interesting property of approaching the
finite value gσ L2 /2, as θ → 0.
(b) In Fig. 2.45, the base of the rectangle has length 2R(1 − cos θ ). Its mass is
therefore 2σ RL(1 − cos θ ). Equating the weight with the upward component of
the normal forces, 2N sin θ , gives N = σ gRL(1 − cos θ )/ sin θ . The horizontal
component of N is therefore
σ gRL(1 − cos θ ) cos θ
.
(2.27)
N cos θ =
sin θ
This equals zero for both θ = π/2 and θ = 0 (because 1 − cos θ ≈ θ 2 /2 goes
to zero faster than sin θ ≈ θ, for small θ). Taking the derivative to find where it
reaches a maximum, we obtain (using sin2 θ = 1 − cos2 θ ),
N cos θ =
N
R
L
θ
Rcosθ
Fig. 2.45
cos3 θ − 2 cos θ + 1 = 0.
(2.28)
Fortunately, there is an easy root of this cubic equation, namely cos θ = 1,
which we know is not the maximum. Dividing through by the factor (cos θ − 1)
gives cos2 θ + cos θ − 1 = 0. The roots of this quadratic equation are
√
−1 ± 5
.
(2.29)
cos θ =
2
We must choose the plus sign, because we need | cos θ | ≤ 1. So our answer is
cos θ ≈ 0.618, which is the inverse of the golden ratio. The angle θ is ≈ 51.8◦ .
(c) In Fig. 2.46, the length of the hypotenuse shown is R sec θ , so the radius of the
top circle is R(sec θ − 1). Its mass is therefore σ π R2 (sec θ − 1)2 . Equating
the weight with the upward component of the normal forces, 2N sin θ , gives
N = σ gπ R2 (sec θ − 1)2 /(2 sin θ ). The horizontal component of N is therefore
Rsecθ
θ
R
N cos θ =
σ gπ R2 cos θ
2 sin θ
1
−1
cos θ
2
=
σ gπ R2 (1 − cos θ )2
.
2 sin θ cos θ
(2.30)
This equals zero when θ = 0 (using cos θ ≈ 1 − θ 2 /2 and sin θ ≈ θ , for
small θ). For θ → π/2, it behaves like 1/ cos θ, which goes to infinity. In this
limit, N points almost vertically, but its magnitude is so large that the horizontal
component still approaches infinity.
Fig. 2.46
2.8. Hanging chain
T(x+dx)
dx
(dm)g
T(x)
Fig. 2.47
(a) The key fact to note is that the horizontal component, Tx , of the tension is the
same throughout the chain. This is true because the net horizontal force on any
subpart of the chain must be zero. Label the constant value as Tx ≡ C.
Let the shape of the chain be described by the function y(x). Since the tension
points along the chain at all points (see Problem 2.12), its components satisfy
Ty /Tx = y , which gives Ty = Cy . In other words, Ty is proportional to the
slope of the chain.
Now consider a little piece of the chain, with endpoints at x and x + dx, as
shown in Fig. 2.47. The difference in the Ty values at the endpoints is what
balances the weight of the little piece, (dm)g. The length of the piece is ds =
dx 1 + y 2 , so if ρ is the density, we have
dTy = (ρ ds)g = ρg dx 1 + y 2
=⇒
dTy
= ρg 1 + y 2 .
dx
(2.31)
2.5 Solutions
43
Using the Ty = Cy result from above, this becomes Cy = ρg 1 + y 2 . Letting
z ≡ y , we can separate variables and integrate to obtain
√
dz
1 + z2
ρg dx
C
=
=⇒
sinh−1 z =
ρgx
+ A,
C
(2.32)
where A is a constant of integration. We can make this look a little cleaner if we
define constants α and a such that α ≡ ρg/C and a ≡ A/α. We then obtain
sinh−1 z = α(x + a)
=⇒
z = sinh α(x + a).
(2.33)
Recalling that z ≡ dy/dx, we can integrate again to obtain
1
cosh α(x + a) + h.
(2.34)
α
The shape of the chain is therefore a hyperbolic cosine function. The constant
h isn’t too important, because it depends simply on where we pick the y = 0
height. Furthermore, we can eliminate the need for the constant a if we pick
x = 0 to be where the lowest point of the chain is (or where it would be, in the
case where the slope is always nonzero). In this case, using Eq. (2.34), we see
that y (0) = 0 implies a = 0, as desired. We then have (ignoring the constant h)
the nice simple result,
y(x) =
1
cosh(αx).
(2.35)
α
(b) The constant α can be determined from the locations of the endpoints and the
length of the chain. As stated in the problem, the position of the chain may be
described by giving (1) the horizontal distance d between the two endpoints,
(2) the vertical distance λ between the two endpoints, and (3) the length of
the chain, as shown in Fig. 2.48. Note that it isn’t obvious what the horizontal
distances between the ends and the minimum point (which we have chosen as
the x = 0 point) are. If λ = 0, then these distances are d/2, by symmetry. But
otherwise, they aren’t so clear.
If we let the left endpoint be located at x = −x0 , then the first of the above
three facts says that the right endpoint is located at x = d − x0 . We now have
two unknowns, x0 and α. The second fact tells us that (we’ll take the right end
to be higher than the left end, without loss of generality)
y(x) =
y(d − x0 ) − y(−x0 ) = λ.
(2.36)
And the third fact gives, using Eq. (2.35),
=
d−x0
−x0
1 + y 2 dx =
1
sinh(αx)
α
d−x0
−x0
,
(2.37)
where we have used (d/du) cosh u = sinh u, and 1 + sinh2 u = cosh2 u, and
cosh u = sinh u. Writing out Eqs. (2.36) and (2.37) explicitly, we have
cosh α(d − x0 ) − cosh(−αx0 ) = αλ,
sinh α(d − x0 ) − sinh(−αx0 ) = α .
(2.38)
We can eliminate x0 by taking the difference of the squares of these two
equations. Using the hyperbolic identities cosh2 u − sinh2 u = 1 and
cosh u cosh v − sinh u sinh v = cosh(u − v ), we obtain
2 cosh(αd) − 2 = α 2 (
2
− λ2 ).
(2.39)
This is the desired equation that determines α. Given d, λ, and , we can numerically solve for α. Using a “half-angle” formula, you can show that Eq. (2.39)
may also be written as
2 sinh(αd/2) = α
2
− λ2 .
(2.40)
d
λ
l
-x0
x=0
Fig. 2.48
d-x0
44
Statics
Remark: Let’s check a couple limits. If λ = 0 and = d (that is, the chain
forms a horizontal straight line), then Eq. (2.40) becomes 2 sinh(αd/2) = αd.
The solution to this is α = 0, which does indeed correspond to a horizontal
straight line, because for small α, we can use cosh ≈ 1 + 2 /2 to say that the
y(x) in Eq. (2.35) behaves like αx2 /2 (up to an additive constant), which varies
slowly with x for small α. Another limit is where is much larger than both d
and λ. In this case, Eq. (2.40) becomes 2 sinh(αd/2) ≈ α . The solution to this
is a large α (or more precisely, α
1/d), which corresponds to a “droopy”
chain, because the y(x) in Eq. (2.35) varies rapidly with x for large α. ♣
2.9. Hanging gently
We must first find the mass of the chain by calculating its length. Then we must
determine the slope of the chain at the supports, so we can find the components of
the force there. Using the given information, y(x) = (1/α) cosh(αx), the slope of the
chain as a function of x is
y =
cosh(αx)
sinh(αx)
F
1
chain
Fig. 2.49
1
cosh(αx) = sinh(αx).
α
(2.41)
The total length is therefore (using 1 + sinh2 z = cosh2 z)
=
θ
d
dx
d
−d
1 + y 2 dx =
d
−d
cosh(αx) =
2
sinh(αd).
α
(2.42)
The weight of the rope is W = ρ g, where ρ is the mass per unit length. Each support
applies a vertical force of W /2. So this equals F sin θ , where F is the magnitude
of the force at each support, and θ is the angle it makes with the horizontal. Since
tan θ = y (d) = sinh(αd), we see from Fig. 2.49 that sin θ = tanh(αd). Therefore,
F=
W
1
ρg sinh(αd)
ρg
1
·
=
·
=
cosh(αd).
sin θ 2
tanh(αd)
α
α
(2.43)
Taking the derivative of this (as a function of α), and setting the result equal to zero to
find the minimum, gives tanh(αd) = 1/(αd). This must be solved numerically. The
result is
αd ≈ 1.1997 ≡ η.
(2.44)
So α is given by α = η/d, and the shape of the chain that requires the minimum F is
thus
ηx
d
.
(2.45)
y(x) ≈ cosh
η
d
From Eqs. (2.42) and (2.44), the length of the chain is = (2d/η) sinh(η) ≈ (2.52)d.
To further get an idea of what the chain looks like, we can calculate the ratio of the
height, h, to the width, 2d.
y(d) − y(0)
cosh(η) − 1
h
=
=
≈ 0.338.
2d
2d
2η
(2.46)
We can also calculate the angle of the rope at the supports, using tan θ = sinh(αd).
This gives tan θ = sinh η, and so θ ≈ 56.5◦ .
Remark: We can also ask what shape the chain should take in order to minimize
the horizontal or vertical component of F. The vertical component, Fy , is simply half
the weight, so we want the shortest possible chain, namely a horizontal one (which
requires an infinite F). This corresponds to α = 0. The horizontal component, Fx ,
equals F cos θ. From Fig. 2.49, we see that cos θ = 1/ cosh(αd). Therefore, Eq. (2.43)
gives Fx = ρg/α. This goes to zero as α → ∞, which corresponds to a chain with
infinite length, that is, a very “droopy” chain. ♣
2.5 Solutions
2.10. Mountain climber
Cheap lasso: We will take advantage of the fact that a cone is “flat,” in the sense
that we can make one out of a piece of paper, without crumpling the paper. Cut the
cone along a straight line emanating from the peak and passing through the knot of the
lasso, and roll the cone flat onto a plane. Call the resulting figure, which is a sector of
a circle, S (see Fig. 2.50). If the cone is very sharp, then S looks like a thin “pie piece.”
If the cone is very wide, with a shallow slope, then S looks like a pie with a piece taken
out of it. Points on the straight-line boundaries of the sector S are identified with each
other. Let P be the location of the lasso’s knot. Then P appears on each straight-line
boundary, at equal distances from the tip of S. Let β be the angle of the sector S.
The key to this problem is to realize that the path of the lasso’s loop must be a
straight line on S, as shown by the dotted line in Fig. 2.50. This is true because the rope
takes the shortest distance between two points because there is no friction, and rolling
the cone onto a plane doesn’t change distances. But a straight line between the two
identified points P is possible if and only if the sector S is smaller than a semicircle.
The condition for a climbable mountain is therefore β < 180◦ .
What is this condition, in terms of the angle of the peak, α? Let C denote a crosssectional circle of the mountain, a distance d (measured along the cone) from the top.
(We are considering this circle for geometrical convenience. It is not the path of the
lasso; see the remark below.) A semicircular S implies that the circumference of C
equals πd. This then implies that the radius of C equals d/2. Therefore,
sin(α/2) <
1
d/2
=
d
2
=⇒
α < 60◦ .
(2.47)
This is the condition under which the mountain is climbable. In short, having α < 60◦
guarantees that there is a loop around the cone with shorter length than the distance
straight to the peak and back.
Remark: When viewed from the side, the rope will appear perpendicular to the side
of the mountain at the point opposite the lasso’s knot. A common mistake is to assume
that this implies that the climbable condition is α < 90◦ . This is not the case, because
the loop does not lie in a plane. Lying in a plane, after all, would imply an elliptical
loop. But the loop must certainly have a kink in it where the knot is, because there
must exist a vertical component to the tension there to hold the climber up. If we had
posed the problem with a planar, triangular mountain, then the condition would have
been α < 90◦ . ♣
Deluxe lasso: If the mountain is very steep, the climber can slide down the mountain
by means of the loop growing larger. If the mountain has a shallow slope, the climber
can slide down by means of the loop growing smaller. The only situation in which the
climber doesn’t slide down is the one where the change in position of the knot along
the mountain is exactly compensated by the change in length of the loop.
Roll the cone onto a plane as we did in the cheap-lasso case. In terms of the sector S in
a plane, the above condition requires that if we move P a distance up (or down) along
the mountain, the distance between the identified points P must decrease (or increase)
by . In Fig. 2.50, we must therefore have an equilateral triangle, so β = 60◦ .
What peak-angle α does this correspond to? As in part the cheap-lasso case, let C be
a cross-sectional circle of the mountain, a distance d (measured along the cone) from
the top. Then β = 60◦ implies that the circumference of C equals (π/3)d. This then
implies that the radius of C equals d/6. Therefore,
sin(α/2) =
d/6
1
=
d
6
=⇒
α ≈ 19◦ .
(2.48)
This is the condition under which the mountain is climbable. We see that there is
exactly one angle for which the climber can climb up along the mountain. The cheap
45
β
P
Fig. 2.50
P
46
Statics
lasso is therefore much more useful than the fancy deluxe lasso, assuming, of course,
that you want to use it for climbing mountains, and not, say, for rounding up cattle.
Remark: Another way to see the β = 60◦ result is to note that the three directions
of rope emanating from the knot must all have the same tension, because the deluxe
lasso is one continuous piece of rope. They must therefore have 120◦ angles between
themselves (to provide zero net force on the massless knot). This implies that β = 60◦
in Fig. 2.50. ♣
N=4
P
Further remarks: For each type of lasso, we can also ask the question: For what
angles can the mountain be climbed if the lasso is looped N times around the top of
the mountain? The solution here is similar to that above.
For the cheap lasso, roll the cone N times onto a plane, as shown in Fig. 2.51 for
N = 4. The resulting figure, SN , is a sector of a circle divided into N equal sectors,
each representing a copy of the cone. As above, SN must be smaller than a semicircle.
The circumference of the circle C (defined above) must therefore be less than πd/N .
Hence, the radius of C must be less than d/2N . Thus,
P
P
P
P
Fig. 2.51
sin(α/2) <
F
1
d/2N
=
d
2N
=⇒
α < 2 sin−1
sin(α/2) =
1
d/6N
=
d
6N
=⇒
α = 2 sin−1
1
.
6N
♣
(2.50)
2.11. Equality of torques
The proof by induction is as follows. Assume that we have shown that a force F
applied at a distance d is equivalent to a force kF applied at a distance d/k, for all
integers k up to n − 1. We now want to show that the statement holds for k = n.
Consider the situation in Fig. 2.52. Forces F are applied at the ends of a stick, and
forces 2F/(n − 1) are applied at the j /n marks (for 1 ≤ j ≤ n − 1). The stick doesn’t
rotate (by symmetry), and it doesn’t translate (because the net force is zero). Consider
the stick to have a pivot at the left end. Replacing the interior forces by their equivalent
ones at the /n mark (see Fig. 2.52) gives a total force there equal to
...
F
Fig. 2.52
2F
2F
1 + 2 + 3 + · · · + (n − 1) =
n−1
n−1
n(n − 1)
2
= nF.
(2.51)
We therefore see that a force F applied at a distance is equivalent to a force nF
applied at a distance /n, as was to be shown.
We can now show that Claim 2.1 holds, for arbitrary distances a and b (see Fig. 2.53).
Consider the stick to be pivoted at its left end, and let be a tiny distance (small
compared with a). Then a force F3 at a distance a is equivalent to a force F3 (a/ ) at
a distance .10 But a force F3 (a/ ) at a distance is equivalent to a force F3 (a/ )
( /(a + b)) = F3 a/(a + b) at a distance (a + b). This equivalent force at the distance
(a + b) must cancel the force F2 there, because the stick is motionless. Therefore, we
have F3 a/(a + b) = F2 , which proves the claim.
F2
a
(2.49)
For the deluxe lasso, again roll the cone N times onto a plane. From the original
reasoning above, we must have N β = 60◦ . The circumference of C must therefore be
π d/3N , and so its radius must be d/6N . Therefore,
F
F1
1
.
2N
b
F3
2.12. Direction of the tension
Consider an infinitesimal piece of the rope, and look at the torque around one end.
Any forces acting at this end provide no torque around it. If the tension at the other
Fig. 2.53
10
Technically, we can use the reasoning in the previous paragraph to say this only if a/ is an integer,
but since a/ is very large, we can simply pick the closest integer to it, and there will be negligible
error.
2.5 Solutions
end is directed at a finite angle away from the direction of the rope, then this produces
a certain torque. But this torque can’t be canceled by the much smaller torque from
the tiny gravitational force, because this force is proportional to the length of the tiny
piece. Therefore, the tension must point along the rope. It actually points along the
direction of the rope at the end of the little piece it acts on, which isn’t quite along the
direction of the rope at the end we’re considering torques around, because the rope
bends (assuming it’s not vertical). So the tension ends up producing a very small torque
which cancels the very small torque from gravity.
This argument doesn’t work for a rigid stick, because the stick can produce finite
torques around the end of a piece via forces at that end, because the end is really
a cross section of finite size. There is a shearing action in the stick, and the large
shearing forces act with tiny lever arms (relative to, say, a point at the middle of the
cross section) to produce finite torques.
2.13. Find the force
In Fig. 2.54, let the supports at the ends exert forces F1 and F2 , and let the support in
the interior exert a force F. Then
F1 + F2 + F = Mg.
47
F1
a
F
F2
b
Mg
(2.52)
Balancing torques around the left and right ends gives, respectively,
Fig. 2.54
a+b
,
2
(2.53)
a+b
,
Fb + F1 (a + b) = Mg
2
where we have used the fact that the stick can be treated like a point mass at its center.
Note that the equation for balancing the torques around the center of mass is redundant;
it is obtained by taking the difference of the two previous equations and then dividing
by 2. And balancing torques around the middle pivot also takes the form of a linear
combination of these equations, as you can show.
It appears as though we have three equations and three unknowns, but we really have
only two equations, because the sum of Eqs. (2.53) gives Eq. (2.52). Therefore, since
we have two equations and three unknowns, the system is underdetermined. Solving
Eqs. (2.53) for F1 and F2 in terms of F, we see that any forces of the form
Fa + F2 (a + b) = Mg
(F1 , F, F2 ) =
Mg
Fb
Mg
Fa
−
, F,
−
2
a+b
2
a+b
(2.54)
are possible. In retrospect, it makes sense that the forces are not determined. By
changing the height of the new support an infinitesimal distance, we can make F be
anything from 0 up to Mg(a + b)/2b, which is when the stick comes off the left support
(assuming b ≥ a).
2.14. Leaning sticks
Let Ml be the mass of the left stick, and let Mr be the mass of the right stick. Then
Ml /Mr = tan θ. Let N and Ff be the normal and friction forces between the sticks
(see Fig. 2.55). Ff has a maximum value of µN . Balancing the torques on the left
stick (around the contact point with the ground) gives N = (Ml g/2) sin θ . Balancing
the torques on the right stick (around the contact point with the ground) gives Ff =
(Mr g/2) cos θ . The condition Ff ≤ µN is therefore
Mr cos θ ≤ µMl sin θ
=⇒
tan2 θ ≥
1
,
µ
(2.55)
where we have used Ml /Mr = tan θ . This answer checks in the two extremes: In the
limit µ → 0, we see that θ must be very close to π/2, which makes sense. And in the
limit µ → ∞ (that is, very sticky sticks), we see that θ can be very small, which also
makes sense.
N
Ff
θ
Fig. 2.55
48
Statics
2.15. Supporting a ladder
Let F be the desired force. F must be directed along the stick, because otherwise
there would be a net torque on the (massless) stick relative to the pivot at its right
end, and this would contradict the fact that it is at rest. Look at torques on the ladder
around the pivot at its bottom. The gravitational force provides a clockwise torque of
Mg(L/2) cos θ, and the force F from the stick provides a counterclockwise torque of
F( / tan θ). Equating these two torques gives
F=
MgL
sin θ .
2
(2.56)
Remarks: F goes to zero as θ → 0, as it should.11 And F increases to MgL/2
as θ → π/2, which isn’t so obvious (the required torque from the stick is very
small, but the lever arm is also very small). However, in the special case where
the ladder is exactly vertical, no force is required. You can see that our calculations above are not valid in this case, because we divided by cos θ, which is zero
when θ = π/2.
The normal force at the pivot of the stick (which equals the vertical component of
F, because the stick is massless) is equal to MgL sin θ cos θ/2 . This has a maximum
value of MgL/4 at θ = π/4. ♣
x0
2.16. Balancing the stick
Let the stick go off to infinity in the positive x direction, and let it be cut at x = x0 .
Then the pivot point is located at x = x0 + (see Fig. 2.56). Let the density be ρ(x).
The condition that the total gravitational torque relative to x0 + be zero is
x0 + l
Fig. 2.56
τ=
∞
ρ(x) x − (x0 + ) g dx = 0.
(2.57)
x0
We want this to equal zero for all x0 , so the derivative of τ with respect to x0 must be
zero. τ depends on x0 through both the limits of integration and the integrand. In taking
the derivative, the former dependence requires finding the value of the integrand at
the x0 limit, while the latter dependence requires taking the derivative of the integrand
with respect to x0 , and then integrating. (To derive these two contributions, just replace
x0 with x0 + dx0 and expand things to first order in dx0 .) We obtain
0=
dτ
= g ρ(x0 ) − g
dx0
∞
ρ(x) dx.
(2.58)
x0
Taking the derivative of this equation with respect to x0 gives ρ (x0 ) = −ρ(x0 ). The
solution to this is (rewriting the arbitrary x0 as x)
ρ(x) = Ae−x/ .
R
(2.59)
We therefore see that the density decreases exponentially with x. The smaller is, the
quicker it falls off. Note that the density at the pivot is 1/e times the density at the left
end. And you can show that 1 − 1/e ≈ 63 % of the mass is contained between the left
end and the pivot.
T
r
θ
2.17. The spool
(a) Let Ff be the friction force the ground provides. Balancing the horizontal forces
on the spool gives (see Fig. 2.57)
T cos θ = Ff .
Ff
11
Fig. 2.57
(2.60)
For θ → 0, we would need to lengthen the ladder with a massless extension, because the stick
would have to be very far to the right to remain perpendicular to the ladder.
2.5 Solutions
49
Balancing torques around the center of the spool gives
Tr = Ff R.
(2.61)
These two equations imply
r
.
(2.62)
R
The niceness of this result suggests that there is a quicker way to obtain it. And
indeed, we see from Fig. 2.58 that cos θ = r/R is the angle that causes the line
of the tension to pass through the contact point on the ground. Since gravity
and friction provide no torque around this point, the total torque around it is
therefore zero, and the spool remains at rest.
(b) The normal force from the ground is
cos θ =
N = Mg − T sin θ.
T
θ
R
(2.63)
Using Eq. (2.60), the statement Ff ≤ µN becomes
T cos θ ≤ µ(Mg − T sin θ)
=⇒
T ≤
µMg
,
cos θ + µ sin θ
(2.64)
r θ
θ
Fig. 2.58
where θ is given in Eq. (2.62).
(c) The maximum value of T is given in (2.64). This depends on θ , which in turn
depends on r. We want to find the r that minimizes this maximum T . Taking the
derivative with respect to θ , we find that the θ that maximizes the denominator
in Eq. (2.64) is given by tan θ0 = µ. You can then show that the value of T for
this θ0 is
T0 =
µMg
1 + µ2
.
To find the corresponding r, we can use Eq. (2.62) to write tan θ =
The relation tan θ0 = µ then yields
r0 =
R
1 + µ2
.
(2.65)
√
R2 − r 2 /r.
(2.66)
This is the r that yields the smallest upper bound on T . In the limit µ = 0, we
have θ0 = 0, T0 = 0, and r0 = R. And in the limit µ = ∞, we have θ0 = π/2,
T0 = Mg, and r0 = 0.
2.18. Stick on a circle
Let N be the normal force between the stick and the circle, and let Ff be the friction
force between the ground and the circle (see Fig. 2.59). Then we immediately see that
the friction force between the stick and the circle is also Ff , because the torques from
the two friction forces on the circle must cancel. We’ve drawn all forces as acting
on the circle. By Newton’s third law, N and Ff act in the opposite directions on the
stick at its top end.
Looking at torques on the stick around the point of contact with the ground, we
have Mg( /2) cos θ = N , where M = ρ is the mass of the stick, and is its length.
Therefore, N = (ρ g/2) cos θ . Balancing the horizontal forces on the circle gives
N sin θ = Ff + Ff cos θ, so we have
Ff =
N sin θ
ρ g sin θ cos θ
=
.
1 + cos θ
2(1 + cos θ )
(2.67)
But from Fig. 2.59 we have = R/ tan(θ/2). Using the identity tan(θ/2) = sin θ/(1+
cos θ), we finally obtain
Ff =
1
ρgR cos θ .
2
(2.68)
Ff
l
N
θ/2
θ/2
R
Ff
Fig. 2.59
50
Statics
In the limit θ → π/2, Ff approaches zero, which makes sense. In the limit θ → 0
(which corresponds to a very long stick), the friction force approaches ρgR/2, which
isn’t so obvious.
Ci
R
______
tan θ/2
Ni+1
R
Si
R
θ/2
θ/2
θ/2
θ/2
Ni
2.19. Leaning sticks and circles
Let Si be the ith stick, and let Ci be the ith circle. The normal forces that Ci feels
from Si and Si+1 are equal in magnitude, because these two forces provide the only
horizontal forces on the frictionless circle, so they must cancel. Let Ni be this normal
force.
Look at the torques on Si+1 , relative to the hinge on the ground. The torques come
from Ni , Ni+1 , and the weight of Si+1 . From Fig. 2.60, we see that Ni acts at a point
which is a distance R tan(θ/2) away from the hinge. Since the stick has a length
R/ tan(θ/2), this point is a fraction tan2 (θ/2) up along the stick. Therefore, balancing
the torques on Si+1 gives
R tan θ/2
Fig. 2.60
1
θ
Mg cos θ + Ni tan2 = Ni+1 .
2
2
(2.69)
N0 is by definition zero, so we have N1 = (Mg/2) cos θ (as in the previous problem).
If we successively use Eq. (2.69), we see that N2 equals (Mg/2) cos θ 1 + tan2 (θ/2) ,
and N3 equals (Mg/2) cos θ 1 + tan2 (θ/2) + tan4 (θ/2) , and so on. In general,
Ni =
Mg cos θ
2
1 + tan2
θ
θ
θ
+ tan4 + · · · + tan2(i−1)
2
2
2
.
(2.70)
In the limit i → ∞, we may write this infinite geometric sum in closed form as
N∞ ≡ lim Ni =
i→∞
Mg cos θ
2
1
.
1 − tan2 (θ/2)
(2.71)
Note that this is the solution to Eq. (2.69), with Ni = Ni+1 . So if a limit exists, it must
be this. Using M = ρR/ tan(θ/2), we can rewrite N∞ as
N∞ =
ρRg cos θ
2 tan(θ/2)
1
.
1 − tan2 (θ/2)
(2.72)
The identity cos θ = cos2 (θ/2) − sin2 (θ/2) may then be used to write this as
N∞ =
ρRg cos3 (θ/2)
.
2 sin(θ/2)
(2.73)
Remarks: N∞ goes to infinity for θ → 0, which makes sense, because the sticks are
very long. All of the Ni are essentially equal to half the weight of a stick (in order to
cancel the torque from the weight relative to the pivot). For θ → π/2, we see from
Eq. (2.73) that N∞ approaches ρRg/4, which is not at all obvious; the Ni start off at
N1 = (Mg/2) cos θ ≈ 0, but gradually increase to ρRg/4, which is a quarter of the
weight of a stick. Note that the horizontal force that must be applied to the last circle
far to the right is N∞ sin θ = ρRg cos4 (θ/2). This ranges from ρRg for θ → 0, to
ρRg/4 for θ → π/2. ♣
Chapter 3
Using F = ma
The general goal of classical mechanics is to determine what happens to a given
set of objects in a given physical situation. In order to figure this out, we need to
know what makes the objects move the way they do. There are two main ways
of going about this task. The first one, which you are undoubtedly familiar with,
involves Newton’s laws. This is the subject of the present chapter. The second
one, which is more advanced, is the Lagrangian method. This is the subject of
Chapter 6. It should be noted that each of these methods is perfectly sufficient
for solving any problem, and they both produce the same information in the end.
But they are based on vastly different principles. We’ll talk more about this in
Chapter 6.
3.1
Newton’s laws
In 1687 Newton published his three laws in his Principia Mathematica. These
laws are fairly intuitive, although I suppose it’s questionable to attach the adjective
“intuitive” to a set of statements that weren’t written down until a mere 300 years
ago. At any rate, the laws may be stated as follows.
• First law: A body moves with constant velocity (which may be zero) unless acted on
by a force.
• Second law: The time rate of change of the momentum of a body equals the force acting
on the body.
• Third law: For every force on one body, there is an equal and opposite force on another
body.
We could discuss for days on end the degree to which these statements are
physical laws, and the degree to which they are definitions. Sir Arthur Eddington
once made the unflattering remark that the first law essentially says that “every
particle continues in its state of rest or uniform motion in a straight line except
insofar as it doesn’t.” However, although the three laws might seem somewhat
light on content at first glance, there’s actually more to them than Eddington’s
comment implies. Let’s look at each in turn.1
1
A disclaimer: This section represents my view on which parts of the laws are definitions and which
parts have content. But you should take all of this with a grain of salt. For further reading, see
Anderson (1990), Keller (1987), O’Sullivan (1980), and Eisenbud (1958).
51
52
Using F =ma
First law
One thing this law does is give a definition of zero force. Another thing it does
is give a definition of an inertial frame, which is defined simply as a frame of
reference in which the first law holds; since the term “velocity” is used, we have
to state what frame we’re measuring the velocity with respect to. The first law
does not hold in an arbitrary frame. For example, it fails in the frame of a rotating
turntable.2 Intuitively, an inertial frame is one that moves with constant velocity.
But this is ambiguous, because we have to say what the frame is moving with
constant velocity with respect to. But all this aside, an inertial frame is defined
as the special type of frame in which the first law holds.
So, what we now have are two intertwined definitions of “force” and “inertial
frame.” Not much physical content here. But the important point is that the law
holds for all particles. So if we have a frame in which one free particle moves
with constant velocity, then all free particles move with constant velocity. This
is a statement with content. We can’t have a bunch of free particles moving with
constant velocity while another one is doing a fancy jig.
Second law
Momentum is defined3 to be mv. If m is constant,4 then the second law says that
F = ma,
(3.1)
where a ≡ dv/dt. This law holds only in an inertial frame, which is defined by
the first law.
For things moving free or at rest,
Observe what the first law does best.
It defines a key frame,
“Inertial” by name,
Where the second law then is expressed.
You might think that the second law merely gives a definition of force, but
there is more to it than that. There is a tacit implication in the law that this “force”
is something that has an existence that isn’t completely dependent on the particle
whose “m” appears in the law (more on this in the third law below). A spring
force, for example, doesn’t depend at all on the particle on which it acts. And
the gravitational force, GMm/r 2 , depends partly on the particle and partly on
something else (another mass).
2
3
4
It’s possible to modify things so that Newton’s laws hold in such a frame, provided that we introduce
the so-called “fictitious” forces. But we’ll save this discussion for Chapter 10.
We’re doing everything nonrelativistically here, of course. Chapter 12 gives the relativistic
modification to the mv expression.
We’ll assume in this chapter that m is constant. But don’t worry, we’ll get plenty of practice with
changing mass (in rockets and such) in Chapter 5.
3.1 Newton’s laws
If you feel like just making up definitions, then you can define a new quantity,
G = m2 a. This is a perfectly legal thing to do; you can’t really go wrong
in making a definition (well, unless you’ve already defined the quantity to be
something else). However, this definition is completely useless. You can define
it for every particle in the world, and for any acceleration, but the point is that
the definitions don’t have anything to do with each other. There is simply no
(uncontrived) quantity in this world that gives accelerations in the ratio of 4 to
1 when “acting” on masses m and 2m. The quantity G has nothing to do with
anything except the particle you defined it for. The main thing the second law
says is that there does indeed exist a quantity F that gives the same ma when
acting on different particles. The statement of the existence of such a thing is far
more than a definition.
Along this same line, note that the second law says that F = ma, and not,
for example, F = mv, or F = m d 3 x/dt 3 . In addition to being inconsistent with
the real world, these expressions are inconsistent with the first law. F = mv
would say that a nonzero velocity requires a force, in contrast with the first law.
And F = md 3 x/dt 3 would say that a particle moves with constant acceleration
(instead of constant velocity) unless acted on by a force, also in contrast with the
first law.
As with the first law, it is important to realize that the second law holds for
all particles. In other words, if the same force (for example, the same spring
stretched by the same amount) acts on two particles with masses m1 and m2 , then
Eq. (3.1) says that their accelerations are related by
m2
a1
=
.
a2
m1
(3.2)
This relation holds regardless of what the common force is. Therefore, once
we’ve used one force to find the relative masses of two objects, then we know
what the ratio of their a’s will be when they are subjected to any other force. Of
course, we haven’t really defined mass yet. But Eq. (3.2) gives an experimental
method for determining an object’s mass in terms of a standard (say, 1 kg) mass.
All we have to do is compare its acceleration with that of the standard mass, when
acted on by the same force.
Note that F = ma is a vector equation, so it is really three equations in one.
In Cartesian coordinates, it says that Fx = max , Fy = may , and Fz = maz .
Third law
One thing this law says is that if we have two isolated particles interacting through
some force, then their accelerations are opposite in direction and inversely proportional to their masses. Equivalently, the third law essentially postulates that
53
54
Using F =ma
the total momentum of an isolated system is conserved (that is, independent of
time). To see this, consider two particles, each of which interacts only with the
other particle and nothing else in the universe. Then we have
dptotal
dp1
dp2
=
+
dt
dt
dt
= F1 + F2 ,
(3.3)
where F1 and F2 are the forces acting on m1 and m2 , respectively. This demonstrates that momentum conservation (that is, dptotal /dt = 0) is equivalent to
Newton’s third law (that is, F1 = −F2 ). Similar reasoning holds with more
than two particles, but we’ll save this more general case, along with many other
aspects of momentum, for Chapter 5.
There isn’t much left to be defined via this law, so this is a law of pure content.
It can’t be a definition, anyway, because it’s actually not always valid. It holds
for forces of the “pushing” and “pulling” type, but it fails for the magnetic force,
for example. In that case, momentum is carried off in the electromagnetic field
(so the total momentum of the particles and the field is conserved). But we won’t
deal with fields here. Just particles. So the third law will always hold in any
situation we’ll be concerned with.
The third law contains an extremely important piece of information. It says
that we will never find a particle accelerating unless there’s some other particle accelerating somewhere else. The other particle might be far away, as with
the earth–sun system, but it’s always out there somewhere. Note that if we were
given only the second law, then it would be perfectly possible for a given particle
to spontaneously accelerate with nothing else happening in the universe, as long
as a similar particle with twice the mass accelerated with half the acceleration
when placed in the same spot, etc. This would all be fine, as far as the second law
goes. We would say that a force with a certain value is acting at the point, and
everything would be consistent. But the third law says that this is simply not the
way the world (at least the one we live in) works. In a sense, a force without a
counterpart seems somewhat like magic, whereas a force with an equal and opposite counterpart has a “cause and effect” nature, which seems (and apparently is)
more physical.
In the end, however, we shouldn’t attach too much significance to Newton’s
laws, because although they were a remarkable intellectual achievement and
work spectacularly for everyday physics, they are the laws of a theory that is only
approximate. Newtonian physics is a limiting case of the more correct theories
of relativity and quantum mechanics, which are in turn limiting cases of yet more
correct theories. The way in which particles (or waves, or strings, or whatever)
interact on the most fundamental level surely doesn’t bear any resemblance to
what we call forces.
3.2 Free-body diagrams
3.2
55
Free-body diagrams
The law that allows us to be quantitative is the second law. Given a force, we can
apply F = ma to find the acceleration. And knowing the acceleration, we can
determine the behavior of a given object (that is, the position and velocity), provided that we are given the initial position and velocity. This process sometimes
takes a bit of work, but there are two basic types of situations that commonly
arise.
• In many problems, all you are given is a physical situation (for example, a block
resting on a plane, strings connecting masses, etc.), and it is up to you to find all the
forces acting on all the objects, using F = ma. The forces generally point in various
directions, so it’s easy to lose track of them. It therefore proves useful to isolate the
objects and draw all the forces acting on each of them. This is the subject of the present
section.
• In other problems, you are given the force explicitly as a function of time, position,
or velocity, and the task immediately becomes the mathematical one of solving the
F = ma ≡ m¨x equation (we’ll just deal with one dimension here). These differential
equations can be difficult (or impossible) to solve exactly. They are the subject of
Section 3.3.
Let’s consider here the first of these two types of scenarios, where we are presented with a physical situation and we must determine all the forces involved.
The term free-body diagram is used to denote a diagram with all the forces
drawn on a given object. After drawing such a diagram for each object in the
setup, we simply write down all the F = ma equations they imply. The result
will be a system of linear equations in various unknown forces and accelerations, for which we can then solve. This procedure is best understood through an
example.
Example (A plane and masses): Mass M1 is held on a plane with inclination
angle θ, and mass M2 hangs over the side. The two masses are connected by a
massless string which runs over a massless pulley (see Fig. 3.1). The coefficient of
kinetic friction between M1 and the plane is µ. M1 is released from rest. Assuming that
M2 is sufficiently large so that M1 gets pulled up the plane, what is the acceleration
of the masses? What is the tension in the string?
Solution: The first thing to do is draw all the forces on the two masses. These are
shown in Fig. 3.2. The forces on M2 are gravity and the tension. The forces on M1 are
gravity, friction, the tension, and the normal force. Note that the friction force points
down the plane, because we are assuming that M1 moves up the plane.
Having drawn all the forces, we can now write down all the F = ma equations.
When dealing with M1 , we could break things up into horizontal and vertical components, but it is much cleaner to use the components parallel and perpendicular to the
M1
µ
M2
θ
Fig. 3.1
T
N
T
f
θ
M1 g
M2 g
Fig. 3.2
56
Using F =ma
plane.5 These two components of F = ma, along with the vertical F = ma equation
for M2 , give
T − f − M1 g sin θ = M1 a,
N − M1 g cos θ = 0,
(3.4)
M2 g − T = M2 a,
where we have used the fact that the two masses accelerate at the same rate (and
we have defined the positive direction for M2 to be downward). We have also used
the fact that the tension is the same at both ends of the string, because otherwise there
would be a net force on some part of the string which would then undergo infinite
acceleration, because it is massless.
There are four unknowns in Eq. (3.4) (namely T , a, N , and f ), but only three
equations. Fortunately, we have a fourth equation: f = µN , because we are assuming
that M1 is in fact moving, so we can use the expression for kinetic friction. Using this
in the second equation above gives f = µM1 g cos θ. The first equation then becomes
T − µM1 g cos θ − M1 g sin θ = M1 a. Adding this to the third equation leaves us with
only a, so we find
a=
g(M2 − µM1 cos θ − M1 sin θ)
M1 + M2
=⇒
T =
M1 M2 g(1 + µ cos θ + sin θ)
.
M1 + M2
(3.5)
Note that in order for M1 to in fact accelerate upward (that is, a > 0), we must have
M2 > M1 (µ cos θ + sin θ). This is clear from looking at the forces along the plane.
Remark: If we instead assume that M1 is sufficiently large so that it slides down the
plane, then the friction force points up the plane, and we find (as you can check),
a=
g(M2 + µM1 cos θ − M1 sin θ )
,
M1 + M2
and
T =
M1 M2 g(1 − µ cos θ + sin θ )
.
M1 + M2
(3.6)
In order for M1 to in fact accelerate downward (that is, a < 0), we must have M2 <
M1 (sin θ − µ cos θ). Therefore, the range of M2 for which the system doesn’t accelerate
(that is, it just sits there, assuming that it started at rest) is
M1 (sin θ − µ cos θ ) ≤ M2 ≤ M1 (sin θ + µ cos θ ).
(3.7)
If µ is very small, then M2 must essentially be equal to M1 sin θ if the system is to
be static. Equation (3.7) also implies that if tan θ ≤ µ, then M1 won’t slide down,
even if M2 = 0. ♣
In problems like the one above, it’s clear which things you should pick as the
objects you’re going to draw forces on. But in other problems, where there are
5
When dealing with inclined planes, it’s usually the case that one of these two coordinate systems
works much better than the other. Sometimes it isn’t clear which one, but if things get messy with
one system, you can always try the other.
3.2 Free-body diagrams
57
various different subsystems you can choose, you must be careful to include all
the relevant forces on a given subsystem. Which subsystems you want to pick
depends on what quantities you’re trying to find. Consider the following example.
Example (Platform and pulley): A person stands on a platform-and-pulley system, as shown in Fig. 3.3. The masses of the platform, person, and pulley6 are M , m,
and µ, respectively.7 The rope is massless. Let the person pull up on the rope so that
she has acceleration a upward. (Assume that the platform is somehow constrained to
stay level, perhaps by having the ends run along some rails.) Find the tension in the
rope, the normal force between the person and the platform, and the tension in the
rod connecting the pulley to the platform.
Solution: To find the tension in the rope, we simply want to let our subsystem be
the whole system (except the ceiling). If we imagine putting the system in a black
box (to emphasize the fact that we don’t care about any internal forces within the
system), then the forces we see “protruding” from the box are the three weights (Mg,
mg, and µg) downward, and the tension T upward. Applying F = ma to the whole
system gives
T − (M + m + µ)g = (M + m + µ)a
=⇒
T = (M + m + µ)(g + a).
(3.8)
To find the normal force N between the person and the platform, and also the tension
f in the rod connecting the pulley to the platform, it is not sufficient to consider the
system as a whole. This is true because these forces are internal forces to this system,
so they won’t show up in any F = ma equations (which involve only external forces
to a system). So we must consider subsystems:
• Let’s apply F = ma to the person. The forces acting on the person are gravity, the
normal force from the platform, and the tension from the rope (pulling downward
on her hand). So we have
N − T − mg = ma.
(3.9)
• Now apply F = ma to the platform. The forces acting on the platform are gravity,
the normal force from the person, and the force upward from the rod. So we have
f − N − Mg = Ma.
(3.10)
• Now apply F = ma to the pulley. The forces acting on the pulley are gravity, the
force downward from the rod, and twice the tension in the rope (because it pulls
6
7
Assume that the pulley’s mass is concentrated at its center, so that we don’t have to worry about
any rotational dynamics (the subject of Chapter 8).
My apologies for using µ as a mass here, since it usually denotes a coefficient of friction. Alas,
there are only so many symbols for “m.”
a
µ
M
Fig. 3.3
m
Using F =ma
58
up on both sides). So we have
2T − f − µg = µa.
(3.11)
Note that if we add up the three previous equations, we obtain the F = ma equation
in Eq. (3.8), as should be the case, because the whole system is the sum of the three
above subsystems. Equations (3.9)–(3.11) are three equations in the three unknowns,
T , N , and f . Their sum yields the T in (3.8), and then Eqs. (3.9) and (3.11) give,
respectively, as you can show,
N = (M + 2m + µ)(g + a),
T
and
f = (2M + 2m + µ)(g + a).
(3.12)
T
f
Remarks: You can also obtain these results by considering subsystems different from the
ones we chose above. For example, you can choose the pulley-plus-platform subsystem,
etc. But no matter how you choose to break up the system, you will need to produce three
independent F = ma equations in order to solve for the three unknowns, T , N , and f .
In problems like this one, it’s easy to forget to include certain forces, such as the second T
in Eq. (3.11). The safest thing to do is to always isolate each subsystem, draw a box around
it, and then draw all the forces that “protrude” from the box. In other words, draw the
free-body diagram. Figure 3.4 shows the free-body diagram for the subsystem consisting
of only the pulley. ♣
µg
Fig. 3.4
Another class of problems, similar to the above example, goes by the name
of Atwood’s machines. An Atwood’s machine is the name used for any system
consisting of a combination of masses, strings, and pulleys.8 In general, the
pulleys and strings can have mass, but we’ll deal only with massless ones in
this chapter. As we’ll see in the following example, there are two basic steps in
solving an Atwood’s problem: (1) write down all the F = ma equations, and
(2) relate the accelerations of the various masses by noting that the length of the
string(s) doesn’t change, a fact that we call “conservation of string.”
Example (Atwood’s machine): Consider the pulley system in Fig. 3.5, with
masses m1 and m2 . The strings and pulleys are massless. What are the accelerations
of the masses? What is the tension in the string?
m1
Solution: The first thing to note is that the tension T is the same everywhere
throughout the massless string, because otherwise there would be an infinite acceleration of some part of the string. It then follows that the tension in the short string
connected to m2 is 2T . This is true because there must be zero net force on the massless right pulley, because otherwise it would have infinite acceleration. The F = ma
m2
Fig. 3.5
8
George Atwood (1746–1807) was a tutor at Cambridge University. He published the description
of the first of his machines in Atwood (1784). For a history of Atwood’s machines, see Greenslade
(1985).
3.2 Free-body diagrams
59
equations for the two masses are therefore (with upward taken to be positive)
T − m1 g = m1 a1 ,
(3.13)
2T − m2 g = m2 a2 .
We now have two equations in the three unknowns, a1 , a2 , and T . So we need one
more equation. This is the “conservation of string” fact, which relates a1 and a2 .
If we imagine moving m2 and the right pulley up a distance d, then a length 2d
of string has disappeared from the two parts of the string touching the right pulley.
This string has to go somewhere, so it ends up in the part of the string touching m1
(see Fig. 3.6). Therefore, m1 goes down by a distance 2d. In other words, y1 =
−2y2 , where y1 and y2 are measured relative to the initial locations of the masses.
Taking two time derivatives of this statement gives our desired relation between
a1 and a2 ,
a1 = −2a2 .
(3.14)
m1
2m2 − 4m1
,
4m1 + m2
a2 = g
2m1 − m2
,
4m1 + m2
T =
3m1 m2 g
.
4m1 + m2
(3.15)
Remarks: There are all sorts of limits and special cases that we can check here. A couple
are: (1) If m2 = 2m1 , then Eq. (3.15) gives a1 = a2 = 0, and T = m1 g. Everything is
m1 , then Eq. (3.15) gives a1 = 2g, a2 = −g, and T = 3m1 g. In
at rest. (2) If m2
this case, m2 is essentially in free fall, while m1 gets yanked up with acceleration 2g. The
value of T is exactly what is needed to make the net force on m1 equal to m1 (2g), because
m2 .
T − m1 g = 3m1 g − m1 g = m1 (2g). You can check the case where m1
For the more general case where there are N masses instead of two, the “conservation
of string” statement is a single equation that relates all N accelerations. It is most easily
obtained by imagining moving N − 1 of the masses, each by an arbitrary amount, and then
seeing what happens to the last mass. Note that these arbitrary motions undoubtedly do
not correspond to the actual motions of the masses. This is fine; the single “conservation
of string” equation has nothing to do with the N F = ma equations. The combination of
all N + 1 equations is needed to constrain the motions down to a unique set. ♣
In the problems and exercises for this chapter, you will encounter some strange
Atwood’s setups. But no matter how complicated they get, there are only two
things you need to do to solve them, as mentioned above: write down the F = ma
equations for all the masses (which may involve relating the tensions in various
strings), and then relate the accelerations of the masses, using “conservation of
string.”
It may seem, with the angst it can bring,
That an Atwood’s machine’s a cruel thing.
But you just need to say
That F is ma,
And use conservation of string!
d
2d
m2
Combining this with Eq. (3.13), we can now solve for a1 , a2 , and T . The result is
a1 = g
d
Fig. 3.6
60
Using F =ma
3.3
Solving differential equations
Let’s now consider the type of problem where we are given the force as a function
of time, position, or velocity, and our task is to solve the F = ma ≡ m¨x differential equation to find the position, x(t), as a function of time.9 In what follows,
we will develop a few techniques for solving differential equations. The ability
to apply these techniques dramatically increases the number of systems we can
understand.
It’s also possible for the force F to be a function of higher derivatives of x,
in addition to the quantities t, x, and v ≡ x˙ . But these cases don’t arise much, so
we won’t worry about them. The F = ma differential equation we want to solve
is therefore (we’ll just work in one dimension here)
m¨x = F(t, x, v).
(3.16)
In general, this equation cannot be solved exactly for x(t).10 But for most of
the problems we’ll deal with, it can be solved. The problems we’ll encounter
will often fall into one of three special cases, namely, where F is a function of t
only, or x only, or v only. In all of these cases, we must invoke the given initial
conditions, x0 ≡ x(t0 ) and v0 ≡ v(t0 ), to obtain our final solutions. These initial
conditions will appear in the limits of the integrals in the following discussion.11
Note: You may want to just skim the following page and a half, and then refer
back as needed. Don’t try to memorize all the different steps. We present them
only for completeness. The whole point here can basically be summarized by
saying that sometimes you want to write x¨ as dv/dt, and sometimes you want to
write it as v dv/dx (see Eq. (3.20)). Then you “simply” have to separate variables
and integrate. We’ll go through the three special cases, and then we’ll do some
examples.
• F is a function of t only: F = F(t).
Since a = d 2 x/dt 2 , we just need to integrate F = ma twice to obtain x(t). Let’s do this
in a very systematic way, to get used to the general procedure. First, write F = ma as
m
9
dv
= F(t).
dt
(3.17)
In some setups, such as in Problem 3.11, the force isn’t given, so you have to figure out what it is.
But the main part of the problem is still solving the resulting differential equation.
10
You can always solve for x(t) numerically, to any desired accuracy. This topic is discussed in
Section 1.4.
11
It is no coincidence that we need two initial conditions to completely specify the solution to
our second-order (meaning the highest derivative of x that appears is the second one) F = m¨x
differential equation. It is a general result (which we’ll just accept here) that the solution to an
nth-order differential equation has n free parameters, which are determined by the initial conditions.
3.3 Solving differential equations
Then separate variables and integrate both sides to obtain12
v (t)
m
v0
t
dv =
t0
F(t ) dt .
(3.18)
We have put primes on the integration variables so that we don’t confuse them with the
limits of integration, but in practice we usually don’t bother with them. The integral of
dv is just v , so Eq. (3.18) yields v as a function of t, that is, v(t). We can then separate
variables in dx/dt ≡ v(t) and integrate to obtain
x(t)
x0
dx =
t
t0
v(t ) dt .
(3.19)
This yields x as a function of t, that is, x(t). This procedure might seem like a cumbersome way to simply integrate something twice. That’s because it is. But the technique
proves more useful in the following case.
• F is a function of x only: F = F(x).
We will use
a=
dv
dx dv
dv
=
=v
dt
dt dx
dx
(3.20)
dv
= F(x).
dx
(3.21)
to write F = ma as
mv
Now separate variables and integrate both sides to obtain
v (x)
m
v0
v dv =
x
x0
F(x ) dx .
(3.22)
The integral of v is v 2 /2, so the left-hand side involves the square of v(x). Taking
the square root, this gives v as a function of x, that is, v(x). Separating variables in
dx/dt ≡ v(x) then yields
x(t) dx
x0
v(x )
=
t
dt .
(3.23)
t0
Assuming that we can do the integral on the left-hand side, this equation gives t as a
function of x. We can then (in principle) invert the result to obtain x as a function of t,
that is, x(t). The unfortunate thing about this case is that the integral in Eq. (3.23) might
not be doable. And even if it is, it might not be possible to invert t(x) to produce x(t).
12
If you haven’t seen such a thing before, the act of multiplying both sides by the infinitesimal
quantity dt might make you feel a bit uneasy. But it is in fact quite legal. If you wish, you can
imagine working with the small (but not infinitesimal) quantities v and t, for which it is
certainly legal to multiply both sides by t. Then you can take a discrete sum over many t
intervals, and then finally take the limit t → 0, which results in the integral in Eq. (3.18).
61
62
Using F =ma
• F is a function of v only: F = F(v).
Write F = ma as
m
dv
= F(v).
dt
(3.24)
Separate variables and integrate both sides to obtain
v (t) dv
m
v0
=
F(v )
t
dt .
(3.25)
t0
Assuming that we can do this integral, it yields t as a function of v, and hence (in
principle) v as a function of t, that is, v(t). We can then integrate dx/dt ≡ v(t) to obtain
x(t) from
x(t)
x0
dx =
t
t0
v(t ) dt .
(3.26)
Note: In this F = F(v) case, if we want to find v as a function of x, v(x), then we
should write a as v(dv/dx) and integrate
m
v (x) v dv
v0
F(v )
=
x
dx .
(3.27)
x0
We can then obtain x(t) from Eq. (3.23), if desired.
When dealing with the initial conditions, we have chosen to put them in the
limits of integration above. If you wish, you can perform the integrals without
any limits, and just tack on a constant of integration to your result. The constant
is then determined by the initial conditions.
Again, as mentioned above, you do not have to memorize the above three
procedures, because there are variations, depending on what you’re given and
what you want to solve for. All you have to remember is that x¨ can be written
as either dv/dt or v dv/dx. One of these will get the job done (namely, the one
that makes only two of the three variables, t, x, and v, appear in your differential
equation). And then be prepared to separate variables and integrate as many times
as needed.13
a is dv by dt.
Is this useful? There’s no guarantee.
If it leads to “Oh, heck!” s,
Take dv by dx,
And then write down its product with v.
13
We want only two of the variables to appear in the differential equation because the goal is to
separate variables and integrate, and because equations have only two sides. If equations were
triangles, it would be a different story.
3.3 Solving differential equations
Example (Gravitational force): A particle of mass m is subject to a constant
force F = −mg. The particle starts at rest at height h. Because this constant force
falls into all of the above three categories, we should be able to solve for y(t) in two
ways:
(a) Find y(t) by writing a as dv/dt.
(b) Find y(t) by writing a as v dv/dy.
Solution:
(a) F = ma gives dv/dt = −g. Multiplying by dt and integrating yields v =
−gt + A, where A is a constant of integration.14 The initial condition v(0) = 0
gives A = 0. Therefore, dy/dt = −gt. Multiplying by dt and integrating yields
y = −gt 2 /2 + B. The initial condition y(0) = h gives B = h. Therefore,
1
y = h − gt 2 .
2
(3.28)
(b) F = ma gives v dv/dy = −g. Separating variables and integrating yields
v 2 /2 = −gy + C. The initial condition v(h) = 0 gives v 2 /2 = −gy + gh.
Therefore, v ≡ dy/dt = − 2g(h − y). We have chosen the negative square
root because the particle is falling. Separating variables gives
dy
h−y
= − 2g
dt.
(3.29)
√
This yields 2 h − y = 2g t, where we have used the initial condition y(0) =
h. Hence, y = h − gt 2 /2, as in part (a). In part (b) here, we essentially derived
conservation of energy, as we’ll see in Chapter 5.
Example (Dropped ball): A beach ball is dropped from rest at height h. Assume
that the drag force15 from the air takes the form of Fd = −βv. Find the velocity and
height as a function of time.
Solution: For simplicity in future formulas, let’s write the drag force as Fd =
−βv ≡ −mαv (otherwise we’d have a bunch of 1/m’s floating around). Taking
upward to be the positive y direction, the force on the ball is
F = −mg − mαv.
14
15
(3.30)
We’ll do this example by adding on constants of integration which are then determined by the
initial conditions. We’ll do the following example by putting the initial conditions in the limits of
integration.
The drag force is roughly proportional to v as long as the speed is fairly small (say, less than
10 m/s). For large speeds (say, greater than 100 m/s), the drag force is roughly proportional to
v 2 . But these approximate cutoffs depend on various things, and in any event there is a messy
transition region between the two cases.
63
64
Using F =ma
Note that v is negative here, because the ball is falling, so the drag force points
upward, as it should. Writing F = m dv/dt and separating variables gives
v (t)
0
t
dv
=−
dt .
g + αv
0
(3.31)
Integration yields ln(1 + αv/g) = −αt. Exponentiation then gives
v(t) = −
g
1 − e−αt .
α
(3.32)
Writing dy/dt ≡ v(t), and then separating variables and integrating to obtain y(t),
yields
y(t)
h
dy = −
g t
1 − e−αt
α 0
dt .
(3.33)
.
(3.34)
Therefore,
y(t) = h −
g
α
t−
1
1 − e−αt
α
Remarks:
1. Let’s look at some limiting cases. If t is very small (more precisely, if αt
1), then
we can use e−x ≈ 1 − x + x2 /2 to make approximations to leading order in t. You
can show that Eq. (3.32) gives v (t) ≈ −gt. This makes sense, because the drag force
is negligible at the start, so the ball is essentially in freefall. And likewise you can
show that Eq. (3.34) gives y(t) ≈ h − gt 2 /2, which is again the freefall result.
We can also look at large t. In this case, e−αt is essentially equal to zero, so
Eq. (3.32) gives v (t) ≈ −g/α. (This is the “terminal velocity.” Its value makes
sense, because it is the velocity for which the total force, −mg − mα v , vanishes.)
And Eq. (3.34) gives y(t) ≈ h − (g/α)t + g/α 2 . Interestingly, we see that for large
t, g/α 2 is the distance our ball lags behind another ball that started out already at the
terminal velocity, −g/α.
2. You might think that the velocity in Eq. (3.32) doesn’t depend on m, since no m’s
appear. However, there is an m hidden in α. The quantity α (which we introduced
just to make our formulas look a little nicer) was defined by Fd = −β v ≡ −mα v .
But the quantity β ≡ mα is roughly proportional to the cross-sectional area, A, of
the ball. Therefore, α ∝ A/m. Two balls of the same size, one made of lead and one
made of styrofoam, have the same A but different m’s. So their α’s are different, and
they fall at different rates.
If we have a solid ball with density ρ and radius r, then α ∝ A/m ∝ r 2 /(ρr 3 ) =
1/ρr. For large dense objects in a thin medium such as air, the quantity α is small,
so the drag effects are not very noticeable over short times (because if we include
the next term in the expansion for v , we obtain v (t) ≈ −gt + αgt 2 /2). Large dense
objects therefore all fall at roughly the same rate, with an acceleration essentially
equal to g. But if the air were much thicker, then all the α’s would be larger, and
maybe it would have taken Galileo a bit longer to come to his conclusions.
What would you have thought, Galileo,
If instead you dropped cows and did say, “Oh!
3.4 Projectile motion
To lessen the sound
Of the moos from the ground,
They should fall not through air, but through mayo!”16
3.4
♣
Projectile motion
Consider a ball thrown through the air, not necessarily vertically. We will neglect
air resistance in the following discussion. Things get a bit more complicated
when this is included, as Exercise 3.53 demonstrates.
Let x and y be the horizontal and vertical positions, respectively. The force
in the x direction is Fx = 0, and the force in the y direction is Fy = −mg. So
F = ma gives
x¨ = 0,
and
y¨ = −g.
(3.35)
Note that these two equations are “decoupled.” That is, there is no mention
of y in the equation for x¨ , and vice versa. The motions in the x and y directions are therefore completely independent. The classic demonstration of the
independence of the x and y motions is the following. Fire a bullet horizontally (or, preferably, just imagine firing a bullet horizontally), and at the same
time drop a bullet from the height of the gun. Which bullet will hit the ground
first? (Neglect air resistance, the curvature of the earth, etc.) The answer is that
they will hit the ground at the same time, because the effect of gravity on the
two y motions is exactly the same, independent of what is going on in the x
direction.
If the initial position and velocity are (X , Y ) and (Vx , Vy ), then we can easily
integrate Eq. (3.35) to obtain
x˙ (t) = Vx ,
and
y˙ (t) = Vy − gt.
(3.36)
Integrating again gives
x(t) = X + Vx t,
and
1
y(t) = Y + Vy t − gt 2 .
2
(3.37)
These equations for the speeds and positions are all you need to solve a projectile
problem.
16
It’s actually much more likely that Galileo obtained his “all objects fall at the same rate in a
vacuum” result by rolling balls down planes than by dropping balls from the Tower of Pisa; see
Adler and Coulter (1978). So I suppose this limerick is relevant only in the approximation of the
proverbial spherical cow.
65
66
Using F =ma
Example (Throwing a ball):
(a) For a given initial speed, at what inclination angle should a ball be thrown
so that it travels the maximum horizontal distance by the time it returns to the
ground? Assume that the ground is horizontal, and that the ball is released from
ground level.
(b) What is the optimal angle if the ground is sloped upward at an angle β (or
downward, if β is negative)?
Solution:
(a) Let the inclination angle be θ, and let the initial speed be V . Then the horizontal
speed is always Vx = V cos θ, and the initial vertical speed is Vy = V sin θ.
The first thing we need to do is find the time t in the air. We know that the
vertical speed is zero at time t/2, because the ball is moving horizontally at
the highest point. So the second of Eqs. (3.36) gives Vy = g(t/2). Therefore,
t = 2Vy /g. 17 The first of Eqs. (3.37) tells us that the horizontal distance
traveled is d = Vx t. Using t = 2Vy /g in this gives
d=
2Vx Vy
V 2 sin 2θ
V 2 (2 sin θ cos θ)
=
.
=
g
g
g
(3.38)
The sin 2θ factor has a maximum at θ = π/4. The maximum horizontal
distance traveled is then dmax = V 2 /g.
Remark: For θ = π/4, you can show that the maximum height achieved is V 2 /4g.
This is half the maximum height of V 2 /2g (as you can show) if the ball is thrown
straight up. Note that any possible distance you might want to find in this problem
must be proportional to V 2 /g, by dimensional analysis. The only question is what
the numerical factor is. ♣
(b) As in part (a), the first thing we need to do is find the time t in the air. If the
ground is sloped at an angle β, then the equation for the line of the ground is
y = (tan β)x. The path of the ball is given in terms of t by
x = (V cos θ)t,
and
1
y = (V sin θ)t − gt 2 ,
2
(3.39)
where θ is the angle of the throw, as measured with respect to the horizontal
(not the ground). We must solve for the t that makes y = (tan β)x, because this
gives the time when the path of the ball intersects the line of the ground. Using
Eq. (3.39), we find that y = (tan β)x when
t=
17
2V
(sin θ − tan β cos θ).
g
(3.40)
Alternatively, the time of flight can be found from the second of Eqs. (3.37), which says that the
ball returns to the ground when Vy t = gt 2 /2. We will have to use this second strategy in part (b),
where the trajectory is not symmetric around the maximum.
3.4 Projectile motion
(There is, of course, also the solution t = 0.) Plugging this into the expression
for x in Eq. (3.39) gives
x=
2V 2
(sin θ cos θ − tan β cos2 θ).
g
(3.41)
We must now maximize this value for x, which is equivalent to maximizing
the distance along the slope. Setting the derivative with respect to θ equal to
zero, and using the double-angle formulas, sin 2θ = 2 sin θ cos θ and cos 2θ =
cos2 θ − sin2 θ , we find tan β = − cot 2θ . This can be rewritten as tan β =
− tan(π/2 − 2θ ). Therefore, β = −(π/2 − 2θ), so we have
θ=
1
π
β+
.
2
2
(3.42)
In other words, the throwing angle should bisect the angle between the ground
and the vertical.
Remarks:
1. For β ≈ π/2, we have θ ≈ π/2, as should be the case. For β = 0, we have
θ = π/4, as we found in part (a). And for β ≈ −π/2, we have θ ≈ 0, which makes
sense.
2. A quicker method of obtaining the time in Eq. (3.40) is the following. Consider the set
of tilted axes parallel and perpendicular to the ground; let these be the x and y axes,
respectively. The initial velocity in the y direction is V sin(θ −β), and the acceleration
in this direction is g cos β. The time in the air is twice the time it takes the ball to reach
the maximum “height” above the ground (measured in the y direction), which occurs
when the velocity in the y direction is instantaneously zero. The total time is therefore
2V sin(θ − β)/(g cos β), which you can show is equivalent to the time in Eq. (3.40).
Note that the g sin β acceleration in the x direction is irrelevant in calculating this
time. In the present example, using these tilted axes doesn’t save a huge amount of
time, but in some situations (see Exercise 3.50) the tilted axes can save you a lot
of grief.
3. An interesting fact about the motion of the ball in the maximum-distance case is that
the initial and final velocities are perpendicular to each other. The demonstration of
this is the task of Problem 3.16.
4. Substituting the value of θ from Eq. (3.42) into Eq. (3.41), you can show (after a bit
of algebra) that the maximum distance traveled along the tilted ground is
d=
V 2 /g
x
=
.
cos β
1 + sin β
(3.43)
Solving for V , we have V 2 = g(d + d sin β). This can be interpreted as saying that
the minimum speed at which a ball must be thrown in order to pass√over a wall of
height h, at a distance L away on level ground, is given by V 2 = g( L2 + h2 + h).
This checks in the limits of h → 0 and L → 0.
5. A compilation of many other projectile results can be found in Buckmaster
(1985). ♣
67
Using F =ma
68
Along with the bullet example mentioned above, another classic example of
the independence of the x and y motions is the “hunter and monkey” problem.
In it, a hunter aims an arrow (a toy one, of course) at a monkey hanging from
a branch in a tree. The monkey, thinking he’s being clever, tries to avoid the
arrow by letting go of the branch right when he sees the arrow released. The
unfortunate consequence of this action is that he in fact will get hit, because
gravity acts on both him and the arrow in the same way; they both fall the same
distance relative to where they would have been if there were no gravity. And
the monkey would get hit in such a case, because the arrow is initially aimed at
him. You can work this out in Exercise 3.44, in a more peaceful setting involving
fruit.
If a monkey lets go of a tree,
The arrow will hit him, you see,
Because both heights are pared
By a half gt 2
From what they would be with no g.
3.5
r
y
Motion in a plane, polar coordinates
When dealing with problems where the motion lies in a plane, it is often convenient to work with polar coordinates, r and θ . These are related to the Cartesian
coordinates by (see Fig. 3.7)
x = r cos θ ,
θ
and
y = r sin θ .
(3.44)
x
Fig. 3.7
Depending on the problem, either Cartesian or polar coordinates are easier to
use. It is usually clear from the setup which is better. For example, if the problem
involves circular motion, then polar coordinates are a good bet. But to use polar
coordinates, we need to know what Newton’s second law looks like when written
in terms of them. Therefore, the goal of the present section is to determine what
F = ma ≡ m¨r looks like when written in terms of polar coordinates.
At a given position r in the plane, the basis vectors in polar coordinates are
ˆ which is a unit
rˆ , which is a unit vector pointing in the radial direction; and θ,
vector pointing in the counterclockwise tangential direction. In polar coordinates,
a general vector may be written as
r = r rˆ .
(3.45)
Since the goal of this section is to find r¨ , we must, in view of Eq. (3.45), get
a handle on the time derivative of rˆ . And we’ll eventually need the derivative
ˆ too. In contrast with the fixed Cartesian basis vectors (ˆx and yˆ ), the polar
of θ,
ˆ change as a point moves around in the plane. We can
basis vectors (ˆr and θ)
3.5 Motion in a plane, polar coordinates
69
find r˙ˆ and θ˙ˆ in the following way. In terms of the Cartesian basis, Fig. 3.8
shows that
rˆ = cos θ xˆ + sin θ yˆ ,
θˆ = − sin θ xˆ + cos θ yˆ .
y
θ
(3.46)
r˙ˆ = − sin θ θ˙ xˆ + cos θ θ˙ yˆ ,
(3.47)
Using Eq. (3.46), we arrive at the nice clean expressions,
ˆ
r˙ˆ = θ˙ θ,
and
˙
θˆ = −θ˙ rˆ .
(3.48)
These relations are fairly evident if you look at what happens to the basis vectors
as r moves a tiny distance in the tangential direction. Note that the basis vectors
do not change as r moves in the radial direction. We can now start differentiating
Eq. (3.45). One derivative gives (yes, the product rule works fine here)
ˆ
r˙ = r˙ rˆ + r r˙ˆ = r˙ rˆ + r θ˙ θ.
(3.49)
This makes sense, because r˙ is the velocity in the radial direction, and r θ˙ is the
velocity in the tangential direction, often written as rω (where ω ≡ θ˙ is the
angular velocity, or “angular frequency”).18 Differentiating Eq. (3.49) then gives
˙
r¨ = r¨ rˆ + r˙ r˙ˆ + r˙ θ˙ θˆ + r θ¨ θˆ + r θ˙ θˆ
ˆ + r˙ θ˙ θˆ + r θ¨ θˆ + r θ(−
˙ θ˙ rˆ )
= r¨ rˆ + r˙ (θ˙ θ)
2
ˆ
= (¨r − r θ˙ )ˆr + (r θ¨ + 2˙r θ˙ )θ.
(3.50)
Finally, equating m¨r with F ≡ Fr rˆ + Fθ θˆ gives the radial and tangential
forces as
2
Fr = m(¨r − r θ˙ ),
Fθ = m(r θ¨ + 2˙r θ˙ ).
(3.51)
(See Exercise 3.67 for a slightly different derivation of these equations.) Let’s
look at each of the four terms on the right-hand sides of Eqs. (3.51).
18
sin θ
r
θ
Taking the time derivative of these equations gives
˙
θˆ = − cos θ θ˙ xˆ − sin θ θ˙ yˆ .
θ
For r θ˙ to be the tangential velocity, we must measure θ in radians and not degrees. Then rθ is by
definition the position along the circumference, so r θ˙ is the velocity along the circumference.
Fig. 3.8
cos θ
x
Using F =ma
70
• The m¨r term is quite intuitive. For radial motion, it simply states that F = ma along the
radial direction.
• The mr θ¨ term is also quite intuitive. For circular motion, it states that F = ma along
the tangential direction, because r θ¨ is the second derivative of the distance rθ along the
circumference.
2
• The −mr θ˙ term is also fairly clear. For circular motion, it says that the radial force
2
is −m(r θ˙ ) /r = −mv 2 /r, which is the familiar force that causes the centripetal acceleration, v 2 /r. See Problem 3.20 for an alternate (and quicker) derivation of this v 2 /r
result.
• The 2m˙r θ˙ term isn’t so obvious. It is associated with the Coriolis force. There are
various ways to look at this term. One is that it exists in order to keep angular momentum
conserved. We’ll have a great deal to say about the Coriolis force in Chapter 10.
β
Example (Circular pendulum): A mass hangs from a massless string of length
. Conditions have been set up so that the mass swings around in a horizontal circle,
with the string making a constant angle β with the vertical (see Fig. 3.9). What is the
angular frequency, ω, of this motion?
l
m
Solution: The mass travels in a circle, so the horizontal radial force must be
Fr = mr θ˙ 2 ≡ mrω2 (with r = sin β), directed radially inward. The forces on
the mass are the tension in the string, T , and gravity, mg (see Fig. 3.10). There is no
acceleration in the vertical direction, so F = ma in the vertical and radial directions
gives, respectively,
Fig. 3.9
T cos β − mg = 0,
T
T sin β = m( sin β)ω2 .
β
(3.52)
Solving for ω gives
ω=
mg
g
.
cos β
(3.53)
√
If β ≈ 90◦ , then ω → ∞, which makes sense. And if β ≈ 0, then ω ≈ g/ , which
happens to equal the frequency of a plane pendulum of length . The task of Exercise
3.60 is to explain why.
Fig. 3.10
3.6
Problems
Section 3.2: Free-body diagrams
m1 m2
Fig. 3.11
3.1. Atwood’s machine *
A massless pulley hangs from a fixed support. A massless string connecting two masses, m1 and m2 , hangs over the pulley (see Fig. 3.11).
Find the acceleration of the masses and the tension in the string.
3.6 Problems
71
3.2. Double Atwood’s machine **
A double Atwood’s machine is shown in Fig. 3.12, with masses m1 , m2 ,
and m3 . Find the accelerations of the masses.
3.3. Infinite Atwood’s machine ***
Consider the infinite Atwood’s machine shown in Fig. 3.13. A string
passes over each pulley, with one end attached to a mass and the other
end attached to another pulley. All the masses are equal to m, and all
the pulleys and strings are massless. The masses are held fixed and then
simultaneously released. What is the acceleration of the top mass? (You
may define this infinite system as follows. Consider it to be made of
N pulleys, with a nonzero mass replacing what would have been the
(N + 1)th pulley. Then take the limit as N → ∞.)
m1
m 2 m3
Fig. 3.12
3.4. Line of pulleys *
N +2 equal masses hang from a system of pulleys, as shown in Fig. 3.14.
What are the accelerations of all the masses?
m
m
...
3.5. Ring of pulleys **
Consider the system of pulleys shown in Fig. 3.15. The string (which
is a loop with no ends) hangs over N fixed pulleys that circle around
the underside of a ring. N masses, m1 , m2 , . . . , mN , are attached to
N pulleys that hang on the string. What are the accelerations of all the
masses?
m
Fig. 3.13
3.6. Sliding down a plane **
(a) A block starts at rest and slides down a frictionless plane inclined
at an angle θ . What should θ be so that the block travels a given
horizontal distance in the minimum amount of time?
(b) Same question, but now let there be a coefficient of kinetic friction
µ between the block and the plane.
3.7. Sliding sideways on a plane ***
A block is placed on a plane inclined at an angle θ . The coefficient of
friction between the block and the plane is µ = tan θ . The block is given
a kick so that it initially moves with speed V horizontally along the plane
(that is, in the direction perpendicular to the direction pointing straight
down the plane). What is the speed of the block after a very long time?
3.8. Moving plane ***
A block of mass m is held motionless on a frictionless plane of mass
M and angle of inclination θ (see Fig. 3.16). The plane rests on a frictionless horizontal surface. The block is released. What is the horizontal
acceleration of the plane?
N=3
Fig. 3.14
....
....
....
mN
m1
....
m2
Fig. 3.15
m
M
θ
Fig. 3.16
72
Using F =ma
Section 3.3: Solving differential equations
3.9. Exponential force *
A particle of mass m is subject to a force F(t) = ma0 e−bt . The initial
position and speed are zero. Find x(t).
3.10. −kx force **
A particle of mass m is subject to a force F(x) = −kx, with k > 0. The
initial position is x0 , and the initial speed is zero. Find x(t).
3.11. Falling chain **
A chain with length is held stretched out on a frictionless horizontal
table, with a length y0 hanging down through a hole in the table. The
chain is released. As a function of time, find the length that hangs down
through the hole (don’t bother with t after the chain loses contact with
the table). Also, find the speed of the chain right when it loses contact
with the table.19
3.12. Throwing a beach ball ***
A beach ball is thrown upward with initial speed v0 . Assume that the
drag force from the air is Fd = −mαv. What is the speed of the ball,
vf , right before it hits the ground? (An implicit equation is sufficient.)
Does the ball spend more time or less time in the air than it would if it
were thrown in vacuum?
3.13. Balancing a pencil ***
Consider a pencil that stands upright on its tip and then falls over. Let’s
idealize the pencil as a mass m sitting at the end of a massless rod of
length .20
(a) Assume that the pencil makes an initial (small) angle θ0 with the
vertical, and that its initial angular speed is ω0 . The angle will
eventually become large, but while it is small (so that sin θ ≈ θ ),
what is θ as a function of time?
(b) You might think that it should be possible (theoretically, at least)
to make the pencil balance for an arbitrarily long time, by making the initial θ0 and ω0 sufficiently small. However, it turns out
that due to Heisenberg’s uncertainty principle (which puts a constraint on how well we can know the position and momentum of
19
20
Assume that the hole is actually a short frictionless tube bent into a gradual right angle, so that
the chain’s horizontal momentum doesn’t cause it to overshoot the hole. For a description of what
happens in a similar problem when this constraint is removed, see Calkin (1989).
It actually involves only a trivial modification to do the problem correctly using the moment of
inertia and the torque. But the point-mass version is quite sufficient for the present purposes.
3.6 Problems
a particle), it is impossible to balance the pencil for more than a
certain amount of time. The point is that you can’t be sure that
the pencil is initially both at the top and at rest. The goal of this
problem is to be quantitative about this. The time limit is sure to
surprise you.
Without getting into quantum mechanics, let’s just say that
the uncertainty principle says (up to factors of order 1) that
x p ≥ , where = 1.05 · 10−34 J s is Planck’s constant.
The implications of this are somewhat vague, but we’ll just take
it to mean that the initial conditions satisfy ( θ0 )(m ω0 ) ≥ .
With this constraint, your task is to find the maximum time it
can take your θ (t) solution in part (a) to become of order 1. In
other words, determine (roughly) the maximum time the pencil
can balance. Assume m = 0.01 kg, and = 0.1 m.
Section 3.4: Projectile motion
3.14. Maximum trajectory area *
A ball is thrown at speed v from zero height on level ground. At
what angle should it be thrown so that the area under the trajectory
is maximum?
3.15. Bouncing ball *
A ball is thrown straight upward so that it reaches a height h. It falls
down and bounces repeatedly. After each bounce, it returns to a certain
fraction f of its previous height. Find the total distance traveled, and
also the total time, before it comes to rest. What is its average speed?
3.16. Perpendicular velocities **
In the maximum-distance case in part (b) of the example in Section
3.4, show that the initial and final velocities are perpendicular to each
other.21
3.17. Throwing a ball from a cliff **
A ball is thrown with speed v from the edge of a cliff of height h. At
what inclination angle should it be thrown so that it travels the maximum
horizontal distance? What is this maximum distance? Assume that the
ground below the cliff is horizontal.
21
You can grind through this problem and explicitly find the final angle, but there’s a quicker way.
This quicker method makes use of the conservation-of-energy statement that the difference in the
squares of the initial and final speeds depends only on the change in height (the relation happens
to be vi2 − vf2 = 2gh, but you don’t need this actual expression). Hint: Consider the reverse path.
73
Using F =ma
74
3.18. Redirected motion **
A ball is dropped from rest at height h above level ground, and it bounces
off a surface at height y (with no loss in speed). The surface is inclined
so that the ball bounces off at an angle θ with respect to the horizontal.
What should y and θ be so that the ball travels the maximum horizontal
distance by the time it hits the ground?
3.19. Maximum trajectory length ***
A ball is thrown at speed v from zero height on level ground. Let θ0 be
the angle at which the ball should be thrown so that the length of the
trajectory is maximum. Show that θ0 satisfies
sin θ0 ln
1 + sin θ0
cos θ0
= 1.
(3.54)
You can show numerically that θ0 ≈ 56.5◦ .
Section 3.5: Motion in a plane, polar coordinates
3.20. Centripetal acceleration *
Show that the magnitude of the acceleration of a particle moving in a circle at constant speed is v 2 /r. Do this by drawing the position and velocity
vectors at two nearby times, and then making use of some similar
triangles.
3.21. Vertical acceleration **
A bead rests at the top of a fixed frictionless hoop of radius R that lies in
a vertical plane. The bead is given a tiny push so that it slides down and
around the hoop. At what points on the hoop is the bead’s acceleration
vertical?22 What is this vertical acceleration? Note: We haven’t studied
conservation of energy yet, but use the fact that the bead’s speed after it
has fallen a height h is given by v = 2gh.
(top view)
3.22. Circling around a pole **
A mass, which is free to move on a horizontal frictionless surface, is
attached to one end of a massless string that wraps partially around a
frictionless vertical pole of radius r (see the top view in Fig. 3.17). You
hold on to the other end of the string. At t = 0, the mass has speed v0 in
the tangential direction along the dotted circle of radius R shown. Your
task is to pull on the string so that the mass keeps moving along the
r
R
hand
Fig. 3.17
22
One such point is the bottom of the hoop. Another point is technically the top, where a ≈ 0. Find
the other two more interesting points (one on each side).
3.7 Exercises
75
dotted circle. You are required to do this in such a way that the string
remains in contact with the pole at all times. (You will have to move
your hand around the pole, of course.) What is the speed of the mass as
a function of time? There is a special value of the time; what is it and
why is it special?
3.23. A force Fθ = m˙r θ˙ **
Consider a particle that√feels an angular force only, of the form Fθ =
˙ Show that r˙ = A ln r + B, where A and B are constants of
m˙r θ.
integration, determined by the initial conditions. (There’s nothing all
that physical about this force. It simply makes the F = ma equations
solvable.)
3.7
Exercises
m
m/2
m/4
...
3.24. Free particle ***
Consider a free particle in a plane. With Cartesian coordinates, it is easy
to use F = ma to show that the particle moves in a straight line. The
task of this problem is to demonstrate this result in a much more cumbersome way, using polar coordinates and Eq. (3.51). More precisely,
show that cos θ = r0 /r for a free particle, where r0 is the radius at
closest approach to the origin, and θ is measured with respect to this
radius.
m/2 n-1
?
Fig. 3.18
Section 3.2: Free-body diagrams
3.25. A peculiar Atwood’s machine
(a) The Atwood’s machine in Fig. 3.18 consists of n masses, m, m/2,
m/4, . . . , m/2n−1 . All the pulleys and strings are massless. Put a
mass m/2n−1 at the free end of the bottom string. What are the
accelerations of all the masses?
(b) Remove the mass m/2n−1 (which was arbitrarily small, for very
large n) that was attached in part (a). What are the accelerations
of all the masses, now that you’ve removed this infinitesimal
piece?
M
m1
m2
Fig. 3.19
3.26. Keeping the mass still *
In the Atwood’s machine in Fig. 3.19, what should M be, in terms of
m1 and m2 , so that it doesn’t move?
3.27. Atwood’s 1 *
Consider the Atwood’s machine in Fig. 3.20. It consists of three pulleys,
a short piece of string connecting one mass to the bottom pulley, and
a continuous long piece of string that wraps twice around the bottom
m
2m
Fig. 3.20
Using F =ma
76
side of the bottom pulley, and once around the top side of the top two
pulleys. The two masses are m and 2m. Assume that the parts of the string
connecting the pulleys are essentially vertical. Find the accelerations of
the masses.
3.28. Atwood’s 2 *
Consider the Atwood’s machine in Fig. 3.21, with two masses m. The
axle of the bottom pulley has two string ends attached to it, as shown.
Find the accelerations of the masses.
m
m
3.29. Atwood’s 3 *
Consider the Atwood’s machine in Fig. 3.22, with masses m, 2m, and
3m. Find the accelerations of the masses.
Fig. 3.21
m
3m
3.31. Atwood’s 5 **
Consider the Atwood’s machine in Fig. 3.24. The two shaded pulleys
have mass m, and the string slides frictionlessly along all of the pulleys
(so you don’t have to worry about any rotational motion). Find the
accelerations of the two shaded pulleys.
2m
Fig. 3.22
m
3.32. Atwood’s 6 **
Consider the Atwood’s machine in Fig. 3.25. Find the accelerations of
the masses. (This is a strange one.)
3.33. Accelerating plane **
A block of mass m rests on a plane inclined at an angle θ . The coefficient
of static friction between the block and the plane is µ. The plane is
accelerated to the right with acceleration a (which may be negative);
see Fig. 3.26. For what range of a does the block remain at rest with
respect to the plane? In terms of µ, there are two special values of θ;
what are they, and why are they special?
m
Fig. 3.23
m
m
Fig. 3.24
3.30. Atwood’s 4 **
Consider theAtwood’s machine in Fig. 3.23 (and also on the front cover).
If the number of pulleys that have string passing beneath them is N
instead of the 3 shown, find the accelerations of the masses.
3.34. Accelerating cylinders **
Three identical cylinders are arranged in a triangle as shown in Fig. 3.27,
with the bottom two lying on the ground. The ground and the cylinders
are frictionless. You apply a constant horizontal force (directed to the
right) on the left cylinder. Let a be the acceleration you give to the
system. For what range of a will all three cylinders remain in contact
with each other?
3.7 Exercises
3.35. Leaving the sphere **
A small mass rests on top of a fixed sphere of radius R. The coefficient
of friction is µ. The mass is given a sideways kick that produces an
initial angular speed ω0 . Let θ be the angle down from the top of the
sphere. In terms of θ and its derivatives, what is the tangential F = ma
equation? Depending on the value of ω0 , the mass either comes to rest
on the sphere or flies off it. If g = 10 m/s2 , R = 1 m, and µ = 1,
write a program to numerically determine the minimum ω0 for which
the mass leaves the sphere. For this cutoff case, give the angle at which
the mass loses contact, and describe (roughly) what the plot of θ˙ versus
θ looks like. See Prior and Mele (2007) for the exact solution for θ˙ in
terms of θ.
77
m
m
Fig. 3.25
3.36. Comparing the times ***
A block of mass m is projected up along the surface of a plane inclined
at an angle θ. The initial speed is v0 , and the coefficients of static and
kinetic friction are both equal to µ. The block reaches a highest point
and then slides back down to the starting point.
(a) Show that for the block to in fact slide back down instead of
remaining at rest at the highest point, tan θ must be greater than µ.
(b) Assuming that tan θ > µ, is the total up and down time longer
or shorter than the total time it would take if the plane were
frictionless? Or does the answer depend on what θ and µ are?
(c) Assuming that tan θ > µ, show that for a given θ, the value of µ
that yields the minimum total time is given by µ ≈ (0.397) tan θ .
(You will need to solve something numerically.) This minimum
time turns out to be about 90% of the time it would take if the
plane were frictionless.
a
3.38. kx force **
A particle of mass m is subject to a force F(x) = kx, with k > 0. The
initial position is x0 , and the initial speed is zero. Find x(t).
Section 3.4: Projectile motion
3.39. Equal distances *
At what angle should a ball be thrown so that its maximum height equals
the horizontal distance traveled?
m
θ
Fig. 3.26
Section 3.3: Solving differential equations
3.37. −bv2 force *
A particle of mass m is subject to a force F(v) = −bv 2 . The initial
position is zero, and the initial speed is v0 . Find x(t).
µ
F
Fig. 3.27
Using F =ma
78
3.40. Redirected motion *
A ball is dropped from rest at height h. At height y, it bounces off a
surface with no loss in speed. The surface is inclined at 45◦ , so the ball
bounces off horizontally. What should y be so that the ball travels the
maximum horizontal distance? What is this maximum distance?
3.41. Throwing in the wind *
A ball is thrown horizontally to the right, from the top of a vertical
cliff of height h. A wind blows horizontally to the left, and assume
(simplistically) that the effect of the wind is to provide a constant force
to the left, equal in magnitude to the weight of the ball. How fast should
the ball be thrown so that it lands at the foot of the cliff?
3.42. Throwing in the wind again *
A ball is thrown eastward across level ground. A wind blows horizontally
to the east, and assume (simplistically) that the effect of the wind is to
provide a constant force to the east, equal in magnitude to the weight of
the ball. At what angle θ should the ball be thrown so that it travels the
maximum horizontal distance?
3.43. Increasing gravity *
At t = 0 on the planet Gravitus Increasicus, a projectile is fired with
speed v0 at an angle θ above the horizontal. This planet is a strange
one, in that the acceleration due to gravity increases linearly with time,
starting with a value of zero when the projectile is fired. In other words,
g(t) = βt, where β is a given constant. What horizontal distance does
the projectile travel? What should θ be to maximize this distance?
3.44. Newton’s apple *
Newton is tired of apples falling on his head, so he decides to throw a
rock at one of the larger and more formidable-looking apples positioned
directly above his favorite sitting spot. Forgetting all about his work on
gravitation, he aims the rock directly at the apple (see Fig. 3.28). To his
surprise, the apple falls from the tree just as he releases the rock. Show,
by calculating the rock’s height when it reaches the horizontal position
of the apple, that the rock hits the apple.23
Fig. 3.28
u
v
3.45. Colliding projectiles *
Two balls are fired from ground level, a distance d apart. The right one is
fired vertically with speed v (see Fig. 3.29). You wish to simultaneously
d
Fig. 3.29
23
This problem suggests a way in which William Tell and his son might survive their ordeal if they
were plopped down, with no time to practice, on a planet with an unknown gravitational constant
(provided that the son weren’t too short or that g weren’t too big).
3.7 Exercises
79
fire the left one at the appropriate velocity u so that it collides with the
right ball when they reach their highest point. What should u be (give
the horizontal and vertical components)? Given d, what should v be so
that the speed u is minimum?
3.46. Equal tilts *
A plane tilts down at an angle θ below the horizontal. On this plane, a
projectile is fired with speed v at an angle θ above the horizontal, as
shown in Fig. 3.30. What is the distance, d, along the plane that the
projectile travels? What is d in the limit θ → 90◦ ? What θ yields the
maximum horizontal distance?
3.47. Throwing at a wall *
You throw a ball with speed v0 at a vertical wall, a distance away. At
what angle should you throw the ball so that it hits the wall as high as
possible? Assume < v02 /g (why?).
3.48. Firing a cannon **
A cannon, when aimed vertically, is observed to fire a ball to a maximum
height of L. Another ball is then fired with this same speed, but with the
cannon aimed up along a plane of length L, inclined at an angle θ, as
shown in Fig. 3.31. What should θ be so that the ball travels the largest
horizontal distance, d, by the time it returns to the height of the top of
the plane?
3.49. Perpendicular and horizontal **
A plane is inclined at an angle θ below the horizontal. A person throws
a ball with speed v0 from the surface of the plane. How far down along
the plane does the ball hit, if the person throws the ball (a) perpendicular
to the plane? (b) horizontally?
3.50. Cart, ball, and plane **
A cart is held at rest on an inclined plane. A tube is positioned in the
cart with its axis perpendicular to the plane. The cart is released, and
at some later time a ball is fired from the tube. Will the ball eventually land back in the tube? Hint: Choose your coordinate system
wisely.
3.51. Perpendicular to plane **
A hill is sloped downward at an angle β with respect to the horizontal.
A projectile is fired with an initial velocity perpendicular to the hill.
When it eventually lands on the hill, let its velocity make an angle θ
with respect to the horizontal. What is θ? What β yields the minimum
value of θ ? What is this minimum θ ?
v
θ
θ
d
Fig. 3.30
L
θ
Fig. 3.31
d
Using F =ma
80
3.52. Increasing distance **
(a) What is the maximum angle at which you can thrown a ball so
that its distance from you never decreases during its flight?
(b) This maximum angle equals the minimum θ from Exercise 3.51.
Explain why this is true. (It isn’t necessary to have done that
exercise.)
3.53. Projectile with drag ***
A ball is thrown with speed v0 at an angle θ . Let the drag force from the
air take the form Fd = −βv ≡ −mαv.
(a) Find x(t) and y(t).
(b) Assume that the drag coefficient takes the value that makes the
magnitude of the initial drag force equal to the weight of the
ball. If your goal is to have x be as large as possible when
y achieves its maximum value (you don’t care what this maximum
value actually is), show that θ should satisfy sin θ =
√
( 5 − 1)/2, which just happens to be the inverse of the golden
ratio.
Section 3.5: Motion in a plane, polar coordinates
3.54. Low-orbit satellite
What is the speed of a satellite whose orbit is just above the earth’s
surface? Give the numerical value.
3.55. Weight at the equator *
A person stands on a scale at the equator. If the earth somehow stopped
spinning but kept its same shape, would the reading on the scale increase
or decrease? By what fraction?
ω
θ
Fig. 3.32
R
3.56. Banking an airplane *
An airplane flies at speed v in a horizontal circle of radius R. At what
angle should the plane be banked so that you don’t feel like you are
getting flung to the side in your seat? At this angle, what is your apparent
weight (that is, what is the normal force from the seat)?
3.57. Rotating hoop *
A bead lies on a frictionless hoop of radius R that rotates around a vertical
diameter with constant angular frequency ω, as shown in Fig. 3.32. What
should ω be so that the bead maintains the same position on the hoop,
at an angle θ with respect to the vertical? There is a special value of ω;
what is it, and why is it special?
3.7 Exercises
81
3.58. Swinging in circles *
A large number of masses are attached by strings of various lengths to a
point on the ceiling. All of the masses swing around in horizontal circles
of various radii with the same frequency ω (one such circle is drawn
in Fig. 3.33). If you take a picture (from the side) of the setup at an
instant when all the masses lie in the plane of the paper (as shown for
four masses), what does the “curve” formed by the masses look like?
3.59. Swinging triangle *
Two masses m are attached to two vertices of an equilateral triangle
made of three massless rods of length . A pivot is located at the third
vertex, and the triangle is free to swing back and forth in a vertical plane,
as shown in Fig. 3.34. If it is initially released from rest when one of the
rods is vertical (as shown), find the tensions in all three rods (and specify
tension or compression), and also the accelerations of the masses, at the
instant right after it is released.
3.60. Circular and plane pendulums *
Consider the circular pendulum in the example in Section 3.5. Let the
x-y plane be the horizontal plane of the circle. For small β, what (approximately) is the Fx component of the force on the mass when it is at the
position (x, y) on the circle?
Consider now a standard plane pendulum where the mass swings back
and forth in the vertical plane containing the x axis. Let the maximum
angle of the pendulum be the same small angle β. 24 What (approximately) is the Fx component of the force on the mass in terms of its x
coordinate?
Your two results should be the same, which means that the x motions
of the two systems are the same, because when they are each at their
maximum x value ( sin β), they have the same x speed (zero), and we
just showed that they always have the same x acceleration, independent
of any y motion. So the frequencies of the two systems must be equal.
√
(We’ll see in Section 4.2 that the frequency of a plane pendulum is g/ ,
in agreement with this observation.)
3.61. Rolling wheel *
If you paint a dot on the rim of a rolling wheel, the coordinates of the
dot may be written as25
(x, y) = (Rθ + R sin θ, R + R cos θ).
24
25
(3.55)
It’s actually not required that this be the same angle, as long as it’s small. See Problem 3.10.
This follows from writing (x, y) as (Rθ , R) + (R sin θ, R cos θ ). The first term here is the position
of the center of the wheel, and the second term is the position of the dot relative to the center,
where θ is measured clockwise from the top.
ω
Fig. 3.33
pivot
l
g
l
m
l
m
Fig. 3.34
Using F =ma
82
The path of the dot is called a cycloid. Assume that the wheel is rolling
at constant speed, which implies θ = ωt.
(a) Find v(t) and a(t) of the dot.
(b) At the instant the dot is at the top of the wheel, what is the radius
of curvature of its path? The radius of curvature is defined to be
the radius of the circle that matches up with the path locally at a
given point. Hint: You know v and a.
3.62. Radius of curvature **
A projectile is fired at speed v0 and angle θ . What is the radius of
curvature (defined in Exercise 3.61) of the parabolic motion
(a) at the top?
(b) at the beginning?
(c) At what angle should the projectile be fired so that the radius of
curvature at the top equals half the maximum height, as shown in
Fig. 3.35?
Fig. 3.35
3.63. Driving on tilted ground **
A driver encounters a large tilted parking lot, where the angle of the
ground with respect to the horizontal is θ . The driver wishes to drive in
a circle of radius R at constant speed. The coefficient of friction between
the tires and the ground is µ.
side point
θ
(a) What is the largest speed the driver can have if he wants to avoid
slipping?
(b) What is the largest speed the driver can have, assuming he is
concerned only with whether or not he slips at one of the “side”
points on the circle (that is, halfway between the top and bottom
points; see Fig. 3.36)?
Fig. 3.36
3.64. Car on a banked track **
A car travels around a circular banked track of radius R. The
angle of the bank is θ , and the coefficient of friction between the
tires and the track is µ. For what range of speeds does the car
not slip?
3.65. Horizontal acceleration **
A bead rests at the top of a fixed frictionless hoop of radius R that lies in
a vertical plane. The bead is given a tiny push so that it slides down and
around the hoop. At what points on the hoop is the bead’s acceleration
horizontal? Note: We haven’t studied conservation of energy yet, but
use the fact that the bead’s speed after it has fallen a height h is given
by v = 2gh.
3.7 Exercises
83
3.66. Maximum horizontal force **
A bead rests at the top of a fixed frictionless hoop of radius R that
lies in a vertical plane. The bead is given a tiny push so that it slides
down and around the hoop. Consider the horizontal component of the
force from the hoop on the bead. At what points on the hoop does this
component achieve a local maximum or minimum? As in Exercise 3.65,
use v = 2gh.
3.67. Derivation of Fr and Fθ *
In Cartesian coordinates, a general vector takes the form,
r = xxˆ + yyˆ = r cos θ xˆ + r sin θ yˆ .
(3.56)
Derive Eq. (3.51) by taking two derivatives of this expression for r, and
then using Eq. (3.46) to show that the result can be written in the form
ˆ the vectors xˆ and yˆ do not change
of Eq. (3.50). Note that unlike rˆ and θ,
with time.
3.68. A force Fθ = 3m˙r θ˙ **
Consider a particle that feels
√ an angular force only, of the form Fθ =
3m˙r θ˙ . Show that r˙ = ± Ar 4 + B, where A and B are constants of
integration, determined by the initial conditions. Also, show that if the
particle starts with θ˙ = 0 and r˙ > 0, it reaches r = ∞ in a finite time.
(As in Problem 3.23, there’s nothing all that physical about this force.
It simply makes the F = ma equations solvable.)
3.69. A force Fθ = 2m˙r θ˙ **
Consider a particle that feels an angular force only, of the form Fθ =
2m˙r θ˙ . Show that r = Aeθ + Be−θ , where A and B are constants of
integration, determined by the initial conditions. (This force is actually
a physical one. If you put a bead on a stick and swing the stick around
one end at a constant rate, then the normal force from the stick happens
to be 2m˙r θ˙ .26 )
3.70. Stopping on a cone **
When viewed from the side, the cone in Fig. 3.37 subtends an angle 2θ
at its tip. A block of mass m is connected to the tip by a massless string
and moves in a horizontal circle of radius R around the surface. If the
26
Depending on what is meant by “physical,” the forces in Exercise 3.68 and Problem 3.23 might
also be considered to be physical. They correspond to putting a bead on a stick and swinging the
stick around with angular speeds proportional to the bead’s r or 1/r, respectively (as is evident
from the values of θ˙ in the solutions). This can also be deduced from τ = dL/dt, but we won’t
get to torque until Chapter 8.
R
Fig. 3.37
m
84
Using F =ma
initial speed is v0 , and if the coefficient of kinetic friction between the
block and the cone is µ, how much time does it take the block to stop?
(The answer is a little messy, but there are some limits you can check
that will make you feel better about it.)
3.71. Motorcycle circle ***
A motorcyclist wishes to travel in a circle of radius R on level ground.
The coefficient of friction between the tires and the ground is µ. The
motorcycle starts at rest. What is the minimum distance it must travel in
order to achieve its maximum allowable speed, that is, the speed above
which it will skid out of the circular path?27 Solve this in two ways:
(a) Write down the radial and tangential F = ma equations (you’ll
want to write a as v dv/dx), and then demand that the magnitude
of the friction force equals µmg in the optimal case. Take it from
there.
(b) Let the friction force make an angle β(t) with respect to the tangential direction. Write down the radial and tangential F = ma
equations (you’ll want to write a as dv/dt), and then take the
derivative of the radial equation. Take it from there (this is the
slick way).
3.8
Solutions
3.1. Atwood’s machine
Let T be the tension in the string, and let a be the acceleration of m1 (with upward
taken to be positive). Then −a is the acceleration of m2 . So the F = ma equations are
T − m1 g = m1 a,
and
T − m2 g = m2 (−a).
(3.57)
Solving these two equations for a and T gives
a=
(m2 − m1 )g
,
m2 + m1
and
T =
2m1 m2 g
.
m2 + m1
(3.58)
Remarks: As a double-check, a has the correct limits when m2
m1 , m1
m2 ,
and m2 = m1 (namely a ≈ g, a ≈ −g, and a = 0, respectively). As far as T goes, if
m2 , then T ≈ 2m1 g. This is
m1 = m2 ≡ m, then T = mg, as it should. And if m1
correct, because it makes the net upward force on m1 equal to m1 g, which means that
its acceleration is g upward, which is consistent with the fact that m2 is essentially in
free fall. ♣
3.2. Double Atwood’s machine
Let the tension in the lower string be T . Then the tension in the upper string is 2T (by
balancing the forces on the bottom pulley). The three F = ma equations are therefore
(with all the a’s taken to be positive upward)
2T − m1 g = m1 a1 ,
27
T − m2 g = m2 a2 ,
T − m3 g = m3 a3 .
This problem can be traced to an old edition of the Russian magazine Kvant.
(3.59)
3.8 Solutions
And conservation of string says that the acceleration of m1 is
a1 = −
a2 + a 3
2
.
(3.60)
This follows from the fact that the average position of m2 and m3 moves the same
distance as the bottom pulley, which in turn moves the same distance (but in the
opposite direction) as m1 . We now have four equations in the four unknowns, a1 , a2 ,
a3 , and T . With a little work, we can solve for the accelerations,
a1 = g
4m2 m3 − m1 (m2 + m3 )
,
4m2 m3 + m1 (m2 + m3 )
a2 = −g
4m2 m3 + m1 (m2 − 3m3 )
,
4m2 m3 + m1 (m2 + m3 )
a3 = −g
4m2 m3 + m1 (m3 − 3m2 )
.
4m2 m3 + m1 (m2 + m3 )
(3.61)
Remarks: There are many limits we can check here. A couple are: (1) If m2 = m3 =
m1 /2, then all the a’s are zero, which is correct. (2) If m3 is much less than both m1 and
m2 , then a1 = −g, a2 = −g, and a3 = 3g. To understand this 3g, convince yourself
that if m1 and m2 go down by d, then m3 goes up by 3d.
Note that a1 can be written as
a1 = g
4m2 m3
− m1
m2 + m3
4m2 m3
+ m1 .
m2 + m3
(3.62)
In view of the result for a in Eq. (3.58) in Problem 3.1, we see that as far as m1 is
concerned here, the m2 , m3 pulley system acts just like a mass of 4m2 m3 /(m2 + m3 ).
This has the expected properties of equaling zero when either m2 or m3 is zero, and
♣
equaling 2m if m2 = m3 ≡ m.
3.3. Infinite Atwood’s machine
First solution: If the strength of gravity on the earth were multiplied by a factor
η, then the tension in all of the strings in the Atwood’s machine would likewise be
multiplied by η. This is true because the only way to produce a quantity with the units
of tension (that is, force) is to multiply a mass by g. Conversely, if we put the Atwood’s
machine on another planet and discover that all of the tensions are multiplied by η,
then we know that the gravity there must be ηg.
Let the tension in the string above the first pulley be T . Then the tension in the string
above the second pulley is T /2 (because the pulley is massless). Let the downward
acceleration of the second pulley be a2 . Then the second pulley effectively lives in a
world where gravity has strength g − a2 . Consider the subsystem of all the pulleys
except the top one. This infinite subsystem is identical to the original infinite system
of all the pulleys. Therefore, by the arguments in the above paragraph, we must have
T /2
T
=
,
g
g − a2
(3.63)
which gives a2 = g/2. But a2 is also the acceleration of the top mass, so our answer
is g/2.
Remarks: You can show that the relative acceleration of the second and third pulleys
is g/4, and that of the third and fourth is g/8, etc. The acceleration of a mass far down
in the system therefore equals g(1/2 + 1/4 + 1/8 + · · · ) = g, which makes intuitive
sense.
Note that T = 0 also makes Eq. (3.63) true. But this corresponds to putting a mass
of zero at the end of a finite pulley system (see the following solution). ♣
85
Using F =ma
86
Second solution:
as
as
m
m1
m2
Consider the following auxiliary problem.
Problem: Two setups are shown in Fig. 3.38. The first contains a hanging mass m.
The second contains two masses, m1 and m2 , hanging over a pulley. Let both supports
have acceleration as downward. What should m be, in terms of m1 and m2 , so that the
tension T in the top string is the same in both cases?
Answer:
In the first case, we have
mg − T = mas .
Fig. 3.38
(3.64)
In the second case, let a be the acceleration of m2 relative to the support (with downward
taken to be positive). Then we have
T
T
m1 g − = m1 (as − a), and m2 g − = m2 (as + a).
(3.65)
2
2
If we define g ≡ g − as , then we may write the above three equations as
T
T
mg = T , m1 g − = −m1 a, m2 g − = m2 a.
(3.66)
2
2
Eliminating a from the last two of these equations gives T = 4m1 m2 g /(m1 + m2 ).
Using this value of T in the first equation then gives
4m1 m2
m=
.
(3.67)
m1 + m2
Note that the value of as is irrelevant. We effectively have a fixed support in a world
where the acceleration due to gravity is g (see Eq. (3.66)), and the desired m can’t
depend on g , by dimensional analysis. This auxiliary problem shows that for any as
the two-mass system in the second case can equivalently be treated like a mass m,
given by Eq. (3.67), as far as the upper string is concerned.
Now let’s look at our infinite Atwood’s machine. Assume that the system has N pulleys,
where N → ∞. Let the bottom mass be x. Then the auxiliary problem shows that the
bottom two masses, m and x, can be treated like an effective mass f (x), where
4mx
4x
f (x) =
=
.
(3.68)
m+x
1 + (x/m)
We can then treat the combination of the mass f (x) and the next m as an effective
mass f (f (x)). These iterations can be repeated, until we finally have a mass m and a
mass f (N −1) (x) hanging over the top pulley. So we must determine the behavior of
f N (x), as N → ∞. This behavior is clear if we look at the plot of f (x) in Fig. 3.39.
We see that x = 3m is a fixed point of f (x). That is, f (3m) = 3m. This plot shows
that no matter what x we start with, the iterations approach 3m (unless we start at
x = 0, in which case we remain there). These iterations are shown graphically by the
directed lines in the plot. After reaching the value f (x) on the curve, the line moves
horizontally to the x value of f (x), and then vertically to the value f (f (x)) on the curve,
and so on. Therefore, since f N (x) → 3m as N → ∞, our infinite Atwood’s machine
is equivalent to (as far as the top mass is concerned) just two masses, m and 3m. You
can then quickly show that the acceleration of the top mass is g/2. Note that as far
as the support is concerned, the whole apparatus is equivalent to a mass 3m. So 3mg
is the upward force exerted by the support.
3.4. Line of pulleys
Let m be the common mass, and let T be the tension in the string. Let a be the
acceleration of the end masses, and let a be the acceleration of the other N masses,
with upward taken to be positive. These N accelerations are indeed all equal, because
the same net force acts on all of the internal N masses, namely 2T upwards and mg
downwards. The F = ma equations for the end and internal masses are, respectively,
T − mg = ma,
and
2T − mg = ma .
(3.69)
3.8 Solutions
y
y=x
4m
3m
y = f (x)
2m
m
x
2m
m
3m
4m
5m
Fig. 3.39 Problem 3.3, second solution
But the string has fixed length. Therefore,
N (2a ) + a + a = 0.
(3.70)
The “2” here comes from the fact that if one of the inside masses goes up by a distance
d, then a length 2d of string has disappeared and must therefore appear somewhere
else (namely, in the outer two segments). Eliminating T from Eq. (3.69) gives a =
2a + g. Combining this with Eq. (3.70) then gives
a=−
Ng
,
2N + 1
and
a =
g
.
2N + 1
(3.71)
Remarks: For N = 1, we have a = −g/3 and a = g/3. For larger N , a increases
in magnitude and approaches −g/2 as N → ∞, and a decreases in magnitude and
approaches zero as N → ∞. The signs of a and a in Eq. (3.71) may be surprising.
You might think that if, say, N = 100, then these 100 masses will “win” out over the
two end masses, so that the N masses will fall. But this is not correct, because there
are many (2N , in fact) tensions acting up on the N masses. They do not behave like
a mass Nm hanging below one pulley. In fact, two masses of m/2 on the ends will
balance any number N of masses m in the interior (with the help of the upward forces
from the top row of pulleys).
♣
3.5. Ring of pulleys
Let T be the tension in the string. Then F = ma for mi gives
2T − mi g = mi ai ,
(3.72)
with upward taken to be positive. The ai ’s are related by the fact that the string has
fixed length, which implies that the sum of the displacements of all the masses is zero.
In other words,
a1 + a2 + · · · + aN = 0.
(3.73)
87
88
Using F =ma
If we divide Eq. (3.72) by mi , and then add the N such equations together and use
Eq. (3.73), we find that T is given by
2T
1
1
1
+
+ ··· +
m1
m2
mN
− Ng = 0.
(3.74)
Therefore,
T =
NMg
,
2
1
1
1
1
≡
+
+ ··· +
M
m1
m2
mN
where
(3.75)
is the so-called reduced mass of the system. Substituting this value for T into (3.72)
gives
ai = g
NM
−1 .
mi
(3.76)
Remark: A few special cases are: If all the masses are equal, then all the ai = 0.
If mk = 0 (and all the others are not zero), then ak = (N − 1)g, and all the other
ai = −g. If N − 1 of the masses are equal and much smaller than the remaining one,
mk , then mk ≈ −g, and all the other ai ≈ g/(N − 1). ♣
3.6. Sliding down a plane
(a) The component of gravity along the plane is g sin θ . The acceleration in the
horizontal direction is therefore ax = (g sin θ ) cos θ. Our goal is to maximize
ax . By taking the derivative, or by noting that sin θ cos θ = (sin 2θ )/2, we obtain
θ = π/4. The maximum ax is then g/2.
(b) The normal force from the plane is mg cos θ , so the kinetic friction force is
µmg cos θ . The acceleration along the plane is therefore g(sin θ − µ cos θ ), and
so the acceleration in the horizontal direction is ax = g(sin θ − µ cos θ ) cos θ .
We want to maximize this. Setting the derivative equal to zero gives
(cos2 θ − sin2 θ ) + 2µ sin θ cos θ = 0 =⇒ cos 2θ + µ sin 2θ = 0
=⇒ tan 2θ = −
1
.
µ
(3.77)
For µ → 0, this gives the π/4 result in part (a). For µ → ∞, we obtain θ ≈ π/2,
which makes sense.
Remarks: The time to travel a horizontal distance d is obtained from ax t 2 /2 =
d. In part (a), this gives a minimum time of 2 d/g. In part (b), you can show
that the maximum ax is (g/2) 1 + µ2 − µ , which then leads to a minimum
1/2
time of 2 d/g 1 + µ2 + µ
. This has the correct µ → 0 limit, and it
behaves like 2 2µd/g for µ → ∞. ♣
3.7. Sliding sideways on a plane
The normal force from the plane is N = mg cos θ . Therefore, the friction force on the
block is µN = (tan θ )(mg cos θ ) = mg sin θ. This force acts in the direction opposite
to the motion. The block also feels the gravitational force of mg sin θ pointing down
the plane.
Because the magnitudes of the friction force and the gravitational force along the
plane are equal, the acceleration along the direction of motion equals the negative of
the acceleration in the direction down the plane. Therefore, in a small increment of
time, the speed that the block loses along its direction of motion exactly equals the
speed that it gains in the direction down the plane. Letting v be the total speed of the
3.8 Solutions
block, and letting vy be the component of the velocity in the direction down the plane,
we therefore have
v + vy = C,
(3.78)
where C is a constant. C is given by its initial value, which is V + 0 = V . The final
value of C is Vf + Vf = 2Vf (where Vf is the final speed of the block), because the
block is essentially moving straight down the plane after a very long time. Therefore,
2Vf = V
=⇒
Vf =
V
.
2
(3.79)
3.8. Moving plane
Let N be the normal force between the block and the plane. Note that we cannot assume
that N = mg cos θ, because the plane recoils. We can see that N = mg cos θ is in fact
incorrect, because in the limiting case where M = 0, we have no normal force at all.
The various F = ma equations (vertical and horizontal for the block, and horizontal
for the plane) are
mg − N cos θ = may ,
N sin θ = max ,
(3.80)
N sin θ = MAx ,
where we have chosen the positive directions for ay , ax , and Ax to be downward,
rightward, and leftward, respectively. There are four unknowns here: ax , ay , Ax , and
N , so we need one more equation. This fourth equation is the constraint that the block
remains in contact with the plane. The horizontal distance between the block and its
starting point on the plane is (ax + Ax )t 2 /2, and the vertical distance is ay t 2 /2. The
ratio of these distances must equal tan θ if the block is to remain on the plane (imagine
looking at things in the frame of the plane). Therefore, we must have
ay
= tan θ.
(3.81)
ax + Ax
Using Eq. (3.80) to solve for ay , ax , and Ax in terms of N , and then plugging the results
into Eq. (3.81), gives
g−
N
m
N
m
sin θ +
cos θ
N
M
sin θ
= tan θ
=⇒
N = g sin θ tan θ
1
1
+
m
M
+
cos θ
m
−1
.
(3.82)
(In the limit M → ∞, this reduces to N = mg cos θ , as it should.) Having found N ,
the third of Eqs. (3.80) gives Ax , which may be written as
Ax =
mg sin θ cos θ
N sin θ
.
=
M
M + m sin2 θ
(3.83)
Remarks:
1. √
For given M and m, you can show that the angle θ0 that maximizes Ax is tan θ0 =
M /(M + m). If M
m, then θ0 ≈ 0; this makes sense, because the plane gets
squeezed out very fast. If M
m, then θ0 ≈ π/4; this is consistent with the π/4
result from Problem 3.6(a).
2. In the limit M
m, Eq. (3.83) gives Ax ≈ g/ tan θ. This makes sense, because
m falls essentially straight down with acceleration g, and the plane gets squeezed
out to the left.
3. In the limit M
m, Eq. (3.83) gives Ax ≈ g(m/M ) sin θ cos θ. The correctness
of this is more transparent if we instead look at ax = (M /m)Ax ≈ g sin θ cos θ.
Since the plane is essentially at rest in this limit, this value of ax implies that the
acceleration of m along the plane equals ax / cos θ ≈ g sin θ , as expected. ♣
89
90
Using F =ma
3.9. Exponential force
F = ma gives x¨ = a0 e−bt . Integrating this with respect to time gives v (t) =
−a0 e−bt/b + A. Integrating again gives x(t) = a0 e−bt /b2 + At + B. The initial condition v (0) = 0 gives −a0 /b + A = 0 =⇒ A = a0 /b. And the initial condition x(0) = 0
gives a0 /b2 + B = 0 =⇒ B = −a0 /b2 . Therefore,
x(t) = a0
e−bt
1
t
+ − 2
b2
b
b
.
(3.84)
For t → ∞ (more precisely, for bt → ∞), v approaches a0 /b, and x approaches
a0 (t/b−1/b2 ). We see that the particle eventually lags a distance a0 /b2 behind another
particle that starts at the same position but moves with the constant speed v = a0 /b.
For t ≈ 0 (more precisely, for bt ≈ 0), we can expand e−bt in its Taylor series to
obtain x(t) ≈ a0 t 2 /2. This makes sense, because the exponential factor in the force is
essentially equal to 1, so we essentially have a constant force with constant acceleration.
3.10. −kx force
This is simply a Hooke’s-law spring force, which we’ll see much more of in Chapter 4.
F = ma gives −kx = mv d v /dx. Separating variables and integrating yields
−
x
v
kx dx =
x0
mv d v
1 2 1 2
1
kx − kx = mv 2 .
2 0 2
2
=⇒
0
(3.85)
Solving for v ≡ dx/dt and then separating variables and integrating again gives
x
x0
dx
x02
− x2
t
=±
0
k
dt.
m
(3.86)
You can look up this integral, or you can solve it with a trig substitution. Letting
x ≡ x0 cos θ gives dx = −x0 sin θ dθ , and so we have
θ
0
−x0 sin θ dθ
k
=±
t
x0 sin θ
m
=⇒
k
t.
m
θ =∓
(3.87)
From the definition of θ, the solution for x(t) is therefore
x(t) = x0 cos
k
t .
m
(3.88)
We see that the particle oscillates back and forth sinusoidally. It completes a full
oscillation when the argument of the cosine increases by 2π. So the period of the
motion is T = 2π m/k, which interestingly is independent of x0 . It increases with m
and decreases with k, as expected.
3.11. Falling chain
Let the density of the chain be ρ, and let y(t) be the length hanging down through the
hole at time t. Then the total mass is ρ , and the mass hanging below the hole is ρy.
The net downward force on the chain is (ρy)g, so F = ma gives
g
(3.89)
ρgy = (ρ )¨y =⇒ y¨ = y.
At this point, there are two ways we can proceed:
First method: Since we have a function whose second derivative is proportional to
itself, a good bet for the solution is an exponential function. And indeed, a quick check
shows that the solution is
y(t) = Aeαt + Be−αt ,
where α ≡
g
.
(3.90)
3.8 Solutions
Taking the derivative of this to obtain y˙ (t), and using the given information that y˙ (0) =
0, we find A = B. Using y(0) = y0 , we then find A = B = y0 /2. So the length that
hangs below the hole is
y0 αt
e + e−αt ≡ y0 cosh(αt).
2
(3.91)
αy0 αt
e − e−αt ≡ αy0 sinh(αt).
2
(3.92)
y(t) =
And the speed is
y˙ (t) =
The time T that satisfies y(T ) = is given by = y0 cosh(αT ). Using sinh x =
cosh2 x − 1, we find that the speed of the chain right when it loses contact with the
table is
y˙ (T ) = αy0 sinh(αT ) = α
2
− y02 ≡
1 − η02 ,
g
(3.93)
where η0 ≡ y0 / is√the initial fraction hanging below the hole. If η0 ≈ 0, then the
speed at time T is g (this quickly follows from conservation of energy, which is
the subject of Chapter 5). Also, you can show that Eq. (3.91) implies that T goes to
infinity logarithmically as η0 → 0.
Second method: Write y¨ as v d v /dy in Eq. (3.89), and then separate variables and
integrate to obtain
v
y
v d v = α2
y dy
where α ≡
=⇒
v 2 = α 2 (y2 − y02 ),
y0
0
(3.94)
√
g/ . Now write v as dy/dt and separate variables again to obtain
y
dy
y2 − y02
y0
t
=α
dt.
(3.95)
0
The integral on the left-hand side is cosh−1 (y/y0 ), so we arrive at y(t) = y0 cosh(αt),
in agreement with Eq. (3.91). The solution then proceeds as above. However, an easier
way to obtain the final speed with this method is to simply use the result for v in
Eq. (3.94). This tells us that the speed of the chain when it leaves the table (that is,
when y = ) is v = α
2
− y02 , in agreement with Eq. (3.93).
3.12. Throwing a beach ball
On both the way up and the way down, the total force on the ball is
F = −mg − mα v .
(3.96)
On the way up, v is positive, so the drag force points downward, as it should. And on
the way down, v is negative, so the drag force points upward. Our strategy for finding
vf will be to produce two different expressions for the maximum height h, and then
equate them. We’ll find these two expressions by considering the upward and then
the downward motion of the ball. In doing so, we will need to write the acceleration
of the ball as a = v d v /dy. For the upward motion, F = ma gives
−mg − mα v = mv
dv
dy
h
=⇒
dy = −
0
0
v0
v dv
.
g + αv
(3.97)
where we have taken advantage of the fact that the speed of the ball at the top is zero.
Writing v /(g + α v ) as [1 − g/(g + α v )]/α, the integral yields
h=
v0
α
−
g
α v0
ln 1 +
α2
g
.
(3.98)
91
92
Using F =ma
Now consider the downward motion. Let vf be the final speed, which is a positive
quantity. The final velocity is then the negative quantity, −vf . Using F = ma, we
obtain
0
−vf
dy = −
h
0
v dv
.
g + αv
(3.99)
Performing the integration (or just replacing the v0 in Eq. (3.98) with −vf ) gives
h=−
vf
α
−
g
α vf
ln 1 −
α2
g
.
(3.100)
Equating the expressions for h in Eqs. (3.98) and (3.100) gives an implicit equation
for vf in terms of v0 ,
v0 + vf =
g + α v0
g − α vf
g
ln
α
.
(3.101)
Remarks: In the limit of small α (more precisely, in the limit α v0 /g
1), we can
use ln(1 + x) = x − x2 /2 + · · · to obtain approximate values for h in Eqs. (3.98) and
(3.100). The results are, as expected,
h≈
v02
,
2g
and
h≈
vf2
2g
.
(3.102)
We can also make approximations for large α (or large α v0 /g). In this limit, the
log term in Eq. (3.98) is negligible, so we obtain h ≈ v0 /α. And Eq. (3.100) gives
vf ≈ g/α, because the argument of the log must be very small in order to give a very
large negative number, which is needed to produce a positive h on the left-hand side.
There is no way to relate vf and h in this limit, because the ball quickly reaches the
terminal velocity of −g/α (which is the velocity that makes the net force equal to zero),
independent of h. ♣
Let’s now find the times it takes for the ball to go up and to go down. We’ll present
two methods for doing this.
First method: Let T1 be the time for the upward path. If we write the acceleration
of the ball as a = d v /dt, then F = ma gives −mg − mα v = m d v /dt. Separating
variables and integrating yields
T1
0
dt = −
0
v0
dv
g + αv
=⇒
T1 =
1
α v0
ln 1 +
α
g
.
(3.103)
In a similar manner, we find that the time T2 for the downward path is
T2 = −
1
α vf
ln 1 −
α
g
.
(3.104)
Therefore,
T1 + T2 =
1
ln
α
g + α v0
g − α vf
=
v0 + vf
g
,
(3.105)
where we have used Eq. (3.101). This result is shorter than the time in vacuum (namely
2v0 /g) because vf < v0 .
Second method: The very simple form of Eq. (3.105) suggests that there is a cleaner
way of deriving it. And indeed, if we integrate m d v /dt = −mg − mα v with respect to
time on the way up, we obtain −v0 = −gT1 − αh (because v dt = h). Likewise, if
we integrate m d v /dt = −mg − mα v with respect to time on the way down, we obtain
−vf = −gT2 + αh (because v dt = −h). Adding these two results gives Eq. (3.105).
Note that this procedure works only because the drag force is proportional to v .
3.8 Solutions
Remark: The fact that the time here is shorter than the time in vacuum isn’t obvious.
On one hand, the ball doesn’t travel as high in air as it would in vacuum, so you might
think that T1 + T2 < 2v0 /g. But on the other hand, the ball moves slower in air on
the way down, so you might think that T1 + T2 > 2v0 /g. It isn’t obvious which effect
wins, without doing a calculation.28 For any α, you can use Eq. (3.103) to show that
T1 < v0 /g. But T2 is harder to get a handle on, because it is given in terms of vf . But in
the limit of large α, the ball quickly reaches terminal velocity, so we have T2 ≈ h/vf .
Using the results from the previous remark, this becomes T2 ≈ (v0 /α)/(g/α) =
v0 /g. Interestingly, this equals the downward (and upward) time for a ball thrown
in vacuum. ♣
3.13. Balancing a pencil
(a) The component of gravity in the tangential direction is mg sin θ ≈ mgθ. There¨ which may be written as
fore, the tangential F = ma equation is mgθ = m θ,
θ¨ = (g/ )θ . The general solution to this equation is29
θ(t) = Aet/τ + Be−t/τ ,
where τ ≡
/g.
(3.106)
The constants A and B are found from the initial conditions,
θ(0) = θ0
=⇒
A + B = θ0 ,
˙
θ(0)
= ω0
=⇒
(A − B)/τ = ω0 .
(3.107)
Solving for A and B, and then plugging the results into Eq. (3.106) gives
θ(t) =
1
1
(θ0 + ω0 τ ) et/τ + (θ0 − ω0 τ ) e−t/τ .
2
2
(3.108)
√
(b) The constants A and B will turn out to be small (they will each be of order
).
Therefore, by the time the positive exponential has increased enough to make
θ of order 1, the negative exponential will have become negligible. We will
therefore ignore this term from here on. In other words,
1
(3.109)
(θ0 + ω0 τ ) et/τ .
2
The goal is to keep θ small for as long as possible. Hence, we want to minimize
the coefficient of the exponential, subject to the uncertainty-principle constraint,
( θ0 )(m ω0 ) ≥ . This constraint gives ω0 ≥ /(m 2 θ0 ). Therefore,
θ(t) ≈
θ(t) ≥
1
τ
θ0 +
2
m 2 θ0
et/τ .
(3.110)
Taking the derivative with respect to θ0 to minimize the coefficient, we find that
the minimum value occurs at θ0 =
τ/m 2 . Substituting this back into Eq.
(3.110) gives
θ(t) ≥
τ
m
2
et/τ .
(3.111)
Setting θ ≈ 1, and then solving for t gives (using τ ≡
t≤
28
29
1
4
g
ln
m2 3 g
2
.
√
/g)
(3.112)
For a similar setup where it again isn’t obvious (for good reason) which effect wins, see
Exercise 3.36.
If you want, you can derive this by separating variables and integrating. The solution is essentially
the same as in the second method presented in the solution to Problem 3.11.
93
94
Using F =ma
With the given values, m = 0.01 kg and = 0.1 m, along with g = 10 m/s2 and
= 1.06 · 10−34 J s, we obtain
1
t ≤ 0.1 s ln(9 · 1061 ) ≈ 3.5 s.
(3.113)
4
No matter how clever you are, and no matter how much money you spend on
the newest cutting-edge pencil balancing equipment, you can never get a pencil
to balance for more than about four seconds.
Remarks:
1. The smallness of this answer is quite amazing. It is remarkable that a
quantum effect on a macroscopic object can produce an everyday value
for a time scale. Basically, the point is that the fast exponential growth of
θ (which gives rise to the log in the final result for t) wins out over the
smallness of , and produces a result for t of order 1. When push comes to
shove, exponential effects always win.
√
/g term.
2. The above value for t depends strongly on and g, through the
But the dependence on m, , and g in the log term is very weak, because
is so small. If m is increased by a factor of 1000, for example, the result for
t increases by only about 10%. This implies that any factors of order 1 that
we neglected throughout this problem are completely irrelevant. They will
appear in the argument of the log term, and will therefore have negligible
effect.
3. Note that dimensional analysis, which is generally √
a very powerful tool,
/g has dimensions
won’t get you too far in this problem. The quantity
of time, and the quantity η ≡ m2 3 g/ 2 is dimensionless (it is the only
such quantity), so the balancing time must take the form,
t≈
g
f (η),
(3.114)
where f is some function. If the leading term in f were a power (even, for
example, a square root), then t would essentially be infinite (t ≈ 1030 s ≈
1022 years for the square root). But f in fact turns out to be a log (which
you can’t know without solving the problem), which completely cancels
out the smallness of , reducing an essentially infinite time down to a few
seconds. ♣
3.14. Maximum trajectory area
Let θ be the angle at which the ball is thrown. Then the coordinates are given by
x = (v cos θ)t and y = (v sin θ )t − gt 2 /2. The total time in the air is 2(v sin θ )/g, so
the area under the trajectory, A = y dx, is
2v sin θ/g
2v 4
sin3 θ cos θ .
3g 2
0
0
(3.115)
√
Taking the derivative of this, we find that the maximum occurs
when tan θ = 3,
√
that is, when θ = 60◦ . The maximum area is then Amax = 3v 4 /8g 2 . Note that by
dimensional analysis we know that the area, which has dimensions of distance squared,
must be proportional to v 4 /g 2 .
xmax
y dx =
(v sin θ )t −
gt 2
2
(v cos θ dt) =
3.15. Bouncing ball
The ball travels 2h during the first up-and-down journey. It travels 2hf during the
second, then 2hf 2 during the third, and so on. Therefore, the total distance traveled is
D = 2h 1 + f + f 2 + f 3 + · · · =
2h
.
1−f
(3.116)
3.8 Solutions
95
The time it takes to fall down during the first up-and-down is obtained from
h = gt 2 /2. Therefore, the time for the first up-and-down equals 2t = 2 2h/g.
Likewise, the time for the second up-and-down equals 2 2(hf )/g. Each successive
up-and-down time decreases by a factor of f , so the total time is
T =2
2h
2h
1
1 + f 1/2 + f 1 + f 3/2 + · · · = 2
·
.
g
g 1− f
(3.117)
The average speed is therefore
gh/2
D
=
.
T
1+ f
(3.118)
Remark: The average speed for f ≈ 1 is roughly half of the average speed for f ≈ 0.
This may seem counterintuitive, because in the f ≈ 0 case the ball slows down far
more quickly than in the f ≈ 1 case. But the f ≈ 0 case consists of essentially only
one bounce, and the average speed for that one bounce is the largest of any bounce.
Both D and T are smaller for f ≈ 0 than for f ≈ 1, but T is smaller by a larger
factor. ♣
P'''
3.16. Perpendicular velocities
In the maximum-distance case, let vi be the initial speed of the ball, and let vf be the
final speed right before it hits the plane (so vf = vi2 − 2gh, where h is the final height
of the ball). Let the parabolic path be labeled P, and let the beginning and ending points
be A and B, as shown in Fig. 3.40.
Consider the question, “Given an initial speed vf , at what inclination angle should a
ball be thrown down the plane from point B so that it travels the maximum distance?”
The answer is that it should be thrown along the same path P, tracing out the path
backward. This is certainly a possible physical trajectory (reversing time still leads to
a solution to F = ma for projectile motion), and we claim that it does indeed yield the
maximum distance. This can be seen in the following way.
Assume (in search of a contradiction) that with initial speed vf , the maximum distance down the plane is obtained via some other path P that lands farther down the
plane, as shown in Fig. 3.40. Then if we decrease the initial speed vf by an appropriate
amount, we can have the ball land at point A via some path P (not shown, lest the
figure get too cluttered). From the conservation-of-energy result, the final speed at A
in this case is less than vi . But if we simply reverse the motion along path P , we see
that we can get from A to B by using an initial speed that is less than vi . So if we then
increase the speed up to vi , we can hit a point above B on the plane via some other path
P , contradicting the fact that B was the endpoint of the maximum-distance trajectory
starting at A with initial speed vi . This contradiction shows that the maximum distance
down the plane, starting at B with speed vf , must in fact be attained via the path P.
Now, from the example in Section 3.4, we know that in the maximum-distance case,
the throwing angle bisects the angle between the ground and the vertical. We therefore
have the situation shown in Fig. 3.41, because the same path gives the maximum
distance for both the upward and downward throws. Since 2α + 2γ = 180◦ , we see
that α + γ = 90◦ , as we wanted to show.
3.17. Throwing a ball from a cliff
Let the inclination angle be θ . Then the horizontal speed is vx = v cos θ , and the initial
vertical speed is vy = v sin θ . The time it takes for the ball to hit the ground is given
by h + (v sin θ)t − gt 2 /2 = 0. Therefore,
t=
v
g
sin θ +
sin2 θ + β
,
where β ≡
2gh
v2
.
(3.119)
P
B
P'
A
Fig. 3.40
γ
α
α
vi
Fig. 3.41
γ
vf
96
Using F =ma
(The “−” solution for t from the quadratic formula corresponds to the ball being thrown
backward down through the cliff.) The horizontal distance traveled is d = (v cos θ )t,
which gives
d=
v2
sin2 θ + β
cos θ sin θ +
g
.
(3.120)
We want to maximize this function of θ . Taking the derivative, multiplying through
by
sin2 θ + β, and setting the result equal to zero, gives
(cos2 θ − sin2 θ ) sin2 θ + β = sin θ β − (cos2 θ − sin2 θ ) .
cos2 θ
(3.121)
Using
= 1 − sin θ , and then squaring and simplifying this equation, gives an
optimal angle of
2
sin θmax = √
1
≡
2+β
1
.
2 + 2gh/v 2
(3.122)
Plugging this into Eq. (3.120), and simplifying, gives a maximum distance of
dmax =
v2
g
1+β ≡
v2
1+
g
2gh
v2
.
(3.123)
If h = 0, then θmax = π/4 and dmax = v 2 /g, in agreement with the example in Section
3.4. If h → ∞ or v → 0, then θmax ≈ 0, which makes sense.
Remark: If we make use of conservation of energy (discussed in Chapter 5), it turns
out that the final speed of the ball when it hits the ground is vf = v 2 + 2gh. The
maximum distance in Eq. (3.123) can therefore be written as (with vi ≡ v being the
initial speed)
dmax =
vi vf
g
.
(3.124)
Note that this is symmetric in vi and vf , as it must be, because we could imagine
the trajectory running backward. Also, it equals zero if vi is zero, and it reduces to
v 2 /g on level ground, as it should. We can also write the angle θ in Eq. (3.122)
in terms of vf (instead of h). You can show that the result is tan θ = vi /vf . This
implies that the initial and final velocities are perpendicular to each other, because
running the trajectory backward interchanges vi and vf , which means that the product
of the tangents of the two angles equals 1. This is basically the same result as in
Problem 3.16. ♣
3.18. Redirected motion
First solution: We will use the results of Problem 3.17, namely Eqs. (3.123) and
(3.122), which say that an object projected from height y at speed v travels a maximum
horizontal distance of
dmax =
v2
g
1+
2gy
v2
,
(3.125)
.
(3.126)
and the optimal angle yielding this distance is
sin θ =
1
2 + 2gy/v 2
In the problem at hand, the object is dropped from height h, so the kinematic relation
2g(h − y). Plugging this into
Eq. (3.125) tells us that the maximum horizontal distance, as a function of y, is
vf2 = vi2 + 2ad gives the speed at height y as v =
dmax (y) = 2 h(h − y).
(3.127)
3.8 Solutions
This is maximum when y = 0 (assuming we can’t have negative y), in which case the
distance is dmax = 2h. Equation (3.126) gives the associated optimal angle as θ = 45◦ .
Second solution: Assume that the greatest distance d0 is obtained when the surface
is at y = y0 . We will show that y0 must be 0. We will do this by assuming y0 = 0 and
explicitly constructing a situation that yields a greater distance.
Let P be the point where the ball finally hits the ground after it bounces off the
surface at height y0 . Consider a second scenario where a ball is dropped from height
h directly above P. The speed of this ball at P will be the same as the speed of the
original ball at P. This follows from conservation of energy. (Or, since we haven’t
covered energy yet, you can use the kinematic relation vf2 = vi2 + 2ad for the y speeds
of the two balls; apply it in two steps for the first ball.)
Now imagine putting a surface at P at the appropriate angle so that the second ball
bounces off in the direction from which the first ball came. Then the second ball will
travel backward along the parabolic trajectory of the first one. But this means that
after the second ball gets to the location of the platform at y0 (which we have now
removed), it will cover more horizontal distance before it hits the ground. We have
therefore constructed a setup in which the ball travels farther horizontally than in our
proposed maximal case. So the optimal setup must have y0 = 0, in which case the
example in Section 3.4 says that the optimal angle is θ = 45◦ . If we want the ball to
go even farther, we can simply dig a (wide enough) hole in the ground and have the
ball bounce from the bottom of the hole.
3.19. Maximum trajectory length
Let θ be the angle at which the ball is thrown. Then the coordinates are given by
x = (v cos θ)t and y = (v sin θ)t − gt 2 /2. The ball reaches its maximum height at
t = v sin θ/g, so the length of the trajectory is
v sin θ/g
L=2
0
v sin θ/g
=2
dx
dt
2
2
dy
dt
+
dt
(v cos θ)2 + (v sin θ − gt)2 dt
0
v sin θ/g
= 2v cos θ
1 + tan θ −
0
gt
v cos θ
2
dt.
(3.128)
Letting z ≡ tan θ − gt/v cos θ, we obtain
L=−
2v 2 cos2 θ
g
0
tan θ
1 + z 2 dz.
(3.129)
We can look up this integral, or we can derive it by making a z ≡ sinh α substitution.
The result is
L=
=
2v 2 cos2 θ 1
·
z 1 + z 2 + ln z +
g
2
v2
g
sin θ + cos2 θ ln
sin θ + 1
cos θ
1 + z2
tan θ
0
.
(3.130)
As an intermediate check, you can verify that L = 0 when θ = 0, and L = v 2 /g when
θ = 90◦ . Taking the derivative of Eq. (3.130) to find the maximum, we obtain
0 = cos θ − 2 cos θ sin θ ln
+ cos2 θ
cos θ
1 + sin θ
1 + sin θ
cos θ
cos2 θ + (1 + sin θ ) sin θ
.
cos2 θ
(3.131)
97
Using F =ma
98
y
This reduces to
θ = 45
path
.
(3.132)
You can show numerically that the solution for θ is θ0 ≈ 56.5◦ .
x
Fig. 3.42
v2
v1
r2
Remark: A few possible trajectories are shown in Fig. 3.42. Using the standard result
that θ = 45◦ provides the maximum horizontal distance, it follows from the figure
that the θ0 that yields the maximum trajectory length must satisfy θ0 ≥ 45◦ . The exact
angle, however, requires the above detailed calculation. ♣
3.20. Centripetal acceleration
The position and velocity vectors at two nearby times are shown in Fig. 3.43. Their
differences, r ≡ r2 − r1 and v ≡ v2 − v1 , are shown in Fig. 3.44. The angle
between the v’s is the same as the angle between the r’s, because each v makes a right
angle with the corresponding r. Therefore, the triangles in Fig. 3.44 are similar, so we
have
| v|
v
r1
=
| r|
,
r
where r ≡ |r| and v ≡ |v|. Dividing through by
1
v
v
1
=
t
r
r
t
=⇒
|a|
v
=
(3.133)
t gives
|v|
r
=⇒
a=
v2
r
.
(3.134)
We have assumed that t is infinitesimal here, which allows us to get rid of the
in favor of instantaneous quantities.
Fig. 3.43
v2
θ
’s
3.21. Vertical acceleration
Let θ be the angle down from the top of the hoop. The tangential acceleration is
at = g sin θ , and the radial acceleration is ar = v 2 /R = 2gh/R. But the height fallen
is h = R − R cos θ, so we have
∆v
v1
ar =
2gR(1 − cos θ )
= 2g(1 − cos θ ).
R
(3.135)
We want the total acceleration to be vertical, which means that we want the horizontal
components of at and ar in Fig. 3.45 to cancel. That is, at cos θ = ar sin θ. This gives
r2
∆r
θ
(g sin θ) cos θ = 2g(1 − cos θ ) sin θ
=⇒
sin θ = 0 or cos θ = 2/3. (3.136)
The sin θ = 0 root corresponds to the top and bottom of the hoop (θ = 0 and θ = π ).
So we want the cos θ = 2/3 =⇒ θ ≈ ±48.2◦ root. The vertical acceleration is the
sum of the vertical components of at and ar , so
r1
Fig. 3.44
ay = at sin θ + ar cos θ = (g sin θ ) sin θ + 2g(1 − cos θ ) cos θ
R
θ
= g(sin2 θ + 2 cos θ − 2 cos2 θ ).
√
Using cos θ = 2/3, and hence sin θ = 5/3, we have
θ
θ
ar
at
Fig. 3.45
1 + sin θ
cos θ
1 = sin θ ln
ay = g
2
4
5
+2· −2·
9
3
9
= g.
(3.137)
(3.138)
Remark: The reason for this nice answer is the following. If there is no horizontal
acceleration, then the normal force from the hoop must have no horizontal component.
In other words, N sin θ = 0. Therefore, either sin θ = 0 (which gives the top and
bottom solutions of θ = 0 and θ = π), or N = 0, which means that there is no normal
force, so the bead feels only gravity, so it’s in freefall with ay = g.
3.8 Solutions
If we want, we can use this N = 0 requirement as the starting point for a second
solution. Using the ar from above, the radial F = ma equation is mg cos θ − N =
2mg(1 − cos θ), with positive N defined to point outward. Setting N = 0 gives the
desired result, cos θ = 2/3. ♣
3.22. Circling around a pole
Let F be the tension in the string. At the mass, the angle between the string and the
radius of the dotted circle is θ = sin−1 (r/R). In terms of θ , the radial and tangential
F = ma equations are
F cos θ =
mv 2
,
R
F sin θ = mv˙ .
and
(3.139)
Dividing these two equations gives tan θ = (Rv˙ )/v 2 . Separating variables and
integrating gives
v
dv
v0
v2
=
tan θ
R
t
dt
1
=⇒
v0
0
−
1
v
=
1
v (t) =
=⇒
(tan θ )t
R
v0
−
−1
(tan θ )t
R
.
(3.140)
The speed v becomes infinite when
t=T ≡
R
v0 tan θ
.
(3.141)
This means that you can keep the mass moving in the desired circle only up to time
T . After that, it is impossible. (Of course, it will become impossible, for all practical
purposes, long before v becomes infinite.) The total distance, d = v dt, is infinite,
because this integral diverges (barely, like a log) as t approaches T .
3.23. A force Fθ = m˙r θ˙
With the given force, Eq. (3.51) becomes
2
0 = m(¨r − r θ˙ ),
and
m˙r θ˙ = m(r θ¨ + 2˙r θ˙ ).
(3.142)
The second of these equations gives −˙r θ˙ = r θ¨ . Therefore,
θ¨
dt = −
θ˙
r˙
dt
r
=⇒
ln θ˙ = − ln r + C
=⇒
θ˙ =
D
,
r
(3.143)
where D = eC is a constant of integration, determined by the initial conditions. Substituting this value of θ˙ into the first of Eqs. (3.142), and then multiplying through by
r˙ and integrating, gives
r¨ = r
D
r
2
=⇒
Therefore,
r˙ =
where A ≡
2D2
r˙
dt
r
r¨ r˙ dt = D2
=⇒
r˙ 2
= D2 ln r + E. (3.144)
2
√
A ln r + B ,
(3.145)
and B ≡ 2E.
3.24. Free particle
For zero force, Eq. (3.51) gives
2
r¨ = r θ˙ ,
and
r θ¨ = −2˙r θ˙ .
(3.146)
Separating variables in the second equation and integrating yields
θ¨
dt = −
θ˙
2˙r
dt
r
=⇒
ln θ˙ = −2 ln r + C
=⇒
θ˙ =
D
,
r2
(3.147)
99
100
Using F =ma
where D = eC is a constant of integration, determined by the initial conditions.30
Substituting this value of θ˙ into the first of Eqs. (3.146), and then multiplying through
by r˙ and integrating, gives
r¨ = r
2
D
r2
=⇒
r¨ r˙ dt = D2
r˙
dt
r3
r˙ 2
D2
= − 2 + E. (3.148)
2
2r
=⇒
We want r˙ = 0 when r = r0 , which implies that E = D2 /2r02 . Therefore,
r˙ = V 1 −
r02
,
r2
(3.149)
where V ≡ D/r0 . Separating variables and integrating gives
r˙r dt
r2
− r02
=
V dt
=⇒
r 2 − r02 = Vt
=⇒
r=
r02 + (Vt)2 ,
(3.150)
where the constant of integration is zero, because we have chosen t = 0 to correspond
to r = r0 . Plugging this value for r into the θ˙ = D/r 2 ≡ Vr0 /r 2 result in Eq. (3.147)
gives
dθ =
Vr0 dt
r02 + (Vt)2
=⇒
θ = tan−1
Vt
r0
=⇒
cos θ =
r0
r02
.
+ (Vt)2
(3.151)
Finally, combining this with the result for r in Eq. (3.150) gives cos θ = r0 /r, as
desired.
30
The statement that r 2 θ˙ is constant is simply the statement of conservation of angular momentum,
because r 2 θ˙ = r(r θ˙ ) = rvθ , where vθ is the tangential velocity. More on this in Chapters 7
and 8.
Chapter 4
Oscillations
In this chapter we will discuss oscillatory motion. The simplest examples of such
motion are a swinging pendulum and a mass on a spring, but it is possible to make
a system more complicated by introducing a damping force and/or an external
driving force. We will study all of these cases.
We are interested in oscillatory motion for two reasons. First, we study it
because we can study it. This is one of the few systems in physics where we can
solve the motion exactly. There’s nothing wrong with looking under the lamppost
every now and then. Second, oscillatory motion is ubiquitous in nature, for
reasons that will become clear in Section 5.2. If there was ever a type of physical
system worthy of study, this is it. We’ll start off by doing some necessary math
in Section 4.1. And then in Section 4.2 we’ll show how the math is applied to the
physics.
4.1
Linear differential equations
A linear differential equation is one in which x and its time derivatives enter only
through their first powers. An example is 3¨x + 7˙x + x = 0. An example of a
nonlinear differential equation is 3¨x + 7˙x2 + x = 0. If the right-hand side of the
equation is zero, then we use the term homogeneous differential equation. If the
right-hand side is some function of t, as in the case of 3¨x − 4˙x = 9t 2 − 5, then
we use the term inhomogeneous differential equation. The goal of this chapter
is to learn how to solve linear differential equations, both homogeneous and
inhomogeneous. These come up again and again in physics, so we had better
find a systematic method of solving them.
The techniques that we will use are best learned through examples, so let’s
solve a few differential equations, starting with some simple ones. Throughout
this chapter, x is understood to be a function of t. Hence, a dot denotes time
differentiation.
Example 1 (x˙ = ax): This is a very simple differential equation. There are two
ways (at least) to solve it.
101
102
Oscillations
First method: Separate variables to obtain dx/x = a dt, and then integrate to
obtain ln x = at + c, where c is a constant of integration. Then exponentiate to obtain
x = Aeat ,
(4.1)
where A ≡ ec . A is determined by the value of x at, say, t = 0.
Second method: Guess an exponential solution, that is, one of the form x =
Aeαt . Substitution into x˙ = ax immediately gives α = a. Therefore, the solution is
x = Aeat . Note that we can’t solve for A, due to the fact that our differential equation
is homogeneous and linear in x (translation: A cancels out). A is determined by the
initial condition.
This method may seem a bit silly, and somewhat cheap. But as we will see below,
guessing these exponential functions (or sums of them) is actually the most general
thing we can try, so the method is indeed quite general.
Remark: Using this method, you might be concerned that although we have found one
solution, we might have missed another one. But the general theory of differential equations
says that a first-order linear equation has only one independent solution (we’ll just accept
this fact here). So if we find one solution, then we know that we’ve found the whole
thing. ♣
Example 2 (¨x = ax): If a is negative, then we’ll see that this equation describes
the oscillatory motion of, say, a spring. If a is positive, then it describes exponentially
growing or decaying motion. There are two ways (at least) to solve this equation.
First method: We can use the separation-of-variables method from Section 3.3
here, because our system is one in which the force depends only on the position x. But
this method is rather cumbersome, as you found if you did Problem 3.10 or Exercise
3.38. It will certainly work, but in the case where our equation is linear in x, there is
a much simpler method:
Second method: As in the first example above, we can guess a solution of the
form x(t) = Aeαt and then find out what α must be. Again, we can’t solve for A,
√
because it cancels out. Plugging Aeαt into x¨ = ax gives α = ± a. We have therefore
found two solutions. The most general solution is an arbitrary linear combination of
these,
√
√
x(t) = Ae a t + Be− a t ,
(4.2)
which you can quickly check does indeed work. A and B are determined by the initial
conditions. As in the first example above, you might be concerned that although
we have found two solutions to the equation, we might have missed others. But the
general theory of differential equations says that our second-order linear equation has
only two independent solutions. Therefore, having found two independent solutions,
we know that we’ve found them all.
Very Important Remark: The fact that the sum of two different solutions is again a solution to our equation is a monumentally important property of linear differential equations.
4.1 Linear differential equations
This property does not hold for nonlinear differential equations, for example x¨ 2 = bx,
because the act of squaring after adding the two solutions produces a cross term which
destroys the equality, as you should check (see Problem 4.1). This property is called the
principle of superposition. That is, superposing two solutions yields another solution. In
other words, linearity leads to superposition. This fact makes theories that are governed by
linear equations much easier to deal with than those that are governed by nonlinear ones.
General Relativity, for example, is based on nonlinear equations, and solutions to most
General Relativity systems are extremely difficult to come by.
For equations with one main condition
(Those linear), you have permission
To take your solutions,
With firm resolutions,
And add them in superposition. ♣
Let’s say a little more about the solution in Eq. (4.2). If a is negative, then it is
helpful to define a ≡ −ω2 , where ω is a real number. The solution then becomes
x(t) = Aeiωt + Be−iωt . Using eiθ = cos θ + i sin θ , this can be written in terms of
trig functions, if desired. Various ways of writing the solution are:
x(t) = Aeiωt + Be−iωt ,
x(t) = C cos ωt + D sin ωt,
x(t) = E cos(ωt + φ1 ),
(4.3)
x(t) = F sin(ωt + φ2 ).
Depending on the specifics of a given system, one of the above forms will work better
than the others. The various constants in these expressions are related to each other.
For example, C = E cos φ1 and D = −E sin φ1 , which follow from the cosine sum
formula. Note that there are two free parameters in each of the above expressions for
x(t). These parameters are determined by the initial conditions (say, the position and
velocity at t = 0). In contrast with these free parameters, the quantity ω is determined
by the particular physical system we’re dealing with. For example, we’ll see that for
a spring, ω = k/m, where k is the spring constant. ω is independent of the initial
conditions.
If a is positive, then let’s define a ≡ α 2 , where α is a real number. The solution in
Eq. (4.2) then becomes x(t) = Aeαt + Be−αt . Using eθ = cosh θ + sinh θ, this can
be written in terms of hyperbolic trig functions, if desired. Various ways of writing
the solution are:
x(t) = Aeαt + Be−αt ,
x(t) = C cosh αt + D sinh αt,
x(t) = E cosh(αt + φ1 ),
(4.4)
x(t) = F sinh(αt + φ2 ).
Again, the various constants are related to each other. If you are unfamiliar with the
hyperbolic trig functions, a few facts are listed in Appendix A.
103
104
Oscillations
Although the solution in Eq. (4.2) is completely correct for both signs of a, it’s
generally more illuminating to write the negative-a solutions in either the trig forms
or the e±iωt exponential form where the i’s are explicit.
The usefulness of our method of guessing exponential solutions cannot be overemphasized. It may seem somewhat restrictive, but it works. The examples in the
remainder of this chapter should convince you of this.
This is our method, essential,
For equations we solve, differential.
It gets the job done,
And it’s even quite fun.
We just try a routine exponential.
Example 3 (¨x +2γ x˙ + ax = 0): This will be our last mathematical example, and
then we’ll start doing some physics. As we’ll see later, this example pertains to a
damped harmonic oscillator. We’ve put a factor of 2 in the coefficient of x˙ here to
make some later formulas look nicer. The force in this example (if we switch from
math to physics for a moment) is −2γ x˙ − ax (times m), which depends on both v
and x. Our methods of Section 3.3 therefore don’t apply; we’re not going to be able
to use separation of variables here. This leaves us with only our method of guessing
an exponential solution, Aeαt . So let’s see what it gives us. Plugging x(t) = Aeαt into
the given equation, and canceling the nonzero factor of Aeαt , gives
α 2 + 2γ α + a = 0.
The solutions for α are −γ ±
solution to our equation is
(4.5)
γ 2 − a. Call these α1 and α2 . Then the general
x(t) = Aeα1 t + Beα2 t
√
√
2
2
= e−γ t Aet γ −a + Be−t γ −a .
(4.6)
If γ 2 − a < 0, then we can write this in terms of sines and cosines, so we have
oscillatory motion that decreases in time due to the e−γ t factor (or it increases, if
γ < 0, but this is rarely physical). If γ 2 − a > 0, then we have exponential motion.
We’ll talk more about these different possibilities in Section 4.3.
In the first two examples above, the solutions were fairly clear. But in the present
case, you’re not apt to look at the above solution and say, “Oh, of course. It’s obvious!”
So our method of trying solutions of the form Aeαt isn’t looking so silly anymore.
In general, if we have an nth order homogeneous linear differential equation,
d nx
d n−1 x
dx
+
c
+ · · · + c1 + c0 x = 0,
n−1
n
n−1
dt
dt
dt
(4.7)
4.2 Simple harmonic motion
105
then our strategy is to guess an exponential solution, x(t) = Aeαt , and to then (in
theory) solve the resulting nth order equation, α n + cn−1 α n−1 + · · · + c1 α + c0 =
0, for α, to obtain the solutions α1 , . . . , αn . The general solution for x(t) is then
the superposition,
x(t) = A1 eα1 t + A2 eα2 t + · · · + An eαn t ,
(4.8)
where the Ai are determined by the initial conditions. In practice, however, we
will rarely encounter differential equations of degree higher than 2. (Note: if
some of the αi happen to be equal, then Eq. (4.8) is not valid, so a modification
is needed. We will encounter such a situation in Section 4.3.)
4.2
Simple harmonic motion
Let’s now do some real live physical problems. We’ll start with simple harmonic
motion. This is the motion undergone by a particle subject to a force F(x) = −kx.
The classic system that undergoes simple harmonic motion is a mass attached
to a massless spring, on a frictionless table (see Fig. 4.1). A typical spring has
a force of the form F(x) = −kx, where x is the displacement from equilibrium
(see Section 5.2 for the reason behind this). This is “Hooke’s law,” and it holds as
long as the spring isn’t stretched or compressed too far. Eventually this expression
breaks down for any real spring. But if we assume a −kx force, then F = ma
gives −kx = m¨x, or
x¨ + ω2 x = 0,
where ω ≡
k
.
m
(4.9)
This is simply the equation we studied in Example 2 in the previous section.
From Eq. (4.3), the solution may be written as
x(t) = A cos(ωt + φ).
(4.10)
This trig solution shows that the system oscillates back and forth forever in time.
ω is the angular frequency. If t increases by 2π/ω, then the argument of the
cosine increases by 2π , so the position and velocity are back to what they were.
The period (the time for one complete cycle) is therefore T = 2π/ω = 2π m/k.
The frequency in cycles per second (hertz) is ν = 1/T = ω/2π . The constant
A (or rather the absolute value of A, if A is negative) is the amplitude, that is,
the maximum distance the mass gets from the origin. Note that the velocity as a
function of time is v(t) ≡ x˙ (t) = −Aω sin(ωt + φ).
The constants A and φ are determined by the initial conditions. If, for example,
x(0) = 0 and x˙ (0) = v, then we must have A cos φ = 0 and −Aω sin φ = v.
Hence, φ = π/2, and so A = −v/ω (or φ = −π/2 and A = v/ω, but this
leads to the same solution). Therefore, we have x(t) = −(v/ω) cos(ωt + π/2).
This looks a little nicer if we write it as x(t) = (v/ω) sin(ωt). It turns out that
k
m
Fig. 4.1
106
Oscillations
if the facts you’re given are the initial position and velocity, x0 and v0 , then the
x(t) = C cos ωt + D sin ωt expression in Eq. (4.3) usually works best, because
(as you can verify) it yields the nice clean results, C = x0 and D = v0 /ω.
Problem 4.3 gives another setup that involves initial conditions.
θ
l
m
Example (Simple pendulum): Another classic system that undergoes (approximately) simple harmonic motion is the simple pendulum, that is, a mass that hangs
on a massless string and swings in a vertical plane. Let be the length of the string,
and let θ (t) be the angle the string makes with the vertical (see Fig. 4.2). Then the
gravitational force on the mass in the tangential direction is −mg sin θ. So F = ma
in the tangential direction gives
Fig. 4.2
¨
−mg sin θ = m( θ).
(4.11)
The tension in the string combines with the radial component of gravity to produce the
radial acceleration, so the radial F = ma equation serves only to tell us the tension,
which we won’t need here.
We will now enter the realm of approximations and assume that the amplitude of
the oscillations is small. Without this approximation, the problem cannot be solved
in closed form. Assuming θ is small, we can use sin θ ≈ θ in Eq. (4.11) to obtain
θ¨ + ω2 θ = 0,
where ω ≡
g
.
(4.12)
Therefore,
θ(t) = A cos(ωt + φ),
(4.13)
where A and φ are determined by the initial conditions. So the pendulum undergoes
√
simple harmonic motion with a frequency of g/ . The period is therefore T =
√
2π/ω = 2π /g. The true motion is arbitrarily close to this, for sufficiently small
amplitudes. Exercise 4.23 deals with the higher-order corrections to the motion in the
case where the amplitude is not small.
There will be many occasions throughout your physics education where you
will plow through a calculation and then end up with a simple equation of the
form z¨ + ω2 z = 0, where ω2 is a positive quantity that depends on various
parameters in the problem. When you encounter such an equation, you should
jump for joy, because without any more effort you can simply write down the
answer: the solution for z must be of the form z(t) = A cos(ωt + φ). No matter
how complicated the system looks at first glance, if you end up with an equation
that looks like z¨ + ω2 z = 0, then you know that the system undergoes simple
harmonic motion with a frequency equal to the square root of the coefficient of z,
no matter what that coefficient
is. If you end up with z¨ + (zucchini)z = 0, then
√
the frequency is ω = zucchini (well, as long as the zucchini is positive and has
the dimensions of inverse time squared).
4.3 Damped harmonic motion
4.3
107
Damped harmonic motion
Consider a mass m attached to the end of a spring with spring constant k. Let
the mass be subject to a drag force proportional to its velocity, Ff = −bv (the
subscript f here stands for “friction”; we’ll save the letter d for “driving” in the
next section); see Fig. 4.3. Why do we study this Ff = −bv damping force? Two
reasons: First, it is linear in x, which will allow us to solve for the motion. And
second, it is a perfectly realistic force; an object moving at a slow speed through
a fluid generally experiences a drag force proportional to its velocity. Note that
this Ff = −bv force is not the force that a mass would feel if it were placed on
a table with friction. In that case the drag force would be (roughly) constant.
Our goal in this section is to solve for the position as a function of time. The
total force on the mass is F = −b˙x − kx. So F = m¨x gives
x¨ + 2γ x˙ + ω2 x = 0,
k
v
m
Ff
Fig. 4.3
(4.14)
where 2γ ≡ b/m, and ω2 ≡ k/m. This is conveniently the same equation we
already solved in Example 3 in Section 4.1 (with a → ω2 ). Now, however, we
have the physical restrictions that γ > 0 and ω2 > 0. Letting 2 ≡ γ 2 − ω2 for
simplicity, we may write the solution in Eq. (4.6) as
x(t) = e−γ t Ae
t
+ Be−
t
,
where
≡
γ 2 − ω2 .
(4.15)
There are three cases to consider.
Case 1: Underdamping (
2
< 0)
If 2 < 0, then γ < ω. Since is imaginary, let us define the real number
ω˜ ≡ ω2 − γ 2 , so that = iω.
˜ Equation (4.15) then gives
˜
˜
+ Be−iωt
x(t) = e−γ t Aeiωt
≡ e−γ t C cos(ωt
˜ + φ).
(4.16)
These two forms are equivalent. Using eiθ = cos θ + i sin θ , the constants in
Eq. (4.16) are related by A + B = C cos φ and A − B = iC sin φ. Note that in a
physical problem, x(t) is real, so we must have A∗ = B, where the star denotes
complex conjugation. The two constants A and B, or the two constants C and φ,
are determined by the initial conditions.
Depending on the given problem, one of the expressions in Eq. (4.16) will
inevitably work better than the other. Or perhaps one of the other forms in Eq. (4.3)
(times e−γ t ) will be the most useful one. The cosine form makes it apparent that
the motion is harmonic motion whose amplitude decreases in time, due to the
e−γ t factor. A plot of such motion is shown in Fig. 4.4.1 The frequency of the
1
To be precise, the amplitude doesn’t decrease exactly like Ce−γ t , as Eq. (4.16) suggests, because
Ce−γ t describes the envelope of the motion, and not the curve that passes through the extremes of
x
Ce
t
x(t)
Fig. 4.4
108
Oscillations
motion, ω˜ =
oscillator.
ω2 − γ 2 , is smaller than the natural frequency ω of the undamped
Remarks: If γ is very small (more precisely, if γ
ω), then ω˜ ≈ ω, which makes sense
because we almost have an undamped oscillator. If γ is very close to ω, then ω˜ ≈ 0. So the
oscillations are very slow (more precisely, ω˜
ω). Of course, for very small ω˜ it’s hard to even
tell that the oscillations exist, because they will damp out on a time scale of order 1/γ ≈ 1/ω,
which is short compared with the long time scale of the oscillations, 1/ω.
˜
♣
x
Case 2: Overdamping (
x(t)
If
2
> 0, then γ > ω.
2
> 0)
is real (and taken to be positive), so Eq. (4.15) gives
x(t) = Ae−(γ −
t
Fig. 4.5
)t
+ Be−(γ +
)t
.
(4.17)
There is no oscillatory motion in this case; see Fig. 4.5. Since γ >
≡
γ 2 − ω2 , both of the exponents are negative, so the motion goes to zero for
large t. This had better be the case, because a real spring is certainly not going
to have the motion fly off to infinity. If we had obtained a positive exponent
somehow, we’d know we had made a mistake.
Remarks: If γ is just slightly larger than ω, then
≈ 0, so the two terms in (4.17) are
ω
roughly equal, and we essentially have exponential decay, according to e−γ t . If γ
(that is, strong damping), then
≈ γ , so the first term in (4.17) dominates (it has the less
negative exponent), and we essentially have exponential decay according to e−(γ − )t . We can
be somewhat quantitative about this by approximating as
≡
γ 2 − ω2 = γ 1 − ω2 /γ 2 ≈ γ (1 − ω2 /2γ 2 ).
(4.18)
Hence, the exponential behavior goes like e−ω t/2γ . Because γ
ω, this is slow decay (that
is, slow compared with t ∼ 1/ω), which makes sense if the damping is very strong. The mass
slowly creeps back to the origin, as in the case of a weak spring immersed in molasses. ♣
2
Case 3: Critical damping (
2
= 0)
If 2 = 0, then γ = ω. Equation (4.14) therefore becomes x¨ + 2γ x˙ + γ 2 x = 0.
In this special case, we have to be careful in solving our differential equation. The
solution in Eq. (4.15) is not valid, because in the procedure leading to Eq. (4.6),
the roots α1 and α2 are equal (to −γ ), so we have really found only one solution,
e−γ t . We’ll just invoke here the result from the theory of differential equations
that says that in this special case, the other solution is of the form te−γ t .
Remark: You should check explicitly that te−γ t solves the equation x¨ + 2γ x˙ + γ 2 x = 0. Or if
you want to, you can derive it in the spirit of Problem 4.2. In the more general case where there
are n identical roots in the procedure leading to Eq. (4.8) (call them all α), the n independent
solutions to the differential equation are t k eαt , for 0 ≤ k ≤ (n − 1). But more often than not,
there are no repeated roots, so you don’t have to worry about this. ♣
the motion. You can show that the amplitude in fact decreases like Ce−γ t cos(tan−1 (γ /ω)).
˜ This
is the expression for the curve that passes through the extremes; see Castro (1986). But for small
damping (γ
ω), this is essentially equal to Ce−γ t . And in any event, it is proportional to e−γ t .
4.4 Driven (and damped) harmonic motion
Our solution is therefore of the form,
x(t) = e−γ t (A + Bt).
x
(4.19)
The exponential factor eventually wins out over the Bt term, so the motion goes
to zero for large t (see Fig. 4.6).
If we are given a spring with a fixed ω, and if we look at the system for
different values of γ , then critical damping (when γ = ω) is the case where
the motion converges to zero in the quickest way (which is like e−ωt ). This is
true because in the underdamped case (γ < ω), the envelope of the oscillatory
motion goes like e−γ t , which goes to zero slower than e−ωt , because γ < ω.
And in the overdamped case (γ > ω), the dominant piece is the e−(γ − )t term.
And as you can verify, if γ > ω then γ − ≡ γ − γ 2 − ω2 < ω, so this
motion also goes to zero slower than e−ωt . Critical damping is very important in
many real systems, such as screen doors and shock absorbers, where the goal is
to have the system head to zero (without overshooting and bouncing around) as
fast as possible.
4.4
Driven (and damped) harmonic motion
Before we examine driven harmonic motion, we must learn how to solve a new
type of differential equation. How can we solve something of the form
x¨ + 2γ x˙ + ax = C0 eiω0 t ,
(4.20)
where γ , a, ω0 , and C0 are given quantities? This is an inhomogeneous differential
equation, due to the term on the right-hand side. It’s not very physical, because
the right-hand side is complex, but let’s not worry about that for now. Equations
of this sort come up again and again, and fortunately there’s a straightforward
(although sometimes messy) method for solving them. As usual, the method
involves making a reasonable guess, plugging it in, and seeing what condition
comes out. Since we have the eiω0 t sitting on the right-hand side of Eq. (4.20),
let’s guess a solution of the form x(t) = Aeiω0 t . A will depend on ω0 , among
other things, as we will see. Plugging this guess into Eq. (4.20) and canceling the
nonzero factor of eiω0 t , we obtain
(−ω02 )A + 2γ (iω0 )A + aA = C0 .
(4.21)
Solving for A, we find that our solution for x is
x(t) =
−ω02
C0
eiω0 t .
+ 2iγ ω0 + a
109
(4.22)
Note the differences between this technique and the one in Example 3 in
Section 4.1. In that example, the goal was to determine the α in x(t) = Aeαt .
x(t) = e-γt (A+Bt)
t
Fig. 4.6
110
Oscillations
And there was no way to solve for A; the initial conditions determined A. But in
the present technique, the ω0 in x(t) = Aeiω0 t is a given quantity, and the goal
is to solve for A in terms of the given constants. Therefore, in the solution in
Eq. (4.22), there are no free constants to be determined by the initial conditions.
We’ve found one particular solution, and we’re stuck with it. The term particular
solution is used for Eq. (4.22).
With no freedom to adjust the solution in Eq. (4.22), how can we satisfy an
arbitrary set of initial conditions? Fortunately, Eq. (4.22) does not represent the
most general solution to Eq. (4.20). The most general solution is the sum of
our particular solution in Eq. (4.22), plus the “homogeneous” solution we found
in Eq. (4.6). This sum is certainly a solution, because the solution in Eq. (4.6)
was explicitly constructed to yield zero when plugged into the left-hand side of
Eq. (4.20). Therefore, tacking it on to our particular solution doesn’t change the
equality in Eq. (4.20), because the left side is linear. The principle of superposition
has saved the day. The complete solution to Eq. (4.20) is therefore
x(t) = e−γ t Aet
√
γ 2 −a
+ Be−t
√
γ 2 −a
+
−ω02
C0
eiω0 t ,
+ 2iγ ω0 + a
(4.23)
where A and B are determined by the initial conditions.
With superposition in mind, it’s clear what the strategy should be if we have
a slightly more general equation to solve, for example,
x¨ + 2γ x˙ + ax = C1 eiω1 t + C2 eiω2 t .
(4.24)
Simply solve the equation with only the first term on the right. Then solve the
equation with only the second term on the right. Then add the two solutions. And
then add on the homogeneous solution from Eq. (4.6). We are able to apply the
principle of superposition because the left-hand side of Eq. (4.24) is linear.
Finally, let’s look at the case where we have many such terms on the right-hand
side, for example,
N
x¨ + 2γ x˙ + ax =
Cn eiωn t .
(4.25)
n=1
We need to solve N different equations, each with only one of the N terms on
the right-hand side. Then we add up all the solutions, and then we add on the
homogeneous solution from Eq. (4.6). If N is infinite, that’s fine; we just have
to add up an infinite number of solutions. This is the principle of superposition
at its best.
Remark: The previous paragraph, combined with a basic result from Fourier analysis, allows
us to solve (in principle) any equation of the form
x¨ + 2γ x˙ + ax = f (t).
(4.26)
4.4 Driven (and damped) harmonic motion
111
Fourier analysis says that any (nice enough) function f (t) can be decomposed into its Fourier
components,
f (t) =
∞
−∞
g(ω)eiωt dω.
(4.27)
In this continuous sum, the functions g(ω) (times dω) take the place of the coefficients Cn in
Eq. (4.25). So if Sω (t) is the solution for x(t) when there is only the term eiωt on the right-hand
side of Eq. (4.26) (that is, Sω (t) is the solution given in Eq. (4.22), without the C0 factor), then
the principle of superposition tells us that the complete particular solution to Eq. (4.26) is
x(t) =
∞
−∞
g(ω)Sω (t) dω.
(4.28)
Finding the coefficients g(ω) is the hard part (or rather, the messy part), but we won’t get into
that here. We won’t do anything with Fourier analysis in this book, but it’s nevertheless good
to know that it is possible to solve (4.26) for any function f (t). Most of the functions we’ll
consider will be nice functions like cos ω0 t, which has a very simple Fourier decomposition,
namely cos ω0 t = 12 (eiω0 t + e−iω0 t ). ♣
Let’s now do a physical example.
v
k
Example (Damped and driven spring): Consider a spring with spring constant
k. A mass m at the end of the spring is subject to a drag force proportional to its
velocity, Ff = −bv. The mass is also subject to a driving force, Fd (t) = Fd cos ωd t
(see Fig. 4.7). What is its position as a function of time?
Solution: The force on the mass is F(x, x˙ , t) = −b˙x −kx+Fd cos ωd t. So F = ma
gives
x¨ + 2γ x˙ + ω2 x = F cos ωd t
=
F iωd t
e
+ e−iωd t .
2
(4.29)
where 2γ ≡ b/m, ω2 ≡ k/m, and F ≡ Fd /m. Note that there are two different
frequencies here, ω and ωd , which need not have anything to do with each other.
Equation (4.22), along with the principle of superposition, tells us that our particular
solution is
xp (t) =
F/2
−ωd2 + 2iγ ωd + ω2
eiωd t +
F/2
−ωd2 − 2iγ ωd + ω2
e−iωd t . (4.30)
The complete solution is the sum of this particular solution and the homogeneous
solution from Eq. (4.15).
Let’s now eliminate the i’s in Eq. (4.30) (which we had better be able to do,
because x must be real), and write x in terms of sines and cosines. Getting the i’s
out of the denominators (by multiplying both the numerator and the denominator by
the complex conjugate of the denominator), and using eiθ = cos θ + i sin θ, we find,
m
Ff
Fig. 4.7
Fd (t)
112
Oscillations
after a little work,
⎛
⎜
xp (t) = ⎝
⎞
F ω2 − ωd2
ω2 − ωd2
2
+ 4γ 2 ωd2
⎛
⎟
⎜
⎠ cos ωd t + ⎝
⎞
2Fγ ωd
ω2 − ωd2
2
+ 4γ 2 ωd2
⎟
⎠ sin ωd t.
(4.31)
Remarks: If you want, you can solve Eq. (4.29) by taking the real part of the solution to
Eq. (4.20) (with C0 → F), that is, the x(t) in Eq. (4.22). This is true because if we take
the real part of Eq. (4.20), we obtain
d
d2
Re(x) + 2γ
Re(x) + a Re(x) = Re C0 eiω0 t
dt 2
dt
= C0 cos(ω0 t).
(4.32)
In other words, if x satisfies Eq. (4.20) with a C0 eiω0 t on the right-hand side, then Re(x)
satisfies it with a C0 cos(ω0 t) on the right. At any rate, it’s clear that the real part of the
solution in Eq. (4.22) (with C0 → F) does indeed give the result in Eq. (4.31), because
in Eq. (4.30) we simply took half of a quantity plus its complex conjugate, which is the
real part.
If you don’t like using complex numbers, another way of solving Eq. (4.29) is to keep it in
the form with the cos ωd t on the right, and guess a solution of the form A cos ωd t+B sin ωd t,
and then solve for A and B (this is the task of Problem 4.8). The result is Eq. (4.31). ♣
We can now write Eq. (4.31) in a very simple form. If we define
ω2 − ωd2
R≡
2
+ (2γ ωd )2 ,
(4.33)
then we can rewrite Eq. (4.31) as
xp (t) =
≡
F
R
ω2 − ωd2
R
cos ωd t +
2γ ωd
sin ωd t
R
F
cos(ωd t − φ),
R
(4.34)
R
=
(ω 2
-ω 2
d )2
+ (
2γ
ω
d
)2
where φ (the phase) is defined by
φ
ω2-ωd2
cos φ =
ω2 − ωd2
R
2γωd
,
sin φ =
2γ ωd
R
=⇒
tan φ =
2γ ωd
ω2 − ωd2
.
(4.35)
The triangle describing the angle φ is shown in Fig. 4.8. Note that 0 ≤ φ ≤ π,
because the sin φ in Eq. (4.35) is greater than or equal to zero. See the end of this
section for more discussion of φ.
Recalling the homogeneous solution in Eq. (4.15), we can write the complete
solution to Eq. (4.29) as
Fig. 4.8
x(t) =
F
cos(ωd t − φ) + e−γ t Ae t + Be− t .
R
(4.36)
4.4 Driven (and damped) harmonic motion
The constants A and B are determined by the initial conditions. If there is any damping
at all in the system (that is, if γ > 0), then the homogeneous part of the solution
goes to zero for large t, and we are left with only the particular solution. In other
words, the system approaches a definite x(t), namely xp (t), independent of the initial
conditions.
Resonance
The amplitude of the motion given in Eq. (4.34) is proportional to
1
=
R
1
ω2 − ωd2
2
.
(4.37)
+ (2γ ωd )2
Given ωd and γ , this is maximum when ω = ωd . Given ω and γ , it is maximum
when ωd = ω2 − 2γ 2 , as you can show in Exercise 4.29. But for weak damping
(that is, γ
ω, which is usually the case we are concerned with), this reduces
to ωd ≈ ω also. The term resonance is used to describe the situation where the
amplitude of the oscillations is as large as possible. It is quite reasonable that
this is achieved when the driving frequency equals the frequency of the spring.
But what is the value of the phase φ at resonance? Using Eq. (4.35), we see that
φ satisfies tan φ ≈ ±∞ when ωd ≈ ω. Therefore, φ = π/2 (it is indeed π/2,
and not −π/2, because the sin φ in Eq. (4.35) is positive), and the motion of the
particle lags the driving force by a quarter of a cycle at resonance. For example,
when the particle moves rightward past the origin (which means it has a quarter
of a cycle to go before it hits the maximum value of x), the force is already at its
maximum. And when the particle makes it out to the maximum value of x, the
force is already back to zero.
The fact that the force is maximum when the particle is moving fastest makes
sense from an energy point of view.2 If you want the amplitude to become large,
then you need to give the system as much energy as you can. That is, you must
do as much work as possible on the system. And in order to do as much work
as possible, you should have your force act over as large a distance as possible,
which means that you should apply your force when the particle is moving fastest,
that is, as it speeds past the origin. And similarly, you don’t want to waste your
force when the particle is barely moving near the endpoints of its motion. In short,
v is the derivative of x and therefore a quarter cycle ahead of x (which is a general
property of a sinusoidal function, as you can show). Since we want the force to
be in phase with v at resonance (by the above energy argument), we see that the
force is also a quarter cycle ahead of x.
Resonance has a large number of extremely important applications (both
wanted and unwanted) in the real world. On the desirable side, resonance makes
2
Energy is one of the topics of the next chapter, so you may want to come back and read this
paragraph after reading that.
113
114
Oscillations
it possible to have a relaxing day at the beach at the Bay of Fundy, talking to
a friend over your cell phone while pushing a child on a swing at low tide. On
the undesirable side, your ride home on that newly discovered “washboard” dirt
road will be annoyingly bumpy at a certain speed, and any attempt to take your
mind off the discomfort by turning up the radio will result only in certain parts
of your car rattling in perfect sync (well, actually 90◦ out of phase) with the bass
line of your formerly favorite song.3
The phase φ
Equation (4.35) gives the phase of the motion as
tan φ =
2γ ωd
,
ω2 − ωd2
(4.38)
where 0 ≤ φ ≤ π. Let’s look at a few cases for ωd (not necessarily at resonance)
and see what the resulting phase φ is. Using Eq. (4.38), we have:
• If ωd ≈ 0 (or more precisely, if γ ωd
ω2 − ωd2 ), then φ ≈ 0. This means that the
motion is in phase with the force. The mathematical reason for this is that if ωd ≈ 0,
then both x¨ and x˙ are small, because they are proportional to ωd2 and ωd , respectively.
Therefore, the first two terms in Eq. (4.29) are negligible, so we end up with x ∝ cos ωd t.
In other words, the phase is zero.
The physical reason is that since there is essentially no acceleration, the net force is
always essentially zero. This means that the driving force always essentially balances
the spring force (that is, the two forces are 180◦ out of phase), because the damping
force is negligible (since x˙ ∝ ωd ≈ 0). But the spring force is 180◦ out of phase with
the motion (because of the minus sign in F = −kx). Therefore, the driving force is in
phase with the motion.
• If ωd ≈ ω, then φ ≈ π/2. This is the case of resonance, discussed above.
• If ωd ≈ ∞ (or more precisely, if γ ωd
ωd2 − ω2 ), then φ ≈ π. The mathematical
reason for this is that if ωd ≈ ∞, then the x¨ term in Eq. (4.29) dominates, so we have
x¨ ∝ cos ωd t. Therefore, x¨ is in phase with the force. But x is 180◦ out of phase with
x¨ (this is a general property of a sinusoidal function), so x is 180◦ out of phase with
the force.
The physical reason is that if ωd ≈ ∞, then the mass hardly moves, because from
Eq. (4.37) we see that the amplitude is proportional to 1/ωd2 . This amplitude then implies
that the velocity is proportional to 1/ωd . Therefore, both x and v are always small. But
if x and v are always small, then the spring and damping forces can be ignored. So
we basically have a mass that feels only one force, the driving force. But we already
understand very well a situation where a mass is subject to only one oscillating force:
a mass on a spring. The mass in our setup can’t tell if it’s being driven by an oscillating
3
Some other examples of resonance that are often cited are in fact not actually examples of resonance,
but rather of “negative damping” (also known as positive feedback). Musical instruments fall into
this category, as does the well-known Tacoma Narrows bridge failure. For a detailed discussion of
this issue, see Billah and Scanlan (1991) and also Green and Unruh (2006).
4.5 Coupled oscillators
driving force, or being pushed and pulled by an oscillating spring force. They both feel
the same. Therefore, both phases must be the same. But in the spring case, the minus
sign in F = −kx tells us that the force is 180◦ out of phase with the motion. Hence, the
same result holds in the ωd ≈ ∞ case.
Another special case for the phase occurs when γ = 0 (no damping), for
which we have tan φ = ±0, depending on the sign of ω2 − ωd2 . So φ is either
0 or π. The motion is therefore either exactly in phase or out of phase with the
driving force, depending on which of ω or ωd is larger.
4.5
Coupled oscillators
In the previous sections, we dealt with only one function of time, x(t). What if
we have two functions of time, say x(t) and y(t), that are related by a pair of
“coupled” differential equations? For example, we might have
2¨x + ω2 (5x − 3y) = 0,
2¨y + ω2 (5y − 3x) = 0.
(4.39)
For now, let’s not worry about how these equations might arise. Let’s just try
to solve them (we’ll do a physical example later in this section). We’ll assume
ω2 > 0 here, although this isn’t necessary. We’ll also assume there aren’t any
damping or driving forces, although a few of the problems and exercises for this
chapter deal with these additions. We call the above equations “coupled” because
there are x’s and y’s in both of them, and it isn’t immediately obvious how to
separate them to solve for x and y. There are two methods (at least) for solving
these equations.
First method: Sometimes it is easy, as in this case, to find certain linear combinations of the given equations for which nice things happen. If we take the sum,
we find
(¨x + y¨ ) + ω2 (x + y) = 0.
(4.40)
This equation involves x and y only in the combination of their sum, x + y. With
z ≡ x + y, Eq. (4.40) is just our old friend, z¨ + ω2 z = 0. The solution is
x + y = A1 cos(ωt + φ1 ),
(4.41)
where A1 and φ1 are determined by the initial conditions. We may also take the
difference of Eqs. (4.39), which results in
(¨x − y¨ ) + 4ω2 (x − y) = 0.
(4.42)
This equation involves x and y only in the combination of their difference, x − y.
The solution is
x − y = A2 cos(2ωt + φ2 ).
(4.43)
115
116
Oscillations
Taking the sum and difference of Eqs. (4.41) and (4.43), we find that x(t) and
y(t) are given by
x(t) = B1 cos(ωt + φ1 ) + B2 cos(2ωt + φ2 ),
y(t) = B1 cos(ωt + φ1 ) − B2 cos(2ωt + φ2 ),
(4.44)
where the Bi ’s are half of the Ai ’s. The strategy of this solution was simply
to fiddle around and try to form differential equations that involved only one
combination of the variables. This allowed us to write down the familiar solution
for these combinations, as we did in Eqs. (4.41) and (4.43).
We’ve managed to solve our equations for x and y. However, it turns out that
the more interesting thing we’ve done is produce the equations (4.41) and (4.43).
The combinations (x + y) and (x − y) are called the normal coordinates of the
system. These are the combinations that oscillate with one pure frequency. The
motion of x and y will in general look complicated, and it may be difficult to tell
that the motion is really made up of just the two frequencies in Eq. (4.44). But
if you plot the values of (x + y) and (x − y) as time goes by, for any motion of
the system, then you will find nice sinusoidal graphs, even if x and y are each
behaving in a rather unpleasant manner.
In the above method, it was fairly easy to guess which combinations of Eqs. (4.39) would produce equations involving only one combination
of x and y. But surely there are problems in physics where the guessing isn’t so
easy. What do we do then? Fortunately, there is a fail-safe method for solving for
x and y. It proceeds as follows.
In the spirit of Section 4.1, let’s try a solution of the form x = Aeiαt and
y = Beiαt , which we will write, for convenience, as
Second method:
x
A iαt
=
e .
y
B
(4.45)
It isn’t obvious that there should exist solutions for x and y that have the same t
dependence, but let’s try it and see what happens. We’ve explicitly put the i in
the exponent, but there’s no loss of generality here. If α happens to be imaginary,
then the exponent is real. It’s personal preference whether or not you put the i in.
Plugging our guess into Eqs. (4.39), and dividing through by eiωt , we find
2A(−α 2 ) + 5Aω2 − 3Bω2 = 0,
2B(−α 2 ) + 5Bω2 − 3Aω2 = 0,
(4.46)
or equivalently, in matrix form,
−2α 2 + 5ω2
−3ω2
−3ω2
−2α 2 + 5ω2
A
0
=
.
B
0
(4.47)
4.5 Coupled oscillators
This homogeneous equation for A and B has a nontrivial solution (that is, one
where A and B aren’t both 0) only if the matrix is not invertible. This is true
because if it were invertible, then we could multiply through by the inverse to
obtain (A, B) = (0, 0). When is a matrix invertible? There is a straightforward
(although tedious) method for finding the inverse. It involves taking cofactors,
taking a transpose, and dividing by the determinant. The step that concerns
us here is the division by the determinant, since this implies that the inverse
exists if and only if the determinant is not zero. So we see that Eq. (4.47) has
a nontrivial solution only if the determinant equals zero. Because we seek a
nontrivial solution, we must therefore have
0=
−2α 2 + 5ω2
−3ω2
−3ω2
−2α 2 + 5ω2
= 4α 4 − 20α 2 ω2 + 16ω4 .
(4.48)
This is a quadratic equation in α 2 , and the roots are α = ±ω and α = ±2ω.
We have therefore found four types of solutions. If α = ±ω, then we can plug
this back into Eq. (4.47) to obtain A = B. (Both equations give this same result.
This was essentially the point of setting the determinant equal to zero.) And if
α = ±2ω, then Eq. (4.47) gives A = −B. (Again, the equations are redundant.)
Note that we cannot solve specifically for A and B, but only for their ratio. Adding
up our four solutions according to the principle of superposition, we see that x
and y take the general form (written in vector form for the sake of simplicity and
bookkeeping),
x
1 iωt
1 −iωt
= A1
e + A2
e
y
1
1
+ A3
1
1
e2iωt + A4
e−2iωt .
−1
−1
(4.49)
The four Ai are determined by the initial conditions. We can rewrite Eq. (4.49)
in a somewhat cleaner form. If the coordinates x and y describe the positions of
particles, they must be real. Therefore, A1 and A2 must be complex conjugates,
and likewise for A3 and A4 . If we then define some φ’s and B’s via A∗2 = A1 ≡
(B1 /2)eiφ1 and A∗4 = A3 ≡ (B2 /2)eiφ2 , we may rewrite our solution in the form,
as you can verify,
x
1
1
= B1
cos(ωt + φ1 ) + B2
cos(2ωt + φ2 ),
y
1
−1
(4.50)
where the Bi and φi are real (and are determined by the initial conditions). We
have therefore reproduced the result in Eq. (4.44).
It is clear from Eq. (4.50) that the combinations x + y and x − y (the normal
coordinates) oscillate with the pure frequencies, ω and 2ω, respectively, because
117
118
Oscillations
the combination x + y makes the B2 terms disappear, and the combination x − y
makes the B1 terms disappear.
It is also clear that if B2 = 0, then x = y at all times, and they both oscillate
with frequency ω. And if B1 = 0, then x = −y at all times, and they both
oscillate with frequency 2ω. These two pure-frequency motions are called the
normal modes. They are labeled by the vectors (1, 1) and (1, −1), respectively.
In describing a normal mode, both the vector and the frequency should be stated.
The significance of normal modes will become clear in the following example.
k
k
m
Fig. 4.9
k
m
Example (Two masses, three springs): Consider two masses m, connected to
each other and to two walls by three springs, as shown in Fig. 4.9. The three springs
have the same spring constant k. Find the most general solution for the positions
of the masses as functions of time. What are the normal coordinates? What are the
normal modes?
Solution: Let x1 (t) and x2 (t) be the positions of the left and right masses, respectively, relative to their equilibrium positions. Then the middle spring is stretched
a distance x2 − x1 compared with the stretch at equilibrium. Therefore, the net
force on the left mass is −kx1 + k(x2 − x1 ), and the net force on the right mass
is −kx2 − k(x2 − x1 ). It’s easy to make a mistake in the sign of the second term
in these expressions, but you can check it by, say, looking at the force when x2 is
very big. At any rate, the second term must have the opposite sign in the two expressions, by Newton’s third law. With these forces, F = ma on each mass gives, with
ω2 = k/m,
x¨ 1 + 2ω2 x1 − ω2 x2 = 0,
x¨ 2 + 2ω2 x2 − ω2 x1 = 0.
(4.51)
These are rather friendly looking coupled equations, and we can see that the sum and
difference are the useful combinations to take. The sum gives
(¨x1 + x¨ 2 ) + ω2 (x1 + x2 ) = 0,
(4.52)
(¨x1 − x¨ 2 ) + 3ω2 (x1 − x2 ) = 0.
(4.53)
and the difference gives
The solutions to these equations are the normal coordinates,
x1 + x2 = A+ cos(ωt + φ+ ),
√
x1 − x2 = A− cos( 3ωt + φ− ).
Taking the sum and difference of these normal coordinates, we have
√
x1 (t) = B+ cos(ωt + φ+ ) + B− cos( 3ωt + φ− ),
√
x2 (t) = B+ cos(ωt + φ+ ) − B− cos( 3ωt + φ− ),
(4.54)
(4.55)
4.5 Coupled oscillators
119
where the B’s are half of the A’s. Along with the φ’s, they are determined by the initial
conditions.
Remark: For practice, let’s also derive Eq. (4.55) by using the determinant method.
Letting x1 = Aeiαt and x2 = Beiαt in Eq. (4.51), we see that for there to be a nontrivial
solution for A and B, we must have
0=
−α 2 + 2ω2
−ω2
−ω2
−α 2 + 2ω2
= α 4 − 4α 2 ω2 + 3ω4 .
(4.56)
√
This is a quadratic equation in α 2 , and √
the roots are α = ±ω and α = ± 3ω. If α = ±ω,
then Eq. (4.51) gives A = B. If α = ± 3ω, then Eq. (4.51) gives A = −B. The solutions
for x1 and x2 therefore take the general form
x1
x2
= A1
1 iωt
1 −iωt
e + A2
e
1
1
+ A3
≡ B+
√
√
1
1
e 3iωt + A4
e− 3iωt
−1
−1
√
1
1
cos(ωt + φ+ ) + B−
cos( 3ωt + φ− ),
1
−1
(4.57)
where the last line follows from the same substitutions that led to Eq. (4.50). This expression
is equivalent to Eq. (4.55). ♣
The normal modes are obtained by setting either B− or B+ equal to zero in Eq. (4.55)
or Eq. (4.57). Therefore, the normal modes are (1, 1) and (1, −1). How do we visualize
these? The mode (1, 1) oscillates with frequency ω. In this case (where B− = 0), we
have x1 (t) = x2 (t) = B+ cos(ωt + φ+ ) at all times. So the masses simply oscillate
back and forth in the same manner, as shown in Fig. 4.10. It is clear that such motion
has frequency ω, because as far as the masses are concerned, the middle spring is
effectively not there, so each mass moves under the influence of only one spring, and
therefore has frequency ω.
√
The mode (1, −1) oscillates with√frequency 3ω. In this case (where B+ = 0),
we have x1 (t) = −x2 (t) = B− cos( 3ωt + φ− ) at all times. So the masses oscillate
back and forth with opposite displacements, as shown in Fig. 4.11. It is clear that this
mode should have a frequency larger than that for the other mode, because the middle
spring is stretched (or compressed), so the √
masses feel a larger force. But it takes a
little thought to show that the frequency is 3ω. 4
The normal mode (1, 1) above is associated with the normal coordinate x1 +x2 .
They both involve the frequency ω. However, this association is not due to the
fact that the coefficients of both x1 and x2 in this normal coordinate are equal to 1.
4
√
If you want to obtain this 3ω result without going through all of the above work, just note that
the center of the middle spring doesn’t move. Therefore, it acts like two “half springs,” each with
spring constant 2k (as you can verify). Hence, each mass is effectively attached to a “k” spring and
√
a “2k” spring, yielding a total effective spring constant of 3k. Thus the 3.
Fig. 4.10
Fig. 4.11
120
Oscillations
Rather, it is due to the fact that the other normal mode, namely (x1 , x2 ) ∝ (1, −1),
gives no contribution to the sum x1 + x2 . There are a few too many 1’s floating
around in the above example, so it’s hard to see which results are meaningful and
which results are coincidence. But the following example should clear things up.
Let’s say we solved a problem using the determinant method, and we found the
solution to be
x
3
1
= B1
cos(ω1 t + φ1 ) + B2
cos(ω2 t + φ2 ).
y
2
−5
(4.58)
Then 5x + y is the normal coordinate associated with the normal mode (3, 2),
which has frequency ω1 . (This is true because there is no cos(ω2 t + φ2 ) dependence in the combination 5x + y.) And similarly, 2x − 3y is the normal coordinate
associated with the normal mode (1, −5), which has frequency ω2 (because there
is no cos(ω1 t + φ1 ) dependence in the combination 2x − 3y).
Note the difference between the types of differential equations we solved in
the previous chapter in Section 3.3, and the types we solved throughout this
chapter. The former dealt with forces that did not have to be linear in x or x˙ , but
that had to depend on only x, or only x˙ , or only t. The latter dealt with forces
that could depend on all three of these quantities, but that had to be linear in
x and x˙ .
4.6
Problems
Section 4.1: Linear differential equations
4.1. Superposition
Let x1 (t) and x2 (t) be solutions to x¨ 2 = bx. Show that x1 (t) + x2 (t) is
not a solution to this equation.
4.2. A limiting case *
Consider the equation x¨ = ax. If a = 0, then the solution to x¨ = 0 is
simply x(t) = C + Dt. Show that in the limit a → 0, Eq. (4.2) reduces
to this form. Note: a → 0 is a sloppy way of saying what we mean.
What is the proper way to write this limit?
Section 4.2: Simple harmonic motion
4.3. Increasing the mass **
A mass m oscillates on a spring with spring constant k. The amplitude is d. At the moment (let this be t = 0) when the mass is at
position x = d/2 (and moving to the right), it collides and sticks to
another mass m. The speed of the resulting mass 2m right after the collision is half the speed of the moving mass m right before the collision
4.6 Problems
121
(from momentum conservation, discussed in Chapter 5). What is the
resulting x(t)? What is the amplitude of the new oscillation?
4.4. Average tension **
Is the average (over time) tension in the string of a pendulum larger
or smaller than mg? By how much? As usual, assume that the angular
amplitude A is small.
4.5. Walking east on a turntable **
A person walks at constant speed v eastward with respect to a turntable
that rotates counterclockwise at constant frequency ω. Find the general
expression for the person’s coordinates with respect to the ground (with
the x direction taken to be eastward).
Section 4.3: Damped harmonic motion
4.6. Maximum speed **
A mass on the end of a spring (with natural frequency ω) is released
from rest at position x0 . The experiment is repeated, but now with the
system immersed in a fluid that causes the motion to be overdamped
(with damping coefficient γ ). Find the ratio of the maximum speed in
the former case to that in the latter. What is the ratio in the limit of strong
damping (γ
ω)? In the limit of critical damping?
Section 4.4: Driven (and damped) harmonic motion
4.7. Exponential force *
A particle of mass m is subject to a force F(t) = ma0 e−bt . The initial
position and speed are both zero. Find x(t). (This problem was already
given as Problem 3.9, but solve it here by guessing an exponential
function, in the spirit of Section 4.4.)
4.8. Driven oscillator *
Derive Eq. (4.31) by guessing a solution of the form x(t) = A cos ωd t +
B sin ωd t in Eq. (4.29).
k
k
m
k
2m
Section 4.5: Coupled oscillators
Fig. 4.12
4.9. Unequal masses **
Three identical springs and two masses, m and 2m, lie between two walls
as shown in Fig. 4.12. Find the normal modes.
4.10. Weakly coupled **
Three springs and two equal masses lie between two walls, as shown
in Fig. 4.13. The spring constant, k, of the two outside springs is much
κ
k
m
Fig. 4.13
k
m
122
Oscillations
larger than the spring constant, κ, of the middle spring. Let x1 and x2 be
the positions of the left and right masses, respectively, relative to their
equilibrium positions. If the initial positions are given by x1 (0) = a and
x2 (0) = 0, and if both masses are released from rest, show that x1 and
x2 can be written as (assuming κ
k)
Fd (t)
x1 (t) ≈ a cos (ω + )t cos( t),
x2 (t) ≈ a sin (ω + )t sin( t),
where ω ≡ k/m and
motion looks like.
Fig. 4.14
(4.59)
≡ (κ/2k)ω. Explain qualitatively what the
4.11. Driven mass on a circle **
Two identical masses m are constrained to move on a horizontal hoop.
Two identical springs with spring constant k connect the masses and wrap
around the hoop (see Fig. 4.14). One mass is subject to a driving force
Fd cos ωd t. Find the particular solution for the motion of the masses.
4.12. Springs on a circle ****
(a) Two identical masses m are constrained to move on a horizontal
hoop. Two identical springs with spring constant k connect the
masses and wrap around the hoop (see Fig. 4.15). Find the normal
modes.
(b) Three identical masses are constrained to move on a hoop. Three
identical springs connect the masses and wrap around the hoop
(see Fig. 4.16). Find the normal modes.
(c) Now do the general case with N identical masses and N identical
springs.
Fig. 4.15
4.7
Fig. 4.16
Exercises
Section 4.1: Linear differential equations
4.13. kx force *
A particle of mass m is subject to a force F(x) = kx, with k > 0. What
is the most general form of x(t)? If the particle starts out at x0 , what is
the one special value of the initial velocity for which the particle doesn’t
eventually get far away from the origin?
4.14. Rope on a pulley **
A rope with length L and mass density σ kg/m hangs over a massless
pulley. Initially, the ends of the rope are a distance x0 above and below
their average position. The rope is given an initial speed. If you want
4.7 Exercises
the rope to not eventually fall off the pulley, what should this initial
speed be? (Don’t worry about the issue discussed in Calkin (1989).)
123
k
m
m
Section 4.2: Simple harmonic motion
4.15. Amplitude *
Find the amplitude of the motion given by x(t) = C cos ωt + D sin ωt.
4.16. Angled rails *
Two particles of mass m are constrained to move along two horizontal
frictionless rails that make an angle 2θ with respect to each other. They
are connected by a spring with spring constant k, whose relaxed length
is at the position shown in Fig. 4.17. What is the frequency of oscillations for the motion where the spring remains parallel to the position
shown?
4.17. Effective spring constant *
(a) Two springs with spring constants k1 and k2 are connected in parallel, as shown in Fig. 4.18. What is the effective spring constant,
keff ? In other words, if the mass is displaced by x, find the keff
for which the force equals F = −keff x.
(b) Two springs with spring constants k1 and k2 are connected in
series, as shown in Fig. 4.19. What is the effective spring
constant, keff ?
4.18. Changing k **
Two springs each have spring constant k and equilibrium length . They
are both stretched a distance and attached to a mass m and two walls,
as shown in Fig. 4.20. At a given instant, the right spring constant is
somehow magically changed to 3k (the relaxed length remains ). What
is the resulting x(t)? Take the initial position to be x = 0.
4.19. Removing a spring **
The springs in Fig. 4.21 are at their equilibrium length. The mass oscillates along the line of the springs with amplitude d. At the moment (let
this be t = 0) when the mass is at position x = d/2 (and moving to the
right), the right spring is removed. What is the resulting x(t)? What is
the amplitude of the new oscillation?
2θ
Fig. 4.17
k1
k2
Fig. 4.18
k2
k1
Fig. 4.19
2l
2l
m
k
k
3k
Fig. 4.20
k
k
m
Fig. 4.21
4.20. Springs all over **
(a) A mass m is attached to two springs that have relaxed lengths of
zero. The other ends of the springs are fixed at two points (see
Fig. 4.22). The two spring constants are equal. The mass sits at
its equilibrium position and is then given a kick in an arbitrary
k
Fig. 4.22
m
k
124
Oscillations
direction. Describe the resulting motion. (Ignore gravity, although
you actually don’t need to.)
(b) A mass m is attached to n springs that have relaxed lengths of
zero. The other ends of the springs are fixed at various points in
space (see Fig. 4.23). The spring constants are k1 , k2 , . . . , kn . The
mass sits at its equilibrium position and is then given a kick in an
arbitrary direction. Describe the resulting motion. (Again, ignore
gravity, although you actually don’t need to.)
Fig. 4.23
4.21. Rising up ***
In Fig. 4.24, a mass hangs from a ceiling. A piece of paper is held
up to obscure three strings and two springs; all you see is two other
strings protruding from behind the paper, as shown. How should the
three strings and two springs be attached to each other and to the two
visible strings (different items can be attached only at their endpoints)
so that if you start with the system at its equilibrium position and then
cut a certain one of the hidden strings, the mass will rise up?5
?
4.22. Projectile on a spring ***
A projectile of mass m is fired from the origin at speed v0 and angle θ. It
is attached to the origin by a spring with spring constant k and relaxed
length zero.
Fig. 4.24
(a) Find x(t) and y(t).
(b) Show that for small ω ≡ k/m, the trajectory reduces to normal projectile motion. And show that for large ω, the trajectory
reduces to simple harmonic motion, that is, oscillatory motion
along a line (at least before the projectile smashes back into the
ground). What are the more meaningful statements that should
replace “small ω” and “large ω”?
(c) What value should ω take so that the projectile hits the ground
when it is moving straight downward?
4.23. Corrections to the pendulum ***
(a) For small oscillations, the period of a pendulum is approximately
√
T ≈ 2π /g, independent of the amplitude, θ0 . For finite
oscillations, use dt = dx/v to show that the exact expression
for T is
T =
5
8
g
θ0
0
√
dθ
.
cos θ − cos θ0
(4.60)
Thanks to Paul Horowitz for this extremely cool problem. For more applications of the idea behind
it, see Cohen and Horowitz (1991).
4.7 Exercises
(b) Find an approximation to this T , up to second order in θ02 , in the
following way. Make use of the identity cos φ = 1 − 2 sin2 (φ/2)
to write T in terms of sines (because it’s more convenient to work
with quantities that go to zero as θ → 0). Then make the change of
variables, sin x ≡ sin(θ/2)/ sin(θ0 /2) (you’ll see why). Finally,
expand your integrand in powers of θ0 , and perform the integrals
to show that6
T ≈ 2π
g
1+
θ02
+ ···
16
.
(4.61)
Section 4.3: Damped harmonic motion
4.24. Crossing the origin
Show that an overdamped or critically damped oscillator can cross the
origin at most once.
4.25. Strong damping *
In the strong damping (γ
ω) case discussed in the remark in the
2
overdamping subsection, we saw that x(t) ∝ e−ω t/2γ for large t. Using
the definitions of ω and γ , this can be written as x(t) ∝ e−kt/b , where b
is the coefficient of the damping force. By looking at the forces on the
mass, explain why this makes sense.
4.26. Maximum speed *
A critically damped oscillator with natural frequency ω starts out at
position x0 > 0. What is the maximum initial speed (directed toward
the origin) it can have and not cross the origin?
4.27. Another maximum speed **
An overdamped oscillator with natural frequency ω and damping coefficient γ starts out at position x0 > 0. What is the maximum initial speed
(directed toward the origin) it can have and not cross the origin?
4.28. Ratio of maxima **
A mass on the end of a spring is released from rest at position x0 . The
experiment is repeated, but now with the system immersed in a fluid
that causes the motion to be critically damped. Show that the maximum
speed of the mass in the first case is e times the maximum speed in the
second case.7
6
7
If you like this sort of thing, you can show that the next term in the parentheses is (11/3072)θ04 .
But be careful, this fourth-order correction comes from two terms.
The fact that the maximum speeds differ by a fixed numerical factor follows from dimensional
analysis, which tells us that the maximum speed in the first case must be proportional to ωx0 . And
125
126
Oscillations
Section 4.4: Driven (and damped) harmonic motion
4.29. Resonance
Given ω and γ , show that the R in Eq. (4.33) is minimum when ωd =
ω2 − 2γ 2 (unless this is imaginary, in which case the minimum occurs
at ωd = 0).
4.30. No damping force *
A particle of mass m is subject to a spring force, −kx, and also a driving
force, Fd cos ωd t. But there is no damping force. Find the particular
solution for x(t) by guessing x(t) = A cos ωd t + B sin ωd t. If you write
this in the form C cos(ωd t − φ), where C > 0, what are C and φ? Be
careful about the phase (there are two cases to consider).
k
k
m
m
Fig. 4.25
k
k
m
k
m
k
m
k
m
θ
Fig. 4.27
4.31. Springs and one wall **
Two identical springs and two identical masses are attached to a wall as
shown in Fig. 4.25. Find the
√ normal modes, and show that the frequencies
can be written as k/m( 5 ± 1)/2. This numerical factor is the golden
ratio (and its inverse).
4.32. Springs between walls **
Four identical springs and three identical masses lie between two walls
(see Fig. 4.26). Find the normal modes.
Fig. 4.26
m
Section 4.5: Coupled oscillators
4.33. Beads on angled rails **
Two horizontal frictionless rails make an angle θ with each other, as
shown in Fig. 4.27. Each rail has a bead of mass m on it, and the beads
are connected by a spring with spring constant k and relaxed length zero.
Assume that one of the rails is positioned a tiny distance above the other,
so that the beads can pass freely through the crossing. Find the normal
modes.
4.34. Coupled and damped **
The system in the example in Section 4.5 is modified by immersing it in
a fluid so that both masses feel a damping force, Ff = −bv. Solve for
x1 (t) and x2 (t). Assume underdamping.
4.35. Coupled and driven **
The system in the example in Section 4.5 is modified by subjecting the
left mass to a driving force Fd cos(2ωt), and the right mass to a driving
since γ = ω in the critical-damping case, the damping doesn’t introduce a new parameter, so the
maximum speed has no choice but to again be proportional to ωx0 . But showing that the maximum
speeds differ by the nice factor of e requires a calculation.
4.8 Solutions
force 2Fd cos(2ωt), where ω = k/m. Find the particular solution for
x1 (t) and x2 (t), and explain why your answer makes sense.
4.8
Solutions
4.1. Superposition
The sum x1 + x2 is a solution to x¨ 2 = bx if
d 2 (x1 + x2 )
dt 2
⇐⇒
⇐⇒
2
= b(x1 + x2 )
(¨x1 + x¨ 2 )2 = b(x1 + x2 )
x¨ 12 + 2¨x1 x¨ 2 + x¨ 22 = b(x1 + x2 ).
(4.62)
But x¨ 12 = bx1 and x¨ 22 = bx2 , by assumption. So we are left with the 2¨x1 x¨ 2 term on the
left-hand side, which destroys the equality. (Note that 2¨x1 x¨ 2 can’t be zero, because if
either x¨ 1 or x¨ 2 is identically zero, then either x1 or x2 is also, so we didn’t really have
a solution to begin with.)
4.2. A limiting case
The expression “a → 0” is sloppy because a has units of inverse time squared, and the
number
√Eq. (4.2) reduces to x(t) = C + Dt
√0 has no units. The proper statement is that
1, or equivalently when t
1/ a, which is now a comparison of
when a t
quantities with the same units. The smaller a is, the larger
√ t can be. Therefore, if
1, we can write
“a √→ 0,” then t can basically be anything. Assuming a t
√
e± a t ≈ 1 ± a t, and Eq. (4.2) becomes
√
√
x(t) ≈ A(1 + a t) + B(1 − a t)
√
= (A + B) + a(A − B)t
≡ C + Dt.
(4.63)
C is the initial position, and D is the speed of the particle. If these quantities are of
order 1 in the units chosen, then if we solve for A and
√ B, we see that they must be
roughly negatives of each other, and both of order 1/ a. So if the speed and initial
position are of order 1, then A and B actually diverge in the “a →√0” limit. If a is small
1 won’t hold,
but nonzero, then t will eventually become large enough so that a t
in which case the linear form in Eq. (4.63) won’t be valid.
4.3. Increasing the mass
The first thing we must do is find the velocity of the mass right before the collision. The
motion before the collision looks like x(t) = d cos(ωt + φ), where ω = k/m. The
collision happens at t = 0 (although it actually doesn’t matter what time we plug in
here), so we have d/2 = x(0) = d cos φ, which gives φ = ±π/3. The velocity right
before the collision is therefore
√
v (0) ≡ x˙ (0) = −ωd sin φ = −ωd sin(±π/3) = ∓( 3/2)ωd.
(4.64)
We want the plus sign, because we are told that the mass is moving to the right. Finding
the motion after the collision is now reduced to an initial conditions problem. We have
a mass 2m
√ on a spring with spring constant k, with initial position d/2 and initial
velocity ( 3/4)ωd (half of the result above). In situations where we know the initial
position and velocity, it turns out that the best form to use for x(t) from the expressions
in Eq. (4.3) is
x(t) = C cos ω t + D sin ω t,
(4.65)
127
128
Oscillations
because the initial position at t = 0 is simply C, and the initial velocity at t = 0
is ω D. The initial conditions are therefore easy to apply. We have put a prime on
the frequency in Eq. (4.65) to remind us that it is different from
√ the initial frequency,
because the mass is now 2m. So we have ω = k/2m = ω/ 2. The initial conditions
therefore give
x(0) = d/2 =⇒
√
v (0) = ( 3/4)ωd
C = d/2,
=⇒
√
ω D = ( 3/4)ωd
=⇒
√
D = ( 6/4)d.
(4.66)
Our solution for x(t) is therefore
x(t) =
d
cos ω t +
2
√
6d
sin ω t,
4
k
.
2m
where ω =
(4.67)
To find the amplitude, we must calculate the maximum value of x(t). This is the task
of Exercise 4.15,
√ and the result is that the amplitude of the x(t) = C cos ω t + D sin ω t
motion is A = C 2 + D2 . So we have
A=
d2
6d 2
+
=
4
16
5
d.
8
(4.68)
This is smaller than the original amplitude d, because energy is lost to heat during the
collision (but energy is one of the topics of the next chapter).
4.4. Average tension
Let the length of the pendulum be . We know that the angle θ depends on time
according to
θ (t) = A cos(ωt),
(4.69)
√
where ω = g/ . If T is the tension in the string, then the radial F = ma equation is
T − mg cos θ = m θ˙ 2 . Using Eq. (4.69), this becomes
T = mg cos A cos(ωt) + m
2
− ωA sin(ωt) .
(4.70)
Since A is small, we can use the small-angle approximation cos α ≈ 1 − α 2 /2, which
gives
T ≈ mg 1 −
1 2
A cos2 (ωt) + m ω2 A2 sin2 (ωt)
2
= mg + mgA2 sin2 (ωt) −
1
cos2 (ωt) ,
2
(4.71)
where we have made use of ω2 = g/ . The average value of both sin2 θ and cos2 θ
over one period is 1/2 (you can show this by doing the integrals, or you can just note
that the averages are equal and they add up to 1), so the average value of T is
Tavg = mg +
mgA2
,
4
(4.72)
which is larger than mg, by mgA2 /4. It makes sense that Tavg > mg, because the
average value of the vertical component of T equals mg (because the pendulum has
no net rise or fall over a long period of time), and there is some nonzero contribution
to the magnitude of T from the horizontal component.
4.5. Walking east on a turntable
The velocity of the person with respect to the ground is the sum of v xˆ and u, where
u is the velocity (at the person’s position) of the turntable with respect to the ground.
4.8 Solutions
129
In terms of the angle θ in Fig. 4.28, the velocity components with respect to the ground
are
x˙ = v − u sin θ,
and
y˙ = u cos θ .
east
(4.73)
ω
But u = rω. So we have, using r sin θ = y and r cos θ = x,
x˙ = v − ωy,
and
y˙ = ωx.
(4.74)
u
r
Taking the derivative of the first equation, and then plugging in y˙ from the second,
gives x¨ = −ω2 x. Therefore, x(t) = A cos(ωt + φ). The first equation then quickly
gives y(t), and the result is that the general expression for the person’s position is
(x, y) = A cos(ωt + φ),
A sin(ωt + φ) + v /ω .
v
θ
(4.75)
This describes a circle centered at the point (0, v /ω). The constants A and φ are
determined by the initial x and y values. You can show that
A=
x02 + (y0 − v /ω)2 ,
and
y0 − v /ω
.
x0
tan φ =
Fig. 4.28
(4.76)
Remarks: It turns out that in the frame of the turntable, the person’s path is also a
circle. This can be seen in the following way. Imagine a distant object (say, a star)
located in the eastward direction. In the frame of the turntable, this star rotates clockwise with frequency ω. And in the frame of the turntable, the person’s velocity always
points toward the star. Therefore, the person’s velocity rotates clockwise with frequency ω. And since the magnitude of the velocity is constant, this means that the
person travels clockwise in a circle in the frame of the turntable. From the usual
expression v = rω, we see that this circle has a radius v /ω.
This result leads to another way of showing that the person’s path is a circle as
viewed in the ground frame. In short, when the person’s clockwise circular motion at
speed v with respect to the turntable is combined with the counterclockwise motion of
the turntable with respect to the ground (with the same frequency ω, but in the opposite
direction), the resulting motion of the person with respect to the ground is a circle with
its center at the point (0, v /ω).
The situation is summarized in Fig. 4.29. From the above result (that the path
is a circle in the turntable frame), we may characterize the person’s motion in the
turntable frame by imagining her riding on a merry-go-round that rotates clockwise
with frequency ω with respect to the turntable. This merry-go-round is shown at five
different times in the figure. The effect of the merry-go-round’s clockwise rotational
motion is to cancel the counterclockwise rotation of the turntable, so that the merrygo-round ends up not rotating at all with respect to the ground. Therefore, if the person
(the dot in the figure) starts at the top of the merry-go-round (which she in fact does,
because this corresponds to walking eastward with respect to the turntable), then she
always remains at the top. She therefore travels in a circle that is simply shifted upward
by the vertical radius of the merry-go-round (the dotted lines shown, which have length
v /ω), relative to the center of the turntable. This agrees with the original result. You
can also see from the figure how the values of A and φ in Eq. (4.76) arise. For example,
A is the length of the solid lines shown. ♣
4.6. Maximum speed
For the undamped case, the general form of x is x(t) = C cos(ωt + φ). The initial
condition v (0) = 0 tells us that φ = 0, and then the initial condition x(0) = x0 tells
us that C = x0 . Therefore, x(t) = x0 cos(ωt), and so v (t) = −ωx0 sin(ωt). This has a
maximum magnitude of ωx0 .
Now consider the overdamped case. Equation (4.17) gives the position as
x(t) = Ae−(γ −
)t
+ Be−(γ +
)t
.
(4.77)
east
ω
ω
Fig. 4.29
130
Oscillations
The initial conditions are
x(0) = x0
=⇒
A + B = x0 ,
v (0) = 0
=⇒
−(γ −
)A − (γ +
(4.78)
)B = 0.
Solving these equations for A and B, and then plugging the results into Eq. (4.77),
gives
x0
2
x(t) =
)e−(γ −
(γ +
)t
Taking the derivative to find v (t), and using γ 2 −
−ω2 x0 −(γ −
e
2
v (t) =
)e−(γ +
− (γ −
)t
2
)t
.
(4.79)
= ω2 , gives
− e−(γ +
)t
.
(4.80)
Taking the derivative again, we find that the maximum speed occurs at
tmax =
1
ln
2
γ+
γ−
.
(4.81)
Plugging this into Eq. (4.80), and taking advantage of the logs in the exponentials,
gives
v (tmax ) =
−ω2 x0
γ
exp −
ln
2
2
= −ωx0
γ+
γ−
γ+
γ−
−
γ−
γ+
γ /2
γ−
γ+
.
(4.82)
The desired ratio, R, of the maximum speeds in the two scenarios is therefore
γ+
γ−
R=
1/2
2γ
ω2 /2γ
In the limit of critical damping (γ ≈ ω,
R≈
.
ω), we have
In the limit of strong damping (γ
So the ratio becomes
R≈
γ /2
1+
1−
=
(4.83)
≡
γ 2 − ω2 ≈ γ − ω2 /2γ .
2γ
.
ω
≈ 0), we have, with
(4.84)
/γ ≡ ,
1/2
≈ (1 + 2 )1/2 ≈ e,
(4.85)
in agreement with the result of Exercise 4.28 (the solution to which is much quicker
than the one above, since you don’t need to deal with all the ’s). You can also show
that in these two limits, tmax equals ln(2γ /ω)/γ and 1/γ ≈ 1/ω, respectively.
4.7. Exponential force
F = ma gives x¨ = a0 e−bt . Let’s guess a particular solution of the form x(t) = Ce−bt .
Plugging this in gives C = a0 /b2 . And since the solution to the homogeneous equation
x¨ = 0 is x(t) = At + B, the complete solution for x is
x(t) =
a0 e−bt
+ At + B.
b2
(4.86)
4.8 Solutions
The initial condition x(0) = 0 gives B = −a0 /b2 . And the initial condition v (0) = 0
applied to v (t) = −a0 e−bt /b + A gives A = a0 /b. Therefore,
x(t) = a0
e−bt
1
t
+ − 2
b2
b
b
,
(4.87)
in agreement with Problem 3.9.
4.8. Driven oscillator
Plugging x(t) = A cos ωd t + B sin ωd t into Eq. (4.29) gives
− ωd2 A cos ωd t − ωd2 B sin ωd t
− 2γ ωd A sin ωd t + 2γ ωd B cos ωd t
+ ω2 A cos ωd t + ω2 B sin ωd t = F cos ωd t.
(4.88)
If this is true for all t, the coefficients of cos ωd t on both sides must be equal. And
likewise for sin ωd t. Therefore,
−ωd2 A + 2γ ωd B + ω2 A = F,
(4.89)
−ωd2 B − 2γ ωd A + ω2 B = 0.
Solving this system of equations for A and B gives
A=
F ω2 − ωd2
ω2
2
− ωd2
+ 4γ 2 ωd2
,
B=
2Fγ ωd
ω2
− ωd2
2
+ 4γ 2 ωd2
,
(4.90)
in agreement with Eq. (4.31).
4.9. Unequal masses
Let x1 and x2 be the positions of the left and right masses, respectively, relative to
their equilibrium positions. The forces on the two masses are −kx1 + k(x2 − x1 ) and
−kx2 − k(x2 − x1 ), respectively, so the F = ma equations are
x¨ 1 + 2ω2 x1 − ω2 x2 = 0,
2¨x2 + 2ω2 x2 − ω2 x1 = 0.
(4.91)
The appropriate linear combinations of these equations aren’t obvious, so we’ll use
the determinant method. Letting x1 = A1 eiαt and x2 = A2 eiαt , we see that for there to
be a nontrivial solution for A and B, we must have
0=
−α 2 + 2ω2
−ω2
−ω2
−2α 2 + 2ω2
= 2α 4 − 6α 2 ω2 + 3ω4 .
(4.92)
The roots of this quadratic equation in α 2 are
√
3− 3
(4.93)
≡ ±α2 .
2
√
If α 2 = α12 , then the normal mode is proportional to ( 3 + 1, −1). And if α 2 = α22 ,
√
then the normal mode is proportional to ( 3 − 1, 1). So the normal modes are
√
x1
3+1
=
cos(α1 t + φ1 ), and
x2
−1
(4.94)
√
x1
3−1
=
cos(α2 t + φ2 ).
x2
1
α = ±ω
√
3+ 3
≡ ±α1 ,
2
and
α = ±ω
131
132
Oscillations
Note that these two vectors are not orthogonal (there is no need for them
√ to be). The
normal√coordinates associated with these normal modes are x1 − ( 3 − 1)x2 and
x1 + ( 3 + 1)x2 , respectively, because these are the combinations that make the α2
and α1 frequencies disappear, respectively.
4.10. Weakly coupled
The magnitude of the force in the middle spring is κ(x2 − x1 ), so the F = ma equations
are
m¨x1 = −kx1 + κ(x2 − x1 ),
(4.95)
m¨x2 = −kx2 − κ(x2 − x1 ).
Adding and subtracting these equations gives
m(¨x1 + x¨ 2 ) = −k(x1 + x2 )
=⇒
x1 + x2 = A cos(ωt + φ),
m(¨x1 − x¨ 2 ) = −(k + 2κ)(x1 − x2 )
=⇒
˜
x1 − x2 = B cos(ωt
˜ + φ),
(4.96)
where
ω≡
k
,
m
and
ω˜ ≡
k + 2κ
.
m
(4.97)
The initial conditions are x1 (0) = a, x˙ 1 (0) = 0, x2 (0) = 0, and x˙ 2 (0) = 0. The easiest
way to apply these is to plug them into the normal coordinates in Eq. (4.96), before
solving for x1 (t) and x2 (t). The velocity conditions quickly give φ = φ˜ = 0, and then
the position conditions give A = B = a. Solving for x1 (t) and x2 (t) yields
a
a
x1 (t) = cos(ωt) + cos(ωt),
˜
2
2
(4.98)
a
a
˜
x2 (t) = cos(ωt) − cos(ωt).
2
2
Writing ω and ω˜ as
ω˜ + ω
ω˜ − ω
ω˜ + ω
ω˜ − ω
−
, and ω˜ =
+
,
2
2
2
2
and using the identity cos(α + β) = cos α cos β − sin α sin β, gives
ω=
x1 (t) = a cos
ω˜ + ω
ω˜ − ω
t cos
t ,
2
2
x2 (t) = a sin
ω˜ + ω
t sin
2
ω˜ − ω
t .
2
a cos(εt)
ω˜ ≡
t
where
k + 2κ
=
m
k
κ
2κ
≈ω 1+
1+
m
k
k
≡ω+2 ,
x2 (t) ≈ a sin (ω + )t sin( t),
a sin(εt)
t
(4.101)
≡ (κ/2k)ω = (κ/2m) m/k, we can write x1 and x2 as
x1 (t) ≈ a cos (ω + )t cos( t),
Fig. 4.30
(4.100)
If we now approximate ω˜ as
x1(t)
x2(t)
(4.99)
(4.102)
ω,
as desired. To get an idea of what this motion looks like, let’s examine x1 . Since
the cos( t) oscillation is much slower than the cos((ω + )t) oscillation. Therefore,
cos( t) is essentially constant on the time scale of the cos((ω + )t) oscillations. This
means that for the time span of a few of these oscillations, x1 essentially oscillates with
frequency ω + ≈ ω and amplitude a cos( t). This a cos( t) term is the “envelope”
of the oscillation, as shown in Fig. 4.30, for /ω = 1/10. Initially, the amplitude of
x1 is a, but it decreases to zero when t = π/2. By this time, the amplitude of the x2
4.8 Solutions
oscillation, which is a sin( t), has grown to a. So at t = π/2 , the right mass has all
the motion, and the left mass is at rest. This process keeps repeating. After each time
period of π/2 , the motion of one mass gets transferred to the other. The weaker the
coupling (that is, the spring constant κ) between the masses, the smaller the , and so
the longer the time period.
Remarks: The above reasoning also holds for two pendulums connected by a weak
spring. All the above steps carry through, with the only change being that k is replaced
by mg/ , because the spring force, −kx, is replaced by the tangential gravitational
force, −mg sin θ ≈ −mg(x/ ). So after a time
t=
π
π
=
2
2
2m
κ
k
m
−→
πm
κ
g
,
(4.103)
the pendulum that was initially oscillating is now momentarily at rest, and the other
pendulum has all the motion. Since√
the time scale, Ts , of a single mass on the end of
the weak spring is proportional to m/κ, and since the time scale, Tp , of a simple
√
pendulum is proportional to
/g, we see that the above t is proportional to Ts2 /Tp .
The existence of the “beats” in Fig. 4.30 can be traced to the fact that the expressions
in Eq. (4.98) are linear combinations of sinusoidal functions with two very close
frequencies. The physics here is the same as the physics that produces the beats you
hear when listening to two musical notes of nearly the same pitch, as when tuning
a guitar.8 The time between the zeros of, say, x1 in Fig. 4.30 is π/ , so the angular
frequency of the beats is 2π/(π/ ) = 2 . ♣
4.11. Driven mass on a circle
Label two diametrically opposite points as the equilibrium positions. Let the positions
of the masses relative to these points be x1 and x2 , measured counterclockwise. If the
driving force acts on mass “1,” then the F = ma equations are
m¨x1 + 2k(x1 − x2 ) = Fd cos ωd t,
m¨x2 + 2k(x2 − x1 ) = 0.
(4.104)
To solve these equations, we can treat the driving force as the real part of Fd eiωd t
and try solutions of the form x1 (t) = A1 eiωd t and x2 (t) = A2 eiωd t , and then solve for
A1 and A2 . Or we can try some trig functions. If we take the latter route, we quickly
find that the solutions can’t involve any sine terms (this is due to the fact that there
are no first derivatives of the x’s in Eq. (4.104)). Therefore, the trig functions must
look like x1 (t) = A1 cos ωd t and x2 (t) = A2 cos ωd t. Using either of the two methods,
Eq. (4.104) becomes
−ωd2 A1 + 2ω2 (A1 − A2 ) = F,
−ωd2 A2 + 2ω2 (A2 − A1 ) = 0,
(4.105)
where ω ≡ k/m and F ≡ Fd /m. Solving for A1 and A2 , we find that the desired
particular solution is
x1 (t) =
8
−F 2ω2 − ωd2
ωd2 4ω2 − ωd2
cos ωd t,
x2 (t) =
−2Fω2
cos ωd t.
ωd2 4ω2 − ωd2
(4.106)
If the two frequencies involved aren’t too close to each other, then you can actually hear a faint
note with a frequency equal to the difference of the original frequencies (and possibly some other
notes too, involving various combinations of the frequencies). But this is a different phenomenon
from the above beats; it is due to the nonlinear way in which the ear works. See Hall (1981) for
more details.
133
134
Oscillations
The most general solution is the sum of this particular solution and the homogeneous
solution found in Eq. (4.111) in the solution to Problem 4.12 below.
Remarks:
1. If ωd = 2ω, the amplitudes of the motions go to infinity. This makes sense,
considering that there is no damping, and that the natural frequency of the system
(calculated
√ in Problem 4.12) is 2ω.
2. If ωd = 2ω, then the mass that is being driven doesn’t move. The reason for
this is that the driving force balances the force that the mass feels from the
√ two
springs due to the other mass’s motion. And indeed, you can show that 2ω is
the frequency that one mass moves at if the other√mass is at rest (and thereby
acts essentially like a brick wall). Note that ωd = 2ω is the cutoff between the
masses moving in the same direction or in opposite directions.
3. If ωd → ∞, then both motions go to zero. But x2 is fourth-order small, whereas
x1 is only second-order small.
4. If ωd → 0, then A1 ≈ A2 ≈ −F/2ωd2 , which is very large. The slowly changing
driving force basically spins the masses around in one direction for a while, and
then reverses and spins them around in the other direction. We essentially have the
driving force acting on a mass 2m, and two integrations of Fd cos ωd t = (2m)¨x
show that the amplitude of the motion is F/2ωd2 , as above. Equivalently, you can
calculate the A1 − A2 difference in the ωd → 0 limit to show that the springs
stretch just the right amount to cause there to be a net force of (Fd /2) cos ωd t on
each mass. This leads to the same F/2ωd2 amplitude. ♣
4.12. Springs on a circle
(a) Label two diametrically opposite points as the equilibrium positions. Let
the positions of the masses relative to these points be x1 and x2 , measured
counterclockwise. Then the F = ma equations are
m¨x1 + 2k(x1 − x2 ) = 0,
m¨x2 + 2k(x2 − x1 ) = 0.
(4.107)
The determinant method works here, but let’s just do it the easy way. Adding
the equations gives
x¨ 1 + x¨ 2 = 0,
(4.108)
(¨x1 − x¨ 2 ) + 4ω2 (x1 − x2 ) = 0.
(4.109)
and subtracting them gives
The normal coordinates are therefore
x1 + x2 = At + B,
x1 − x2 = C cos(2ωt + φ).
(4.110)
Solving these two equations for x1 and x2 , and writing the results in vector form,
gives
x1
x2
=
1
1
(At + B) + C
cos(2ωt + φ),
1
−1
(4.111)
where the constants A, B, and C are defined to be half of what they were in
Eq. (4.110). The normal modes are therefore
1
(At + B),
1
x1
x2
=
x1
x2
1
=C
cos(2ωt + φ).
−1
and
(4.112)
4.8 Solutions
The first mode has frequency zero. It corresponds to the masses sliding around
the circle, equally spaced, at constant speed. The second mode has both masses
moving to the left, then both to the right, back and forth. Each mass feels a force
of
√ 4kx (because there are two springs, and each one stretches by 2x), hence the
4 = 2 in the frequency.
(b) Label three equally spaced points as the equilibrium positions. Let the positions of the masses relative to these points be x1 , x2 , and x3 , measured
counterclockwise. Then the F = ma equations are, as you can show,
m¨x1 + k(x1 − x2 ) + k(x1 − x3 ) = 0,
m¨x2 + k(x2 − x3 ) + k(x2 − x1 ) = 0,
(4.113)
m¨x3 + k(x3 − x1 ) + k(x3 − x2 ) = 0.
The sum of all three of these equations definitely gives something nice. Also,
differences between any two of the equations give something useful. But let’s
use the determinant method to get some practice. Trying solutions of the form
x1 = A1 eiαt , x2 = A2 eiαt , and x3 = A3 eiαt , we obtain the matrix equation,
⎛ 2
⎞⎛ ⎞ ⎛ ⎞
−α + 2ω2
−ω2
−ω2
A1
0
2
2
2
2
⎝ −ω
⎠ ⎝A2 ⎠ = ⎝0⎠ .
(4.114)
−α + 2ω
−ω
A3
0
−ω2
−ω2
−α 2 + 2ω2
Setting the determinant equal to zero yields a cubic equation in α 2 . But it’s a
nice cubic equation, with α 2 = 0 as a solution. The other solution is the double
root α 2 = 3ω2 .
The α = 0 root corresponds to A1 = A2 = A3 . That is, it corresponds to the
vector (1, 1, 1). This α = 0 case is the one case where our exponential solution
isn’t really an exponential. But α 2 equalling zero in Eq. (4.114) basically tells
us that we’re dealing with a function whose second derivative is zero, that is, a
linear function At + B. Therefore, the normal mode is
⎛ ⎞ ⎛ ⎞
1
x1
⎝x2 ⎠ = ⎝1⎠ (At + B).
(4.115)
x3
1
This mode has frequency zero. It corresponds to the masses sliding around the
circle, equally spaced, at constant speed.
The two α 2 = 3ω2 roots correspond to a two-dimensional subspace of normal
modes. You can show that any vector
√ of the form (a, b, c) with a + b + c = 0
is a normal mode with frequency 3ω. We will arbitrarily pick the vectors
(0, 1, −1) and (1, 0, −1) as basis vectors for this space. We can then write the
normal modes as linear combinations of the vectors
⎛ ⎞
⎛ ⎞
0
x1
√
⎝x2 ⎠ = C1 ⎝ 1 ⎠ cos( 3ωt + φ1 ),
x3
−1
(4.116)
⎛ ⎞
⎛ ⎞
1
x1
√
⎝x2 ⎠ = C2 ⎝ 0 ⎠ cos( 3ωt + φ2 ).
x3
−1
Remarks: The α 2 = 3ω2 case is very similar to the example in Section 4.5
with two masses and three springs oscillating between two walls. The way we’ve
written the two modes in Eq. (4.116), the first one has the first mass stationary
(so there could be a wall there,
√for all the other two masses know). Similarly for
the second mode. Hence the 3ω result here, as in the example.
135
136
Oscillations
The normal coordinates in this problem are x1 +x2 +x3 (obtained by adding the
three equations in (4.113)), and also any combination of the form ax1 +bx2 +cx3 ,
where a + b + c = 0 (obtained by taking a times the first equation in Eq. (4.113),
plus b times the second, plus c times the third). The three normal coordinates
that correspond to the mode in Eq. (4.115) and the two modes we chose in Eq.
(4.116) are, respectively, x1 + x2 + x3 , x1 − 2x2 + x3 , and −2x1 + x2 + x3 ,
because each of these combinations gets no contribution from the other two
modes (demanding this is how you can derive the coefficients of the xi ’s, up to
an overall constant). ♣
(c) In part (b), when we set the determinant of the matrix in Eq. (4.114) equal to zero,
we were essentially finding the eigenvectors and eigenvalues9 of the matrix,
⎛
⎞
⎛
⎞
2
−1 −1
1 1 1
⎝−1
2
−1⎠ = 3I − ⎝1 1 1⎠ ,
(4.117)
−1 −1
2
1 1 1
where I is the identity matrix. We haven’t bothered writing the common factor
ω2 here, because it doesn’t affect the eigenvectors. As an exercise, you can show
that for the general case of N springs and N masses on a circle, the above matrix
becomes the N × N matrix,
⎛
⎞
1 1 0 0
1
⎜ 1 1 1 0 ··· 0 ⎟
⎜
⎟
⎜ 0 1 1 1
0 ⎟
⎜
⎟
3I − ⎜ 0 0 1 1
(4.118)
0 ⎟ ≡ 3I − M .
⎜
⎟
⎜
⎟
..
.
..
.
⎝
. . ⎠
.
1 0 0 0 ··· 1
In the matrix M , the three consecutive 1’s keep shifting to the right, and they
wrap around cyclicly. We must now find the eigenvectors of M , which will
require being a little clever.
We can guess the eigenvectors and eigenvalues of M if we take a hint from
its cyclic nature. A particular set of things that are rather cyclic are the N th roots
of 1. If β is an N th root of 1, you can verify that (1, β, β 2 , . . . , β N −1 ) is an
eigenvector of M with eigenvalue β −1 + 1 + β. (This general method works for
any matrix where the entries keep shifting to the right. The entries don’t have
to be equal.) The eigenvalues of the entire matrix in Eq. (4.118) are therefore
3 − (β −1 + 1 + β) = 2 − β −1 − β. There are N different N th roots of 1, namely
βn = e2πin/N , for 0 ≤ n ≤ N − 1. So the N eigenvalues are
λn = 2 − e−2π in/N + e2πin/N = 2 − 2 cos(2πn/N )
= 4 sin2 (πn/N ).
(4.119)
The corresponding eigenvectors are
Vn = 1, βn , βn2 , . . . , βnN −1 .
(4.120)
Since the numbers n and N − n yield the same value for λn in Eq. (4.119),
the eigenvalues come in pairs (except for n = 0, and n = N /2 if N is even).
This is fortunate, because we can then form real linear combinations of the
9
An eigenvector v of a matrix M is a vector that gets taken into a multiple of itself when acted
upon by M . That is, M v = λv, where λ is some number (the eigenvalue). This can be rewritten as
(M − λI )v = 0, where I is the identity matrix. By our usual reasoning about invertible matrices,
a nonzero vector v exists only if λ satisfies det |M − λI | = 0.
4.8 Solutions
two corresponding complex eigenvectors given in Eq. (4.120). We see that the
vectors
⎛
⎞
1
⎜
⎟
cos(2π n/N )
⎜
⎟
1
⎜
⎟
cos(4π
n/N
)
+
Vn ≡ (Vn + VN −n ) = ⎜
(4.121)
⎟
⎜
⎟
..
2
⎝
⎠
.
cos 2(N − 1)πn/N
and
⎛
Vn−
⎜
⎜
1
⎜
≡ (Vn − VN −n ) = ⎜
⎜
2i
⎝
0
sin(2π n/N )
sin(4π n/N )
..
.
sin 2(N − 1)πn/N
⎞
⎟
⎟
⎟
⎟
⎟
⎠
(4.122)
both have eigenvalue λn = λN −n (as does any linear combination of these
vectors). For the special case of n = 0, the eigenvector is V0 = (1, 1, 1, . . . , 1)
with eigenvalue λ0 = 0. And for the special case of n = N /2 if N is even, the
eigenvector is VN /2 = (1, −1, 1, . . . , −1) with eigenvalue λN /2 = 4.
Referring back to the N = 3 case in Eq. (4.114), we see that we must take the
square root of the eigenvalues and then multiply by ω to obtain the frequencies
(because it was an α 2 that appeared in the matrix, and because we dropped the
factor of ω2 ). The frequency corresponding to the above two normal modes is
therefore, using Eq. (4.119),
ωn = ω λn = 2ω sin(π n/N ).
(4.123)
For even N , the largest value of the frequency is 2ω, with the masses moving
in alternating equal positive and negative displacements. But for odd N , it is
slightly less than 2ω.
To sum everything up, the N normal modes are the vectors in Eqs. (4.121)
and (4.122), where n runs from 1 up to the greatest integer less than N /2. And
then we have to add on the V0 vector, and also the VN /2 vector if N is even.10
The frequencies are given in Eq. (4.123). Each frequency is associated with two
modes, except the V0 mode and the VN /2 mode if N is even.
Remark: Let’s check our results for N = 2 and N = 3. For N = 2: The values
of n are the two “special” cases of n = 0 and n = N /2 = 1. If n = 0, we have
ω0 = 0 and V0 = (1, 1). If n = 1, we have ω1 = 2ω and V1 = (1, −1). These
results agree with the two modes in Eq. (4.112).
For N = 3: If n = 0, we have ω0 = 0√and V0 = (1, 1, 1), in agreement
with Eq. (4.115). If n = 1, we have ω1 = 3ω, and V1+ = (1, −1/2, −1/2)
and V1− = (0, 1/2, −1/2). These two vectors span the same space we found in
Eq. (4.116). And they have the same frequency as in Eq. (4.116). You can also
find the vectors for N = 4. These are fairly intuitive, so try to write them down
first without using the above results. ♣
10
If you want, you can treat the n = 0 and n = N /2 cases the same as all the others. But in both of
these cases, the V − vector is the zero vector, so you can ignore it. So no matter what route you
take, you will end up with exactly N nontrivial eigenvectors.
137
Chapter 5
Conservation of energy and
momentum
Conservation laws are extremely important in physics. They are enormously
helpful, both quantitatively and qualitatively, in figuring out what is going on in
a physical system. When we say that something is “conserved,” we mean that
it is constant over time. If a certain quantity is conserved, for example, while a
ball rolls around in a valley, or while a group of particles interact, then the possible final motions are greatly restricted. If we can write down enough conserved
quantities (which we are generally able to do, at least for the systems we’ll be
concerned with), then we can restrict the final motions down to just one possibility, and so we have solved our problem. Conservation of energy and momentum
are two of the main conservation laws in physics. A third, conservation of angular
momentum, is discussed in Chapters 7–9.
It should be noted that it isn’t necessary (in principle) to use conservation of
energy and momentum when solving a problem. We’ll derive these conservation
laws from Newton’s laws. Therefore, if you felt like it, you could always (in
theory) simply start with first principles and use F = ma, etc. However, at best,
you would soon grow weary of this approach. And at worst, you would throw
in the towel after finding the problem completely intractable. For example, you
would get nowhere trying to analyze the collision between two shopping carts
(whose contents are free to shift around) by looking at the forces on all the various
objects. But conservation of momentum can quickly give you a great deal of
information. The point of conservation laws is that they make your calculations
much easier, and they also provide a means for getting a good idea of the overall
qualitative behavior of a system.
5.1
Conservation of energy in one dimension
Consider a force, in just one dimension for now, that depends only on position.
That is, F = F(x). If the force acts on a particle of mass m, and if we write a as
v dv/dx, then F = ma becomes
F(x) = mv
138
dv
.
dx
(5.1)
5.1 Conservation of energy in one dimension
We can separate variables here and integrate from a given point x0 where the
velocity is v0 to an arbitrary point x where the velocity is v. The result is
x
x0
F(x ) dx =
v
v0
mv dv
x
=⇒
F(x ) dx =
x0
=⇒
E=
1 2
mv −
2
1 2 1 2
mv − mv0
2
2
x
F(x ) dx ,
(5.2)
x0
where E ≡ mv02 /2. E depends on v0 , so it therefore also depends on the choice
of x0 , because a different x0 would yield (in general) a different v0 . What we’ve
done here is simply follow the procedure in Section 3.3, for a function that
depends only on x. If we now define the potential energy, V (x), as
x
V (x) ≡ −
F(x ) dx ,
(5.3)
x0
then Eq. (5.2) becomes
1 2
mv + V (x) = E.
2
(5.4)
We define the first term here to be the kinetic energy. Since this equation is
true at all points in the particle’s motion, the sum of the kinetic energy and
potential energy is constant. In other words, the total energy is conserved. If a
particle loses (or gains) potential energy, then its speed increases (or decreases).
Common examples of potential energy are kx2 /2 for a Hooke’s-law spring force
(−kx), with x0 chosen to be zero; and mgy for the gravitational force (−mg), with
y0 chosen to be zero.
In Boston, lived Jack as did Jill,
Who gained mgh on a hill.
In their liquid pursuit,
Jill exclaimed with a hoot,
“I think we’ve just climbed a landfill!”
While noting, “Oh, this is just grand,”
Jack tripped on some trash in the sand.
He changed his potential
To kinetic, torrential,
But not before grabbing Jill’s hand.
So that’s what really happened on that hill. People don’t just magically come
“tumbling after” for no reason, of course.
For a particle undergoing a given motion, both E and V (x) depend on the
arbitrary choice of x0 in Eq. (5.3). This implies that E and V (x) have no real
meaning by themselves. Only the difference between E and V (x) is relevant (and
139
140
Conservation of energy and momentum
it equals the kinetic energy); this difference is independent of the choice of x0 .
But in order to be concrete in a given setup, we need to pick an arbitrary x0 , so we
must remember to state which x0 we’ve chosen. For example, it makes no sense
to simply say that the gravitational potential energy of an object at height y above
the ground is − F dy = − (−mg) dy = mgy. We have to say that the potential
energy is mgy with respect to the ground (assuming that our y0 is at ground level).
If we wanted to, we could say that the potential energy is mgy + mg(7 m) with
respect to a point 7 meters below the ground. This is perfectly legitimate, although
a bit unconventional.1 But no matter what reference point we pick, the difference
in the potential energies at the points, say, y = 3 m and y = 5 m equals mg(2 m).
Note that although we introduced the x0 in Eq. (5.2) as a point where the
particle was at some time, the particle in fact need not ever be at the point x0 . For
example, we can throw a ball up at 8 m/s from a height of 5 m while measuring the
gravitational potential energy with respect to a point a kilometer high. The ball is
certainly not going to reach a height of a kilometer, but that’s fine. All that matters
is the difference between E and V (x) (both of which are very much negative here)
throughout the motion, so a constant shift is irrelevant. We’ve simply added on
the same negative quantity to both E and V (x) in Eq. (5.4), compared with the
values someone would measure if the ground were the reference point.
If we take the difference between Eq. (5.4) evaluated at two points, x1 and x2
(or if we just integrate Eq. (5.1) from x1 to x2 ), then we obtain
1 2
1
mv (x2 ) − mv 2 (x1 ) = V (x1 ) − V (x2 )
2
2
=
x2
x1
F(x ) dx ≡ (Work)x1 →x2 .
(5.5)
Here it is clear that only differences in potential energies matter. If we define the
integral in this equation to be the work done on the particle as it moves from x1
to x2 , then we have produced the work–energy theorem:2
Theorem 5.1 The change in a particle’s kinetic energy between points x1
and x2 equals the work done on the particle between x1 and x2 .
If the force points in the same direction as the motion (that is, if the F(x) and
the dx in Eq. (5.5) have the same sign), then the work is positive and the speed
increases. If the force points in the direction opposite to the motion, then the
work is negative and the speed decreases.
1
2
It gets to be a pain to keep repeating “with respect to the ground.” Therefore, whenever anyone talks
about gravitational potential energy in a setup on the surface of the earth, it’s generally understood
that the ground is the reference point. If, on the other hand, the experiment reaches out to distances
far from the earth, then r = ∞ is understood to be the reference point, for reasons of convenience
we will see in the first example below.
In the form stated here, this theorem holds only for a point particle with no internal structure. See
the “Work vs. potential energy” subsection below for the general theorem.
5.1 Conservation of energy in one dimension
Referring back to Eq. (5.4), and assuming we’ve chosen a reference point
x0 for the potential energy (and perhaps we’ve added on a constant to V (x),
just because we felt like it), let’s draw in Fig. 5.1 the V (x) curve and also the
constant E line (which we can determine if we’re given, say, the initial position
and speed). Then the difference between E and V (x) gives the kinetic energy. The
places where V (x) > E are the regions where the particle cannot go. The places
where V (x) = E are the “turning points” where the particle stops and changes
direction. In the figure, the particle is trapped between x1 and x2 , and oscillates
back and forth. The potential V (x) is extremely useful this way because it makes
clear the general properties of the motion.
141
V(x)
E
x1
x2
x
Fig. 5.1
Remark: It may seem silly to introduce a specific x0 as a reference point, considering that
it’s only the differences in the potential (which are independent of x0 ) that have any meaning.
It’s sort of like taking the difference between 17 and 8 by first finding their sizes relative to
5, namely 12 and 3, and then subtracting 3 from 12 to obtain 9. However, since integrals are
harder to do than simple subtractions, it’s advantageous to do the integral once and for all and
thereby label all positions with a definite number V (x), and to then take differences between
the V ’s when needed. ♣
The differential form of Eq. (5.3) is
F(x) = −
dV (x)
.
dx
(5.6)
Given V (x), it is easy to take its derivative to obtain F(x). But given F(x), it
may be difficult (or impossible) to perform the integration in Eq. (5.3) and write
V (x) in closed form. But this is not of much concern. The function V (x) is well
defined (assuming that the force is a function of x only), and if needed it can be
computed numerically to any desired accuracy.
Example (Gravitational potential energy): Consider two point masses, M
and m, separated by a distance r. Newton’s law of gravitation says that the force
between them is attractive and has magnitude GMm/r 2 (we’ll talk more about gravity
in Section 5.4.1). The potential energy of the system at separation r, measured relative
to separation r0 , is therefore
r −GMm
−GMm GMm
+
dr =
,
V (r) − V (r0 ) = −
2
r
r0
r
r0
V(r)
r
(5.7)
_____
V(r) = -GMm
r
where the minus sign in the integrand comes from the attractive nature of the force. A
convenient choice for r0 is ∞, because this makes the second term vanish. It will be
understood from now on that this r0 = ∞ reference point has been chosen. Therefore
(see Fig. 5.2),
Fig. 5.2
−GMm
V (r) =
.
r
(5.8)
142
Conservation of energy and momentum
Example (Gravity near the earth): What is the gravitational potential energy
of a mass m at height y, relative to the ground? We know, of course, that it is
mgy, but let’s do it the hard way. If M is the mass of the earth and R is its radius,
then Eq. (5.8) gives (assuming y
R)
V (R + y) − V (R) =
−GMm −GMm
−GMm
−
=
R+y
R
R
1
−1
1 + y/R
−GMm
(1 − y/R) − 1
R
GMmy
,
=
R2
≈
(5.9)
where we have used the Taylor series approximation for 1/(1 + ) to obtain the second
line. We have also used the fact that a sphere can be treated like a point mass, as far
as gravity is concerned. We’ll prove this in Section 5.4.1.
Using g ≡ GM/R2 , we see that the potential energy difference in Eq. (5.9) equals
mgy. We have, of course, simply gone around in circles here. We integrated in
Eq. (5.7), and then we basically differentiated in Eq. (5.9) by taking the difference
between the forces at nearby points. But it’s good to check that everything works out.
A good way to visualize a potential V (x) is to imagine a ball sliding around in a
valley or on a hill. For example, the potential of a typical spring is V (x) = kx2/2
(this produces the Hooke’s-law force, F(x) = −dV/dx = −kx), and we can get
a decent idea of what is going on if we imagine a valley with height given by
y = x2/2. The gravitational potential of the ball is then mgy = mgx2/2. Choosing
mg = k gives the desired potential. If we then look at the projection of the ball’s
motion on the x axis, it seems like we have constructed a setup identical to the
original spring.
However, although this analogy helps in visualizing the basic properties of
the motion, the two setups are not the same. The details of this fact are left
for Problem 5.7, but the following observation should convince you that they
are indeed different. Let the ball be released from rest in both setups at a large
value of x. Then the force kx due to the spring is very large. But the force in
the x direction on the particle in the valley is only a fraction of mg, namely
(mg sin θ ) cos θ , where θ is the angle of the valley at that point. The setups are
approximately the same, however, for small oscillations near the bottom of the
valley. See Problem 5.7 for more details.
Conservative forces
Given any force (it can depend on x, v, t, and/or whatever), the work it does on
a particle is defined by W ≡ F dx. If the particle starts at x1 and ends up at
x2 , then no matter how it gets there (it might speed up or slow down, or reverse
direction a few times), we can calculate the total work done on it by all the forces
5.1 Conservation of energy in one dimension
in the setup, and then equate the result with the change in kinetic energy, via
Wtotal ≡
x2
x1
Ftotal dx =
x2
m
x1
v dv
1
1
dx = mv22 − mv12 .
dx
2
2
(5.10)
For some forces, the work done is independent of how the particle moves.
A force that depends only on position (in one dimension) has this property,
because the integral in Eq. (5.10) depends only on the endpoints. The W = F dx
integral is the (signed) area under the F vs. x graph, and this area is independent
of how the particle goes from x1 to x2 .
For other forces, the work done depends on how the particle moves. Such is
the case for forces that depend on t or v, because it then matters when or how
quickly the particle goes from x1 to x2 . A common example of such a force is
friction. If you slide a brick across a table from x1 to x2 , then the work done
by friction equals −µmg| x|. But if you slide the brick by wiggling it back
and forth for an hour before you finally reach x2 , then the amount of negative
work done by friction is very large. Since friction always opposes the motion, the
contributions to the W = F dx integral are always negative, so there is never
any cancellation. The result is therefore a large negative number.
The issue with friction is that although the µmg force looks like a constant
force (which is a subset of position-only dependent forces), it actually isn’t. At a
given location, the friction can point to the right or to the left, depending on
which way the particle is moving. Friction is therefore a function of velocity.
True, it’s a function only of the direction of the velocity, but that’s enough to ruin
the position-only dependence.
We now define a conservative force as one for which the work done on a
particle between two given points is independent of how the particle makes the
journey. From the preceding discussion, we know that a one-dimensional force
is conservative if and only if it depends only on x (or is constant).3 The point
we’re leading up to here is that although we can calculate the work done by any
force, it makes sense to talk about the potential energy associated with a force
only if the force is conservative. This is true because we want to be able to label
x
each value of x with a unique number, V (x), given by V (x) = − x0 F dx. If this
integral were dependent on how the particle goes from x0 to x, then it wouldn’t be
well defined, so we wouldn’t know what number to assign to V (x). We therefore
talk about potential energies only if they are associated with conservative forces.
In particular, it makes no sense to talk about the potential energy associated with
a friction force.
A useful fact about the gravitational potential energy, mgz, is that it doesn’t
depend on the path the particle takes, even in two or three dimensions. This
is true because even if the particle moves in a complicated direction, only the
3
In two or three dimensions, however, we will see in Section 7 that a conservative force must satisfy
another requirement, in addition to being dependent only on position.
143
144
Conservation of energy and momentum
vertical z component of the displacement is relevant in calculating the work done
by gravity. If we break the path up into many little pieces, the total work done
by gravity is obtained by adding up the many little −mg(dz) terms. But the sum
of all the dz’s is always equal to the total z, independent of the path. Therefore,
no matter what the particle is doing in the two horizontal directions, the change
in gravitational potential energy is always just mgz. So the gravitational force is
a conservative force in three dimensions. We’ll see in Section 5.3 that this is a
special case of a more general result.
Example (Unwinding string): A mass is connected to one end of a massless
string, the other end of which is connected to a very thin frictionless vertical pole.
The string is initially wound completely around the pole, in a very large number of
tiny horizontal circles, so that the mass touches the pole. The mass is released, and
the string gradually unwinds. What angle does the string make with the pole at the
moment it becomes completely unwound?
Solution: Let be the length of the string, and let θ be the final angle it makes with
the pole. Then the final height of the mass is cos θ below the starting point. So the
mass loses a potential energy of mg( cos θ). Conservation of energy therefore gives
(picking the initial height as y = 0, although this doesn’t matter)
Ki + Vi = Kf + Vf
=⇒ 0 + 0 =
1 2
mv − mg cos θ
2
=⇒ v 2 = 2g cos θ.
(5.11)
There are two unknowns here, v and θ, so we need one more equation. This will be the
radial F = ma equation for the (essentially) final circular motion. Because the pole is
very thin, the motion can always be approximated by a horizontal circle, which very
slowly lowers as time goes by. Because there is essentially no motion in the vertical
direction, the total force in this direction is zero. Therefore, the vertical component
of the tension is essentially mg. The horizontal component is then mg tan θ, so the
F = ma equation for the final circular motion (which has a radius of sin θ) is
mg tan θ =
mv 2
sin θ
=⇒
v 2 = g sin θ tan θ.
Equating our two expressions for v 2 gives tan θ =
this angle is independent of and g.
(5.12)
√
2 =⇒ θ ≈ 54.7◦ . Interestingly,
Work vs. potential energy
When you drop a ball, does its speed increase because the gravitational force is
doing work on it, or because its gravitational potential energy is decreasing? Well,
both (or more precisely, either). Work and potential energy are two different ways
of talking about the same thing (at least for conservative forces). Either method of
5.1 Conservation of energy in one dimension
reasoning gives the correct result. However, be careful not to use both reasonings
and “double count” the effect of gravity on the ball. Which choice of terminology
you use depends on what you call your “system.” Just as with F = ma and freebody diagrams, it is important to label your system when dealing with work and
energy, as we’ll see in the example below.
The work–energy theorem stated in Theorem 5.1 is relevant to one particle.
What if we are dealing with the work done on a system that is composed of
various parts? The general work–energy theorem states that the work done on a
system by external forces equals the change in energy of the system. This energy
may come in the form of (1) overall kinetic energy, (2) internal potential energy,
or (3) internal kinetic energy (heat falls into this category, because it’s simply
the random motion of molecules). So we can write the general work–energy
theorem as
Wexternal =
K+
V+
Kinternal
(5.13)
For a point particle, there is no internal structure, so we have only the first of
the three terms on the right-hand side, in agreement with Theorem 5.1. But to
see what happens when a system has internal structure, consider the following
example.
Example (Raising a book): Assume that you lift a book up at constant speed, so
there is no change in kinetic energy. Let’s see what the general work–energy theorem
says for various choices of the system.
• System = (book): Both you and gravity are external forces, and there is no
change in the energy of the book as a system in itself. So the W–E theorem says
Wyou + Wgrav = 0
⇐⇒
mgh + (−mgh) = 0.
(5.14)
• System = (book + earth): Now you are the only external force. The gravitational
force between the earth and the book is an internal force which produces an
internal potential energy. So the W–E theorem says
Wyou =
Vearth–book
⇐⇒
mgh = mgh.
(5.15)
• System = (book + earth + you): There is now no external force. The internal
energy of the system changes because the earth–book gravitational potential
energy increases, and also because your potential energy decreases. In order to
lift the book, you have to burn some calories from the dinner you ate. So the
W–E theorem says
0=
Vearth–book +
Vyou
⇐⇒
0 = mgh + (−mgh).
(5.16)
145
146
Conservation of energy and momentum
Actually, a human body isn’t 100% efficient, so what really happens here is that
your potential energy decreases by more than mgh, but heat is produced. The
sum of these two changes in energy equals −mgh. So, including an amount η of
energy in the form of heat, we have
0=
⇐⇒
Vearth–book +
Vyou +
0 = mgh + (−mgh − η) + η.
Kinternal
(5.17)
Another contribution to this η heat term can come from, for example, the heat
from friction if you slide the book up a rough wall as you lift it.4
The moral of all this is that you can look at a setup in various ways, depending
on what you pick as your system. Potential energy in one way might show up
as work in another. In practice, it is usually more convenient to work in terms
of potential energy. So for a dropped ball, people usually consider (consciously
or not) gravity to be an internal force in the earth–ball system, as opposed to an
external force on the ball system. In general, “conservation of energy,” commonly
used in setups involving gravity and/or springs, is a straightforward principle to
apply (and you’ll get plenty of practice with it in the problems and exercises for
this chapter). So it turns out that you can usually ignore all these issues about
work and about picking your system.
But let’s look at one more example, just to make sure we’re on the same
page. Consider a car that is braking (but not skidding). The friction force from
the ground on the tires is what causes the car to slow down. But this force
does no work on the car, because the ground isn’t moving; the force acts over
zero distance. So the external work on the left side of Eq. (5.13) is zero. The
right side is therefore also zero. That is, the total energy of the car doesn’t
change. This is indeed true, because although the overall kinetic energy of the
car decreases, there is an equal increase in internal kinetic energy in the form
of heat in the brake pads and discs. In other words, K = − Kinternal , and the
total energy remains constant. The unfortunate fact about this process is that the
energy that goes into heat is lost and can’t be converted back into overall kinetic
energy of the car. It makes much more sense to convert the overall kinetic energy
into some form of internal potential energy (that is, K = − U ), which can
then be converted back into overall kinetic energy. Such is the case with hybrid
cars which convert overall kinetic energy into chemical potential energy in a
battery.
Conversely, when a car accelerates, the friction force from the ground does
no work (because the ground isn’t moving), so the total energy of the car remains
4
In this case we have included the wall as a fourth object in our system, because it might contain
some of the heat. If you want to instead consider the wall as an external object providing a force,
then things get tricky; see Problem 5.6.
5.2 Small oscillations
the same. The internal potential energy of the gasoline (or battery) is converted
into overall kinetic energy (along with some heat and sound). A similar thing
happens when you stand at rest and then start walking. The friction force from
the ground does no work on you, so your total energy remains the same. You’re
simply trading your breakfast for overall kinetic energy (plus some heat). For
further discussion of work, see Mallinckrodt and Leff (1992).
5.2
Small oscillations
Consider an object in one dimension, subject to the potential V (x). Let the object
initially be at rest at a local minimum of V (x), and then let it be given a small
kick so that it moves back and forth around the equilibrium point. What can we
say about this motion? Is it simple harmonic? Does the frequency depend on the
amplitude?
It turns out that for small amplitudes, the motion is indeed simple harmonic,
and the frequency can easily be found, given V (x). To see this, expand V (x) in
a Taylor series around the equilibrium point, x0 ,
V (x) = V (x0 ) + V (x0 )(x − x0 ) +
+
1
V (x0 )(x − x0 )2
2!
1
V (x0 )(x − x0 )3 + · · · .
3!
(5.18)
This looks like a bit of a mess, but we can simplify it greatly. V (x0 ) is an irrelevant
additive constant. We can ignore it because only differences in energy matter (or
equivalently, because F = −dV/dx). And V (x0 ) = 0, by definition of the
equilibrium point. So that leaves us with the V (x0 ) and higher-order terms.
But for sufficiently small displacements, these higher-order terms are negligible
compared with the V (x0 ) term, because they are suppressed by additional powers
of (x − x0 ). So we are left with5
V (x) ≈
1
V (x0 )(x − x0 )2 .
2
(5.19)
But this has exactly the same form as the Hooke’s-law potential, V (x) =
(1/2)k(x − x0 )2 , provided that we let V (x0 ) be our “spring constant” k. Equivalently, the force is F = −dV/dx = −V (x0 )(x − x0 ) ≡ −k(x − x0 ). The
frequency of small oscillations, ω = k/m, is therefore
ω=
5
V (x0 )
.
m
(5.20)
Even if V (x0 ) is much larger than V (x0 ), we can always pick (x − x0 ) small enough so that
the third-order term is negligible. The one case where this is not true is when V (x0 ) = 0. But the
result in Eq. (5.20) is still valid in this case. The frequency ω just happens to be zero, in the limit
of infinitesimal oscillations.
147
148
Conservation of energy and momentum
V(x)
Example: A particle moves under the influence of the potential V (x) =
A/x2 − B/x, where A, B > 0. Find the frequency of small oscillations around the
equilibrium point. This potential is relevant to planetary motion, as we’ll see in
Chapter 7. The rough shape is shown in Fig. 5.3.
x
Solution: The first thing we need to do is calculate the equilibrium point x0 . The
minimum occurs where
2A
B
0 = V (x) = − 3 + 2
x
x
Fig. 5.3
=⇒
x=
2A
≡ x0 .
B
(5.21)
The second derivative of V (x) is
6A 2B
V (x) = 4 − 3 .
x
x
(5.22)
Plugging in x0 = 2A/B, we find
ω=
parabola
V(x)
V (x0 )
=
m
B4
.
8mA3
(5.23)
Equation (5.20) is an important result, because any function V (x) looks basically like a parabola (see Fig. 5.4) in a small enough region around a minimum
(except in the special case where V (x0 ) = 0).
A potential may look quite erratic,
And its study may seem problematic.
But down near a min,
You can say with a grin,
“It behaves like a simple quadratic!”
Fig. 5.4
5.3
Conservation of energy in three dimensions
The concepts of work and potential energy in three dimensions are somewhat
more complicated than in one dimension, but the general ideas are the same.6 As
in the 1-D case, we’ll start with Newton’s second law, which now takes the vector
form, F = ma. And as in the 1-D case, we’ll deal only with forces that depend
only on position, that is, F = F(r), because these are the only ones that have
any chance of being conservative. The F = ma vector equation is shorthand for
three equations analogous to Eq. (5.1), namely mvx (dvx /dx) = Fx , and likewise
for y and z. The Fx here is a function of position, so we should really be writing
6
We’ll invoke a few results from vector calculus here. If you haven’t seen such material before, a
brief review is given in Appendix B.
5.3 Conservation of energy in three dimensions
Fx (x, y, z), but we’ll drop the arguments, lest our expressions get too cluttered.
Multiplying through by dx, etc., in these three equations, and then adding them
together gives
Fx dx + Fy dy + Fz dz = m(vx dvx + vy dvy + vz dvz ).
(5.24)
The left-hand side is the work done on the particle. With dr ≡ (dx, dy, dz),
this work can be written as F · dr (see Appendix B for the definition of the “dot
product”). Using Eq. (B.2), we see that the work can also be written as F|dr| cos θ ,
where θ is the angle between F and dr. Grouping this as (F cos θ)|dr| shows
that the work equals the distance moved times the component of the force along
the displacement. Alternatively, grouping it as F(|dr| cos θ) shows that the work
also equals the magnitude of the force times the component of the displacement
in the direction of the force.
Integrating Eq. (5.24) from the point (x0 , y0 , z0 ) to the point (x, y, z) yields7
E+
x
Fx dx +
x0
y
Fy dy +
y0
z
Fz dz =
z0
1
1
m(vx2 + vy2 + vz2 ) = mv 2 ,
2
2
(5.25)
where E is a constant of integration; it equals mv02 /2, where v0 is the speed at
(x0 , y0 , z0 ). Note that the integrations on the left-hand side depend on the path
in 3-D space that the particle takes in going from (x0 , y0 , z0 ) to (x, y, z), because
the components of F are functions of position. We’ll address this issue below.
In terms of the dot product, Eq. (5.25) can be written in the more compact form,
1 2
mv −
2
r
F(r ) · dr = E.
(5.26)
r0
Therefore, if we define the potential energy V (r) as
V (r) ≡ −
r
F(r ) · dr ,
(5.27)
r0
then we can write
1 2
mv + V (r) = E.
2
(5.28)
In other words, the sum of the kinetic energy and potential energy is constant.
7
We’ve put primes on the integration variables so that we don’t confuse them with the limits of
integration. And as mentioned above, Fx is really Fx (x , y , z ), etc.
149
150
Conservation of energy and momentum
Conservative forces in three dimensions
For a force that depends only on position (as we have been assuming), there is
one complication that arises in 3-D that we didn’t have to worry about in 1-D.
In 1-D, there is only one route that goes from x0 to x. The motion itself may
involve speeding up or slowing down, or backtracking, but the path is always
restricted to be along the line containing x0 and x. But in 3-D, there is an infinite
number of routes that go from r0 to r. In order for the potential V (r) to have
any meaning and to be of any use, it must be well defined. That is, it must be
path-independent. As in the 1-D case, we call the force associated with such
a potential a conservative force. Let’s now see what types of 3-D forces are
conservative.
Theorem 5.2 Given a force F(r), a necessary and sufficient condition for the
potential,
V (r) ≡ −
r
F(r ) · dr ,
(5.29)
r0
to be well defined (that is, to be path-independent) is that the curl of F is zero
everywhere (that is, ∇ × F = 0; see Appendix B for the definition of the curl).8
B
2
1
y
A
x
Fig. 5.5
S
C
2
r
r0
1
Proof: First, let us show that ∇ × F = 0 is a necessary condition for
path-independence. In other words, “If V (r) is path-independent, then ∇ ×
F = 0.” This follows quickly from the discussion of the curl in Appendix B.
We show in Eq. (B.24) that the integral F · dr (that is, the work done) around a
little rectangle in the x-y plane equals the z component of the curl times the area.
If the work done in going from corner A to corner B in Fig. 5.5 is the same for
paths “1” and “2” (as we are assuming), then the round-trip integral around the
rectangle is zero, because one of the paths is being traced out backwards, so it
cancels the contribution from the other path. So path-independence implies that
the round-trip integral F · dr is zero for any arbitrary rectangle in the x-y plane.
Equation (B.24) therefore says that the z component of the curl must be zero
everywhere. Likewise for the y and x components. We have therefore shown that
∇ × F = 0 is a necessary condition for path independence.
Now let us show that it is sufficient. In other words, “If ∇ × F = 0, then
V (r) is path-independent.” The proof of sufficiency follows immediately from
Stokes’ theorem, which is stated in Eq. (B.25). This theorem implies that if
∇ × F = 0 everywhere, then C F · dr = 0 for any closed curve. But Fig. 5.6
shows that traversing the loop C counterclockwise entails traversing path “1” in
the “forward” direction, and then traversing path “2” in the “backward” direction.
Therefore, from the same reasoning as in the previous paragraph, the integrals
8
Fig. 5.6
If the force is infinite at any point, then the proof of sufficiency below (which is based on Stokes’
theorem) isn’t valid, and it turns out that a second condition is required; see Feng (1969). But we
won’t worry about that here.
5.3 Conservation of energy in three dimensions
from r0 to r along paths “1” and “2” are equal. This holds for arbitrary points r0
and r, and arbitrary curves C, so V (r) is path-independent.
Remark: Another way to show that ∇ ×F = 0 is a necessary condition for path-independence
(that is, “If V (r) is path-independent, then ∇ × F = 0”) is the following. If V (r) is pathindependent (and therefore well defined), then it is legal to write down the differential form of
Eq. (5.27), namely
dV (r) = −F(r) · dr ≡ −(Fx dx + Fy dy + Fz dz).
(5.30)
But another expression for dV is
dV (r) =
∂V
∂V
∂V
dx +
dy +
dz.
∂x
∂y
∂z
(5.31)
These two expressions must be equivalent for arbitrary dx, dy, and dz. So we have
(Fx , Fy , Fz ) = −
∂V ∂V ∂V
,
,
∂x ∂y ∂z
=⇒
F(r) = −∇V (r).
(5.32)
In other words, the force is the gradient of the potential. Therefore,
∇ × F = −∇ × ∇V (r) = 0,
(5.33)
because the curl of a gradient is identically zero, as you can verify by using the definition of the
curl in Eq. (B.20) and the fact that partial differentiation is commutative (that is, ∂ 2 V/∂x ∂y =
∂ 2 V/∂y ∂x). ♣
Example (Central force): A central force is defined to be a force that points
radially and whose magnitude depends only on r. That is, F(r) = F(r)ˆr. Show that
a central force is conservative by explicitly showing that ∇ × F = 0.
Solution: The force F may be written as
x y z
, ,
.
r r r
F(x, y, z) = F(r)ˆr = F(r)
(5.34)
Now, as you can verify,
∂ x2 + y2 + z 2
x
∂r
=
= ,
∂x
∂x
r
(5.35)
and similarly for y and z. Therefore, the z component of ∇ × F equals (writing F
for F(r), and F for dF(r)/dr, and making use of the chain rule)
∂Fy
∂Fx
∂(yF/r) ∂(xF/r)
−
=
−
∂x
∂y
∂x
∂y
=
1 ∂r
y ∂r
F
− yF 2
r ∂x
r ∂x
=
yxF
yxF
− 3
r2
r
Likewise for the x and y components.
−
−
x ∂r
1 ∂r
F
− xF 2
r ∂y
r ∂y
xyF
xyF
− 3
r2
r
= 0.
(5.36)
151
152
Conservation of energy and momentum
5.4
5.4.1
Gravity
Newton’s universal law of gravitation
The gravitational force on a point mass m, located a distance r from a point mass
M, is given by Newton’s law of gravitation,
−GMm
,
(5.37)
r2
where the minus sign indicates an attractive force. The numerical value of G
is 6.67 · 10−11 m3 /(kg s2 ). We’ll show how this value is obtained in Section
5.4.2 below.
What is the force if we replace the point mass M by a sphere of radius R and
mass M ? The answer (assuming that the sphere is spherically symmetric, that
is, the density is a function only of r) is that it is still −GMm/r 2 . A sphere acts
just like a point mass at its center, for the purposes of gravity (as long as we’re
considering a mass m outside the sphere). This is an extremely pleasing result,
to say the least. If it were not the case, then the universe would be a far more
complicated place than it is. In particular, the motion of planets and such things
would be much harder to describe.
To demonstrate that spheres behave like points, as far as gravity is concerned,
it turns out to be much easier to calculate the potential energy due to a sphere, and
to then take the derivative to obtain the force, rather than to calculate the force
explicitly.9 So this is the route we will take. It will suffice to demonstrate the
result for a thin spherical shell, because a sphere is the sum of many such shells.
Our strategy for calculating the potential energy at a point P, due to a spherical
shell, will be to slice the shell into rings as shown in Fig. 5.7. Let the radius of
the shell be R, let P be a distance r from the center of the shell, and let the ring
make the angle θ shown. The distance, , from P to the ring is a function of R,
r, and θ . It may be found as follows. In Fig. 5.8, segment AB has length R sin θ ,
and segment BP has length r − R cos θ . So the length in triangle ABP is
F(r) =
l
R
θ
P
r
=
Fig. 5.7
A
R
l
R sin θ
θ
R cos θ B
Fig. 5.8
r - R cos θ
P
(R sin θ )2 + (r − R cos θ)2 =
R2 + r 2 − 2rR cos θ.
(5.38)
What we’ve done here is just prove the law of cosines.
The area of a ring between θ and θ + dθ is its width (which is R dθ) times its
circumference (which is 2π R sin θ ). Letting σ = M/(4π R2 ) be the mass density
per unit area of the shell, we see that the potential energy of a mass m at P due to
a thin ring is −Gmσ (R dθ)(2π R sin θ )/ . This is true because the gravitational
potential energy,
V( ) =
9
−Gm1 m2
,
(5.39)
The reason for this is that the potential energy is a scalar quantity (just a number), whereas the
force is a vector. If we tried to calculate the force, we would have to worry about forces pointing
in all sorts of directions. With the potential energy, we simply have to add up a bunch of numbers.
5.4 Gravity
is a scalar quantity, so the contributions from the little mass pieces simply add.
Every piece of the ring is the same distance from P, and this distance is all that
matters; the direction from P is irrelevant (unlike it would be with the force).
The total potential energy at P is therefore
π
2πσ GR2 m sin θ dθ
√
R2 + r 2 − 2rR cos θ
0
2πσ GRm
=−
R2 + r 2 − 2rR cos θ
r
V (r) = −
π
.
(5.40)
0
The sin θ in the numerator is what makes this integral nice and doable. We must
now consider two cases. If r > R, then we have
V (r) = −
2π σ GRm
G(4πR2 σ )m
GMm
(r + R) − (r − R) = −
=−
,
r
r
r
(5.41)
which is the potential due to a point mass M located at the center of the shell, as
desired. If r < R, then we have
V (r) = −
2π σ GRm
G(4πR2 σ )m
GMm
(r + R) − (R − r) = −
=−
,
r
R
R
(5.42)
which is independent of r. Having found V (r), we can now find F(r) by taking
the negative of the gradient of V . The gradient is just rˆ (d/dr) here, because V is
a function only of r. Therefore,
F(r) = −
F(r) = 0,
GMm
,
r2
if r > R,
(5.43)
if r < R.
These forces are directed radially, of course. A solid sphere is the sum of many
spherical shells, so if P is outside a given sphere, then the force at P is −GMm/r 2 ,
where M is the total mass of the sphere. This result holds even if the shells have
different mass densities (but each one must have uniform density). Note that the
gravitational force between two spheres is the same as if they were both replaced
by point masses. This follows from two applications of our “point mass” result.
Newton looked at the data, numerical,
And then observations, empirical.
He said, “But, of course,
We get the same force
From a point mass and something that’s spherical!”
If P is inside a given sphere, then the only relevant material is the mass inside
a concentric sphere through P, because all the shells outside this region give zero
153
154
Conservation of energy and momentum
P
Fig. 5.9
force, from the second equation in Eq. (5.43). The material “outside” of P is, for
the purposes of gravity, not there.
It isn’t obvious that the force inside a spherical shell is zero. Consider the
point P in Fig. 5.9. A piece of mass dm on the right side of the shell gives a larger
force on P than a piece of mass dm on the left side, due to the 1/r 2 dependence.
But from the figure, there is more mass on the left side than the right side. These
two effects happen to exactly cancel, as you can show in Problem 5.10.
Example (Supporting a tube): Imagine the following unrealistic undertaking.
Drill a narrow tube, with cross-sectional area A, from the surface of the earth down
to the center. Then line the cylindrical wall of the tube with a frictionless coating.
Then fill the tube back up with the dirt (and magma, etc.) that you originally removed.
What force is necessary at the bottom of the tube of dirt (that is, at the center of the
earth) to hold it up? Let the earth’s radius be R, and assume (incorrectly) a uniform
mass density ρ.
Solution: The gravitational force on a mass dm at radius r is effectively due to the
mass inside the radius r (call this Mr ). The mass outside r is effectively not there.
The gravitational force is therefore
Fdm =
G (4/3)πr 3 ρ dm
GMr dm
4
=
= πGρr dm,
3
r2
r2
(5.44)
which we see increases linearly with r. The dirt in the tube between r and r + dr has
volume A dr, so its mass is dm = ρA dr. The total gravitational force on the entire
tube is therefore
F=
Fdm =
=
R 4
0
3
πGρr(ρA dr) =
R
4
πGρ 2 A
r dr
3
0
R2
4
2
πGρ 2 A ·
= πGρ 2 AR2 .
3
2
3
(5.45)
The force at the bottom of the tube must be equal and opposite to this force. In terms
of the mass of the earth, ME = (4/3)πR3 ρ, and the total mass of the tube, Mt = ρAR,
this result can be written as F = GME Mt /2R2 . So the required force is half of what
a scale on the surface of the earth would read if all of the tube’s dirt sat in a lump on
top of it. The reason for this is basically that the force in Eq. (5.44) is linear in r.
An important subtopic of gravity is the tidal force, but because this is most
easily discussed in the context of accelerating frames of reference and fictitious
forces, we’ll postpone it until Chapter 10.
5.4.2
The Cavendish experiment
How do we determine the numerical value of G in Eq. (5.37)? If we can produce
a setup in which we know the values F, M, m, and r, then we can determine G.
5.4 Gravity
The first strategy that comes to mind is to take advantage of the fact that the
gravitational force on an object on the earth’s surface is known to be F = mg.
Combining this with Eq. (5.37) gives g = GME /R2 . We know the values of g
and R,10 so this tells us what the product GME is. However, this information
unfortunately doesn’t help us, because we don’t know what the mass of the earth
is (without knowing G first, which we’re assuming we don’t know yet; see the last
paragraph in this section). For all we know, the mass of the earth might be 10 times
larger than what we think it is, with G being 10 times smaller. The only way to
find G is to use a setup with two known masses. But this then presents the problem
that the resulting force is extremely small. So the task boils down to figuring out
a way to measure the tiny force between two known masses. Henry Cavendish
solved this problem in 1798 by performing an extremely delicate experiment
(which was devised a few years earlier by John Michell, but he died before being
able to perform it).11 The basic idea behind the experiment is the following.
Consider the setup in Fig. 5.10, which shows the top view. A dumbbell with
two masses m on its ends hangs from a very thin wire. The dumbbell is free to twist,
although if it twists, the wire will provide a tiny restoring torque.12 The dumbbell
starts with no twist in the wire, and then two other masses M are placed (fixed)
at the positions shown. These masses produce attractive forces on the dumbbell
masses and cause the dumbbell to twist counterclockwise. The dumbbell will
oscillate back and forth before finally settling down at some tiny angle θ away
from the initial position.
The torque on the dumbbell that arises from the twist in the wire takes the
form of τ = −bθ (with counterclockwise torque taken to be positive), where b
is a constant that depends on the thickness and makeup of the wire. This linear
relation between τ and θ holds for small θ for all the same reasons that the
F = −kx Hooke’s-law result in Section 5.2 holds.
The gravitational force between each pair of masses is GMm/d 2 , where d
is the separation between the centers of the masses in each pair. So the torque
on the dumbbell due to the two gravitational forces is 2(GMm/d 2 ) , where is
half the length of the dumbbell. Demanding that the total torque on the dumbbell be zero gives
2GMm
− bθ = 0
d2
10
11
12
=⇒
G=
bθd 2
.
2Mm
(5.46)
The radius of the earth has been known (at least roughly) since the time of Eratosthenes, about
250 BC. For an interesting way to measure it yourself, see Rawlins (1979).
The purpose of the experiment, as intended by Michell and Cavendish, was actually to measure
the density of the earth, and not G; see Clotfelter (1987). But as we’ll see below, this is equivalent
to measuring G.
We won’t talk about torque until Chapter 8, so you may want to come back and read this section
after that. We’ll invoke some results about rotational dynamics here, but the general setup should
be clear even if you’re not familiar with rotations.
155
(top view)
initial position
dumbbell
wire
(out of page)
Fig. 5.10
M
m
156
Conservation of energy and momentum
We know all the parameters on the right-hand side except b, so if we can determine
that, then we’re done. It’s very difficult to measure b directly with any reasonable
accuracy, because the torque in the wire is so tiny. But fortunately there’s a sneaky
way to determine b that involves taking a page from our playbook on oscillations.
The bread-and-butter equation for rotations is τ = I θ¨ , where I is the moment of
inertia (which we can calculate for the dumbbell); this is the rotational analog of
Newton’s second law, F = m¨x. Now, if the torque takes the form of τ = −bθ
for small θ , then τ = I θ¨ becomes −bθ = I θ¨ . This is a good old simpleharmonic-oscillator equation, so we know that the frequency of the oscillations
is ω = b/I . Therefore, all we need to do is measure the period, T = 2π/ω,
of the oscillations while the dumbbell is settling down, and we can determine b
from b = I ω2 = I (2π/T )2 . (The time T is large, because b is small on the scale
of things, because otherwise there wouldn’t be any noticeable twist in the wire.)
Plugging this value of b into Eq. (5.46) finally gives
G=
4π 2 I θd 2
.
2Mm T 2
(5.47)
The Cavendish experiment is also known as the “weighing the earth” (or
perhaps the “massing the earth”) experiment, because now that we know G (and
also g and R), we can use g = GME /R2 to calculate the mass of the earth, ME . The
only possible way to determine ME (without examining every cubic meter of the
inside of the earth, which is obviously impossible) is to determine G first, as we
have done.13 Interestingly, the resulting value of ME , which is roughly 6·1024 kg,
leads to an average density of the earth of about 5.5 g/cm3 . This is larger than
the density of the earth’s crust and mantle, so we conclude that there must be
something very dense deep down inside the earth. So the Cavendish experiment,
which involves masses hanging from a wire, amazingly tells us something about
the earth’s core!14
5.5
5.5.1
Momentum
Conservation of momentum
Newton’s third law says that for every force there is an equal and opposite force.
In other words, if Fab is the force that particle a feels due to particle b, and if
Fba is the force that particle b feels due to particle a, then Fba = −Fab at all
times. This law has important implications concerning momentum, p ≡ mv.
Consider two particles that interact over a period of time. Assume that they are
isolated from outside forces. From Newton’s second law, F = dp/dt, we see by
integrating this that the total change in a particle’s momentum equals the time
13
14
If you want to determine ME without using g or R (but still using G, of course, because ME appears
only through the combination GME ), see Celnikier (1983).
For a comprehensive discussion of the earth’s core, see Brush (1980).
5.5 Momentum
integral of the force acting on it. That is,
t2
p(t2 ) − p(t1 ) =
F dt.
(5.48)
t1
This integral is called the impulse. If we now invoke the third law, Fba = −Fab ,
we find
pa (t2 ) − pa (t1 ) =
t2
Fab dt = −
t1
t2
Fba dt = − pb (t2 ) − pb (t1 ) .
(5.49)
t1
Therefore,
pa (t2 ) + pb (t2 ) = pa (t1 ) + pb (t1 ).
(5.50)
This is the statement that the total momentum of this isolated system of two
particles is conserved; it does not depend on time. Note that Eq. (5.50) is a vector
equation, so it is really three equations, namely conservation of px , py , and pz .
Remark: Newton’s third law makes a statement about forces. But force is related to momentum
via F = dp/dt. So the third law essentially postulates conservation of momentum. (The “proof”
above in Eq. (5.49) is hardly a proof. It involves one simple integration.) So you might wonder
if momentum conservation is something you can prove, or if it’s something you have to assume,
as we have basically done because we have simply accepted the third law.
The difference between a postulate and a theorem is rather nebulous. One person’s postulate
might be another person’s theorem, and vice versa. You have to start somewhere in your
assumptions. We chose to start with the third law. In the Lagrangian formalism in Chapter 6,
the starting point is different, and momentum conservation is deduced as a consequence of
translational invariance (as we will see). So it looks more like a theorem in that formalism.
But one thing is certain. Momentum conservation for two particles cannot be proved from
scratch for arbitrary forces, because it is not necessarily true. For example, if two charged
particles interact in a certain way through the magnetic fields they produce, then the total
momentum of the two particles might not be conserved. Where is the missing momentum? It is
carried off in the electromagnetic field. The total momentum of the system is indeed conserved,
but the crucial point is that the system consists of the two particles plus the electromagnetic
field. Said in another way, each particle actually interacts with the electromagnetic field, and
not the other particle. Newton’s third law does not necessarily hold for particles subject to such
a force. ♣
Let’s now look at momentum conservation for a system of many particles.
As above, let Fij be the force that particle i feels due to particle j. Then Fij = −Fji
at all times. Assume that the particles are isolated from outside forces. The change
in momentum of the ith particle from t1 to t2 is (we won’t bother writing all the
t’s in the expressions below)
pi =
Fij dt.
(5.51)
j
Therefore, the change in the total momentum of all the particles is (switching the
order of the integration and the sum over i on the right-hand side here)
P≡
pi =
i
Fij dt.
i
j
(5.52)
157
158
Conservation of energy and momentum
But i j Fij = 0 at all times, because for every term Fab there is a term Fba ,
and Fab + Fba = 0 (and also, Faa = 0). The forces all cancel in pairs. Therefore,
the total momentum of an isolated system of particles is conserved.
Example (Snow on a sled): You are riding on a sled that is given an initial push
and slides across frictionless ice. Snow is falling vertically (in the frame of the ice) on
the sled. Assume that the sled travels in tracks that constrain it to move in a straight
line. Which of the following three strategies causes the sled to move the fastest? The
slowest?
A: You sweep the snow off the sled so that it leaves the sled in the direction
perpendicular to the sled’s tracks, as seen by you in the frame of the sled.
B: You sweep the snow off the sled so that it leaves the sled in the direction
perpendicular to the sled’s tracks, as seen by someone in the frame of the ice.
C: You do nothing.
First solution: The sideways motion of the snow after you sweep it off is irrelevant,
because although Newton’s third law says that the swept snow applies a sideways force
on the sled, the normal force from the tracks keeps the sled from sliding off the tracks.
Also, the vertical motion of the snow when it hits the sled is irrelevant, because the
vertical normal force from the tracks keeps the sled from falling through the ground.
We are therefore concerned only with the motion in the direction of the tracks. And
since there are no external forces in this direction on the you/sled/snow system, the
momentum in this direction is conserved.
In general, the speed of the sled can (possibly) change due to two kinds of events:
(1) new snow hitting the sled (and eventually coming to rest with respect to the sled),
and (2) snow being swept off the sled.
Let’s first compare A with C. In strategy A, your sweeping action doesn’t change
the speed of the sled, because you are sweeping the snow directly sideways in your
frame. Since the sideways motion of the snow is irrelevant as far as the forward
momentum goes, you are essentially just reaching out and plopping a ball of snow
on the ice. This snow then simply travels along next to the sled at the same speed;
it might as well be connected to the sled (at least for a moment, until new snow hits
the sled). In strategy C, your (non) action obviously don’t change the speed of the
sled. So in comparing A with C, the departure of snow doesn’t differentiate them. We
therefore need only consider what happens to the sled when new snow hits it. But
this is easy to do: Since the sled is heavier in C than in A, a new snowflake slows it
down less in C than in A. Therefore, C is faster than A.
Now let’s compare B with C. B is faster than C because in B the snowflakes
have zero forward momentum in the end, whereas in C they have nonzero forward
momentum (because they are sitting on the moving sled). The momenta of the two
systems must be the same (equal to the initial momentum of the sled plus you), so
the sled must be moving faster in B.
Therefore, B is faster than C, which is faster than A. As a consistency check, it’s
easy to see that B is faster than A, because in B you must push the snow backward with
5.5 Momentum
159
respect to the sled. So by Newton’s third law, the swept snow in B exerts a forward
force on the sled.
Second solution: In the end, this solution is basically the same as the first one, but
it’s a slightly more systematic way of looking at things: In B, all the snow is moving
slower than the sled; in fact, it is all at rest (at least in the forward direction) with
respect to the ice. In C, all the snow is on the sled. And in A, all the snow is moving
faster than the sled; it is moving along with various forward speeds, ranging from the
initial speed of the sled down to the present speed, depending on when it was swept off.
By conservation of momentum, the total momentum of the sled (including you)
plus the snow at any given time is the same in all three cases. The only consistent
way to combine the facts in the previous paragraph with conservation of momentum
is for the speed of the sled to satisfy B > C > A. This is true because if someone
claimed that C > B, then the slowest object (namely everything, since all the snow
is on the sled) in C would be going faster than the fastest object (the sled) in B; this
would contradict the fact that the momenta of the two systems are equal. Therefore,
we must have B > C. Likewise, if someone claimed that A > C, then the slowest
object (the sled) in A would be going faster than the fastest object (everything) in C;
this would again be a contradiction. Therefore, we must have C > A. Putting these
together gives B > C > A. See Exercise 5.70 for a quantitative treatment of this setup.
5.5.2
Rocket motion
The quantitative application of momentum conservation can get a little tricky
when the mass m is allowed to vary. Such is the case with rockets, because most
of their mass consists of fuel which is eventually ejected.
Let mass be ejected backward with (constant) speed u relative to the rocket.15
Since u is a speed here, it is defined to be positive. This means that the velocity of
the ejected particles is obtained by subtracting u from the velocity of the rocket.
Let the rocket have initial mass M, and let m be the (changing) mass at a later
time. Then the rate of change of the rocket’s mass is dm/dt, which is negative.
So mass is ejected at a rate |dm/dt| = −dm/dt, which is positive. In other words,
during a small time dt, a negative mass dm gets added to the rocket, and a positive
mass (−dm) gets shot out the back. (If you wanted, you could define dm to be
positive, and then subtract it from the rocket’s mass, and have dm get shot out the
back. Either way is fine.) It may sound silly, but the hardest thing about rocket
motion is picking a sign for these quantities and sticking with it.
Consider a moment when the rocket has mass m and speed v. Then at a time
dt later (see Fig. 5.11), the rocket has mass m + dm and speed v + dv, while the
15
Just to emphasize, u is the speed with respect to the rocket. It wouldn’t make sense to say “relative
to the ground,” because the rocket’s engine shoots out the matter relative to itself, and the engine
has no way of knowing how fast the rocket is moving with respect to the ground.
m
v
-dm
m+dm
v-u
Fig. 5.11
v+dv
160
Conservation of energy and momentum
exhaust has mass (−dm) and speed v − u (which may be positive or negative,
depending on the relative sizes of v and u). There are no external forces, so the
total momentum at each of these times must be equal. Therefore,
mv = (m + dm)(v + dv) + (−dm)(v − u).
(5.53)
Ignoring the second-order term dm dv, this simplifies to m dv = −u dm. Dividing
by m and integrating from t1 to t2 gives
v2
v1
dv = −
m2
u
m1
dm
m
=⇒
v2 − v1 = u ln
m1
.
m2
(5.54)
For the case where the initial mass is M and the initial speed is 0, we have
v = u ln(M/m). Note that we haven’t assumed anything about dm/dt in this
derivation. There is no need for it to be constant; it can change in any way it
wants. The only thing that matters (assuming that M and u are given) is the final
mass m. In the special case where dm/dt happens to be constant (call it −η, where
η is positive), we have v(t) = u ln[M/(M − ηt)].
The log in the result in Eq. (5.54) is not very encouraging. If the mass of
the metal in the rocket is m, and if the mass of the fuel is 9m, then the final
speed is only u ln 10 ≈ (2.3)u. If the mass of the fuel is increased by a factor
of 11 up to 99m (which is probably not even structurally possible, given the
amount of metal required to hold it),16 then the final speed only doubles to
u ln 100 = 2(u ln 10) ≈ (4.6)u. How do you make a rocket go significantly
faster? Exercise 5.69 deals with this question.
Remark: If you want, you can solve this rocket problem by using force instead of conservation
of momentum. If a chunk of mass (−dm) is ejected out the back, then its momentum changes
by u dm (which is negative). Therefore, because force equals the rate of change in momentum,
the force on this chunk is u dm/dt. By Newton’s third law, the remaining part of the rocket then
feels a force of −u dm/dt (which is positive). This force accelerates the remaining part of the
rocket, so F = ma gives −u dm/dt = m d v /dt,17 which is equivalent to the m d v = −u dm
result above.
We see that this rocket problem can be solved by using either force or conservation of
momentum. In the end, these two strategies are really the same, because the latter was derived
from F = dp/dt. But the philosophies behind the approaches are somewhat different. The
choice of strategy depends on personal preference. In an isolated system such as a rocket,
conservation of momentum is usually simpler. But in a problem involving an external force,
you have to use F = dp/dt. You’ll get lots of practice with F = dp/dt in the problems for this
section and also in Section 5.8. Note that we used both F = dp/dt and F = ma in this second
solution to the rocket problem. For further discussion of which expression to use in a given
situation, see Appendix C. ♣
16
17
The space shuttle’s external fuel tank, just by itself, has a fuel-to-container mass ratio of only
about 20.
Whether we use m or m + dm here for the mass of the rocket doesn’t matter. Any differences are
of second order.
5.6 The center of mass frame
5.6
5.6.1
161
The center of mass frame
Deﬁnition
When talking about momentum, it is understood that a certain frame of reference
has been chosen. After all, the velocities of the particles have to be measured
with respect to some coordinate system. Any inertial (that is, nonaccelerating)
frame is legal to pick, but we will see that there is one particular reference frame
that is often advantageous to use.
Consider a frame S and another frame S that moves at constant velocity u
with respect to S (see Fig. 5.12). Given a system of particles, the velocity of the
ith particle in S is related to its velocity in S by
vi = vi + u.
y'
u
y
S'
(5.55)
S
This relation implies that if momentum is conserved during a collision in frame
S , then it is also conserved in frame S. This is true because both the initial and
final momenta of the system in S are increased by the same amount, ( mi )u,
compared with what they are in S .18
Let us therefore consider the unique frame in which the total momentum of a
system of particles is zero. This is called the center of mass frame, or CM frame.
If the total momentum is P ≡ mi vi in frame S, then the CM frame is the frame
S that moves with velocity
u=
with respect to S, where M ≡
Eq. (5.55) to write
P =
mi vi =
P
≡
M
mi vi
M
(5.56)
mi is the total mass. This follows from using
mi vi −
P
M
= P − P = 0.
(5.57)
The CM frame is extremely useful. Physical processes are generally much more
symmetrical in this frame, and this makes the results more transparent. The CM
frame is sometimes called the “zero momentum” frame. But the “center of mass”
name is commonly used because the center of mass of the particles doesn’t move
in the CM frame, for the following reason. The position of the center of mass is
defined by
RCM ≡
18
mi ri
.
M
(5.58)
Alternatively, nowhere in our earlier derivation of momentum conservation did we say what frame
we were using. We assumed only that the frame wasn’t accelerating. If it were accelerating, then
F would not equal ma. We will see in Chapter 10 how F = ma is modified in a noninertial frame.
But there’s no need to worry about that here.
Fig. 5.12
x
x'
162
Conservation of energy and momentum
This is the location of the pivot upon which a rigid system would balance, as
we’ll see in Chapter 8. The fact that the CM doesn’t move with respect to the CM
frame follows from the fact that the derivative of RCM is the velocity of the CM
frame in Eq. (5.56). The center of mass may therefore be chosen as the origin of
the CM frame.
If we take two derivatives of Eq. (5.58), we obtain
M aCM ≡
mi ai =
Fi = Ftotal .
(5.59)
So as far as the acceleration of the CM goes, we can treat the system of particles
like a point mass at the CM, and then just apply F = ma to this point mass. Since
the internal forces cancel in pairs, we need only consider the external forces when
calculating Ftotal .
Along with the CM frame, the other frame that people generally work with
is the lab frame. There is nothing at all special about this frame. It is simply
the frame (assumed to be inertial) in which the conditions of the problem are
given. Any inertial frame can be called the “lab frame.” Solving problems often
involves switching back and forth between the lab and CM frames. For example,
if the final answer is requested in the lab frame, then you may want to transform
the given information from the lab frame to the CM frame where things are more
obvious, and then transform back to the lab frame to give the answer.
m
v
Example (Two masses in 1-D): A mass m with speed v approaches a stationary
mass M (see Fig. 5.13). The masses bounce off each other without any loss in total
energy. What are the final velocities of the particles? Assume that the motion takes
place in 1-D.
M
Fig. 5.13
Solution: Doing this problem in the lab frame would require a potentially messy
use of conservation of energy (see the example in Section 5.7.1). But if we work in
the CM frame, things are much easier. The total momentum in the lab frame is mv,
so the CM frame moves to the right with speed mv/(m + M ) ≡ u with respect to the
lab frame. Therefore, in the CM frame, the velocities of the two masses are
vm = v − u =
Mv
,
m+M
and
vM = 0 − u = −
mv
.
m+M
(5.60)
As a double-check, the difference in the velocities is v, and the ratio of the speeds is
M /m, which makes the total momentum zero.
The important point to realize now is that in the CM frame, the two particles must
simply reverse their velocities after the collision (assuming that they do indeed hit
each other). This is true because the speeds must still be in the ratio M/m after the
collision, in order for the total momentum to remain zero. Therefore, the speeds must
either both increase or both decrease. But if they do either of these, then energy is not
conserved.19
19
So we did have to use conservation of energy in this CM-frame solution. But it was far less messy
than it would have been in the lab frame.
5.6 The center of mass frame
If we now go back to the lab frame by adding the CM velocity of mv/(m + M )
to the two new velocities of −M v/(m + M ) and mv/(m + M ), we obtain final lab
velocities of
vm =
(m − M )v
,
m+M
and
vM =
2mv
.
m+M
(5.61)
Remark: If m = M , then the left mass stops, and the right mass picks up a velocity of v
(this should be familiar to pool players). If M
m, then the left mass bounces back with
M,
velocity ≈ −v , and the right mass hardly moves (it’s essentially a brick wall). If m
then the left mass keeps plowing along with velocity ≈ v , and the right mass picks up a
velocity of ≈ 2v . This 2v is an interesting result (it is clearer if you consider things in the
frame of the heavy mass m, which is essentially the CM frame), and it leads to some neat
effects, such as in Problem 5.23. ♣
5.6.2
Kinetic energy
Given a system of particles, the relation between the total kinetic energy in two
different frames is not very enlightening in general. But if one of the frames is
the CM frame, then the relation turns out to be quite nice. Let S be the CM
frame, which moves at constant velocity u with respect to another frame S. Then
the velocities of the particles in the two frames are related by vi = vi + u. The
kinetic energy in the CM frame is
KCM =
1
2
mi |vi |2 .
(5.62)
And the kinetic energy in frame S is
1
2
1
=
2
1
=
2
KS =
mi |vi + u|2
mi (vi · vi + 2vi · u + u · u)
mi |vi |2 + u ·
1
= KCM + Mu2 ,
2
1
mi vi + |u|2
2
mi
(5.63)
where M is the total mass of the system, and where we have used i mi vi = 0,
by definition of the CM frame. Therefore, the K in any frame equals the K in the
CM frame, plus the K of the whole system treated like a point mass M located
at the CM, which moves with velocity u. An immediate corollary of this fact is
that if the K is conserved in a collision in one frame (which implies that KCM
is conserved, because conservation of momentum says that the CM speed u is
the same before and after the collision), then it is conserved in any other frame
(because again, the u relevant to that frame is the same before and after the
collision).
163
164
Conservation of energy and momentum
5.7
Collisions
There are two basic types of collisions among particles, namely elastic ones (in
which kinetic energy is conserved), and inelastic ones (in which kinetic energy
is lost). In any collision, the total energy is conserved, but in inelastic collisions
some of this energy goes into the form of heat (that is, relative motion of the
molecules inside the particles) instead of showing up in the net translational
motion of the particles.20
We’ll deal mainly with elastic collisions here, although some situations are
inherently inelastic, as we’ll see in Section 5.8. For inelastic collisions where
it is stated that a certain fraction, say 20%, of the kinetic energy is lost, only a
trivial modification to the procedure is required. To solve any elastic collision
problem, we just have to write down the conservation of energy and momentum
equations, and then solve for whatever variables we want to find.
5.7.1
One-dimensional motion
Let’s first look at one-dimensional motion. To see the general procedure, we’ll
solve the example from Section 5.6.1 again.
m
v
M
Example (Two masses in 1-D, again): A mass m with speed v approaches a
stationary mass M (see Fig. 5.14). The masses bounce off each other elastically. What
are the final velocities of the particles? Assume that the motion takes place in 1-D.
Fig. 5.14
Solution: Let vf and Vf be the final velocities of the masses. Then conservation
of momentum and energy give, respectively,
mv + 0 = mvf + MVf ,
1
1
1 2
mv + 0 = mvf2 + MVf2 .
2
2
2
(5.64)
We must solve these two equations for the two unknowns vf and Vf . Solving for Vf
in the first equation and substituting into the second gives
mv 2 = mvf2 + M
m2 (v − vf )2
,
M2
=⇒
0 = (m + M )vf2 − 2mvvf + (m − M )v 2 ,
=⇒
0 = (m + M )vf − (m − M )v (vf − v).
(5.65)
One solution is vf = v, but this isn’t the one we’re concerned with. It is of course
a solution, because the initial conditions certainly satisfy conservation of energy and
20
We’ll use the terminology where “kinetic energy” refers to the overall translational energy of a
particle. That is, we’ll exclude heat from this definition, even though heat is just the relative kinetic
energy of molecules inside a particle.
5.7 Collisions
momentum with the initial conditions (a fine tautology indeed). If you want, you can
view vf = v as the solution where the particles miss each other. The fact that vf = v
is always a root can often save you a lot of quadratic-formula trouble.
The vf = v(m − M )/(m + M ) root is the one we want. Plugging this vf back into
the first of Eqs. (5.64) to obtain Vf gives
vf =
(m − M )v
,
m+M
and
Vf =
2mv
,
m+M
(5.66)
in agreement with Eq. (5.61).
This solution was somewhat of a pain, because it involved a quadratic
equation. The following theorem is extremely useful because it offers a way to
avoid the hassle of quadratic equations when dealing with 1-D elastic collisions.
Theorem 5.3 In a 1-D elastic collision, the relative velocity of the two particles
after the collision is the negative of the relative velocity before the collision.
Proof: Let the masses be m and M . Let vi and Vi be the initial velocities, and
let vf and Vf be the final velocities. Conservation of momentum and energy give
mvi + MVi = mvf + MVf ,
1 2 1
1
1
mv + MVi2 = mvf2 + MVf2 .
2 i
2
2
2
(5.67)
Rearranging these yields
m(vi − vf ) = M (Vf − Vi ),
m(vi2 − vf2 ) = M (Vf2 − Vi2 ).
(5.68)
Dividing the second equation by the first gives vi + vf = Vi + Vf . Therefore,
vi − Vi = −(vf − Vf ),
(5.69)
as we wanted to show. In taking the quotient of these two equations, we have lost
the vf = vi and Vf = Vi solution. But as stated in the above example, this is the
trivial solution.
This is a splendid theorem. It has the quadratic energy-conservation statement
built into it. Hence, using this theorem along with momentum conservation (both
of which are linear equations and thus easy to deal with) gives the same information as the standard combination of Eqs. (5.67). Another quick proof is the
following. It is fairly easy to see that the theorem is true in the CM frame (as
we argued in the example in Section 5.6.1), so it is therefore true in any frame,
because it involves only differences in velocities.
165
166
Conservation of energy and momentum
5.7.2
Two-dimensional motion
Let’s now look at the more general case of two-dimensional motion. Threedimensional motion is just more of the same, so we’ll confine ourselves to 2-D.
Everything is basically the same as in 1-D, except that there is one more momentum equation, and one more variable to solve for. This is best seen through an
example.
m
Example (Billiards): Abilliard ball with speed v approaches an identical stationary
one. The balls bounce off each other elastically, in such a way that the incoming one
gets deflected by an angle θ (see Fig. 5.15). What are the final speeds of the balls?
What is the angle φ at which the stationary ball is deflected?
m
v
vf
φ
Solution: Let vf and Vf be the final speeds of the balls. Then conservation of px ,
py , and E give, respectively,
m
θ
mv = mvf cos θ + mVf cos φ,
Vf
0 = mvf sin θ − mVf sin φ,
m
(5.70)
1
1
1 2
mv = mvf2 + mVf2 .
2
2
2
Fig. 5.15
We must solve these three equations for the three unknowns vf , Vf , and φ. There are
various ways to do this. Here’s one. Eliminate φ by adding the squares of the first
two equations (after putting the vf terms on the left-hand side) to obtain
v 2 − 2vvf cos θ + vf2 = Vf2 .
(5.71)
Now eliminate Vf by combining this with the third equation to obtain21
vf = v cos θ.
(5.72)
Vf = v sin θ.
(5.73)
The third equation then yields
The second equation then gives m(v cos θ) sin θ = m(v sin θ) sin φ, which implies
cos θ = sin φ (or θ = 0, which corresponds to no collision). Therefore,
φ = 90◦ − θ.
(5.74)
In other words, the balls bounce off at right angles with respect to each other. This
fact is well known to pool players. Problem 5.19 gives another (cleaner) way to
demonstrate this result. Note that we needed to specify one of the four quantities, vf ,
Vf , θ , φ (we chose θ ), because we have only three equations. Intuitively, we can’t
expect to solve for all four of these quantities, because we can imagine one ball hitting
21
Another solution is vf = 0. In this case, φ must equal zero, and θ is not well defined. This is
simply the 1-D motion in the example in Section 5.6.1.
5.8 Inherently inelastic processes
the other at various distances away from directly head-on, which will cause the balls
to be deflected at various angles.
As we saw in the 1-D example in Section 5.6.1, collisions are often much
easier to deal with in the CM frame. Using the same reasoning (conservation
of p and E) that we used in that example, we conclude that in 2-D (or 3-D) the
final speeds of two elastically colliding particles must be the same as the initial
speeds. The only degree of freedom in the CM frame is the angle of the line
containing the final (oppositely directed) velocities. This simplicity in the CM
frame invariably provides for a cleaner solution than the lab frame yields. A good
example of this is Exercise 5.81, which gives yet another way to derive the above
right-angle billiard result.
5.8
Inherently inelastic processes
There is a nice class of problems where the system has inherently inelastic properties, even if it doesn’t appear so at first glance. In such a problem, no matter
how you try to set it up, there will be inevitable kinetic energy loss that shows
up in the form of heat. Total energy is conserved, of course, since heat is simply
another form of energy. But the point is that if you try to write down a bunch of
(1/2)mv 2 ’s and conserve their sum, then you’re going to get the wrong answer.
The following example is the classic illustration of this type of problem.
Example (Sand on conveyor belt): Sand drops vertically (from a negligible
height) at a rate σ kg/s onto a moving conveyor belt.
(a) What force must you apply to the belt in order to keep it moving at a constant
speed v?
(b) How much kinetic energy does the sand gain per unit time?
(c) How much work do you do per unit time?
(d) How much energy is lost to heat per unit time?
Solution:
(a) Your force equals the rate of change in momentum. If we let m be the combined
mass of the conveyor belt plus the sand on the belt, then
F=
dp
d(mv)
dv
dm
=
=m
+
v = 0 + σ v,
dt
dt
dt
dt
(5.75)
where we have used the fact that v is constant.
(b) The kinetic energy gained per unit time is
d
dt
mv 2
2
=
dm
dt
v2
2
=
σ v2
.
2
(5.76)
167
168
Conservation of energy and momentum
(c) The work done by your force per unit time is
d(Work)
F dx
=
= Fv = σ v 2 ,
dt
dt
(5.77)
where we have used Eq. (5.75).
(d) If work is done at a rate σ v 2 , and kinetic energy is gained at a rate σ v 2 /2, then
the “missing” energy must be lost to heat at a rate σ v 2 − σ v 2 /2 = σ v 2 /2.
In this example, it turned out that exactly the same amount of energy was lost
to heat as was converted into kinetic energy of the sand. There is an interesting
and simple way to see why this is true. In the following explanation, we’ll just
deal with one particle of mass M that falls onto the conveyor belt, for simplicity.
In the lab frame, the mass gains a kinetic energy of M v 2/2 by the time it
finally comes to rest with respect to the belt, because the belt moves at speed v.
Now look at things in the conveyor belt’s reference frame. In this frame, the mass
comes flying in with an initial kinetic energy of M v 2/2, and then it eventually
slows down and comes to rest on the belt. Therefore, all of the M v 2/2 energy
is converted to heat. And since the heat is the same in both frames, this is the
amount of heat in the lab frame, too.
We therefore see that in the lab frame, the equality of the heat loss and the
gain in kinetic energy is a consequence of the obvious fact that the belt moves at
the same rate with respect to the lab (namely v) as the lab moves with respect to
the belt (also v).
In the solution to the above example, we did not assume anything about the
nature of the friction force between the belt and the sand. The loss of energy
to heat is an unavoidable result. You might think that if the sand comes to rest
on the belt very “gently” (over a long period of time), then you can avoid the
heat loss. This is not the case. In that scenario, the smallness of the friction force
is compensated by the fact that the force must act over a very large distance.
Likewise, if the sand comes to rest on the belt very abruptly, then the largeness
of the friction force is compensated by the smallness of the distance over which
it acts. No matter how you set things up, the work done by the friction force is
the same nonzero quantity.
In other problems such as the following one, it is fairly clear that the process
is inelastic. But the challenge is to correctly use F = dp/dt instead of F = ma,
because F = ma will get you into trouble due to the changing mass.
Example (Chain on a scale): An “idealized” (see the comments following this
example) chain with length L and mass density σ kg/m is held such that it hangs
vertically just above a scale. It is then released. What is the reading on the scale, as a
function of the height of the top of the chain?
5.8 Inherently inelastic processes
First solution: Let y be the height of the top of the chain, and let F be the desired
force applied by the scale. The net force on the entire chain is F −(σ L)g, with upward
taken to be positive. The momentum of the entire chain (which just comes from the
moving part) is (σ y)˙y. Note that this is negative, because y˙ is negative. Equating the
net force on the entire chain with the rate of change in its momentum gives
F − σ Lg =
d(σ y˙y)
dt
= σ y¨y + σ y˙ 2 .
(5.78)
The part of the chain that is still above the scale is in free fall. Therefore, y¨ = −g.
And conservation of energy gives y˙ = 2g(L − y), because the chain has fallen a
distance L − y. Plugging these into Eq. (5.78) gives
F = σ Lg − σ yg + 2σ (L − y)g
= 3σ (L − y)g,
(5.79)
which happens to be three times the weight of the chain already on the scale. This
answer for F has the expected property of equaling zero when y = L, and also the
interesting property of equaling 3(σ L)g right before the last bit touches the scale.
Once the chain is completely on the scale, the reading suddenly drops down to the
weight of the chain, namely (σ L)g.
If you used conservation of energy to do this problem and assumed that all of the
lost potential energy goes into the kinetic energy of the moving part of the chain, then
you would obtain a speed of infinity for the last infinitesimal part of the chain to hit
the scale. This is certainly incorrect, and the reason is that there is inevitable heat loss
that arises when the pieces of the chain inelastically smash into the scale.
Second solution: The normal force from the scale is responsible for doing two
things. It holds up the part of the chain that already lies on the scale, and it also
changes the momentum of the atoms that are suddenly brought to rest when they hit
the scale. The first of these two parts of the force is simply the weight of the chain
already on the scale, which is Fweight = σ (L − y)g.
To find the second part of the force, we need to find the change in momentum, dp,
of the part of the chain that hits the scale during a given time dt. The amount of mass
that hits the scale in a time dt is dm = σ |dy| = σ |˙y| dt = −σ y˙ dt, since y˙ is negative.
This mass initially has velocity y˙ , and then it is abruptly brought to rest. Therefore,
the change in its momentum is dp = 0−(dm)˙y = σ y˙ 2 dt, which is positive. The force
required to cause this change in momentum is
Fdp/dt =
But as in the first solution, we have y˙ =
the scale is
dp
= σ y˙ 2 .
dt
(5.80)
2g(L − y). Therefore, the total force from
F = Fweight + Fdp/dt = σ (L − y)g + 2σ (L − y)g
= 3σ (L − y)g.
(5.81)
169
170
Conservation of energy and momentum
Note that Fdp/dt = 2Fweight (until the chain is completely on the scale), independent
of y.
In this example, we assumed that the chain was “ideal,” in the sense that it was
completely flexible, infinitesimally thin, and unstretchable. The simplest model
that satisfies these criteria is a series of point masses connected by short massless
strings. But in the above example, the strings actually don’t even matter. You
could instead start with many little unconnected point masses held in a vertical
line, with the bottom one just above the scale. If you then dropped all of them
simultaneously, they would successively smash into the scale in the same manner
as if they were attached by little strings; the tension in the strings would all be
zero. However, even though the strings aren’t necessary in this chain and scale
example, there are many setups involving idealized chains where they are in fact
necessary, because a tension is required in them. This is evident in many of the
problems and exercises for this chapter, as you will see.
An interesting fact is that even with the above definition of an ideal
chain, there are some setups (in contrast with the one above) for which it
is impossible to specify how the system behaves, without being given more
information. This information involves the relative size of two specific length
scales, as we’ll see below. To illustrate this, consider the two following
scenarios for the setup in Problem 5.28 (see Fig. 5.16), where a vertical
ideal chain is dropped with its bottom end attached to the underneath of a
support.
hand
L
Fig. 5.16
....
....
Fig. 5.18
....
Fig. 5.17
• First scenario (energy nonconserving): Let the spacing between the point masses in
our ideal chain be large compared with the horizontal span of the bend in the chain at
its bottom; see Fig. 5.17. Then the system is for all practical purposes one dimensional.
Each of the masses stops abruptly when it reaches the bend. This stoppage is a completely
inelastic collision in the same way it was in the above example with the chain falling on
the scale. Note that at any point in time, the bend consists of a massless piece of string
folded back along itself (or perhaps it consists of one of the masses, if we happen to be
looking at it right when a mass stops). There is no tension in this bottom piece of string
(if there were, then the massless bend would have an infinite acceleration upward), so
there is no tension pulling down the part of the chain on the left side of the bend. The
left part of the chain is therefore in freefall.
• Second scenario (energy conserving): Let the spacing between the point masses in
our ideal chain be small compared with the horizontal span of the bend in the chain
at its bottom; see Fig. 5.18. The system is now inherently two dimensional, and the
masses are essentially continuously distributed along the chain, as far as the bend is
concerned. This has the effect of allowing each mass to gradually come to rest, so
there is no abrupt inelastic stopping like there was in the first scenario. Each mass
5.8 Inherently inelastic processes
in the bend keeps the same distance from its two neighbors, whereas in the first scenario the mass that has just stopped soon sees the next mass fly directly past it before
abruptly coming to rest. The process in this second scenario is elastic; no energy is lost
to heat.
The basic difference between the two scenarios is whether or not there is slack
in any of the strings in the bend. If there is, then the relative speed between a pair
of masses changes abruptly at some point, which means that the relative kinetic
energy of the masses goes into damped (perhaps very overdamped) vibrational
motion in the connecting string, which then decays into the random motion
of heat.22
If no energy is lost to heat in the second scenario, then you might think that the
last infinitesimal piece of the chain will have an infinite speed. However, there
isn’t one last piece of the chain. When the left part of the chain has disappeared
and we are left with only the bend and the right part, the small (but nonzero)
bend is the last “piece,” and it ends up swinging horizontally with a large speed.
This then drags the whole chain to the side in a very visible motion (which can
be traced to the horizontal force from the support), at which point we have a very
noticeably two-dimensional system. The initial potential energy of the chain ends
up as kinetic energy of the final wavy side-to-side motion.
A consequence of energy conservation in the second scenario is that for a given
height fallen, the left part of the chain will be moving faster than the left part in
the first scenario. In other words, the left part in the second scenario accelerates
downward faster than the freefall g. But although this result follows quickly
from energy considerations, it isn’t so obvious in terms of a force argument.
Apparently there exists a tension at the left end of the bend in the second scenario
that drags down the left part of the chain to give it an acceleration greater than
g. A qualitative way of seeing why a tension exists there is the following. A tiny
piece of the chain that enters the bend from the left part slows down as it gradually
joins the fixed right part of the chain. There must therefore be an upward force on
this tiny piece. This upward force can’t occur at the bottom of the piece, because
any tension there pulls down on it. The force must therefore occur at the top of
the piece. In other words, there is a tension at this point, and so by Newton’s third
law this tension pulls down on the left part of the chain, thereby causing it to
accelerate faster than g. One of the tasks of Problem 5.29 is to find the tensions
at the two ends of the bend.
There is a simple way to demonstrate the existence of a tension that pulls
on the free part of the chain. The following setup is basically the falling-chain
setup without gravity, but it still has all the essential parts. Place a rope on a
22
If the strings were ideal springs with weak spring constants, then the energy would keep changing
back and forth between potential energy of the springs and kinetic energy of the masses, causing
the masses to bounce around and possibly run into each other. But we’re assuming that the strings
in our ideal chain are essentially very rigid overdamped springs.
171
172
Conservation of energy and momentum
(fairly smooth) table, in the shape of a very thin “U” so that it doubles back
along itself. Then quickly yank on one of the ends, in the direction away from
the bend. You will find that the other end moves backwards, in the direction
opposite to the motion of your hand, toward the bend (at least until the bend
reaches it and drags it forward). There must therefore exist a tension in the rope
to drag the other end backwards. But there’s no need to take my word for it – all
you need is a piece of rope. This effect is essentially the same as the one (in a
simplified version, since the rope here has constant density) that leads to the crack
of a whip.
Note that a perfectly flexible thin rope, with its continuous mass distribution,
does indeed behave elastically like the second scenario above. The continuous
rope may be thought of as a series of point masses with infinitesimal separation, and so this separation is much smaller than the small (but finite) length
of the bend. As long as the thickness of the rope is much smaller than the
length of the bend, every piece of the rope slows to a stop gradually in the
original falling-chain setup, or starts up from rest gradually in the preceding
“U” setup. So there is no heat loss in either setup from abrupt changes in
motion.
Returning to the falling-chain system with one of our ideal chains, you
might think that if the bend is made really small, so that the system looks onedimensional, then it should behave inelastically like the first scenario above.
However, the only relevant fact is whether the bend is smaller than the spacing
between the point masses in our ideal chain. The word “small” is meaningless,
of course, because we are talking about the length of the bend, which is a dimensionful quantity. It makes sense only to use the word “smaller,” that is, to compare
one length with another. The other length here is the spacing between the masses.
If the length of the bend is large compared with this spacing, then no matter what
the actual length of the bend is, the system behaves elastically like the second
scenario above.
So which of the two scenarios better describes a real chain? Details of an
actual experiment involving a falling chain are given in Calkin and March (1989).
The results show that a real chain behaves basically like the chain in the second
scenario above, at least until the final part of the motion. In other words, it is
energy conserving, and the left part accelerates faster than g. 23
Having said all this, it turns out that the energy-conserving second scenario
leads to complicated issues in problems (such as the numerical integration in
Problem 5.29), so for all the problems and exercises in this chapter (with the
exception of Problem 5.29), we’ll assume that we’re dealing with the inelastic
first scenario.
23
Spur-of-the-moment (but still plenty convincing) experiments were also performed by Wes
Campbell in the physics laboratory of John Doyle at Harvard.
5.9 Problems
5.9
173
Problems
Section 5.1 Conservation of energy in one dimension
5.1. Minimum length *
The shortest configuration of string joining three given points is the one
shown in the first setup in Fig. 5.19, where all three angles are 120◦ . 24
Explain how you could experimentally prove this fact by cutting three
holes in a table and making use of three equal masses attached to the ends
of strings, the other ends of which are connected as shown in the second
setup in Fig. 5.19.
120
120
120
m
m
m
5.2. Heading to zero *
A particle moves toward x = 0 under the influence of a potential
V (x) = −A|x|n , where A > 0 and n > 0. The particle has barely enough
energy to reach x = 0. For what values of n will it reach x = 0 in a
finite time?
Fig. 5.19
5.3. Leaving the sphere *
A small mass rests on top of a fixed frictionless sphere. The mass is given
a tiny kick and slides downward. At what point does it lose contact with
the sphere?
5.4. Pulling the pucks **
(a) A massless string of length 2 connects two hockey pucks that
lie on frictionless ice. A constant horizontal force F is applied to
the midpoint of the string, perpendicular to it (see Fig. 5.20). By
calculating the work done in the transverse direction, find how
much kinetic energy is lost when the pucks collide, assuming they
stick together.
(b) The answer you obtained above should be very clean and nice.
Find the slick solution that makes it transparent why the answer is
so nice.
5.5. Constant y˙ **
A bead, under the influence of gravity, slides down a frictionless wire
whose height is given by the function y(x). Assume that at position
(x, y) = (0, 0), the wire is vertical and the bead passes this point with
a given speed v0 downward. What should the shape of the wire be (that
is, what is y as a function of x) so that the vertical speed remains v0 at all
times? Assume that the curve heads toward positive x.
24
If the three points form a triangle that has an angle greater than 120◦ , then the string simply passes
through the point where that angle is. We won’t worry about this case.
l
F
l
Fig. 5.20
174
Conservation of energy and momentum
5.6. Dividing the heat ***
A block rests on a table where the coefficient of kinetic friction is µk .
You pull the block at a constant speed across the table by applying a
force µk N . Consider a period of time during which the block moves a
distance d. How much work is done on the block? On the table? How
much does each object heat up? Is it possible to answer these questions?
Hint: You’ll have to make some sort of model, however crude, for the
way that friction works.
height = V(x)
y
x
Fig. 5.21
5.7. V (x) vs. a hill ***
A bead, under the influence of gravity, slides along a frictionless wire
whose height is given by the function V (x), as shown in Fig. 5.21. Find
an expression for the bead’s horizontal acceleration, x¨ . (It can depend on
whatever quantities you need it to depend on.) You should find that the
result is not the same as the x¨ for a particle moving in one dimension
in the potential mgV (x), in which case x¨ = −gV . But if you grab hold of
the wire, is there any way you can move it so that the bead’s x¨ is equal to
the x¨ = −gV result for the one-dimensional potential mgV (x)?
Section 5.2: Small oscillations
5.8. Hanging mass
The potential for a mass hanging from a spring is V (y) = ky2/2 + mgy,
where y = 0 corresponds to the position of the spring when nothing
is hanging from it. Find the frequency of small oscillations around the
equilibrium point.
5.9. Small oscillations *
A particle moves under the influence of the potential V (x) = −Cxn e−ax .
Find the frequency of small oscillations around the equilibrium point.
Section 5.4.1: Gravity
P
5.10. Zero force inside a sphere *
Show that the gravitational force inside a spherical shell is zero by showing that the pieces of mass at the ends of the thin cones in Fig. 5.22 give
canceling forces at point P.
5.11. Escape velocity *
Fig. 5.22
(a) Find the escape velocity (that is, the velocity above which a
particle escapes to r = ∞) for a particle on a spherical planet
of radius R and mass M . What is the numerical value for the
earth? The moon? The sun?
5.9 Problems
(b) Approximately how small must a spherical planet be in order for
a human to be able to jump off? Assume a density roughly equal
to the earth’s.
5.12. Ratio of potentials **
Consider a cube of uniform mass density. Find the ratio of the gravitational potential energy of a mass at a corner to that of the same mass
at the center. Hint: There’s a slick way that doesn’t involve any messy
integrals.
5.13. Through the hole **
(a) A hole of radius R is cut out from an infinite flat sheet with mass
density σ per unit area. Let L be the line that is perpendicular to
the sheet and that passes through the center of the hole. What is
the force on a mass m that is located on L, at a distance x from the
center of the hole? Hint: Consider the plane to consist of many
concentric rings.
(b) If a particle is released from rest on L, very close to the center of
the hole, show that it undergoes oscillatory motion, and find the
frequency of these oscillations.
(c) If a particle is released from rest on L, at a distance x from the
sheet, what is its speed when it passes through the center of the
hole? What is your answer in the limit x
R?
Section 5.5.1: Momentum
5.14. Snowball *
A snowball is thrown against a wall. Where does its momentum go?
Where does its energy go?
5.15. Propelling a car **
For some odd reason, you decide to throw baseballs at a car of mass
M that is free to move frictionlessly on the ground. You throw the
balls at the back of the car at speed u, and they leave your hand at a
mass rate of σ kg/s (assume the rate is continuous, for simplicity). If
the car starts at rest, find its speed and position as a function of time,
assuming that the balls bounce elastically directly backward off the back
window.
5.16. Propelling a car again **
Do the previous problem, except now assume that the back window is
open, so that the balls collect inside the car.
175
176
Conservation of energy and momentum
5.17. Leaky bucket **
At t = 0, a massless bucket contains a mass M of sand. It is connected to
a wall by a massless spring with constant tension T (that is, independent
of length).25 See Fig. 5.23. The ground is frictionless, and the initial
distance to the wall is L. At later times, let x be the distance from the
wall, and let m be the mass of sand in the bucket. The bucket is released,
and on its way to the wall, it leaks sand at a rate dm/dx = M/L. In other
words, the rate is constant with respect to distance, not time; and it ends
up empty right when it reaches the wall. Note that dx is negative, so dm
is also.
T
Fig. 5.23
(a) What is the kinetic energy of the (sand in the) bucket, as a function
of x? What is its maximum value?
(b) What is the magnitude of the momentum of the bucket, as a
function of x? What is its maximum value?
5.18. Another leaky bucket ***
Consider the setup in Problem 5.17, but now let the sand leak at a rate
proportional to the bucket’s acceleration. That is, dm/dt = b¨x. Note that
x¨ is negative, so dm is also.
(a) Find the mass as a function of time, m(t).
(b) Find v(t) and x(t) during the time when the bucket contains a
nonzero amount of sand. Also find v(m) and x(m). What is the
speed right before all the sand leaves the bucket (assuming it
hasn’t hit the wall yet)?
(c) What is the maximum value of the bucket’s kinetic energy,
assuming it is achieved before it hits the wall?
(d) What is the maximum value of the magnitude of the
bucket’s momentum, assuming it is achieved before it hits
the wall?
(e) For what value of b does the bucket become empty right when it
hits the wall?
Section 5.7: Collisions
5.19. Right angle in billiards *
A billiard ball collides elastically with an identical stationary one. Use
the fact that mv 2 /2 may be written as m(v · v)/2 to show that the angle
25
You can construct a constant-tension spring with a regular Hooke’s-law spring in the following
way. Pick the spring constant to be very small, and stretch the spring a very large distance. Have
it pass through a hole in the wall, with its other end bolted down a large distance to the left of the
wall. Any changes in the bucket’s position will yield a negligible change in the spring force.
5.9 Problems
177
between the resulting trajectories is 90◦ . Hint: Take the dot product of
the conservation of momentum equation with itself.
5.20. Bouncing and recoiling **
A ball of mass m and initial speed v0 bounces back and forth between
a fixed wall and a block of mass M , with M
m; see Fig. 5.24. The
block is initially at rest. Assume that the ball bounces elastically and
instantaneously. The coefficient of kinetic friction between the block
and the ground is µ. There is no friction between the ball and the ground.
What is the speed of the ball after the nth bounce off the block? How far
does the block eventually move? How much total time does the block
actually spend in motion? Work in the approximation where M
m,
and assume that the distance to the wall is large enough so that the block
comes to rest by the time the next bounce occurs.
m v0
M
µ
Fig. 5.24
5.21. Drag force on a sheet **
A sheet of mass M moves with speed V through a region of space
that contains particles of mass m and speed v. There are n of these
particles per unit volume. The sheet moves in the direction of its normal.
Assume m
M , and assume that the particles do not interact with
each other.
(a) If v
V , what is the drag force per unit area on the sheet?
(b) If v
V , what is the drag force per unit area on the sheet?
Assume, for simplicity, that the component of every particle’s
velocity in the direction of the sheet’s motion is exactly ±v/2.26
5.22. Drag force on a cylinder **
A cylinder of mass M and radius R moves with speed V through a region
of space that contains particles of mass m that are at rest. There are n
of these particles per unit volume. The cylinder moves in a direction
perpendicular to its axis. Assume m
M , and assume that the particles
do not interact with each other. What is the drag force per unit length on
the cylinder?
B2
B1
h
5.23. Basketball and tennis ball **
(a) A tennis ball with a small mass m2 sits on top of a basketball with
a large mass m1 (see Fig. 5.25). The bottom of the basketball
is a height h above the ground, and the bottom of the tennis
ball is a height h + d above the ground. The balls are dropped.
To what height does the tennis ball bounce? Note: Work in the
26
In reality, the velocities are randomly distributed, but this idealization actually gives the correct
answer because the average speed in any direction is |vx | = v/2, as you can show.
Fig. 5.25
178
Conservation of energy and momentum
B4
B3
B2
n=4
B1
h
Fig. 5.26
approximation where m1 is much larger than m2 , and assume that
the balls bounce elastically. Also assume, for the sake of having
a nice clean problem, that the balls are initially separated by a
small distance, and that the balls bounce instantaneously.
(b) Now consider n balls, B1 , . . . , Bn , having masses m1 , m2 , . . . , mn
(with m1
m2
···
mn ), standing in a vertical stack (see
Fig. 5.26). The bottom of B1 is a height h above the ground, and
the bottom of Bn is a height h + above the ground. The balls are
dropped. In terms of n, to what height does the top ball bounce?
Note: Make assumptions and approximations similar to the ones
in part (a).
If h = 1 meter, what is the minimum number of balls needed for
the top one to bounce to a height of at least 1 kilometer? To reach
escape velocity? Assume that the balls still bounce elastically
(which is a bit absurd here), and ignore wind resistance, etc., and
assume that is negligible.
5.24. Maximal deflection ***
A mass M collides with a stationary mass m. If M < m, then it is possible for M to bounce directly backward. However, if M > m, then there
is a maximal angle of deflection of M . Show that this maximal angle
equals sin−1 (m/M ). Hint: It is possible to do this problem by working
in the lab frame, but you can save yourself a lot of time by considering what happens in the CM frame, and then shifting back to the lab
frame.
Section 5.8: Inherently inelastic processes
Note: In the problems involving chains in this section (with the exception
of Problem 5.29), we’ll assume that the chains are of the type described
in the first scenario near the end of Section 5.8.
5.25. Colliding masses *
A mass M , initially moving at speed V , collides and sticks to a mass m,
initially at rest. Assume M
m, and work in this approximation. What
are the final energies of the two masses, and how much energy is lost to
heat, in:
(top view)
hand
v
L
Fig. 5.27
(a) The lab frame?
(b) The frame in which M is initially at rest?
5.26. Pulling a chain **
A chain with length L and mass density σ kg/m lies straight on a frictionless horizontal surface. You grab one end and pull it back along itself,
in a parallel manner (see Fig. 5.27). Assume that you pull it at constant
5.9 Problems
179
speed v. What force must you apply? What is the total work that you
do, by the time the chain is straightened out? How much energy is lost
to heat, if any?
5.27. Pulling a chain again **
A chain with mass density σ kg/m lies in a heap on the floor. You
grab an end and pull horizontally with constant force F. What is the
position of the end of the chain, as a function of time, while it is unravelling? Assume that the chain is greased, so that it has no friction with
itself.
5.28. Falling chain **
A chain with length L and mass density σ kg/m is held in the position
shown in Fig. 5.28, with one end attached to a support. Assume that
only a negligible length of the chain starts out below the support. The
chain is released. Find the force that the support applies to the chain, as
a function of time.
5.29. Falling chain (energy conserving) ***
Consider the setup in the previous problem, but now let the chain be of
the type in the second scenario described in Section 5.8. Show that the
total time it takes the chain to straighten out is approximately 85% of
the time it would take if the left part were in freefall (as it was in the
previous problem); you will need to solve something numerically. Also,
show that the tension at the left end of the infinitesimal bend equals the
tension at the right end at all times.27
5.30. Falling from a table ***
(a) A chain with length L lies in a straight line on a frictionless table,
except for a very small piece at one end which hangs down through
a hole in the table. This piece is released, and the chain slides down
through the hole. What is the speed of the chain at the instant it
loses contact with the table? (See Footnote 3.19.)
(b) Answer the same question, but now let the chain lie in a heap on a
table, except for a very small piece at one end which hangs down
through the hole. Assume that the chain is greased, so that it has
no friction with itself. Which of these two scenarios has the larger
final speed?
27
The “ends” of the bend actually aren’t well defined, because the chain is at least a little bit curved
everywhere. But since we’re assuming that the horizontal span of the chain is very small, we can
define the height of the bend to be, say, 100 times this horizontal span, and this height is still
negligible compared with the total height of the chain.
hand
L
Fig. 5.28
180
Conservation of energy and momentum
5.31. The raindrop ****
Assume that a cloud consists of tiny water droplets suspended (uniformly
distributed, and at rest) in air, and consider a raindrop falling through
them. What is the acceleration of the raindrop? Assume that the raindrop is initially of negligible size and that when it hits a water droplet,
the droplet’s water gets added to it. Also, assume that the raindrop is
spherical at all times.
5.10
Exercises
Section 5.1: Conservation of energy in one dimension
5.32. Cart in a valley
A cart containing sand starts at rest and then rolls, without any energy
loss to friction, down into a valley and then up a hill on the other side.
Let the initial height be h1 , and let the final height attained on the other
side be h2 . If the cart leaks sand along the way, how does h2 compare
with h1 ?
5.33. Walking on an escalator
An escalator moves downward at constant speed. You walk up the escalator at this same speed, so that you remain at rest with respect to the
ground. In the ground frame, are you doing any work?
5.34. Lots of work
If you push on a wall with your hand, you don’t do any work, because
your hand doesn’t move. But in the reference frame of a person moving
past you (from front to back), you do in fact do work, because your
hand moves. And since the person’s speed can be made arbitrarily large,
you can do an arbitrarily large amount of work in the person’s frame.
It therefore seems like you should quickly use up your dinner from the
night before and become very hungry. But you don’t. Why not?
5.35. Spring energy
Using the explicit form of the position of a mass on the end of a spring,
x(t) = A cos(ωt + φ), verify that the total energy is conserved.
5.36. Damping work *
A damped oscillator (with m¨x = −kx − b˙x) has initial position x0 and
speed v0 . After a long time, it will essentially be at rest at the origin.
Therefore, by the work–energy theorem, the work done by the damping
force must equal −kx02 /2 − mv02 /2. Verify that this is true. Hint: It’s
rather messy to explicitly find x˙ in terms of the initial conditions and
5.10 Exercises
181
then calculate the desired integral. An easier way is to use the F = ma
equation to rewrite the x˙ in your integral.
5.37. Heading to infinity *
A particle moves under the influence of a potential V (x) = −A|x|n ,
where A > 0 and n > 0. It starts at a positive value of x with a
velocity that points in the positive x direction. For what values of n
will the particle reach infinity in a finite time? You may assume that
E > 0, although this isn’t necessary. (You can compare this exercise with
Problem 5.2.)
5.38. Work in different frames *
An object, initially at rest, is subject to a force that gives it constant
acceleration a for time t. Verify explicitly that W = K in (a) the lab
frame, and (b) a frame moving to the left at speed V .
5.39. Roller coaster *
A roller coaster car starts at rest and coasts down a frictionless track.
It encounters a vertical loop of radius R. How much higher than the top
of the loop must the car start if it is to remain in contact with the track
at all times?
L
5.40. Pendulum and peg *
A pendulum of length L is held with its string horizontal, and then
released. The string runs into a peg a distance d below the pivot, as
shown in Fig. 5.29. What is the smallest value of d for which the string
remains taut at all times?
5.41. Circling around a cone *
A fixed hollow frictionless cone is positioned with its tip pointing down.
A particle is released from rest on the inside surface. After it has slid part
way down to the tip, it bounces elastically off a platform. The platform
is positioned at a 45◦ angle along the surface of the cone, so the particle
ends up being deflected horizontally along the surface (in other words,
into the page in Fig. 5.30). If the resulting motion of the particle is a
horizontal circle around the cone, what is the ratio of the initial height
of the particle to the height of the platform?
5.42. Hanging spring *
A massless spring with spring constant k hangs vertically from a ceiling,
initially at its relaxed length. A mass m is then attached to the bottom
and is released.
(a) Calculate the potential energy V of the system, as a function of
the height y (which is negative), relative to the initial position.
d
Fig. 5.29
platform
Fig. 5.30
182
Conservation of energy and momentum
(b) Find y0 , the point at which the potential energy is minimum. Make
a rough plot of V (y).
(c) Rewrite the potential energy as a function of z ≡ y − y0 . Explain
why your result shows that a hanging spring can be considered
to be a spring in a world without gravity, provided that the new
equilibrium point, y0 , is now called the “relaxed” length of the
spring.
m
k
µ
θ
Fig. 5.31
(a) You move the block down the plane, compressing the spring.
What is the maximum compression distance of the spring (relative
to the relaxed length it has when nothing is attached to it) that
allows the block to remain at rest when you let go of it?
(b) Assume that the block is at the maximum compression you found
in part (a). At a given instant, you somehow cause the plane to
become frictionless, and the block gets pushed up along the plane.
What must the relation between θ and the original µ be, so that
the block reaches its maximum height when the spring is at its
relaxed length?
5.44. Spring and friction **
A spring with spring constant k stands vertically, and a mass m is placed
on top of it. The mass is slowly lowered to its equilibrium position.
With the spring held at this compression length, the system is rotated to
a horizontal position. The left end of the spring is attached to a wall, and
the mass is placed on a table with a coefficient of friction (both kinetic
and static) of µ = 1/8; see Fig. 5.32. The mass is released.
m
k
k
m
µ =1/8
Fig. 5.32
5.43. Removing the friction *
A block of mass m is supported by a spring on an inclined plane, as shown
in Fig. 5.31. The spring constant is k, the plane’s angle of inclination is
θ , and the coefficient of static friction between the block and the plane
is µ.
(a) What is the initial compression of the spring?
(b) How much does the maximal stretch (or compression) of the
spring decrease after each half-oscillation?
(c) How many times does the mass oscillate back and forth before
coming to rest?
5.45. Keeping contact **
A frictionless circle of radius R is made out of a strip of metal and held
fixed in a vertical plane. A massless spring with spring constant k has one
end attached to the bottom point on the inside surface of the circle, and the
other end attached to a mass m. The spring is compressed to zero length,
5.10 Exercises
183
with the mass touching the inside surface of the circle at the bottom.
(Whatever negligible length of spring remains is essentially horizontal.)
The spring is then released, and the mass gets pushed initially to the
right and then up along the circle; the setup at a random later time is
shown in Fig. 5.33. Let be the equilibrium length of the spring. What
is the minimum value of for which the mass remains in contact with
the circle at all times?
5.46. Spring and hoop **
A fixed hoop of radius R stands vertically. A spring with spring constant
k and relaxed length of zero is attached to the top of the hoop.
(a) A block of mass m is attached to the unstretched spring and
dropped from the top of the hoop. If the resulting motion of the
mass is a linear vertical oscillation between the top and bottom
points on the hoop, what is k?
(b) The block is now removed from the spring, and the spring is
stretched and connected to a bead, also of mass m, at the bottom
of the hoop, as shown in Fig. 5.34. The bead is constrained to
move along the hoop. It is given a rightward kick and acquires an
initial speed v0 . Assuming that it moves frictionlessly, how does
its speed depend on its position along the hoop?
5.47. Constant x˙ **
A bead, under the influence of gravity, slides along a frictionless wire
whose height is given by the function y(x). Assume that at position
(x, y) = (0, 0), the wire is horizontal and the bead passes this point with
a given speed v0 to the right. What should the shape of the wire be (that
is, what is y as a function of x) so that the horizontal speed remains v0
at all times? One solution is simply y = 0. Find the other.28
5.48. Over the pipe **
A frictionless cylindrical pipe with radius r is positioned with its axis
parallel to the ground, at height h. What is the minimum speed at which
a ball must be thrown (from ground level) in order to make it over the
pipe? Consider two cases: (a) the ball is allowed to touch the pipe, and
(b) the ball is not allowed to touch the pipe.
5.49. Pendulum projectile **
A pendulum is held with its string horizontal and is then released. The
mass swings down, and then on its way back up, the string is cut when
28
Solve this exercise in the spirit of Problem 5.5, that is, by solving a differential equation. Once
you get the answer, you’ll see that you could have just written it down without any calculations,
based on your knowledge of a certain kind of physical motion.
m
k
Fig. 5.33
Part (b):
R
k
m
v0
Fig. 5.34
184
Conservation of energy and momentum
θ
cut
5.50. Centered projectile motion **
A mass is attached to one end of a massless string, the other end of which
is attached to a fixed support. The mass swings around in a vertical circle
as shown in Fig. 5.36. Assuming that the mass has the minimum speed
necessary at the top of the circle to keep the string from going slack, at
what location should you cut the string so that the resulting projectile
motion of the mass has its maximum height located directly above the
center of the circle?
Fig. 5.35
cut
θ
5.51. Beads on a hoop **
Two beads of mass m are initially at rest at the top of a frictionless hoop
of mass M and radius R, which stands vertically on the ground. The
beads are given tiny kicks, and they slide down the hoop, one to the
right and one to the left, as shown in Fig. 5.37. What is the largest value
of m/M for which the hoop never rises up off the ground?
Fig. 5.36
m
m
M
Fig. 5.37
m
M
µ=1
Fig. 5.38
it makes an angle θ with the vertical; see Fig. 5.35. What should θ be so
that the mass travels the largest horizontal distance by the time it returns
to the height it had when the string was cut?
5.52. Stationary bowl ***
A hemispherical bowl of mass M rests on a table. The inside surface
of the bowl is frictionless, while the coefficient of friction between the
bottom of the bowl and the table is µ = 1. A particle of mass m is
released from rest at the top of the bowl and slides down into it, as
shown in Fig. 5.38. What is the largest value of m/M for which the
bowl never slides on the table? Hint: The angle you’re concerned with
is not 45◦ .
5.53. Leaving the hemisphere ****
A point particle of mass m sits at rest on top of a frictionless hemisphere
of mass M , which rests on a frictionless table. The particle is given a
tiny kick and slides down the (recoiling) hemisphere. At what angle θ
(measured from the top of the hemisphere) does the particle lose contact
with the hemisphere? In answering this question for m = M , it is sufficient for you to produce an equation that θ must satisfy (it’s a cubic).
However, for the special case of m = M , the equation can be solved
without too much difficulty; find the angle in this case.
5.54. Tetherball ****
A small ball is attached to a massless string of length L, the other end
of which is attached to a very thin pole. The ball is thrown so that it
initially travels in a horizontal circle, with the string making an angle
θ0 with the vertical. As time goes on, the string wraps itself around the
5.10 Exercises
185
pole. Assume that (1) the pole is thin enough so that the length of string
in the air decreases very slowly, so that the ball’s motion may always be
approximated as a circle, and (2) the pole has enough friction so that the
string does not slide on the pole, once it touches it. Show that the ratio
of the ball’s final speed (right before it hits the pole) to initial speed is
vf /vi = sin θ0 .
Section 5.4: Gravity
5.55. Projectile between planets *
Two planets of mass M and radius R are at rest (somehow) with respect
to each other, with their centers a distance 4R apart. You wish to fire
a projectile from the surface of one planet to the other. What is the
minimum firing speed for which this is possible?
m
θ θ
5.56. Spinning quickly *
Consider a planet with uniform mass density ρ. If the planet rotates
too fast, it will fly apart. Show that the minimum period of rotation is
given by
T =
L/2
L/2
3π
.
Gρ
What is the minimum T if ρ = 5.5 g/cm3 (the average density of the
earth)?
5.57. A cone **
(a) A particle of mass m is located at the tip of a hollow cone (such
as an ice cream cone without the ice cream) with surface mass
density σ . The slant height of the cone is L, and the half-angle at
the vertex is θ. What can you say about the gravitational force on
the mass m due to the cone?
(b) If the top half of the cone is removed and thrown away (see
Fig. 5.39), what is the gravitational force on the mass m due to
the remaining part of the cone? For what angle θ is this force
maximum?
Fig. 5.39
a)
R
L
5.58. Sphere and cones **
(a) Consider a thin hollow fixed spherical shell of radius R and surface mass density σ . Aparticle initially at rest falls in from infinity.
What is its speed when it reaches the center of the shell? Assume
that a tiny hole has been cut in the shell to let the particle through;
see Fig. 5.40(a).
R
b)
Fig. 5.40
186
Conservation of energy and momentum
(b) Consider two hollow fixed cones (such as ice cream cones without the ice cream), arranged as shown in Fig. 5.40(b). They have
base radius R, slant height L, and surface mass density σ . A particle initially at rest falls in from infinity, along the perpendicular
bisector line shown. What is its speed when it reaches the tip of
the cones?
5.59. Ratio of potentials **
Consider the following two systems: (1) a mass m is placed at a corner of
a flat square sheet of mass M , and (2) a mass m is placed at the center of
a flat square sheet of mass M . What is the ratio of the potential energies
of m in the two systems? Hint: Find A and B in the suggestive relations
in Fig. 5.41. You’ll need to use a scaling argument to find B.
= A
= B
5.60. Solar escape velocity **
What is the minimum initial velocity (with respect to the earth) required
for an object to escape from the solar system?29 Take the orbital motion
of the earth into account (but ignore the rotation of the earth, and ignore
the other planets). You are free to choose (wisely) the firing direction.
Make the (good) approximation that the process occurs in two separate
steps: first the object escapes from the earth, and then it escapes from
the sun (starting at the radius of the earth’s orbit). Some useful quantities
are given in the solution to Problem 5.11; also, the orbital speed of the
earth is about 30 km/s. Hint: a common incorrect answer is 13.5 km/s.
Fig. 5.41
5.61. Spherical shell **
(a) A spherical shell of mass M has inner radius R1 and outer radius
R2 . A particle of mass m is located a distance r from the center of
the shell. Calculate (and make a rough plot of) the force on m, as
a function of r, for 0 ≤ r ≤ ∞.
(b) If the mass m is dropped from r = ∞ and falls down through the
shell (assume that a tiny hole has been drilled in it), what will its
speed be at the center of the shell? You can let R2 = 2R1 in this
part of the problem, to keep things from getting too messy. Give
your answer in terms of R ≡ R1 .
later
R
5.62. Orbiting stick **
Consider a planet of mass M and radius R. A very long stick of length
2R extends from just above the surface of the planet out to a radius 3R.
If initial conditions have been set up so that the stick moves in a circular
orbit while always pointing radially (see Fig. 5.42), what is the period
2R
Fig. 5.42
29
This problem is discussed in Hendel and Longo (1988).
5.10 Exercises
187
of this orbit? How does this period compare with the period of a satellite
in a circular orbit of radius 2R?
5.63. Speedy travel **
A straight tube is drilled between two points on the earth, as shown
in Fig. 5.43. An object is dropped into the tube. What is the resulting
motion? How long does it take to reach the other end? Ignore friction,
and assume (incorrectly) that the density of the earth is constant (ρ =
5.5 g/cm3 ).
5.64. Mine shaft **
(a) If the earth had constant density, the gravitational force would
decrease linearly with radius as you descend in a mine shaft; see
Eq. (5.44). However, the density of the earth is not constant, and
in fact the gravitational force increases as you descend. Show that
the general condition under which this is true is ρc < (2/3)ρavg ,
where ρavg is the average density of the earth, and ρc is the density
of the crust at the surface. (The values for the earth are ρc ≈
3 g/cm3 and ρavg ≈ 5.5 g/cm3 .) See Zaidins (1972).
(b) A similar problem to the one in part (a), which actually turns
out to be exactly the same, is the following. Consider a large
flat horizontal sheet of material with density ρ and thickness
x. Show that the gravitational force (from the earth plus the
sheet) just below the sheet is larger than the force just above it
if ρ < (2/3)ρavg , where ρavg is the average density of the earth.
A sheet of wood (with a density roughly equal to that of water)
satisfies this inequality, but a sheet of gold doesn’t. A result from
Problem 5.13 will be useful here.
(c) Assuming that the density of a planet is a function of radius only,
what should ρ(r) look like if you want the gravitational force to
be independent of the depth in a mine shaft, all the way down to
the center of the planet?
5.65. Space elevator **
(a) Let the earth’s radius be R, its average density be ρ, and its angular
frequency of rotation be ω. Show that if a satellite is to remain
above the same point on the equator at all times, then it must
travel in a circle of radius ηR, where
η3 =
4π Gρ
.
3ω2
What is the numerical value of η?
(5.82)
Fig. 5.43
188
Conservation of energy and momentum
(b) Instead of a satellite, consider a long rope with uniform mass
density extending radially from just above the surface of the earth
out to a radius η R. 30 Show that if the rope is to remain above the
same point on the equator at all times, then η must be given by
η2+η =
8π Gρ
.
3ω2
(5.83)
What is the numerical value of η ? Where does the tension in the
rope achieve its maximum value? Hint: no messy calculations
required.
5.66. Force from a straight wire ***
A particle of mass m is placed a distance away from an infinitely long
straight wire with mass density σ kg/m. Show that the force on the
particle is F = 2Gσ m/ . Do this in two ways:
(a) Integrate along the wire the contributions to the force.
(b) Integrate along the wire the contributions to the potential, and then
differentiate to obtain the force. You will find that the potential due
to the infinite wire is infinite,31 but you can escape this difficulty
by letting the wire have a large but finite length, then finding the
potential and force, and then letting the length go to infinity.
5.67. Maximal gravity ***
Given a point P in space, and given a piece of malleable material of
constant density, how should you shape and place the material in order
to create the largest possible gravitational field at P?
Section 5.5: Momentum
5.68. Maximum P and E of a rocket *
A rocket that starts at rest with mass M ejects exhaust at a given speed
u. What is the mass of the rocket (including unused fuel) when its
momentum is maximum? What is the mass when its energy is maximum?
5.69. Speedy rockets *
Assume that it is impossible to build a structurally sound container that
can hold fuel of more than, say, nine times its mass (the actual limit
is higher than this, but let’s use this number just to be concrete). It
would then seem like the limit for the speed of a rocket is u ln 10, from
Eq. (5.54). How can you build a rocket that goes faster than this?
30
31
Any proposed space elevator wouldn’t have uniform mass density. But this simplified problem
still gives a good idea of the general features. For more on the space elevator, see Aravind (2007).
There’s nothing bad about this. All that matters as far as the force is concerned is differences in
the potential, and these differences are finite.
5.10 Exercises
5.70. Snow on a sled, quantitative **
Consider the setup in the example in Section 5.5.1. At t = 0, let the mass
of the sled (including you) be M , and let its speed be V0 . If the snow
hits the sled at a rate of σ kg/s, find the speed as a function of time for
the three cases.
5.71. Leaky bucket ***
Consider the setup in Problem 5.17, but now let the sand leak at a rate
dm/dt = −bM . In other words, the rate is constant with respect to time,
not distance. We’ve factored out an M here, just to make the calculations
a little nicer.
(a) Find v(t) and x(t) during the time when the bucket contains a
nonzero amount of sand.
(b) What is the maximum value of the bucket’s kinetic energy,
assuming it is achieved before it hits the wall?
(c) What is the maximum value of the magnitude of the bucket’s
momentum, assuming it is achieved before it hits the wall?
(d) For what value of b does the bucket become empty right when it
hits the wall?
5.72. Throwing a brick ***
A brick is thrown from ground level, at an angle θ with respect to the
(horizontal) ground. Assume that the long face of the brick remains
parallel to the ground at all times, and that there is no deformation in the
ground or the brick when the brick hits the ground. If the coefficient of
friction between the brick and the ground is µ, what should θ be so that
the brick travels the maximum total horizontal distance before finally
coming to rest? Assume that the brick doesn’t bounce. Hint: The brick
slows down when it hits the ground. Think in terms of impulse.
Section 5.7: Collisions
5.73. A one-dimensional collision *
Consider the following one-dimensional collision. A mass 2m moves to
the right, and a mass m moves to the left, both with speed v. They collide
elastically. Find their final lab-frame velocities. Solve this by:
(a) Working in the lab frame.
(b) Working in the CM frame.
5.74. Perpendicular vectors *
A moving mass m collides elastically with a stationary mass 2m. Let
their resulting velocities be v1 and v2 , respectively. Show that v2 must
be perpendicular to 2v1 + v2 . Hint: See Problem 5.19.
189
190
Conservation of energy and momentum
5.75. Three pool balls *
A pool ball with initial speed v is aimed right between two other pool
balls, as shown in Fig. 5.44. If the two right balls leave the (elastic) collision with equal speeds, find the final velocities of all three
balls.
v
Fig. 5.44
F
A
E
v
D
B
C
Fig. 5.45
5.76. Seven pool balls **
Seven pool balls are situated at rest as shown in Fig. 5.45. The middle ball
suddenly somehow acquires a speed v to the right. Assume that starting
with ball A, the balls spiral out an infinitesimal amount. So A is closer to
the center ball than B is, and B is closer than C is, etc. This means that the
center ball collides with A first, then it gets deflected into B, and then it
gets deflected into C, and so on. But all the collisions happen in the blink
of an eye. What will the center ball’s velocity be after it collides (elastically) with all six balls? (You can use the results from the example in
Section 5.7.2.)
5.77. Midair collision **
A ball is held and then released. At the instant it is released, an identical
ball, moving horizontally with speed v, collides elastically with it and is
deflected at an upward angle. What is the maximum horizontal distance
the latter ball can travel by the time it returns to the height of the collision?
(You can use the results from the example in Section 5.7.2.)
5.78. Maximum number of collisions **
N identical balls are constrained to move in one dimension. If you are
allowed to pick their initial velocities, what is the maximum number
of collisions you can arrange for the balls to have among themselves?
Assume that the collisions are elastic.
θ
Fig. 5.46
5.79. Triangular room **
A ball is thrown against a wall of a very long triangular room which has
vertex angle θ . The initial direction of the ball is parallel to the angle
bisector (see Fig. 5.46). How many (elastic) bounces does the ball make?
Assume that the walls are frictionless.
5.80. Equal angles **
(a) Amass 2m moving at speed v0 collides elastically with a stationary
mass m. If the two masses scatter at equal (nonzero) angles with
respect to the incident direction, what is this angle?
(b) What is the largest number that the above “2” can be replaced
with, if you want it to be possible for the masses to scatter at
equal angles?
5.10 Exercises
191
5.81. Right angle in billiards **
A billiard ball collides elastically with an identical stationary one. By
looking at the collision in the CM frame, show that the angle between
the resulting trajectories in the lab frame is 90◦ . (We proved this result
by working in the lab frame in the example in Section 5.7.2.)
5.82. Equal vx ’s **
A mass m moving with speed v in the x direction collides elastically with
a stationary mass nm, where n is some number. After the collision, it is
observed that both masses have equal x components of their velocities.
What angle does the velocity of mass nm make with the x axis? (This
can be solved by working in the lab frame or the CM frame, but the CM
solution is slick.)
5.83. Maximum vy **
A mass M moving in the positive x direction collides elastically with
a stationary mass m. The collision is not necessarily head-on, so the
masses may come off at angles, as shown in Fig. 5.47. Let θ be the
angle of m’s resulting motion. What should θ be so that m has the largest
possible speed in the y direction? Hint: Think about what the collision
should look like in the CM frame.
5.84. Bouncing between rings **
Two fixed circular rings, in contact with each other, stand in a vertical
plane. A ball bounces elastically back and forth between the rings (see
Fig. 5.48). Assume that initial conditions have been set up so that the
ball’s motion forever lies in one parabola. Let this parabola hit the rings
at an angle θ from the horizontal. Show that if you want the magnitude
of the change in the horizontal component of the ball’s momentum
at
√
each bounce to be maximum, then you should pick cos θ = ( 5 − 1)/2,
which just happens to be the inverse of the golden ratio.
5.85. Bouncing between surfaces **
Consider the following generalization of the previous exercise. A ball
bounces back and forth between a surface defined by f (x) and its reflection across the y axis (see Fig. 5.49). Assume that initial conditions have
been set up so that the ball’s motion forever lies in one parabola, with the
contact points located at ±x0 . For what function f (x) is the magnitude
of the change in the horizontal component of the ball’s momentum at
each bounce independent of x0 ?
5.86. Drag force on a sphere **
A sphere of mass M and radius R moves with speed V through a region
of space that contains particles of mass m that are at rest. There are n of
M
m
M
θ
m
Fig. 5.47
θ
Fig. 5.48
f (-x)
-x0
Fig. 5.49
f (x)
x0
192
Conservation of energy and momentum
these particles per unit volume. Assume m
M , and assume that the
particles do not interact with each other. What is the drag force on the
sphere?
Total mass M
m
Fig. 5.50
5.87. Balls in a semicircle ***
N identical balls lie equally spaced in a semicircle on a frictionless
horizontal table, as shown. The total mass of these balls is M . Another
ball of mass m approaches the semicircle from the left, with the proper
initial conditions so that it bounces (elastically) off all N balls and finally
leaves the semicircle, heading directly to the left (see Fig. 5.50).
(a) In the limit N → ∞ (so the mass of each ball in the semicircle,
M /N , goes to zero), find the minimum value of M /m that allows
the incoming ball to come out heading directly to the left. Hint:
You’ll need to do Problem 5.24 first.
(b) In the minimum M /m case found in part (a), show that the ratio
of m’s final speed to initial speed equals e−π .
5.88. Block and bouncing ball ****
A block with large mass M slides with speed V0 on a frictionless table
toward a wall. It collides elastically with a ball with small mass m, which
is initially at rest at a distance L from the wall. The ball slides toward the
wall, bounces elastically, and then proceeds to bounce back and forth
between the block and the wall.
(a) How close does the block come to the wall?
(b) How many times does the ball bounce off the block, by the time
the block makes its closest approach to the wall?
Assume M
m, and give your answers to leading order in m/M .
Section 5.8: Inherently inelastic processes
Note: In the exercises involving chains in this section, we’ll assume that
the chains are of the type described in the first scenario near the end of
Section 5.8.
5.89. Slowing down, speeding up *
A plate of mass M moves horizontally with initial speed v on a frictionless table. A mass m is dropped vertically onto it and soon comes to rest
with respect to it. How much energy is required to bring the system back
up to speed v? Explain intuitively your answer in the M
m limit.
5.90. Pulling a chain back **
A chain with length L and mass density σ kg/m lies outstretched on
a frictionless horizontal table. You grab one end and pull it back along
5.10 Exercises
193
itself, in a parallel manner, as shown in Fig. 5.51. If your hand starts from
rest and has constant acceleration a, what is your force at the moment
right before the chain is straightened out?
5.91. Falling chain **
A chain with length L and mass density σ kg/m is held in a heap, and
you grab an end that protrudes a tiny bit out of the top. The chain is
then released. As a function of time, what is the force that your hand
must apply to keep the top end of the chain motionless? Assume that
the chain has no friction with itself, so that the remaining part of the
heap is always in freefall. The setup at a random later time is shown in
Fig. 5.52.
5.92. Pulling a chain down **
A chain with mass density σ kg/m lies in a heap at the edge of a table.
One end of the chain initially sticks out an infinitesimal distance from
the heap. You grab this end and accelerate it downward with acceleration
a. Assume that there is no friction of the chain with itself as it unravels.
As a function of time, what force does your hand apply to the chain?
Find the value of a that makes your force always equal to zero. (In other
words, find the a with which the chain naturally falls.)
(top view)
hand
a
L
Fig. 5.51
heap
Fig. 5.52
5.93. Raising a chain **
A chain with length L and mass density σ kg/m lies in a heap on the
floor. You grab one end of the chain and pull upward with a force such
that the chain moves at constant speed v. What is the total work you do,
by the time the chain is completely off the floor? How much energy is
lost to heat, if any? Assume that the chain is greased, so that it has no
friction with itself.
5.94. Downhill dustpan **
A plane inclined at an angle θ is covered with dust. An essentially massless dustpan on wheels is released from rest and rolls down the plane,
gathering up dust. The density of dust in the path of the dustpan is
σ kg/m. What is the acceleration of the dustpan?
5.95. Heap and block **
A chain with mass density σ kg/m lies in a heap on the floor, with one
end attached to a block of mass M . The block is given a sudden kick
and instantly acquires a speed V0 . Let x be the distance traveled by the
block. In terms of x, what is the tension in the chain, just to the right
of the heap; that is, at the point P shown in Fig. 5.53? There is no
friction in this problem; none with the floor, and none in the chain with
itself.
V0
m
M
(start)
P
x
(later)
Fig. 5.53
M
194
Conservation of energy and momentum
k
σ
L
heap
5.96. Touching the floor ****
A chain with mass density σ kg/m hangs from a spring with spring
constant k. In the equilibrium position, a length L is in the air, and the
bottom part of the chain lies in a heap on the floor; see Fig. 5.54. The
chain is raised by a very small distance, b, and then released. What is
the amplitude of the oscillations, as a function of time?
Assume that (1) L
b, (2) the chain is very thin, so that the size of
the heap on the floor is very small compared with b, (3) the length of
the chain in the initial heap is larger than b, so that some of the chain
always remains in contact with the floor, and (4) there is no friction of
the chain with itself inside the heap.
Fig. 5.54
5.11
Solutions
5.1. Minimum length
Drop the masses through the three holes, and let the system reach its equilibrium
position. The equilibrium position is the one with the lowest potential energy of the
masses, that is, the one with the most string hanging below the table. In other words, it
is the one with the least string lying on the table. This is the desired minimum-length
configuration.
What are the angles at the vertex of the string? The tensions in all three strings are
equal to mg, because they are holding up the masses. The vertex of the string is in
equilibrium, so the net force on it must be zero. This implies that each string must
bisect the angle formed by the other two. Therefore, the angles between the strings
must all be 120◦ .
5.2. Heading to zero
The energy of the particle is E = mv 2/2 − A|x|n . The given information
√ tells us that
v = 0 when x = 0. Therefore, E = 0, which then implies that v = − 2Axn /m (we’ll
assume x > 0; the x < 0 case works the same). We have chosen the minus sign because
the particle is heading toward the origin. Writing v as dx/dt and separating variables
gives
0
x0
dx
2A
=−
xn/2
m
T
0
dt = −T
2A
,
m
(5.84)
where x0 is the initial position and T is the time to reach the origin. The integral on the
left is finite if and only if n/2 < 1. Therefore, the condition that T is finite is n < 2.
Remark: If 0 < n < 1, then V (x) has a cusp at x = 0 (infinite slope on either side),
so it’s clear that T is finite. If n = 1, then the slope is a finite constant, so it’s also clear
than T is finite. If n > 1, then the slope of V (x) is zero at x = 0, so it’s not obvious
what happens with T . But the above calculation shows that n = 2 is the value where
T becomes infinite.
The particle therefore takes a finite time to reach the top of a triangle or the curve
−Ax3/2 . But it takes an infinite time to reach the top of a parabola, cubic, etc. A circle
looks like a parabola at the top, so T is infinite in that case also. In fact, any nice
polynomial function V (x) requires an infinite T to reach a local maximum, because
the Taylor series starts at order two (at least) around an extremum. ♣
5.11 Solutions
195
5.3. Leaving the sphere
First solution: Let R be the radius of the sphere, and let θ be the angle of the mass,
measured from the top of the sphere. The radial F = ma equation is
mv 2
,
(5.85)
R
where N is the normal force. The mass loses contact with the sphere when the normal
force becomes zero (that is, when the normal component of gravity is barely large
enough to account for the centripetal acceleration of the mass). Therefore, the mass
loses contact when
mg cos θ − N =
mv 2
= mg cos θ .
R
(5.86)
But conservation of energy gives mv 2 /2 = mgR(1 − cos θ ). Hence, v =
2gR(1 − cos θ). Plugging this into Eq. (5.86) gives
2
(5.87)
=⇒ θ ≈ 48.2◦ .
3
Second solution: Let’s assume (incorrectly) that the mass always stays in contact
with the sphere, and then find the point where the horizontal component of v starts
to decrease, which it of course can’t do, because the normal force doesn’t have a
“backward” component. From above, the horizontal component of v is
cos θ =
vx = v cos θ =
2gR(1 − cos θ ) cos θ .
(5.88)
Taking the derivative of this, we find that the maximum occurs when cos θ = 2/3.
So this is where vx would start to decrease if the mass were constrained to remain on
the sphere. But there is no such constraining force available, so the mass loses contact
when cos θ = 2/3.
5.4. Pulling the pucks
(a) Let θ be defined as in Fig. 5.55. Then the tension in the string is T = F/(2 cos θ ),
because the force on the massless kink in the string must be zero. Consider the
“top” puck. The component of the tension in the y direction is −T sin θ =
−(F/2) tan θ . The work done on the puck by this component is therefore
Wy =
=
=
0
−F tan θ
dy =
2
0
π/2
0
π/2
y
x
T
θ
θ
F
T
Fig. 5.55
−F tan θ
d( sin θ )
2
−F sin θ
dθ
2
F cos θ
2
0
π/2
=
F
.
2
(5.89)
By the work–energy theorem (or equivalently, by separating variables and integrating Fy = mvy d vy /dy), this work equals the value of mvy2 /2 right before the
collision. There are two pucks, so the total kinetic energy lost when they stick
together is twice this quantity (vx doesn’t change during the collision), which
is F .
(b) Consider two systems, A and B (see Fig. 5.56). A is the original setup, while B
starts with θ already at zero. Let the pucks in both systems start simultaneously at
x = 0. As the force F is applied, all four pucks will have the same x(t), because
the same force in the x direction, namely F/2, is applied to every puck at all
times. After the collision, both systems will therefore look exactly the same.
Let the collision of the pucks occur at x = d. At this point, F(d + ) work
has been done on system A, because the center of the string (where the force
system A
F
system B
F
Fig. 5.56
196
Conservation of energy and momentum
is applied) ends up moving a distance more than the masses. However, only
Fd work has been done on system B. Since both systems have the same kinetic
energy after the collision, the extra F work done on system A must be what is
lost in the collision.
Remark: The reasoning in this second solution can be used to solve the problem
in the case where we have a uniform massive rope (so the rope flops down, as in
Fig. 5.57). The center of mass of the rope moves in exactly the same manner as
the position of the two pucks in system B (assuming that the mass of each puck
is chosen to be half the mass of the rope), because the same force F acts on both
systems. You can show that this implies that the force acts over an extra distance
of /2 on the rope, compared with system B, by the time the rope has flopped
into a straight line. From the reasoning above, F /2 of work must therefore be
lost to heat in the rope. ♣
F
Fig. 5.57
5.5. Constant y˙
By conservation of energy, the bead’s speed at any time is given by (note that y is
negative here)
1 2
1
mv + mgy = mv02
2
2
=⇒
v=
v02 − 2gy .
(5.90)
The vertical component of the velocity is y˙ = v sin θ, where θ is the (negative) angle
the wire makes with the horizontal. The slope of the wire is tan θ = dy/dx ≡ y ,
which yields sin θ = y / 1 + y 2 . The requirement y˙ = −v0 , which is equivalent to
v sin θ = −v0 , may therefore be written as
y
v02 − 2gy ·
1+y2
= −v0 .
(5.91)
√
Squaring both sides and solving for y ≡ dy/dx yields dy/dx = −v0 / −2gy.
Separating variables and integrating gives
−2gy dy = −v0
dx
=⇒
(−2gy)3/2
= v0 x,
3g
(5.92)
where the constant of integration has been set to zero, because (x, y) = (0, 0) is a point
on the curve. Therefore,
y=−
block
table
Fig. 5.58
F
(3g v0 x)2/3
.
2g
(5.93)
5.6. Dividing the heat
It turns out that it isn’t possible to answer these questions without being given more
information. The way that the work is divided up between the objects depends on what
their surfaces look like. It’s theoretically possible for one of the objects to gain all the
heat, while the other doesn’t heat up at all.
To understand this, we’ll need to make a model of how friction works. The general
way that friction works is that molecules from one surface rub against molecules from
the other surface. The molecules stretch to the side and then bounce back and vibrate.
This vibrational motion is the kinetic energy associated with heat. The model we’ll
use here will have a bunch of springs with masses on the ends, on both surfaces at the
interface. When the surfaces rub against each other, the masses catch on each other
for a short time (as shown in Fig. 5.58), and then they release, whereupon they vibrate
back and forth on the springs. This is the kinetic energy of the heat we see. Though an
oversimplification, this is basically the way friction works.
Now, if everything is symmetrical between the two objects (that is, if the springs and
masses on one object look like those on the other), then both objects will heat up by
5.11 Solutions
the same amount. But things need not be symmetrical. You can imagine the springs on
one surface being much stiffer (that is, having a much larger k value) than the springs
on the other surface. Or, you can even take the limit where one surface (say, the block)
is made of completely rigid teeth, as shown (for one tooth) in Fig. 5.59. In this case,
only the bottom surface (the table) will end up with heat from vibrational motion.
Is this asymmetrical result consistent with what we obtain from the work–energy
theorem? Well, the net work done on the block is zero, because your pulling force does
Fd positive work (where F = µk N ), while the friction force (the sum of all the forces
from the masses on the little teeth) does Fd negative work. Therefore, since zero net
work is done on the block, its total energy is constant. And since the kinetic energy due
to its motion as a whole is constant, its internal thermal energy must also be constant.
In other words, it doesn’t heat up.
The net work done on the table in this scenario comes from the force from the teeth
on the little masses on the springs. These teeth do Fd positive work on all the little
spring–mass systems, so Fd is the work done on the table. Therefore, its total energy
increases by Fd. And since the kinetic energy due to its motion as a whole is constant
(and zero, since the table is just sitting there), its internal thermal energy must increase
by Fd. In other words, it heats up.
Now consider the reverse situation, where the table has the rigid teeth and the block
has the springs and masses, as shown in Fig. 5.60. The block is now the object that
heats up, because it has the vibrational motion of the masses. And as above, we can
show that this is consistent with the work–energy theorem, as follows. In the present
case, the force from the table’s teeth does no work on the block (because the teeth
aren’t moving), so the net work done on the block is simply the Fd from your pulling,
so it heats up. Likewise, the little masses do no work on the teeth (because the teeth
aren’t moving), so the work done on the table is zero, so it doesn’t heat up.
For the in between case where the spring constants on the two objects are equal, the
net work done on each object is Fd/2, where d is the distance the block moves. This
is true because the two masses in Fig. 5.58 each move half as far as the block moves.
So the work done on the block is Fd − Fd/2 = Fd/2 (this is the positive work done
by you, plus the negative work done by the table’s little masses). And the work done
on the table is Fd/2 (this is the positive work done by the block’s little masses). So the
objects heat up the same amount. For more discussion of the issues in this problem,
see Sherwood (1984).
5.7. V (x) vs. a hill
First solution: Consider the normal force N acting on the bead at a given point. Let
θ be the angle that the tangent to V (x) makes with the horizontal, as shown in Fig. 5.61.
The horizontal F = ma equation is
−N sin θ = m¨x.
197
block
table
Fig. 5.59
block
Fig. 5.60
θ N
V(x)
mg
θ
Fig. 5.61
N cos θ − mg = m¨y
=⇒
N cos θ = mg + m¨y.
(5.95)
Dividing Eq. (5.94) by Eq. (5.95) gives
x¨
.
g + y¨
(5.96)
x¨ = −(g + y¨ )V .
(5.97)
− tan θ =
But tan θ = V (x). Therefore,
We see that this is not equal to −gV . In fact, there is in general no way to construct
a curve with height z(x) that gives the same horizontal motion that a 1-D potential
mgV (x) gives, for all initial conditions. We would need −(g + z¨ )z = −gV , for all
x. But at a given x, the quantities V and z are fixed, whereas z¨ depends on the initial
F
table
(5.94)
The vertical F = ma equation is
F
198
Conservation of energy and momentum
conditions. For example, if there is a bend in the wire, then z¨ will be large if z˙ is large.
And z˙ depends (in general) on how far the bead has fallen.
Equation (5.97) holds the key to constructing a situation that does give the x¨ = −gV
result. All we have to do is get rid of the y¨ term. So here’s what we do. We grab our
y = V (x) wire and move it up and/or down in precisely the manner that makes the bead
stay at the same height with respect to the ground. (Actually, constant vertical speed
would be good enough.) This will make the y¨ term vanish, as desired. The vertical
movement of the curve doesn’t change the slope V at a given value of x, so the θ in
the above derivation is still the same θ .
Note that the quantity y here is the vertical position of the bead. It equals V (x) if the
curve is stationary, but not if the curve is being moved up and down.
Remark: There is one case where x¨ is (approximately) equal to −gV , even when the
wire remains stationary. In the case of small oscillations of the bead near a minimum
of V (x), y¨ is small compared with g. Hence, Eq. (5.97) shows that x¨ is approximately
equal to −gV . Therefore, for small oscillations, it is reasonable to model a particle
in a 1-D potential mgV (x) as a particle sliding in a valley whose height is given
by y = V (x). ♣
Second solution: The component of gravity along the wire is what causes the change
in velocity of the bead. That is,
−g sin θ =
dv
,
dt
(5.98)
where θ is given by
tan θ = V (x)
=⇒
sin θ = √
V
1+V
2
,
cos θ = √
1
1+V
2
.
(5.99)
We are, however, not concerned with the rate of change of v , but rather with the rate
of change of x˙ . In view
√ of this, let us write v in terms of x˙ . Since x˙ = v cos θ , we
have v = x˙ / cos θ = x˙ 1 + V 2 (dots denote d/dt, primes denote d/dx). Therefore,
Eq. (5.98) becomes
−gV
√
1+V
2
=
d
x˙ 1 + V
dt
= x¨ 1 + V
2
2
x˙ V (dV /dt)
+ √
.
1+V 2
(5.100)
Hence, x¨ is given by
x¨ =
−gV
1+V
2
−
x˙ V (dV /dt)
.
1+V 2
(5.101)
We’ll simplify this in a moment, but first a remark.
Remark: A common incorrect solution to this
√problem is the following. The accelthe horizontal
eration along the curve is g sin θ = −g(V / 1 + V 2 ). Calculating
√
component of this acceleration brings in a factor of cos θ = 1/ 1 + V 2 . Therefore,
we might think that
x¨ =
−gV
1+V
2
(incorrect).
(5.102)
We have missed the second term in Eq. (5.101). Where is the mistake? The error is that
we forgot to take into account the possible change in the curve’s slope (Eq. (5.102) is
true for straight lines). We addressed only the acceleration due to a change in speed.
We forgot to consider the acceleration due to a change in the direction of motion (the
term we missed is the one with dV /dt). Intuitively, if we have a sharp enough bend in
5.11 Solutions
the wire, then x˙ can change at an arbitrarily large rate, even if v is roughly constant.
In view of this fact, Eq. (5.102) is definitely incorrect, because it is bounded (by g/2,
in fact). ♣
To simplify Eq. (5.101), note that V ≡ dV/dx = (dV/dt)/(dx/dt) ≡ V˙ /˙x (V˙ is just
the rate of change in the bead’s height). Therefore, the numerator in the second term
on the right-hand side of Eq. (5.101) is
x˙ V
V˙
x˙
dV
d
= x˙ V
dt
dt
= V V¨ − V x¨
x˙ V¨ − V˙ x¨
x˙ 2
= x˙ V
V˙
x˙
= V V¨ − V 2 x¨ .
(5.103)
Substituting this into Eq. (5.101) yields
x¨ = −(g + V¨ )V ,
(5.104)
in agreement with Eq. (5.97), because y = V (x) if the wire is stationary.
Equation (5.104) is valid only for a curve V (x) that remains fixed. If we grab the
wire and start moving it up and down, then the above solution is invalid, because the
starting point, Eq. (5.98), rests on the assumption that gravity is the only force that
does work on the bead. But if we move the wire, then the normal force also does work.
It turns out that for a moving wire, we simply need to replace the V¨ in Eq. (5.104)
by y¨ , which then gives Eq. (5.97). This can be seen by looking at things in the vertically
accelerating frame in which the wire is at rest. We won’t cover accelerating frames
until Chapter 10, so we’ll just invoke here the result that there is an extra fictitious
“translational” force in this accelerating frame, and the consequence of this is that the
bead thinks that it lives in a world where the acceleration from gravity is g + h¨ (if h¨ is
positive, then the bead thinks gravity is larger), where h is the position of your hand
that is accelerating the wire. In this new frame, the wire is at rest, so the above solution
¨ we have x¨ = −(g + h¨ + V¨ )V .
is valid. So with the g in Eq. (5.104) replaced by g + h,
But V¨ (which is the vertical acceleration of the bead with respect to the new frame)
plus h¨ (which is the vertical acceleration of the new frame with respect to the ground)
equals y¨ (which, by our definition in the first solution, is the vertical acceleration of
the bead with respect to the ground). We have therefore reproduced Eq. (5.97).
5.8. Hanging mass
We will calculate the equilibrium point y0 , and then use ω =
derivative of V is
V (y0 )/m. The
V (y) = ky + mg.
(5.105)
Therefore, V (y) = 0 when y = −mg/k ≡ y0 . The second derivative of V is
V (y) = k.
(5.106)
We therefore have
ω=
V (y0 )
=
m
k
.
m
(5.107)
Remark: This is independent of y0 , which makes sense because the only effect of
gravity is to change the equilibrium position. More precisely, if yr is the position
relative to y0 (so that y ≡ y0 + yr ), then the total force as a function of yr is
mg
F(yr ) = −k(y0 + yr ) − mg = −k −
+ yr − mg = −kyr ,
(5.108)
k
so it still looks like a regular spring. (This works only because the spring force is
linear.) Alternatively, you can think in terms of the potential energy; this is the task of
Exercise 5.42. ♣
199
200
Conservation of energy and momentum
5.9. Small oscillations
We will calculate the equilibrium point x0 , and then use ω =
derivative of V is
V (x0 )/m. The
V (x) = −Ce−ax xn−1 (n − ax).
(5.109)
Therefore, V (x) = 0 when x = n/a ≡ x0 . The second derivative of V is
V (x) = −Ce−ax xn−2 (n − 1 − ax)(n − ax) − ax .
(5.110)
Plugging in x0 = n/a simplifies this a bit, and we find
ω=
B'
A'
P
B
Fig. 5.62
A
V (x0 )
=
m
Ce−n nn−1
.
man−2
(5.111)
5.10. Zero force inside a sphere
Let a be the distance from P to piece A, and let b be the distance from P to piece B
(see Fig. 5.62). Draw the “perpendicular” bases of the cones, and call them A and
B . The ratio of the areas of A and B is a2 /b2 . The key point here is that the angle
between the planes of A and A is the same as the angle between B and B . This is true
because the chord between A and B meets the circle at equal angles at its ends. So the
ratio of the areas of A and B is also a2 /b2 . But the gravitational force decreases like
1/r 2 , and this effect exactly cancels the a2 /b2 ratio of the areas. Therefore, the forces
at P due to A and B (which can be treated like point masses, because the cones are
assumed to be thin) are equal in magnitude; and opposite in direction, of course. If we
draw enough cones to cover the whole shell, then the contributions to the force from
little pieces over the whole shell cancel in pairs, so we are left with zero force at P.
This holds for any point P inside the shell.
Remark: Interestingly, the force inside an ellipsoidal shell of constant density (per
volume) is also zero, assuming that the shell is defined to be the region between the
surfaces described by ax2 + by2 + cz 2 = k, for two different values of k. In short,
this is true because an ellipsoid is simply a stretched sphere. In detail: Let the above
spherical shell have some thickness dr (which was irrelevant for the sphere, but it will
be important in the ellipsoid case). From above, we know that the masses at the ends
of the thin cones in the sphere have canceling forces. If we now stretch the sphere into
an ellipsoid (uniformly in each direction, but the factors in the three directions can be
different), then the distances from the end masses to P will still be in the ratio of a to b
(as you can verify). And the end masses will still be in the ratio of a2 to b2 , because both
masses change by the same factor. This is true because every infinitesimal cube in the
end masses has its volume changed by the same factor (namely fx fy fz , where these f ’s
are the stretching factors in each direction). So the masses are still in the ratio of a2 to
b2 , and the same argument about canceling forces holds as in the spherical case. Note
that this zero-force result is not true for an ellipsoid with constant thickness, because
such an object isn’t the result of stretching a spherical shell (because a stretching results
in the ellipsoidal shell being thicker near the ends where it is more “pointy”). ♣
5.11. Escape velocity
(a) The cutoff case is where the particle barely makes it to infinity, that is, where
its speed is zero at infinity. Conservation of energy for this situation gives
GMm
1 2
mv −
= 0 + 0.
2 esc
R
(5.112)
2 /2, must account for the gain in
In other words, the initial kinetic energy, mvesc
potential energy, GMm/R. Therefore,
vesc =
2GM
.
R
(5.113)
5.11 Solutions
In terms of the acceleration,
g = GM/R2 , at the surface of a planet, we can
√
write this as vesc = 2gR. Using M = 4πρR3/3, we can also write it as
vesc = 8π GR2 ρ/3. So for a given density ρ, vesc grows like R. Using the
values of g and R given in Appendix J, we have:
√
For the earth, vesc = 2gR ≈ 2(9.8 m/s2 )(6.4 · 106 m) ≈ 11.2 km/s.
√
For the moon, vesc = 2gR ≈ 2(1.6 m/s2 )(1.7 · 106 m) ≈ 2.3 km/s.
√
For the sun, vesc = 2gR ≈ 2(270 m/s2 )(7.0 · 108 m) ≈ 620 km/s.
Remark: Another reasonable question to ask is: what is the escape velocity
from the sun for an object located
where the earth is (but imagine that the earth
√
isn’t there)? The answer is 2GMS /RES , where RES is the earth–sun distance.
Numerically, this is
2(6.67 · 10−11 m3 /kg s2 )(2 · 1030 kg)/(1.5 · 1011 m) ≈ 42 km/s. (5.114)
If you want to bring the earth back in (but let’s assume it’s at rest and not orbiting)
and find the escape velocity (from both the sun and the earth) from a point on the
earth’s surface, you can’t just add the 42 km/s and 11.2 km/s results. Instead,
you have to take the square root of the sum of the squares. This follows from
Eq. (5.112) and from the fact that potentials simply add. The result is about
43.5 km/s. The task of Exercise 5.60 is to find the escape velocity if the orbital
motion of the earth is included. ♣
(b) To get a rough answer, we’ll assume that the initial speed of a person’s jump
on the small planet is the same as it is on the earth. This probably isn’t quite
true, but it’s close enough for the purposes here. A good jump on the earth is
about a meter. For this jump,√
conservation of energy√gives mv 2 /2 = mg(1 m).
Therefore, v = 2g(1 m) ≈ 20 m/s. So we want 20 m/s = 8πGR2 ρ/3.
Using ρ ≈ 5500 kg/m3 , we find R ≈ 2.5 km. On such a planet, you should tread
lightly.
5.12. Ratio of potentials
Let ρ be the mass density of the cube. Let V cor be the potential energy of a mass m at
the corner of a cube of side , and let V cen be the potential energy of a mass m at the
center of a cube of side . By dimensional analysis,
V cor ∝
G(ρ 3 )m
∝
2
.
(5.115)
Therefore,32
V cor = 4V cor
/2 .
(5.116)
But a cube of side can be built from eight cubes of side /2. So by superposition, we
have
V cen = 8V cor
/2 ,
(5.117)
because the center of the larger cube lies at a corner of the eight smaller cubes (and
because potentials just add). Therefore,
4V cor
V cor
1
/2
=
= .
V cen
8V cor
2
/2
32
(5.118)
In other words, imagine expanding a cube of side /2 to one of side . If we consider corresponding
pieces of the two cubes, then the larger piece has 23 = 8 times the mass of the smaller. But
corresponding distances are twice as big in the large cube as in the small cube. Therefore, the
larger piece contributes 8/2 = 4 times as much to V cor as the smaller piece contributes to V cor
/2 .
201
202
Conservation of energy and momentum
5.13. Through the hole
(a) By symmetry, only the component of the gravitational force perpendicular to
the plane survives. A piece of mass dm at radius r on the plane provides a force
To obtain the component perpendicular to the plane,
equal to Gm(dm)/(r 2 +x2 ). √
we must multiply this by x/ r 2 + x2 . Slicing the plane up into rings with mass
dm = (2πr dr)σ , we find that the total force is
∞
F(x) = −
R
Gm(2πrσ dr)x
= 2πσ Gmx(r 2 + x2 )−1/2
(r 2 + x2 )3/2
2πσ Gmx
= −√
.
R2 + x2
r=∞
r=R
(5.119)
Note that if R = 0 (so that we have a uniform plane without a hole), then
F = −2π σ Gm, which is independent of the distance from the plane.
(b) If x
R, then Eq. (5.119) gives F(x) ≈ −2πσ Gmx/R, so F = ma yields
x¨ +
2πσ G
R
x = 0.
(5.120)
The frequency of small oscillations is therefore
2πσ G
,
R
ω=
(5.121)
which is independent of m.
Remark: For everyday values of R, this is a very small number because G
is so small. Let’s determine the rough size. If the sheet has thickness d, and
if it is made of a material with density ρ (per unit volume), then σ = ρd.
So ω = 2πρdG/R. In the above analysis, we assumed that the sheet was
infinitely thin. In practice, we need d to be much smaller than the amplitude
of the motion. But this amplitude must be much smaller than R in order for
our approximation to hold. So we conclude that d
R. To get a rough upper
bound on ω, let’s pick d/R = 1/10. And let’s make our sheet out of gold (with
ρ ≈ 2 · 104 kg/m3 ). We then find ω ≈ 1 · 10−3 s−1 , which corresponds to
an oscillation about every 100 minutes. For the analogous system consisting of
electrical charges, the frequency is much larger, because the electrical force is
so much stronger than the gravitational force. ♣
(c) Integrating the force in Eq. (5.119) to obtain the potential energy (relative to the
center of the hole) gives
x
V (x) = −
F(x) dx =
0
= 2πσ Gm
R2
+ x2
x
0
x
0
2πσ Gmx dx
√
R2 + x 2
= 2πσ Gm
R2 + x2 − R .
(5.122)
By conservation of energy, the speed at the center of the hole is given by
mv 2 /2 = V (x). Therefore,
v = 2 πσ G
For x
R2 + x2 − R .
√
R this reduces to v ≈ 2 πσ Gx.
(5.123)
Remark: You can also obtain this last result by noting that for large x, the force
in Eq. (5.119) reduces to F = −2πσ Gm. This is constant, so it’s basically just
like a gravitational force F = √
mg , where g ≡ 2πσ G. But we know that in this
familiar case, v = 2g h → 2(2πσ G)x, as above. ♣
5.11 Solutions
5.14. Snowball
All of the snowball’s momentum goes into the earth, which then translates (and rotates)
a tiny bit faster (or slower, depending on which way the snowball was thrown).
What about the energy? Let m and v be the mass and initial speed of the snowball.
Let M and V be the mass and final speed of the earth (with respect to the original rest
frame of the earth). Since m
M , conservation of momentum gives V ≈ mv /M . The
kinetic energy of the earth is therefore
m
mv 2
1
1 2
1
M
= mv 2
(5.124)
mv .
2
M
2
M
2
There is also a rotational kinetic-energy term of the same order of magnitude, but that
doesn’t matter. We see that essentially none of the snowball’s energy goes into the
earth. It therefore must all go into the form of heat, which melts some of the snow
(and/or heats up the wall). This is a general result for a small object hitting a large
object: The large object picks up essentially all of the momentum but essentially none
of the energy (except possibly in the form of heat).
5.15. Propelling a car
Let the speed of the car be v (t). Consider the collision of a ball of mass dm with the car.
In the instantaneous rest frame of the car, the speed of the ball is u − v . In this frame,
the ball reverses velocity when it bounces (because the car is so much more massive),
so its change in momentum is −2(u − v ) dm. This is also the change in momentum
in the lab frame, because the two frames are related by a given speed at any instant.
Therefore, in the lab frame the car gains a momentum of 2(u − v ) dm from each ball
that hits it. The rate of change in momentum of the car (that is, the force) is thus
dp
= 2σ (u − v ),
dt
(5.125)
where σ ≡ dm/dt is the rate at which mass hits the car. σ is related to the given σ by
σ = σ (u − v )/u, because although you throw the balls at speed u, the relative speed
of the balls and the car is only (u − v ). We therefore have
M
dv
2(u − v )2 σ
=
dt
u
v
=⇒
0
2σ
dv
=
(u − v )2
Mu
=⇒
1
1
2σ t
− =
u−v
u
Mu
=⇒
v (t) =
2σ t
M
1+
u
2σ t
M
.
t
dt
0
(5.126)
Note that v → u as t → ∞, as it should. Writing this speed as u 1 − 1/(1 + 2σ t/M ) ,
we can integrate it to obtain the position,
x(t) = ut −
2σ t
Mu
ln 1 +
2σ
M
,
(5.127)
where the constant of integration is zero because x = 0 at t = 0. We see that even
though the speed approaches u, the car will eventually be an arbitrarily large distance
behind an object that moves with constant speed u (for example, pretend that your first
ball misses the car and continues forward at speed u).
5.16. Propelling a car again
We can carry over some of the results from the previous problem. The only change in
the calculation of the force on the car is that since the balls don’t bounce backward,
we don’t pick up the factor of 2 in Eq. (5.125). The force on the car is therefore
m
(u − v )2 σ
dv
,
=
u
dt
(5.128)
203
204
Conservation of energy and momentum
where m(t) is the mass of the car-plus-contents, as a function of time. The main
difference between this problem and the previous one is that this mass m changes
because the balls are collecting inside the car. From the previous problem, the rate at
which mass enters the car is σ = σ (u − v )/u. Therefore,
dm
(u − v )σ
=
.
dt
u
(5.129)
We must now solve the two preceding differential equations. Dividing Eq. (5.128) by
Eq. (5.129), and separating variables, gives33
v
0
dv
=
u−v
m
M
dm
m
=⇒
− ln
u−v
u
= ln
m
M
=⇒
m=
Mu
.
u−v
(5.130)
Note that m → ∞ as v → u, as it should. Substituting this value of m into either
Eq. (5.128) or Eq. (5.129) gives
v
0
dv
=
(u − v )3
t
0
σ dt
Mu2
1
1
σt
− 2 =
2(u − v )2
2u
Mu2
u
v (t) = u −
.
t
1 + 2σ
M
=⇒
=⇒
(5.131)
Note that v → u as t → ∞, as it should. Integrating this speed to obtain the position
gives
x(t) = ut −
Mu
2σ t
Mu
1+
+
,
σ
M
σ
(5.132)
where the constant of integration has been chosen so that x = 0 at t = 0. For a
given t, the v (t) in Eq. (5.131) is smaller than the v (t) for the previous problem in
Eq. (5.126), which is easy to see if the latter is written as u 1 − 1/(1 + 2σ t/M ) . This
makes sense, because in the present problem the balls have less of an effect on v (t),
because (1) they don’t bounce back, and (2) the mass of the car-plus-contents is larger.
5.17. Leaky bucket
(a) First solution: The initial position is x = L. The given rate of leaking implies
that the mass of the bucket at position x is m = M (x/L). Therefore, F = ma
gives −T = (Mx/L)¨x. Writing the acceleration as v d v /dx, and separating
variables and integrating, gives
−
TL
M
x
L
dx
=
x
v
v dv
=⇒
−
0
TL
v2
x
ln
=
.
M
L
2
(5.133)
The kinetic energy at position x is therefore
E=
mv 2
=
2
Mx
L
v2
2
= −Tx ln
x
.
L
(5.134)
In terms of the fraction z ≡ x/L, we have E = −TLz ln z. Setting dE/dz = 0
to find the maximum gives
z=
33
1
e
=⇒
Emax =
TL
.
e
(5.135)
We can also quickly derive this equation by writing down conservation of momentum for the
time interval when a mass dm enters the car: dm u + mv = (m + dm)(v + dv). This yields the
first equality in Eq. (5.130). But we will still need to use one of Eqs. (5.128) and (5.129) in what
follows.
5.11 Solutions
Note that the (fractional) location of Emax is independent of M , T , and L, but its
value depends on T and L. These facts follow from dimensional analysis.
Remark: We began this solution by writing down F = ma, where m is the
mass of the bucket. You may be wondering why we didn’t use F = dp/dt,
where p is the momentum of the bucket. This would certainly give a different
result, because dp/dt = d(mv )/dt = ma + (dm/dt)v . We used F = ma because
at any instant the mass m is what is being accelerated by the force F.
If you want, you can imagine the process occurring in discrete steps: The
force pulls on the mass for a short period of time, then a little piece falls off.
Then the force pulls again on the new mass, then another little piece falls off.
And so on. In this scenario, it is clear that F = ma is the appropriate formula,
because it holds for each step in the process.
It is indeed true that F = dp/dt, if you let F be total force in the problem,
and let p be the total momentum. The tension T is the only horizontal force in
the problem, because we’ve assumed the ground to be frictionless. However,
the total momentum consists of both the sand in the bucket and the sand that has
leaked out and is sliding along on the ground. If we use F = dp/dt, where p is
the total momentum, we obtain (remember that dm/dt is a negative quantity)
−T =
dm
dpleaked
dm
dpbucket
+
= ma +
v + −
v = ma,
dt
dt
dt
dt
(5.136)
as expected. See Appendix C for further discussion of the uses of F = ma and
F = dp/dt. ♣
Second solution: Consider a small time interval during which the bucket
moves from x to x + dx (where dx is negative). The bucket’s kinetic energy
changes by (−T ) dx (this is positive) due to the work done by the spring, and
also changes by a fraction dx/x (this is negative) due to the leaking. Therefore,
dE = −T dx + E dx/x, or
E
dE
= −T + .
(5.137)
dx
x
In solving this differential equation, it is convenient to introduce the variable
y ≡ E/x. Then E = xy + y, where a prime denotes differentiation with respect
to x. Equation (5.137) then becomes xy = −T , which gives
E/x
0
x
dy = −T
L
dx
x
=⇒
E = −Tx ln
x
,
L
(5.138)
as in the first solution.
√
√
(b) From Eq. (5.133), the speed is v = 2TL/M − ln z, where z ≡ x/L.
Therefore, the magnitude of the momentum is
√
(5.139)
p = mv = (Mz)v = 2TLM −z 2 ln z.
Setting dp/dz = 0 to find the maximum gives
1
z= √
e
=⇒
pmax =
TLM
.
e
(5.140)
Note that the (fractional) location of pmax is independent of M, T, and L, but its
value depends on all three. These facts follow from dimensional analysis.
Remark: Emax occurs closer to the wall (that is, at a later time) than pmax . The
reason for this is that v matters more in E = mv 2/2 than it does in p = mv . As
far as E is concerned, it is beneficial for the bucket to lose a little more mass if
it means being able to pick up a little more speed (up to a certain point). ♣
205
206
Conservation of energy and momentum
5.18. Another leaky bucket
(a) F = ma gives −T = m¨x. Combining this with the given dm/dt = b¨x equation
yields m dm = −bT dt. Integration then gives m2/2 = C − bTt. But m = M
when t = 0, so we have C = M 2 /2. Therefore,
m(t) =
M 2 − 2bTt .
(5.141)
This holds for t < M 2 /2bT , provided that the bucket hasn’t hit the wall yet.
(b) The given equation dm/dt = b¨x = b d v /dt integrates to v = m/b + D. But
v = 0 when m = M , so we have D = −M /b. Therefore,
√
m−M
M 2 − 2bTt
M
v (m) =
=⇒ v (t) =
−
.
(5.142)
b
b
b
At the instant right before all the sand leaves the bucket, we have m = 0.
Therefore, v = −M /b at this point. Integrating v (t) to obtain x(t), we find
x(t) =
−(M 2 − 2bTt)3/2
M
M3
− t+L+ 2 ,
2
3b T
b
3b T
(5.143)
where the constant of integration has been chosen so that x = L when t =
0. Solving for t in terms of m from Eq. (5.141), substituting the result into
Eq. (5.143), and simplifying, gives
(M − m)2 (M + 2m)
.
6b2 T
x(m) = L −
(5.144)
(c) Using Eq. (5.142), the kinetic energy is (it’s easier to work in terms of m here)
1 2
1
mv = 2 m(m − M )2 .
2
2b
E=
(5.145)
Taking the derivative dE/dm to find the maximum, we obtain
m=
M
3
=⇒
Emax =
2M 3
.
27b2
(5.146)
(d) Using Eq. (5.142), the momentum is
p = mv =
1
m(m − M ).
b
(5.147)
Taking the derivative to find the maximum magnitude, we obtain
m=
M
2
=⇒
|p|max =
M2
.
4b
(5.148)
(d) We want x = 0 when m = 0, so Eq. (5.144) gives
0=L−
M3
6b2 T
=⇒
b=
M3
.
6TL
(5.149)
This is the only combination of M, T, and L that has the units of b, namely
kg√
s/m. But we needed to do the calculation to find the numerical factor of
1/ 6.
5.19. Right angle in billiards
Let v be the initial velocity, and let v1 and v2 be the final velocities. Since the masses
are equal, conservation of momentum gives v = v1 + v2 . Taking the dot product of
this equation with itself gives
v · v = v1 · v1 + 2v1 · v2 + v2 · v2 .
(5.150)
5.11 Solutions
And conservation of energy gives (dropping the factors of m/2)
v · v = v 1 · v1 + v 2 · v2 .
(5.151)
Taking the difference of these two equations yields
v1 · v2 = 0.
(5.152)
90◦ .
So v1 v2 cos θ = 0, which means that θ =
(Or v1 = 0, which means the incoming
mass stops because the collision is head-on. Or v2 = 0, which means the masses miss
each other.)
5.20. Bouncing and recoiling
Let vi be the speed of the ball after the ith bounce, and let Vi be the speed of the block
right after the ith bounce. Then conservation of momentum gives
mvi = MVi+1 − mvi+1 .
(5.153)
But Theorem 5.3 says that vi = Vi+1 + vi+1 . Solving this system of two linear equations
gives
vi+1 =
(M − m)vi
(1 − )vi
≡
≈ (1 − 2 )vi ,
M +m
1+
where ≡ m/M
the nth bounce is
and
Vi+1 ≈ 2 vi ,
(5.154)
1. This expression for vi+1 implies that the speed of the ball after
vn = (1 − 2 )n v0 ,
where
≡ m/M .
(5.155)
The total distance traveled by the block can be obtained by looking at the work done
by friction. Eventually, the ball has negligible energy, so all of its initial kinetic energy
goes into heat from friction. Therefore, mv02 /2 = Ff d = (µMg)d, which gives
d=
mv02
.
2µMg
(5.156)
To find the total time, we can add up the times, tn , the block moves after each bounce.
Since the product of force and time equals the change in momentum, we have Ff tn =
MVn , and so (µMg)tn = M (2 vn−1 ) = 2M (1 − 2 )n−1 v0 . Therefore,
∞
t=
tn =
n=1
2 v0
µg
∞
(1 − 2 )n =
n=0
v0
1
2 v0
·
=
.
µg 1 − (1 − 2 )
µg
(5.157)
We’ve let the sum go to n = ∞ even though our vn = (1 − 2 )n v0 approximation
eventually breaks down for very large n (because we dropped terms of order 2 in
deriving it). But by the time it breaks down, the terms are negligibly small anyway.
The calculation of d above can also be done by adding up the geometric series of the
distances moved after each bounce.
Remarks: This t = v0 /µg result is much larger than the result obtained in the case
where the ball sticks to the block on the first hit, in which case the answer is t =
mv0 /(µMg). The total time is proportional to the total momentum that the block picks
up, and the t = v0 /µg answer is larger because the wall keeps transferring positive
momentum to the ball, which then gets transferred to the block.
In contrast, d is the same as it would be in the case where the ball sticks to the block
on the first hit. The total distance is proportional to the total energy that the block picks
up, and in both cases the total energy given to the block is mv02 /2. The wall (which is
attached to the very massive earth) transfers essentially no energy to the ball.
m), although
The t = v0 /µg result is independent of the masses (as long as M
it’s not at all intuitively obvious that if we keep the same v0 , but decrease m by a factor
207
208
Conservation of energy and momentum
of 100, that we’ll end up with the same t. The distance d, on the other hand, would
decrease by 100. ♣
5.21. Drag force on a sheet
(a) We will set v = 0 here. When the sheet hits a particle, the particle acquires
a speed of essentially 2V . This follows from Theorem 5.3, or by working in
the frame of the heavy sheet. The momentum of the particle is then 2mV . In
time t, the sheet sweeps through a volume AVt, where A is the area of the
sheet. Therefore, in time t, the sheet hits AVtn particles. The sheet therefore
loses momentum at a rate of dP/dt = −(AVn)(2mV ). But F = dP/dt, so the
magnitude of the drag force per unit area is
F
= 2nmV 2 ≡ 2ρV 2 ,
A
(5.158)
where ρ is the mass density of the particles. We see that the force depends
quadratically on V .
(b) If v
V , the particles now hit the sheet from various directions on both sides,
but we need only consider the particles’ motions along the line of the sheet’s
motion. As stated in the problem, we will assume that all velocities in this
direction are equal to ±v /2. Note that we won’t be able to set V exactly equal
to zero here, because we would obtain a force of zero and miss the lowest-order
effect.
Consider a particle in front of the sheet, moving backward toward the sheet.
The relative speed between the particle and the sheet is v /2 + V . This relative
speed reverses direction during the collision, so the change in momentum of the
particle is 2m(v /2 + V ). We have used the fact that the speed of the heavy sheet
is essentially unaffected by the collision. The rate at which particles collide with
the sheet is A(v /2 + V )(n/2), from the reasoning in part (a). The n/2 factor
comes from the fact that half of the particles move toward the sheet, and half
move away from it.
Now consider a particle behind the sheet, moving forward toward the sheet.
The relative speed between the particle and the sheet is v /2 − V . This relative
speed reverses direction during the collision, so the change in momentum of this
particle is −2m(v /2 − V ). And the rate at which particles collide with the sheet
is A(v /2 − V )(n/2). Therefore, the magnitude of the drag force per unit area is
F
1 dP
= ·
A
A dt
n
(v /2 + V )
=
2
2m(v /2 + V ) +
= 2nmv V ≡ 2ρ v V ,
V θ
θ
cylinder frame
Fig. 5.63
θ
V
n
(v /2 − V )
2
− 2m(v /2 − V )
(5.159)
where ρ is the mass density of the particles. We see that the force depends
linearly on V . The fact that it agrees with the result in part (a) in the case of
v = V is coincidence. Neither result is valid when v = V .
5.22. Drag force on a cylinder
Consider a particle that makes contact with the cylinder at an angle θ with respect
to the line of motion. In the frame of the heavy cylinder (see Fig. 5.63), the particle
comes in with velocity −V and then bounces off with a horizontal velocity component
of V cos 2θ. So in this frame (and hence also in the lab frame), the particle increases
its horizontal momentum by mV (1 + cos 2θ ). The cylinder must therefore lose this
momentum.
The area on the cylinder that lies between θ and θ + dθ sweeps out volume at a
rate (R dθ cos θ)V , where is the length of the cylinder. The cos θ factor here gives
5.11 Solutions
the projection orthogonal to the direction of motion. The force per unit length on the
cylinder (that is, the rate of change in momentum per unit length) is therefore
F
=
π/2
−π/2
n(R dθ cos θ)V
= 2nmRV 2
π/2
−π/2
cos θ(1 − sin2 θ ) dθ
= 2nmRV 2 sin θ −
=
mV (1 + cos 2θ )
1
sin3 θ
3
π/2
−π/2
8
8
nmRV 2 ≡ ρRV 2 ,
3
3
(5.160)
where ρ is the mass density of the particles. Note that the average force per crosssectional area, F/(2R ), equals (4/3)ρV 2 . This is smaller than the result for the sheet
in the previous problem, as it should be, because the particles bounce off somewhat
sideways in the cylinder case.
5.23. Basketball and tennis ball
(a) Right before the basketball hits the ground, both balls move downward with
speed (using mv 2 /2 = mgh)
v=
2gh.
(5.161)
Right after the basketball bounces off the ground, it moves upward with speed
v , while the tennis ball still moves downward with speed v . The relative speed
is therefore 2v . After the balls bounce off each other, the relative speed is still
2v (this follows from Theorem 5.3, or by working in the frame of the heavy
basketball). Since the upward speed of the basketball essentially stays equal to
v , the upward speed of the tennis ball is 2v + v = 3v . By conservation of energy,
it therefore rises to a height of H = d + (3v )2 /(2g). But v 2 = 2gh, so we have
H = d + 9h.
(5.162)
(b) Right before B1 hits the ground, all of the balls move downward with speed
v = 2gh. We will inductively determine the speed of each ball after it bounces
off the one below it. If Bi achieves a speed of vi after bouncing off Bi−1 , then
what is the speed of Bi+1 after it bounces off Bi ? The relative speed of Bi+1 and
Bi (right before they bounce) is v + vi . This is also the relative speed after they
bounce. Therefore, since Bi is still moving upward at essentially speed vi , we
see that the final upward speed of Bi+1 equals (v + vi ) + vi . Thus,
vi+1 = 2vi + v .
(5.163)
Since v1 = v , we obtain v2 = 3v (in agreement with part (a)), and then v3 = 7v ,
and then v4 = 15v , etc. In general,
vn = (2n − 1)v ,
(5.164)
which is easily seen to satisfy Eq. (5.163), with the initial value v1 = v . From
conservation of energy, Bn therefore rises to a height of
H= +
((2n − 1)v )2
= + (2n − 1)2 h.
2g
(5.165)
If h is 1 meter, and if we want this height
to be 1000 meters, then (assuming
√
is not very large) we need 2n − 1 > 1000. Five balls won’t quite do the trick,
209
210
Conservation of energy and momentum
but six will, in which case the height
is almost 4 kilometers. Escape velocity
√
from the earth (which is vesc = 2gR ≈ 11 200 m/s) is reached when
vn ≥ vesc =⇒ (2n − 1) 2gh ≥
2gR
=⇒
n ≥ ln2
R
+1 .
h
(5.166)
With R = 6.4 · 106 m and h = 1 m, we find n ≥ 12. Of course, the elasticity
assumption is absurd in this case, as is the notion that you can find 12 balls with
m2
···
m12 .
the property that m1
5.24. Maximal deflection
First solution: Let’s figure out what the collision looks like in the CM frame. If M has
initial speed V in the lab frame, then the CM moves with speed VCM = MV /(M + m).
The speeds of M and m in the CM frame therefore equal, respectively,
U
θ
u
θ
U
u
Fig. 5.64
Vlab
VCM
U
U = V − VCM =
mV
,
M +m
φ max
MV
____
M+m
mV
____
M+m
MV
.
M +m
(5.167)
In the CM frame, the collision is simple. The particles keep the same speeds, but
simply change their directions (while still moving in opposite directions), as shown
in Fig. 5.64. The angle θ is free to have any value. This scenario clearly satisfies
conservation of energy and momentum, so it must be what happens.
The important point to note is that since θ can have any value, the tip of the U
velocity vector can be located anywhere on a circle of radius U . If we then shift back
to the lab frame, we see that the final velocity of M with respect to the lab frame,
Vlab , is obtained by adding VCM to the vector U, which can point anywhere on the
dotted circle in Fig. 5.65. A few possibilities for Vlab are shown. The largest angle of
deflection is obtained when Vlab is tangent to the dotted circle, in which case we have
the situation shown in Fig. 5.66. The maximum angle of deflection, φmax , is therefore
given by
sin φmax =
Fig. 5.65
u = | − VCM | =
and
U
mV /(M + m)
m
=
=
.
VCM
MV /(M + m)
M
(5.168)
If M < m, then the dotted circle passes to the left of the left vertex of the triangle.
This means that φ can take on any value. In particular, it is possible for M to bounce
directly backward.
Second solution: We’ll work in the lab frame in this solution. Let V and v be the
final speeds, and let φ and γ be the scattering angles of M and m, respectively, in the
lab frame. Then conservation of px , py , and E give
Fig. 5.66
MV = MV cos φ + mv cos γ ,
(5.169)
0 = MV sin φ − mv sin γ ,
(5.170)
1
1
MV 2 = MV
2
2
2
+
1
mv 2 .
2
(5.171)
Putting the φ terms on the left-hand sides of Eqs. (5.169) and (5.170), and then squaring
and adding these equations, gives
M 2 (V 2 + V
Equating this expression for
through by m gives
m2 v 2
2
− 2VV cos φ) = m2 v 2 .
with the one obtained by multiplying Eq. (5.171)
M (V 2 + V
=⇒
(M + m)V
2
(5.172)
2
− 2VV cos φ) = m(V 2 − V 2 )
− (2MV cos φ)V + (M − m)V 2 = 0.
(5.173)
5.11 Solutions
A solution to this quadratic equation in V exists if and only if the discriminant is
non-negative. Therefore, we must have
(2MV cos φ)2 − 4(M + m)(M − m)V 2 ≥ 0
=⇒
m2 ≥ M 2 (1 − cos2 φ)
=⇒
m2 ≥ M 2 sin2 φ
m
≥ sin φ.
M
=⇒
(5.174)
5.25. Colliding masses
(a) By conservation of momentum, the final speed of the combined masses is
MV/(M + m) ≈ (1 − m/M )V , plus higher-order corrections. The final energies
are therefore
m
1
m 1−
2
M
m
1
= M 1−
2
M
1
mV 2 ,
2
2
1
V 2 ≈ MV 2 − mV 2 .
2
2
Em =
EM
V2 ≈
(5.175)
These energies add up to MV 2/2 − mV 2/2, which is mV 2/2 less than the initial
energy of mass M , namely MV 2/2. Therefore, mV 2/2 is lost to heat.
(b) In this frame, mass m has initial speed V , so its initial energy is Ei = mV 2/2. By
conservation of momentum, the final speed of the combined masses is mV/(M +
m) ≈ (m/M )V , plus higher-order corrections. The final energies are therefore
m
1
m
2
M
1
m
= M
2
M
Em =
EM
m 2
Ei ≈ 0,
M
2
m
V2 =
Ei ≈ 0.
M
2
V2 =
(5.176)
This negligible final energy of zero is mV 2/2 less than Ei . Therefore, mV 2 /2 is
lost to heat, in agreement with part (a).
5.26. Pulling a chain
Let x be the distance your hand has moved. Then x/2 is the length of the moving part
of the chain, because the chain gets “doubled up.” The momentum of this moving part
is therefore p = (σ x/2)˙x. The force that your hand applies is found from F = dp/dt,
which gives F = (σ/2)(˙x2 + x¨x). But since v is constant, the x¨ term vanishes. The
change in momentum here is due simply to additional mass acquiring speed v , and not
due to any increase in speed of the part already moving. Hence,
F=
σ v2
,
2
(5.177)
which is constant. Your hand applies this force over a total distance 2L, so the total
work you do is
F(2L) = σ Lv 2 .
(5.178)
The mass of the chain is σ L, so its final kinetic energy is (σ L)v 2/2. This is only half
of the work you do. Therefore, an energy of σ Lv 2/2 is lost to heat. Each atom in the
chain goes abruptly from rest to speed v , and there is no way to avoid heat loss in such
a process. This is clear when viewed in the reference frame of your hand. In this frame,
the chain initially moves at speed v and then eventually comes to rest, piece by piece.
So all of its initial kinetic energy, (σ L)v 2/2, goes into heat.
211
212
Conservation of energy and momentum
5.27. Pulling a chain again
Let x be the position of the end of the chain. The momentum of the chain is then
p = (σ x)˙x. F = dp/dt gives (using the fact that F is constant) Ft = p, so we have
Ft = (σ x)˙x. Separating variables and integrating yields
x
0
t
σ x dx =
Ft dt
=⇒
0
Ft 2
σ x2
=
2
2
=⇒
x = t F/σ .
(5.179)
The position therefore
grows linearly with time. In other words, the speed is constant,
√
and it equals F/σ .
Remark: Realistically, when you grab the chain, there is some small initial value of
x (call it ). The dx integral above now starts at instead of 0, so x takes the form,
√
x = Ft 2 /σ + 2 . If is very small, the speed very quickly√approaches F/σ . Even
if is not small, the position becomes arbitrarily close to t F/σ as t becomes large.
The “head start” of therefore doesn’t help you in the long run. ♣
5.28. Falling chain
First solution: The left part of the chain is in freefall (because we’re dealing with
a chain described by our first scenario), so at time t it is moving at speed gt and has
fallen a distance gt 2/2. The chain gets doubled up below the support, so a length of
only gt 2/4 hangs at rest. Hence, a length of L−gt 2/4 is falling at speed gt. With upward
taken to be positive, the momentum of the entire chain (which just comes from the
moving part, of course) is therefore
p = σ (L − gt 2/4)(−gt) = −σ Lgt + σ g 2 t 3/4.
(5.180)
If Fs is the force from the support, then the net force on the entire chain is Fs − σ Lg.
So F = dp/dt for the entire chain gives
Fs − σ Lg =
d
dt
−σ Lgt +
σ g2t3
4
=⇒
Fs =
3σ g 2 t 2
.
4
(5.181)
√
This result holds until the top of the chain has fallen a distance 2L (at T = 4L/g ).
2
Prior to time T , Fs equals three times the weight, σ (gt /4)g, of the part of the chain
that hangs at rest. After time T , Fs simply equals the total weight of the chain, σ Lg.
So at time T , Fs abruptly drops from 3σ Lg to σ Lg.
Second solution: Fs is responsible for two things: (1) It holds up the part of the chain
that hangs at rest below the support, and (2) it changes the momentum of the atoms
in the chain that are suddenly brought to rest at the kink in the chain. In other words,
Fs = Fweight + Fdp/dt . From the first solution above, we have Fweight = σ (gt 2 /4)g.
Now let’s find Fdp/dt . At time t, the speed of the chain is gt, so in a small time interval
dt, the top of the chain falls a distance (gt) dt. But due to the “doubling up” effect, only
half of this length is brought to rest in the time dt. Therefore, a small piece of mass
σ (1/2)(gt) dt that was moving at speed gt is suddenly brought to rest. The momentum
was (σ/2)(gt)2 dt downward, and then it becomes zero. Hence, dp = +(σ/2)g 2 t 2 dt,
and so Fdp/dt = dp/dt = (σ/2)g 2 t 2 . Therefore,
Fs = Fweight + Fdp/dt =
σ g2t2
3σ g 2 t 2
σ g2t2
+
=
.
4
2
4
(5.182)
5.29. Falling chain (energy conserving)
Let σ be the mass density of the chain, let L be its total length, and let x be the distance
(defined to be positive) that the top of the chain has fallen. For a given x, a piece
of chain with mass σ x and CM position L − x/2 has effectively been replaced by a
thin “U” below the support, with height x/2; so its CM position is −x/4. The loss in
potential energy is therefore (σ x)g(L − x/2) − (σ x)g(−x/4) = σ xg(L − x/4). Since
5.11 Solutions
we are assuming that energy is conserved in this setup, this loss in potential energy
shows up as the gain in kinetic energy of the moving part of the chain. This part has
length L − x/2 (because x/2 is the length hanging below the support), so conservation
of energy gives
1
σ (L − x/2)v 2 = σ xg(L − x/4)
2
=⇒
2gx(L − x/4)
.
L − x/2
v=
(5.183)
This has the expected property of going to infinity for x → 2L. However, the finite
size of the bend becomes relevant when x approaches 2L, so all of the energy never
ends up being concentrated in an infinitely small piece. All speeds therefore remain
finite, as they must.
Writing v as dx/dt in Eq. (5.183), and separating variables and integrating, gives
1
t= √
g
2L
0
L − x/2
dx
2x(L − x/4)
=⇒
L
g
t=
2
0
1 − z/2
dz, (5.184)
2z(1 − z/4)
where we have changed√variables to z ≡ x/L. Numerically integrating this gives a
L/g. Since the freefall time in the previous problem is given
total time of t ≈ (1.694) √
by gt 2 /2 = 2L =⇒ t = 2 L/g, we see that the time in the present energy-conserving
case is about 0.847 times the time in the freefall case.
To find the tension T at the left end of the bend, let’s paint a little dot on the chain
there. We’ll find the acceleration of the falling part above the dot and then use F = ma.
The acceleration of the falling part is a = d v /dt, where v is given in Eq. (5.183). Taking
the derivative, you can show (don’t forget the dx/dt from the chain rule) that a can be
written in the form,
a=g 1+
(x/2)(L − x/4)
.
(L − x/2)2
(5.185)
The F = ma equation for the part of the chain above the dot, with downward taken to
be positive, gives T + mg = ma =⇒ T = m(a − g). Therefore, with m = σ (L − x/2),
we have
T = σ (L − x/2)g
(x/2)(L − x/4)
(L − x/2)2
=
σ gx(L − x/4)
σ v2
=
,
2(L − x/2)
4
(5.186)
where we have used Eq. (5.183). This has the expected properties of equaling zero
when x = 0 and diverging when x → 2L.
To find the tension at the right end of the bend, note that the total upward tension on
the tiny bend equals the rate of change in momentum of the bend (the gravitational force
is negligible). In a small time dt, a length v dt/2 of chain is brought to rest (the factor
of 2 comes from the doubling-up effect). This length goes from moving downward at
speed v to being at rest, so the change in momentum is σ (v dt/2)v upward. Therefore,
dp/dt = σ v 2/2. This must be the total upward force on the bend, and since we found
above that the upward force at the left end is σ v 2/4, there must also be an upward force
at the right end of σ v 2/4. The tensions at the two ends are therefore equal.
5.30. Falling from a table
(a) First solution: Let σ be the mass density of the chain. From conservation of
energy, we know that the chain’s final kinetic energy, which is (σ L)v 2 /2, equals
the loss in potential energy. This loss equals (σ L)(L/2)g, because the center of
mass falls a distance L/2. Therefore,
v=
gL .
(5.187)
This equals the speed obtained by an object that falls a distance L/2. Note that
if the initial piece hanging down through the hole is arbitrarily short, then the
213
214
Conservation of energy and momentum
chain will take an arbitrarily
√ long time to fall down. But the final speed will still
be (arbitrarily close to) gL.
Second solution: Let x be the length that hangs down through the hole. The
gravitational force on this length, which is (σ x)g, is responsible for changing
the momentum of the entire chain, which is (σ L)˙x. Therefore, F = dp/dt gives
(σ x)g = (σ L)¨x, which is simply the F = ma equation. Hence, x¨ = (g/L)x, and
the general solution to this equation is34
x(t) = Aet
√
g/L
+ Be−t
√
g/L
.
(5.188)
Let T be the time for which x(T ) = L. If is very
√ small, then T will be very
large. But for large t (more precisely, for t
L/g ), we may neglect the
negative-exponent term in Eq. (5.188). We then have
x ≈ Aet
√
g/L
=⇒
x˙ ≈ Aet
√
g/L
g/L ≈ x g/L
(for large t).
(5.189)
When x = L, we obtain
x˙ (T ) = L g/L =
gL ,
(5.190)
in agreement with the first solution.
(b) Let σ be the mass density of the chain, and let x be the length that hangs down
through the hole. The gravitational force on this length, which is (σ x)g, is
responsible for changing the momentum of the chain. This momentum is (σ x)˙x,
because only the hanging part is moving. Therefore, F = dp/dt gives
xg = x¨x + x˙ 2 .
(5.191)
Note that F = ma gives the wrong equation, because it neglects the fact that the
amount of moving mass, σ x, is changing. It therefore misses the second term on
the right-hand side of Eq. (5.191). In short, the momentum of the chain increases
because it is speeding up (which gives the x¨x term) and because additional mass
is continually being added to the moving part (which gives the x˙ 2 term, as you
can show).
Let’s now solve Eq. (5.191) for x(t). Since g is the only parameter in the
equation, the solution for x(t) can involve only g’s and t’s.35 By dimensional
analysis, x(t) must then be of the form x(t) = bgt 2 , where b is a numerical
constant to be determined. Plugging this expression for x(t) into Eq. (5.191) and
dividing by g 2 t 2 gives b = 2b2 + 4b2 . Therefore, b = 1/6, and our solution
may be written as
1 g 2
(5.192)
t .
2 3
This is the equation for something that accelerates downward with acceleration
g = g/3. The time the chain takes to fall a distance L is then given by L =
g t 2 /2, which yields t = 2L/g . The final speed is thus
x(t) =
v=gt=
34
35
2Lg =
2gL
.
3
(5.193)
If is the initial value of x, then A = B = /2 satisfies the initial conditions x(0) = and
√
x˙ (0) = 0, in which case we can write x(t) = cosh t g/L . But we won’t need this information
in what follows.
The other dimensionful quantities in the problem, L and σ , do not appear in Eq. (5.191), so they
cannot appear in the solution. Also, the initial position and speed (which in general appear in the
solution for x(t), because Eq. (5.191) is a second-order differential equation) do not appear in this
case, because they are equal to zero.
5.11 Solutions
√
This is smaller than the gL result in part (a). We therefore see that although
the total time for the scenario in part (a) is very large, the final speed in that case
is in fact larger than that in the present scenario. You can show that the speed in
part (a)’s scenario is smaller than the speed in part (b)’s scenario for x less than
2L/3, but larger for x greater than 2L/3.
Remarks: Using Eq. (5.193), you can show that 1/3 of the available potential
energy is lost to heat. This inevitable loss occurs during the abrupt motions that
suddenly bring the atoms from zero to nonzero speed when they join the moving
part of the chain. The use of conservation of energy is therefore not a valid way
to solve part (b). If you used conservation of energy, you would (as you can
verify) obtain an incorrect acceleration of g/2. In view of the above solution
based on F = dp/dt, this g/2 result cannot be correct, because there is simply
not enough downward force in the setup to yield this acceleration. The only
downward force comes from gravity, and we showed above that this leads to an
acceleration of g/3.
If we want to try to get rid of the energy loss and somehow produce an
acceleration of g/2, a plausible idea is to use a continuous piece of rope instead
of the ideal kind of chain we’ve been using. As mentioned near the end of Section
5.8, a rope yields an energy-conserving system. However, from the reasoning on
page 171, there is now a nonzero tension everywhere throughout the rope, even
in the part inside the heap; this is the price we pay for having none of the points
in the rope abruptly go from zero to nonzero speed. This tension then causes the
rope (all of it) in the heap to move. The system is therefore completely different
from the original one with our ideal chain, where it was understood that all the
little strings connecting the ideal point masses in the heap are initially limp with
zero tension; the tension in a given little string becomes nonzero only when it
joins the moving part of the chain. The solution to this new energy-conserving
setup therefore depends on exactly how the rope in the heap is initially situated;
more information is therefore needed to solve the problem. But one thing is
certain: this new setup is definitely not going to yield an acceleration of g/2,
because the conservation-of-energy solution that leads to g/2 is based on the
assumption that the entire loss in potential energy goes exclusively into the gain
in kinetic energy of the part of the rope that has fallen below the table. The fact
that the rope inside the heap is now also moving ruins this assumption. ♣
5.31. The raindrop
Let ρ be the mass density of the raindrop, and let λ be the average mass density in
space of the water droplets. Let r(t), M (t), and v (t) be the radius, mass, and velocity of
the raindrop, respectively. We need three equations to solve for these three unknowns.
The equations we will use are two different expressions for dM/dt, and the F = dp/dt
˙ is obtained by taking the
expression for the raindrop. The first expression for M
derivative of M = (4/3)π r 3 ρ, which gives
˙ = 4π r 2 r˙ ρ
M
= 3M
r˙
.
r
(5.194)
(5.195)
˙ is obtained by noting that the change in M is due to the
The second expression for M
acquisition of water droplets. The raindrop sweeps out volume at a rate given by its
cross-sectional area times its velocity. Therefore,
˙ = π r 2 v λ.
M
(5.196)
The F = dp/dt equation is found as follows. The gravitational force is Mg, and the
momentum is M v . Therefore, F = dp/dt gives
˙ v + M v˙ .
Mg = M
(5.197)
215
216
Conservation of energy and momentum
We now have three equations involving the three unknowns, r, M, and v .36 Our goal
˙ in Eqs.
is to find v˙ . We will do this by first finding r¨ . Equating the expressions for M
(5.194) and (5.196) gives
4ρ
r˙
λ
4ρ
r¨ .
=⇒ v˙ =
λ
Plugging Eqs. (5.195), (5.198), and (5.199) into Eq. (5.197) gives
v=
Mg = 3M
r˙
r
4ρ
r˙ + M
λ
4ρ
r¨ .
λ
(5.198)
(5.199)
(5.200)
Therefore,
gr
˜ = 12˙r 2 + 4r¨r ,
(5.201)
where we have defined g˜ ≡ gλ/ρ, for convenience. The only parameter in Eq. (5.201)
is g.
˜ Therefore, r(t) can depend only on g˜ and t.37 Hence, by dimensional analysis, r
must take the form
r(t) = Agt
˜ 2,
(5.202)
where A is a numerical constant, to be determined. Plugging this expression for r into
Eq. (5.201) gives
g(A
˜ gt
˜ 2 ) = 12(2Agt)
˜ 2 + 4(Agt
˜ 2 )(2Ag)
˜
=⇒
A = 48A2 + 8A2 .
(5.203)
Therefore, A = 1/56, and so r¨ = 2Ag˜ = g/28
˜
= gλ/28ρ. Eq. (5.199) then gives the
acceleration of the raindrop as
g
v˙ = ,
(5.204)
7
independent of ρ and λ. For further discussion of the raindrop problem, see Krane
(1981).
Remark: A common invalid solution to this problem is the following, which (incorrectly) uses conservation of energy: The fact that v is proportional to r˙ (shown in Eq.
(5.198)) means that the volume swept out by the raindrop is a cone. The center of mass
of a cone is 1/4 of the way from the base to the apex (as you can show by integrating
over horizontal circular slices). Therefore, if M is the mass of the raindrop after it has
fallen a height h, then an (incorrect) application of conservation of energy gives
h
1
M v 2 = Mg
2
4
=⇒
v2 =
gh
.
2
(5.205)
Taking the derivative of this (or just using the general result, v 2 = 2ah) gives
g
(incorrect).
(5.206)
v˙ =
4
36
37
Note that we cannot write down the naive conservation-of-energy equation (which would say that
the decrease in the water’s potential energy equals the increase in its kinetic energy), because
mechanical energy is not conserved. The collisions between the raindrop and the droplets are
completely inelastic. The raindrop will, in fact, heat up. See the remark at the end of the solution.
The other dimensionful quantities in the problem, ρ and λ, do not appear in Eq. (5.201), except
through g,
˜ so they cannot appear in the solution. Also, the initial values of r and r˙ (which in
general appear in the solution for r(t), because Eq. (5.201) is a second-order differential equation)
do not appear in this case, because they are equal to zero.
5.11 Solutions
The reason why this solution is invalid is that the collisions between the raindrop and
the droplets are completely inelastic. Heat is generated, and the overall kinetic energy
of the raindrop is smaller than you would otherwise expect.
Let’s calculate how much mechanical energy is lost (and therefore how much the
raindrop heats up) as a function of the height fallen. The loss in mechanical energy is
Elost = Mg
1
h
− M v2 .
4
2
(5.207)
Using v 2 = 2(g/7)h, this becomes
Eint = Elost =
3
Mgh,
28
(5.208)
where Eint is the gain in internal thermal energy. The energy required to heat 1 g of
water by 1 ◦ C is 1 calorie (= 4.18 joules). Therefore, the energy required to heat 1 kg
of water by 1 ◦ C is ≈ 4200 J. In other words,
Eint = 4200 M
T,
(5.209)
where M is measured in kilograms, and T is measured in degrees Celsius. Equations
(5.208) and (5.209) give the increase in temperature as a function of h,
4200 T =
3
gh.
28
(5.210)
How far must the raindrop fall before it starts to boil? If we assume that the water
droplets’ temperature is near freezing, then the height through which the raindrop
must fall to have T = 100 ◦ C is found from Eq. (5.210) to be
h ≈ 400 000 m = 400 km,
(5.211)
which is much larger than the height of the atmosphere. We have, of course, idealized
the problem in a drastic manner. But needless to say, there is no need to worry about
getting burned by the rain. A typical value for h is a few kilometers, which would raise
the temperature by only about one degree. This effect is completely washed out by
many other factors. ♣
217
Chapter 6
The Lagrangian method
In this chapter, we’re going to learn about a whole new way of looking at things.
Consider the system of a mass on the end of a spring. We can analyze this, of
course, by using F = ma to write down m¨x = −kx. The solutions to this equation
are sinusoidal functions, as we well know. We can, however, figure things out
by using another method which doesn’t explicitly use F = ma. In many (in
fact, probably most) physical situations, this new method is far superior to using
F = ma. You will soon discover this for yourself when you tackle the problems
and exercises for this chapter. We will present our new method by first stating its
rules (without any justification) and showing that they somehow end up magically
giving the correct answer. We will then give the method proper justification.
6.1
The Euler–Lagrange equations
Here is the procedure. Consider the following seemingly silly combination of the
kinetic and potential energies (T and V, respectively),
L ≡ T − V.
(6.1)
This is called the Lagrangian. Yes, there is a minus sign in the definition (a plus
sign would simply give the total energy). In the problem of a mass on the end of
a spring, T = m˙x2 /2 and V = kx2 /2, so we have
L=
1 2 1 2
m˙x − kx .
2
2
(6.2)
Now write
d ∂L
dt ∂ x˙
=
∂L
.
∂x
(6.3)
Don’t worry, we’ll show you in Section 6.2 where this comes from. This equation
is called the Euler–Lagrange (E–L) equation. For the problem at hand, we have
∂L/∂ x˙ = m˙x and ∂L/∂x = −kx (see Appendix B for the definition of a partial
derivative), so Eq. (6.3) gives
m¨x = −kx,
218
(6.4)
6.1 The Euler–Lagrange equations
which is exactly the result obtained by using F = ma. An equation such as
Eq. (6.4), which is derived from the Euler–Lagrange equation, is called an
equation of motion.1 If the problem involves more than one coordinate, as most
problems do, we just have to apply Eq. (6.3) to each coordinate. We will obtain
as many equations as there are coordinates. Each equation may very well involve
many of the coordinates (see the example below, where both equations involve
both x and θ ).
At this point, you may be thinking, “That was a nice little trick, but we just
got lucky in the spring problem. The procedure won’t work in a more general
situation.” Well, let’s see. How about if we consider the more general problem
of a particle moving in an arbitrary potential V (x) (we’ll stick to one dimension
for now). The Lagrangian is then
L=
1 2
m˙x − V (x),
2
(6.5)
and the Euler–Lagrange equation, Eq. (6.3), gives
m¨x = −
dV
.
dx
(6.6)
But −dV /dx is the force on the particle. So we see that Eqs. (6.1) and (6.3)
together say exactly the same thing that F = ma says, when using a Cartesian
coordinate in one dimension (but this result is in fact quite general, as we’ll see
in Section 6.4). Note that shifting the potential by a given constant has no effect
on the equation of motion, because Eq. (6.3) involves only the derivative of V .
This is equivalent to saying that only differences in energy are relevant, and not
the actual values, as we well know.
In a three-dimensional setup written in terms of Cartesian coordinates, the
potential takes the form V (x, y, z), so the Lagrangian is
L=
1
m(˙x2 + y˙ 2 + z˙ 2 ) − V (x, y, z).
2
(6.7)
It then immediately follows that the three Euler–Lagrange equations (obtained
by applying Eq. (6.3) to x, y, and z) may be combined into the vector statement,
m¨x = −∇V .
(6.8)
But −∇V = F, so we again arrive at Newton’s second law, F = ma, now in
three dimensions.
Let’s do one more example to convince you that there’s really something
nontrivial going on here.
1
The term “equation of motion” is a little ambiguous. It is understood to refer to the second-order
differential equation satisfied by x, and not the actual equation for x as a function of t, namely
x(t) = A cos(ωt + φ) in this problem, which is obtained by integrating the equation of motion
twice.
219
220
The Lagrangian method
θ
Example (Spring pendulum): Consider a pendulum made of a spring with a
mass m on the end (see Fig. 6.1). The spring is arranged to lie in a straight line
(which we can arrange by, say, wrapping the spring around a rigid massless rod). The
equilibrium length of the spring is . Let the spring have length + x(t), and let its
angle with the vertical be θ(t). Assuming that the motion takes place in a vertical
plane, find the equations of motion for x and θ.
l+x
m
Solution: The kinetic energy may be broken up into the radial and tangential
parts, so we have
Fig. 6.1
T =
1
m x˙ 2 + ( + x)2 θ˙ 2 .
2
(6.9)
The potential energy comes from both gravity and the spring, so we have
1
V (x, θ) = −mg( + x) cos θ + kx2 .
2
(6.10)
The Lagrangian is therefore
L≡T −V =
1
1
m x˙ 2 + ( + x)2 θ˙ 2 + mg( + x) cos θ − kx2 .
2
2
(6.11)
There are two variables here, x and θ. As mentioned above, the nice thing about the
Lagrangian method is that we can just use Eq. (6.3) twice, once with x and once with
θ. So the two Euler–Lagrange equations are
d ∂L
dt ∂ x˙
=
∂L
∂x
=⇒
m¨x = m( + x)θ˙ 2 + mg cos θ − kx,
(6.12)
and
d ∂L
dt ∂ θ˙
=
∂L
∂θ
=⇒
d
m( + x)2 θ˙ = −mg( + x) sin θ
dt
=⇒
m( + x)2 θ¨ + 2m( + x)˙xθ˙ = −mg( + x) sin θ
=⇒
m( + x)θ¨ + 2m˙xθ˙ = −mg sin θ.
(6.13)
Equation (6.12) is simply the radial F = ma equation, complete with the centripetal
acceleration, −( + x)θ˙ 2 . And the first line of Eq. (6.13) is the statement that the
torque equals the rate of change of the angular momentum (this is one of the subjects of Chapter 8). Alternatively, if you want to work in a rotating reference frame,
then Eq. (6.12) is the radial F = ma equation, complete with the centrifugal force,
m( + x)θ˙ 2 . And the third line of Eq. (6.13) is the tangential F = ma equation,
˙ But never mind about this now. We’ll deal
complete with the Coriolis force, −2m˙xθ.
2
with rotating frames in Chapter 10.
2
Throughout this chapter, I’ll occasionally point out torques, angular momenta, centrifugal forces,
and other such things when they pop up in equations of motion, even though we haven’t covered
6.2 The principle of stationary action
Remark: After writing down the E–L equations, it is always best to double-check them by
trying to identify them as F = ma and/or τ = dL/dt equations (once we learn about that).
Sometimes, however, this identification isn’t obvious. And for the times when everything
is clear (that is, when you look at the E–L equations and say, “Oh, of course!”), it is usually
clear only after you’ve derived the equations. In general, the safest method for solving
a problem is to use the Lagrangian method and then double-check things with F = ma
and/or τ = dL/dt if you can. ♣
At this point it seems to be personal preference, and all academic, whether you
use the Lagrangian method or the F = ma method. The two methods produce
the same equations. However, in problems involving more than one variable, it
usually turns out to be much easier to write down T and V , as opposed to writing
down all the forces. This is because T and V are nice and simple scalars. The
forces, on the other hand, are vectors, and it is easy to get confused if they point
in various directions. The Lagrangian method has the advantage that once you’ve
written down L ≡ T − V , you don’t have to think anymore. All you have to do
is blindly take some derivatives.3
When jumping from high in a tree,
Just write down del L by del z.
Take del L by z dot,
Then t-dot what you’ve got,
And equate the results (but quickly!)
But ease of computation aside, is there any fundamental difference between the
two methods? Is there any deep reasoning behind Eq. (6.3)? Indeed, there is …
6.2
The principle of stationary action
Consider the quantity,
S≡
t2
L(x, x˙ , t)dt.
(6.14)
t1
S is called the action. It is a quantity with the dimensions of (Energy)×(Time). S
depends on L, and L in turn depends on the function x(t) via Eq. (6.1).4 Given any
3
4
them yet. I figure it can’t hurt to bring your attention to them. But rest assured, a familiarity with
these topics is by no means necessary for an understanding of what we’ll be doing in this chapter,
so just ignore the references if you want. One of the great things about the Lagrangian method is
that even if you’ve never heard of the terms “torque,” “centrifugal,” “Coriolis,” or even “F = ma”
itself, you can still get the correct equations by simply writing down the kinetic and potential
energies, and then taking a few derivatives.
Well, you eventually have to solve the resulting equations of motion, but you have to do that with
the F = ma method, too.
In some situations, the kinetic and potential energies in L ≡ T − V may explicitly depend on time,
so we have included the “t” in Eq. (6.14).
221
222
The Lagrangian method
y
-g/2
1
t
function x(t), we can produce the quantity S. We’ll deal with only one coordinate,
x, for now.
Integrals like the one in Eq. (6.14) are called functionals, and S is sometimes
denoted by S[x(t)]. It depends on the entire function x(t), and not on just one
input number, as a regular function f (t) does. S can be thought of as a function
of an infinite number of values, namely all the x(t) for t ranging from t1 to t2 . If
you don’t like infinities, you can imagine breaking up the time interval into, say,
a million pieces, and then replacing the integral by a discrete sum.
Let’s now pose the following question: Consider a function x(t), for t1 ≤
t ≤ t2 , which has its endpoints fixed (that is, x(t1 ) = x1 and x(t2 ) = x2 ,
where x1 and x2 are given), but is otherwise arbitrary. What function x(t) yields
a stationary value of S? A stationary value is a local minimum, maximum, or
saddle point.5
For example, consider a ball dropped from rest, and consider the function
y(t) for 0 ≤ t ≤ 1. Assume that we somehow know that y(0) = 0 and
y(1) = −g/2. 6 A number of possibilities for y(t) are shown in Fig. 6.2, and
each of these can (in theory) be plugged into Eqs. (6.1) and (6.14) to generate S.
Which one yields a stationary value of S? The following theorem gives us the
answer.
Fig. 6.2
Theorem 6.1 If the function x0 (t) yields a stationary value (that is, a local
minimum, maximum, or saddle point) of S, then
d ∂L
dt ∂ x˙ 0
=
∂L
.
∂x0
(6.15)
It is understood that we are considering the class of functions whose endpoints
are fixed. That is, x(t1 ) = x1 and x(t2 ) = x2 .
Proof: We will use the fact that if a certain function x0 (t) yields a stationary
value of S, then any other function very close to x0 (t) (with the same endpoint
values) yields essentially the same S, up to first order in any deviations. This is
actually the definition of a stationary value. The analogy with regular functions
is that if f (b) is a stationary value of f , then f (b + ) differs from f (b) only at
second order in the small quantity . This is true because f (b) = 0, so there is
no first-order term in the Taylor series expansion around b.
Assume that the function x0 (t) yields a stationary value of S, and consider the
function
xa (t) ≡ x0 (t) + aβ(t),
5
6
(6.16)
A saddle point is a point where there are no first-order changes in S, and where some of the
second-order changes are positive and some are negative (like the middle of a saddle, of course).
This follows from y = −gt 2 /2, but pretend that we don’t know this formula.
6.2 The principle of stationary action
where a is a number, and where β(t) satisfies β(t1 ) = β(t2 ) = 0 (to keep the
endpoints of the function fixed), but is otherwise arbitrary. When producing the
action S[xa (t)] in (6.14), the t is integrated out, so S is just a number. It depends
on a, in addition to t1 and t2 . Our requirement is that there be no change in S at
first order in a. How does S depend on a? Using the chain rule, we have
∂
∂
S[xa (t)] =
∂a
∂a
t2
t2
L dt =
t1
t1
∂L
dt
∂a
t2
=
t1
∂L ∂xa
∂L ∂ x˙ a
+
∂xa ∂a
∂ x˙ a ∂a
dt.
(6.17)
In other words, a influences S through its effect on x, and also through its effect
on x˙ . From Eq. (6.16), we have
∂xa
= β,
∂a
and
∂ x˙ a
˙
= β,
∂a
(6.18)
so Eq. (6.17) becomes7
∂
S[xa (t)] =
∂a
t2
t1
∂L
∂L
β+
β˙ dt.
∂xa
∂ x˙ a
(6.19)
Now comes the one sneaky part of the proof. We will integrate the second term
by parts (you will see this trick many times in your physics career). Using
∂L
∂L
β−
β˙ dt =
∂ x˙ a
∂ x˙ a
d ∂L
dt ∂ x˙ a
β dt,
(6.20)
Eq. (6.19) becomes
∂
S[xa (t)] =
∂a
t2
t1
∂L
d ∂L
−
∂xa
dt ∂ x˙ a
β dt +
∂L
β
∂ x˙ a
t2
.
(6.21)
t1
But β(t1 ) = β(t2 ) = 0, so the last term (the “boundary term”) vanishes. We now
use the fact that (∂/∂a)S[xa (t)] must be zero for any function β(t), because we
are assuming that x0 (t) yields a stationary value. The only way this can be true
is if the quantity in parentheses above (evaluated at a = 0) is identically equal to
zero, that is,
d ∂L
dt ∂ x˙ 0
7
=
∂L
.
∂x0
(6.22)
Note that nowhere do we assume that xa and x˙ a are independent variables. The partial derivatives
in Eq. (6.18) are very much related, in that one is the derivative of the other. The use of the chain
rule in Eq. (6.17) is still perfectly valid.
223
224
The Lagrangian method
The E–L equation, Eq. (6.3), therefore doesn’t just come out of the blue. It
is a consequence of requiring that the action be at a stationary value. We may
therefore replace F = ma by the following principle.
• The principle of stationary action:
The path of a particle is the one that yields a stationary value of the action.
This principle (also known as Hamilton’s principle) is equivalent to F = ma
because the above theorem shows that if (and only if, as you can show by working
backwards) we have a stationary value of S, then the E–L equations hold. And the
E–L equations are equivalent to F = ma (as we showed for Cartesian coordinates
in Section 6.1, and as we’ll prove for any coordinate system in Section 6.4).
Therefore, “stationary action” is equivalent to F = ma.
If we have a multidimensional setup where the Lagrangian is a function of the
variables x1 (t), x2 (t), . . . , then the above principle of stationary action is still all
we need. With more than one variable, we can now vary the path by varying each
coordinate (or combinations thereof ). The variation of each coordinate produces
an E–L equation which, as we saw in the Cartesian case, is equivalent to an
F = ma equation.
Given a classical mechanics problem, we can solve it with F = ma, or we
can solve it with the E–L equations, which are a consequence of the principle of
stationary action (often called the principle of “least action” or “minimal action,”
but see the fourth remark below). Either method will get the job done. But as
mentioned at the end of Section 6.1, it is often easier to use the latter, because it
avoids the use of force which can get confusing if you have forces pointing in all
sorts of complicated directions.
It just stood there and did nothing, of course,
A harmless and still wooden horse.
But the minimal action
Was just a distraction;
The plan involved no use of force.
Let’s now return to the example of a ball dropped from rest, mentioned above.
The Lagrangian is L = T − V = m˙y2 /2 − mgy, so Eq. (6.22) gives y¨ = −g,
which is simply the F = ma equation (divided through by m), as expected. The
solution is y(t) = −gt 2 /2 + v0 t + y0 , as we well know. But the initial conditions
tell us that v0 = y0 = 0, so our solution is y(t) = −gt 2 /2. You are encouraged
to verify explicitly that this y(t) yields an action that is stationary with respect
to variations of the form, say, y(t) = −gt 2 /2 + t(t − 1), which also satisfies
the endpoint conditions (this is the task of Exercise 6.30). There is, of course,
an infinite number of other ways to vary y(t), but this specific result should help
convince you of the validity of Theorem 6.1.
Note that the stationarity implied by the Euler–Lagrange equation, Eq. (6.22),
is a local statement. It gives information only about nearby paths. It says nothing
6.2 The principle of stationary action
about the global nature of how the action depends on all possible paths. If we find
that a solution to Eq. (6.22) happens to produce a local minimum (as opposed to a
maximum or a saddle), there is no reason to conclude that it is a global minimum,
although in many cases it turns out to be (see Exercise 6.32, for the case of a
thrown ball).
Remarks:
1. Theorem 6.1 is based on the assumption that the ending time, t2 , of the motion is given.
But how do we know this final time? Well, we don’t. In the example of a ball thrown
upward, the total time to rise and fall back to your hand can be anything, depending on the
ball’s initial speed. This initial speed will show up as an integration constant when solving
the E–L equations. The motion must end sometime, and the principle of stationary action
says that for whatever time this happens to be, the physical path has a stationary action.
2. Theorem 6.1 shows that we can explain the E–L equations by the principle of stationary
action. This, however, simply shifts the burden of proof. We are now left with the task
of justifying why we should want the action to have a stationary value. The good news
is that there is a very solid reason for this. The bad news is that the reason involves
quantum mechanics, so we won’t be able to discuss it properly here. Suffice it to say that
a particle actually takes all possible paths in going from one place to another, and each
path is associated with the complex number eiS/ (where = 1.05 · 10−34 J s is Planck’s
constant). These complex numbers have absolute value 1 and are called “phases.” It turns
out that the phases from all possible paths must be added up to give the “amplitude” of
going from one point to another. The absolute value of the amplitude must then be squared
to obtain the probability.8
The basic point, then, is that at a nonstationary value of S, the phases from different
paths differ (greatly, because is so small compared with the typical size of the action for
a macroscopic particle) from one another, which effectively leads to the addition of many
random vectors in the complex plane. These end up canceling each other, yielding a sum
of essentially zero. There is therefore no contribution to the overall amplitude from nonstationary values of S. Hence, we do not observe the paths associated with these S’s. At a
stationary value of S, however, all the phases take on essentially the same value, thereby
adding constructively instead of destructively. There is therefore a nonzero probability
for the particle to take a path that yields a stationary value of S. So this is the path we
observe.
3. But again, the preceding remark simply shifts the burden of proof one step further. We
must now justify why these phases eiS/ should exist, and why the Lagrangian that appears
in S should equal T − V . But here’s where we’re going to stop.
4. The principle of stationary action is sometimes referred to as the principle of “least”
action, but this is misleading. True, it is often the case that the stationary value turns out to
be a minimum value, but it need not be, as we can see in the following example. Consider
a harmonic oscillator which has a Lagrangian equal to
L=
1 2 1 2
m˙x − kx .
2
2
(6.23)
Let x0 (t) be a function that yields a stationary value of the action. Then we know that
x0 (t) satisfies the E–L equation, m¨x0 = −kx0 . Consider a slight variation on this path,
8
This is one of those remarks that is completely useless, because it is incomprehensible to those
who haven’t seen the topic before, and trivial to those who have. My apologies. But this and the
following remarks are definitely not necessary for an understanding of the material in this chapter.
If you’re interested in reading more about these quantum mechanical issues, you should take a
look at Richard Feynman’s book (Feynman, 2006). Feynman was, after all, the one who thought
of this idea.
225
226
The Lagrangian method
x0 (t) + ξ(t), where ξ(t) satisfies ξ(t1 ) = ξ(t2 ) = 0. With this new function, the action
becomes
Sξ =
t2
t1
m 2
k 2
x˙ + 2x˙0 ξ˙ + ξ˙ 2 −
x + 2x0 ξ + ξ 2
2 0
2 0
dt.
(6.24)
The two cross-terms add up to zero, because after integrating the x˙0 ξ˙ term by parts, their
sum is
m˙x0 ξ
t2
t1
t2
−
(m¨x0 + kx0 )ξ dt.
(6.25)
t1
The first term is zero, due to the boundary conditions on ξ(t). The second term is zero,
due to the E–L equation. We’ve basically just reproduced the proof of Theorem 6.1 for
the special case of the harmonic oscillator here.
The terms in Eq. (6.24) involving only x0 give the stationary value of the action
(call it S0 ). To determine whether S0 is a minimum, maximum, or saddle point, we
must look at the difference,
S ≡ Sξ − S0 =
1
2
t2
(mξ˙ 2 − kξ 2 ) dt.
(6.26)
t1
It is always possible to find a function ξ that makes S positive. Simply choose ξ to be
small, but make it wiggle very fast, so that ξ˙ is large. Therefore, it is never the case that
S0 is a maximum. Note that this reasoning works for any potential, not just a harmonic
oscillator, as long as it is a function of position only (that is, it contains no derivatives, as
we always assume).
You might be tempted to use the same line of reasoning to say that it is also always
possible to find a function ξ that makes S negative, by making ξ large and ξ˙ small.
If this were true, then we could put everything together and conclude that all stationary
points are saddle points, for a harmonic oscillator. However, it is not always possible to
make ξ large enough and ξ˙ small enough so that S is negative, due to the boundary
conditions ξ(t1 ) = ξ(t2 ) = 0. If ξ changes from zero to a large value and then back to
zero, then ξ˙ may also have to be large, if the time interval is short enough. Problem 6.6
deals quantitatively with this issue. For now, let’s just recognize that in some cases S0 is
a minimum, in some cases it is a saddle point, and it is never a maximum. “Least action”
is therefore a misnomer.
5. It is sometimes said that nature has a “purpose,” in that it seeks to take the path that
produces the minimum action. In view of the second remark above, this is incorrect.
In fact, nature does exactly the opposite. It takes every path, treating them all on equal
footing. We end up seeing only the path with a stationary action, due to the way the
quantum mechanical phases add. It would be a harsh requirement, indeed, to demand
that nature make a “global” decision (that is, compare paths that are separated by large
distances) and choose the one with the smallest action. Instead, we see that everything
takes place on a “local” scale. Nearby phases simply add, and everything works out
automatically.
When an archer shoots an arrow through the air, the aim is made possible by all the other
arrows taking all the other nearby paths, each with essentially the same action. Likewise,
when you walk down the street with a certain destination in mind, you’re not alone . . .
When walking, I know that my aim
Is caused by the ghosts with my name.
And although I can’t see
Where they walk next to me,
I know they’re all there, just the same.
6.3 Forces of constraint
227
6. Consider a function, f (x), of one variable (for ease of terminology). Let f (b) be a local
minimum of f . There are two basic properties of this minimum. The first is that f (b)
is smaller than all nearby values. The second is that the slope of f is zero at b. From
the above remarks, we see that (as far as the action S is concerned) the first property is
completely irrelevant, and the second one is the whole point. In other words, saddle points
(and maxima, although we showed above that these never exist for S) are just as good as
minima, as far as the constructive addition of the eiS/ phases is concerned.
7. Given that classical mechanics is an approximate theory, while quantum mechanics is
the (more) correct one, it is quite silly to justify the principle of stationary action by
demonstrating its equivalence to F = ma, as we did above. We should be doing it the
other way around. However, because our intuition is based on F = ma, it’s easier to start
with F = ma as the given fact, rather than calling upon the latent quantum-mechanical
intuition hidden deep within all of us. Maybe someday …
At any rate, in more advanced theories dealing with fundamental issues concerning the
tiny building blocks of matter (where actions are of the same order of magnitude as ),
the approximate F = ma theory is invalid, and you have to use the Lagrangian method.
8. When dealing with a system in which a nonconservative force such as friction is present,
the Lagrangian method loses much of its appeal. The reason for this is that nonconservative
forces don’t have a potential energy associated with them, so there isn’t a specific V (x) that
you can write down in the Lagrangian. Although friction forces can in fact be incorporated
in the Lagrangian method, you have to include them in the E–L equations essentially by
hand. We won’t deal with nonconservative forces in this chapter. ♣
6.3
Forces of constraint
A nice thing about the Lagrangian method is that we are free to impose any given
constraints at the beginning of the problem, thereby immediately reducing the
number of variables. This is always done (perhaps without thinking) whenever
a particle is constrained to move on a wire or surface, etc. Often we are concerned not with the exact nature of the forces doing the constraining, but only
with the resulting motion, given that the constraints hold. By imposing the constraints at the outset, we can find the motion, but we can’t say anything about the
constraining forces.
If we want to determine the constraining forces, we must take a different
approach. The main idea of the strategy, as we will show below, is that we must
not impose the constraints too soon. This leaves us with a larger number of
variables to deal with, so the calculations are more cumbersome. But the benefit
is that we are able to find the constraining forces.
Consider the setup of a particle sliding off a fixed frictionless hemisphere of
radius R (see Fig. 6.3). Let’s say that we are concerned only with finding the
equation of motion for θ , and not the constraining force. Then we can write
everything in terms of θ, because we know that the radial distance r is constrained to be R. The kinetic energy is mR2 θ˙ 2 /2, and the potential energy (relative
to the bottom of the hemisphere) is mgR cos θ , so the Lagrangian is
L=
1 2 2
mR θ˙ − mgR cos θ ,
2
(6.27)
R
Fig. 6.3
θ
228
The Lagrangian method
and the equation of motion, via Eq. (6.3), is
θ¨ = ( g/R) sin θ ,
which is equivalent to the tangential F = ma statement.
Now let’s say that we want to find the constraining normal force that the
hemisphere applies to the particle. To do this, let’s solve the problem in a different
way and write things in terms of both r and θ. Also (and here’s the critical step),
let’s be really picky and say that r isn’t exactly constrained to be R, because in
the real world the particle actually sinks into the hemisphere a little bit. This may
seem a bit silly, but it’s really the whole point. The particle pushes and sinks
inward a tiny distance until the hemisphere gets squashed enough to push back
with the appropriate force to keep the particle from sinking in any more (just
consider the hemisphere to be made of lots of little springs with very large spring
constants). The particle is therefore subject to a (very steep) potential arising from
the hemisphere’s force. The constraining potential, V (r), looks something like
the plot in Fig. 6.4. The true Lagrangian for the system is thus
V(r)
r
R
Fig. 6.4
(6.28)
L=
1
m(˙r 2 + r 2 θ˙ 2 ) − mgr cos θ − V (r).
2
(6.29)
(The r˙ 2 term in the kinetic energy will turn out to be insignificant.) The equations of motion obtained from varying θ and r are therefore
mr 2 θ¨ + 2mr˙r θ˙ = mgr sin θ ,
m¨r = mr θ˙ 2 − mg cos θ − V (r).
(6.30)
Having written down the equations of motion, we will now apply the constraint
condition that r = R. This condition implies r˙ = r¨ = 0. (Of course, r isn’t really
equal to R, but any differences are inconsequential from this point onward.) The
first of Eqs. (6.30) then reproduces Eq. (6.28), while the second yields
−
dV
dr
= mg cos θ − mRθ˙ 2 .
(6.31)
r=R
But FN ≡ −dV /dr is the constraint force applied in the r direction, which is
precisely the force we are looking for. The normal force of constraint is therefore
˙ = mg cos θ − mRθ˙ 2 .
FN (θ , θ)
(6.32)
This is equivalent to the radial F = ma equation, mg cos θ − FN = mRθ˙ 2 (which
is certainly a quicker way to find the normal force in the present problem). Note
˙ > 0. If the normal force becomes zero,
that this result is valid only if FN (θ, θ)
then this means that the particle has left the sphere, in which case r no longer
equals R.
6.4 Change of coordinates
Remarks:
1. What if we instead had (unwisely) chosen Cartesian coordinates, x and y, instead of polar
coordinates, r and θ? Since the distance from the particle to the surface of the hemisphere
is η ≡ x2 + y2 − R, we obtain a true Lagrangian equal to
L=
1
m(˙x2 + y˙ 2 ) − mgy − V (η).
2
(6.33)
The equations of motion are (using the chain rule)
m¨x = −
dV ∂η
,
dη ∂x
and
m¨y = −mg −
dV ∂η
.
dη ∂y
(6.34)
We now apply the constraint condition η = 0. Since −dV /dη equals the constraint force
F, you can show that the equations we end up with (namely, the two E–L equations and
the constraint equation) are
m¨x = F
x
,
R
m¨y = −mg + F
y
,
R
and
x2 + y2 − R = 0.
(6.35)
These three equations are sufficient to determine the three unknowns x¨ , y¨ , and F as
functions of the quantities x, x˙ , y, and y˙ . See Exercise 6.37, which should convince
you that polar coordinates are the way to go. In general, the strategy for solving for F
is to take two time derivatives of the constraint equation and then eliminate the second
derivatives of the coordinates by using the E–L equations (this process was trivial in the
polar-coordinate case).
2. You can see from Eq. (6.35) that the E–L equations end up taking the form,
d
dt
∂L
∂ q˙ i
=
∂L
∂η
+F
,
∂qi
∂qi
(6.36)
for each coordinate qi . The quantity η is what appears in the constraint equation, η = 0.
In our hemisphere problem, we had η = r − R in polar coordinates, and η = x2 + y2 −
R in Cartesian coordinates. The E–L equations, combined with the η = 0 condition,
give us exactly the number of equations (N + 1 of them, where N is the number of
coordinates) needed to determine all of the N + 1 unknowns (all the q¨ i , and F) in terms of
the qi and q˙ i .
Writing down the equations in Eq. (6.36) is basically the method of Lagrange multipliers, where the Lagrange multiplier turns out to be the force. But if you’re not familiar with
this method, no need to worry; you can derive everything from scratch using the above
technique involving the steep potential. If you do happen to be familiar with it, then there
might in fact be a need to worry about how you apply it, as the following remark explains.
3. When trying to determine the forces of constraint, you can just start with Eq. (6.36),
without bothering to write down V (η). But you must be careful to make sure that η does
indeed represent the distance the particle is from where it should be. In polar coordinates,
if someone gives you the constraint condition for the hemisphere as 7(r − R) = 0, and
if you use the left-hand side of this as the η in Eq. (6.36), then you will get the wrong
constraint force; it will be too
√ small by a factor of 7. Likewise, in Cartesian coordinates,
writing the constraint as y − R2 − x2 = 0 will give you the wrong force. The best way to
avoid this problem is, of course, to pick one of your variables as the distance the particle
is from where it should be (up to an additive constant, as in the case of r − R = 0). ♣
6.4
Change of coordinates
When L is written in terms of Cartesian coordinates x, y, z, we showed in Section
6.1 that the Euler–Lagrange equations are equivalent to Newton’s F = ma
229
230
The Lagrangian method
equations; see Eq. (6.8). But what about the case where we use polar, spherical,
or some other coordinates? The equivalence of the E–L equations and F = ma
isn’t so obvious. As far as trusting the E–L equations for such coordinates goes,
you can achieve peace of mind in two ways. You can accept the principle of
stationary action as something so beautiful and profound that it simply has to
work for any choice of coordinates. Or, you can take the more mundane road
and show through a change of coordinates that if the E–L equations hold for one
set of coordinates (and we know that they do hold for at least one set, namely
Cartesian coordinates), then they also hold for any other coordinates (of a certain
form, described below). In this section, we will demonstrate the validity of the
E–L equations through the explicit change of coordinates.9
Consider the set of coordinates,
xi :
(x1 , x2 , . . . , xN ).
(6.37)
For example, if N = 6, then x1 , x2 , x3 could be the Cartesian x, y, z coordinates
of one particle, and x4 , x5 , x6 could be the r, θ , φ polar coordinates of a second
particle, and so on. Assume that the E–L equations hold for these variables,
that is,
d ∂L
dt ∂ x˙ i
=
∂L
∂xi
(1 ≤ i ≤ N ).
(6.38)
Consider a new set of variables that are functions of the xi and t,
qi = qi (x1 , x2 , . . . , xN ; t).
(6.39)
We will restrict ourselves to the case where the qi do not depend on the x˙ i . (This is
quite reasonable. If the coordinates depended on the velocities, then we wouldn’t
be able to label points in space with definite coordinates. We’d have to worry
about how the particles were behaving when they were at the points. These would
be strange coordinates indeed.) We can, in theory, invert Eq. (6.39) and express
the xi as functions of the qi and t,
xi = xi (q1 , q2 , . . . , qN ; t).
(6.40)
Claim 6.2 If Eq. (6.38) is true for the xi coordinates, and if the xi and qi are
related by Eq. (6.40), then Eq. (6.38) is also true for the qi coordinates. That is,
d ∂L
dt ∂ q˙ m
9
=
∂L
∂qm
(1 ≤ m ≤ N ).
(6.41)
This calculation is straightforward but a bit messy, so you may want to skip this section and just
settle for the “beautiful and profound” reasoning.
6.4 Change of coordinates
Proof: We have
∂L
=
∂ q˙ m
N
i=1
∂L ∂ x˙ i
.
∂ x˙ i ∂ q˙ m
(6.42)
(Note that if the xi depended on the q˙ i , then we would have to include the
additional term, (∂L/∂xi )(∂xi /∂ q˙ m ). But we have excluded such dependence.)
Let’s rewrite the ∂ x˙ i /∂ q˙ m term. From Eq. (6.40), we have
N
x˙ i =
m=1
∂xi
∂xi
q˙ m +
.
∂qm
∂t
(6.43)
Therefore,
∂ x˙ i
∂xi
=
.
∂ q˙ m
∂qm
(6.44)
Substituting this into Eq. (6.42) and taking the time derivative of both sides gives
d ∂L
dt ∂ q˙ m
N
=
i=1
∂xi
+
∂qm
d ∂L
dt ∂ x˙ i
N
i=1
∂L d ∂xi
∂ x˙ i dt ∂qm
.
(6.45)
In the second term here, it is legal to switch the order of the total derivative, d/dt,
and the partial derivative, ∂/∂qm .
Remark:
In case you have your doubts, let’s prove that this switching is legal.
d
dt
∂xi
∂qm
N
=
k=1
=
∂
∂qk
∂
∂qm
∂xi
∂qm
N
k=1
∂ x˙ i
=
.
∂qm
q˙ k +
∂
∂t
∂xi
∂qm
∂xi
∂xi
q˙ k +
∂qk
∂t
♣
(6.46)
In the first term on the right-hand side of Eq. (6.45), we can use the given
information in Eq. (6.38) and rewrite the (d/dt)(∂L/∂ x˙ i ) term. We then obtain
d ∂L
dt ∂ q˙ m
N
=
i=1
∂L ∂xi
+
∂xi ∂qm
∂L
=
,
∂qm
N
i=1
∂L ∂ x˙ i
∂ x˙ i ∂qm
(6.47)
as we wanted to show.
We have therefore demonstrated that if the Euler–Lagrange equations are true
for one set of coordinates, xi (and they are true for Cartesian coordinates), then
231
232
The Lagrangian method
they are also true for any other set of coordinates, qi , satisfying Eq. (6.39). If
you’re inclined to look at the principle of stationary action with distrust, thinking
that it might be a coordinate-dependent statement, this proof should put you at
ease. The Euler–Lagrange equations are valid in any coordinates.
Note that the above proof did not in any way use the precise form of the
Lagrangian. If L were equal to T + V, or 8T + π V 2 /T , or any other arbitrary
function, our result would still be true: If Eq. (6.38) is true for one set of coordinates, then it is also true for any other set of coordinates qi satisfying Eq. (6.39).
The point is that the only L for which the hypothesis is true at all (that is, for
which Eq. (6.38) holds) is L ≡ T − V (or any constant multiple of this).
Remark: On one hand, it is quite amazing how little we assumed in proving the above claim.
Any new coordinates of the very general form in Eq. (6.39) satisfy the E–L equations, as long
as the original coordinates do. If the E–L equations had, say, a factor of 5 on the right-hand
side of Eq. (6.38), then they would not hold in arbitrary coordinates. To see this, just follow
the proof through with the factor of 5.
On the other hand, the claim is quite believable, if you make an analogy with a function
instead of a functional. Consider the function f (z) = z 2 . This has a minimum at z = 0,
consistent with the fact that df /dz = 0 at z = 0. But let’s now write f in terms of the variable
y defined by, say, z = y4 . Then f (y) = y8 , and f has a minimum at y = 0, consistent with the
fact that df /dy = 0 at y = 0. So f = 0 holds in both coordinates at the corresponding points
y = z = 0. This is the (simplified) analog of the E–L equations holding in both coordinates. In
both cases, the derivative equation describes where the stationary value occurs.
This change-of-variables result may be stated in a more geometrical (and friendly) way. If
you plot a function and then stretch the horizontal axis in an arbitrary manner (which is what
happens when you change coordinates), then a stationary value (that is, one where the slope is
zero) will still be a stationary value after the stretching.10 A picture (or even just the thought of
one) is worth a dozen equations, apparently.
As an example of an equation that does not hold for all coordinates, consider the preceding
example, but with f = 1 instead of f = 0. In terms of z, f = 1 when z = 1/2. And in terms
of y, f = 1 when y = (1/8)1/7 . But the points z = 1/2 and y = (1/8)1/7 are not the same
point. In other words, f = 1 is not a coordinate-independent statement. Most equations are
coordinate dependent. The special thing about f = 0 is that a stationary point is a stationary
point no matter how you look at it. ♣
6.5
6.5.1
Conservation laws
Cyclic coordinates
Consider the case where the Lagrangian does not depend on a certain coordinate
qk . Then
d ∂L
dt ∂ q˙ k
10
=
∂L
=0
∂qk
=⇒
∂L
= C,
∂ q˙ k
(6.48)
There is, however, one exception. A stationary point in one coordinate system might be located at
a kink in another coordinate system, so that f is not defined there. For example, if we had instead
defined y by z = y1/4 , then f (y) = y1/2 , which has an undefined slope at y = 0. Basically, we’ve
stretched (or shrunk) the horizontal axis by a factor of infinity at the origin, and this is a process
that can change a zero slope into an undefined one. But let’s not worry about this.
6.5 Conservation laws
where C is a constant, that is, independent of time. In this case, we say that qk is a
cyclic coordinate, and that ∂L/∂ q˙ k is a conserved quantity (meaning that it doesn’t
change with time). If Cartesian coordinates are used, then ∂L/∂ x˙ k is simply the
momentum, m˙xk , because x˙ k appears only in the the kinetic energy’s m˙xk2 /2 term
(we exclude cases where the potential depends on x˙ k ). We therefore call ∂L/∂ q˙ k
the generalized momentum corresponding to the coordinate qk . And in cases
where ∂L/∂ q˙ k does not change with time, we call it a conserved momentum.
Note that a generalized momentum need not have the units of linear momentum,
as the angular-momentum examples below show.
Example (Linear momentum): Consider a ball thrown through the air. In the
full three dimensions, the Lagrangian is
L=
1
m(˙x2 + y˙ 2 + z˙ 2 ) − mgz.
2
(6.49)
There is no x or y dependence here, so both ∂L/∂ x˙ = m˙x and ∂L/∂ y˙ = m˙y are
constant, as we well know. The fancy way of saying this is that conservation of
px ≡ m˙x arises from spatial translation invariance in the x direction. The fact that the
Lagrangian doesn’t depend on x means that it doesn’t matter if you throw the ball in
one spot, or in another spot a mile down the road. The setup is independent of the x
value. This independence leads to conservation of px . Likewise for py .
Example (Angular and linear momentum in cylindrical coordinates):
Consider a potential that depends only on the distance to the z axis. In cylindrical
coordinates, the Lagrangian is
L=
1
m(˙r 2 + r 2 θ˙ 2 + z˙ 2 ) − V (r).
2
(6.50)
There is no z dependence here, so ∂L/∂ z˙ = m˙z is constant. Also, there is no θ
dependence, so ∂L/∂ θ˙ = mr 2 θ˙ is constant. Since r θ˙ is the velocity in the tangential
˙ is the angular
direction around the z axis, we see that our conserved quantity, mr(r θ),
momentum (discussed in Chapters 7–9) around the z axis. In the same manner as in
the preceding example, conservation of angular momentum around the z axis arises
from rotation invariance around the z axis.
Example (Angular momentum in spherical coordinates): In spherical coordinates, consider a potential that depends only on r and θ. Our convention for spherical
coordinates is that θ is the angle down from the north pole, and φ is the angle around
the equator. The Lagrangian is
L=
1
m(˙r 2 + r 2 θ˙ 2 + r 2 sin2 θ φ˙ 2 ) − V (r, θ).
2
(6.51)
233
234
The Lagrangian method
There is no φ dependence here, so ∂L/∂ φ˙ = mr 2 sin2 θ φ˙ is constant. Since r sin θ is
the distance from the z axis, and since r sin θ φ˙ is the speed in the tangential direction
˙ is the
around the z axis, we see that our conserved quantity, m(r sin θ)(r sin θ φ),
angular momentum around the z axis.
6.5.2
Energy conservation
We will now derive another conservation law, namely conservation of energy.
The conservation of momentum or angular momentum above arose when the
Lagrangian was independent of x, y, z, θ , or φ. Conservation of energy arises
when the Lagrangian is independent of time. This conservation law is different
from those in the above momenta examples, because t is not a coordinate that
the stationary-action principle can be applied to. You can imagine varying the
coordinates x, θ , etc., which are functions of t. But it makes no sense to vary t.
Therefore, we’re going to have to prove this conservation law in a different way.
Consider the quantity
N
E≡
i=1
∂L
q˙ i − L.
∂ q˙ i
(6.52)
E turns out (usually) to be the energy. We’ll show this below. The motivation
for this expression for E comes from the theory of Legendre transforms, but we
won’t get into that here. We’ll just accept the definition in Eq. (6.52) and prove
a very useful fact about it.
Claim 6.3 If L has no explicit time dependence (that is, if ∂L/∂t = 0), then
E is conserved (that is, dE/dt = 0), assuming that the motion obeys the E–L
equations (which it does).
Note that there is one partial derivative and one total derivative in this statement.
Proof: L is a function of the qi , the q˙ i , and possibly t. Making copious use of
the chain rule, we have
d
dE
=
dt
dt
N
=
i=1
N
i=1
dL
∂L
q˙ i −
∂ q˙ i
dt
d ∂L
dt ∂ q˙ i
q˙ i +
∂L
q¨ i −
∂ q˙ i
N
i=1
∂L
∂L
∂L
q˙ i +
q¨ i +
.
∂qi
∂ q˙ i
∂t
(6.53)
There are five terms here. The second cancels with the fourth. And the first
(after using the E–L equation, Eq. (6.3), to rewrite it) cancels with the third. We
6.5 Conservation laws
therefore arrive at the simple result,
dE
∂L
=−
.
dt
∂t
(6.54)
In the event that ∂L/∂t = 0 (that is, there are no t’s sitting on the paper
when you write down L), which is usually the case in the situations we consider (because we generally won’t deal with potentials that depend on time), we
have dE/dt = 0.
Not too many things are constant with respect to time, and the quantity E has
units of energy, so it’s a good bet that it’s the energy. Let’s show this in Cartesian
coordinates (however, see the remark below). The Lagrangian is
L=
1
m(˙x2 + y˙ 2 + z˙ 2 ) − V (x, y, z),
2
(6.55)
E=
1
m(˙x2 + y˙ 2 + z˙ 2 ) + V (x, y, z),
2
(6.56)
so Eq. (6.52) gives
which is the total energy. The effect of the operations in Eq. (6.52) in most cases
is just to switch the sign in front of the potential.
Of course, taking the kinetic energy T and subtracting the potential energy V
to obtain L, and then using Eq. (6.52) to produce E = T + V , seems like a rather
convoluted way of arriving at T + V . But the point of all this is that we used the
E–L equations to prove that E is conserved. Although we know very well from
the F = ma methods in Chapter 5 that the sum T + V is conserved, it’s not fair to
assume that it is conserved in our new Lagrangian formalism. We have to show
that this follows from the E–L equations.
As with the translation and rotation invariance we observed in the examples
in Section 6.5.1, we see that energy conservation arises from time translation
invariance. If the Lagrangian has no explicit t dependence, then the setup looks
the same today as it did yesterday. This fact leads to conservation of energy.
Remark: The quantity E in Eq. (6.52) gives the energy of the system only if the entire system
is represented by the Lagrangian. That is, the Lagrangian must represent a closed system with
no external forces. If the system is not closed, then Claim 6.3 (or more generally, Eq. (6.54)) is
still perfectly valid for the E defined in Eq. (6.52), but this E may simply not be the energy of
the system. Problem 6.8 is a good example of such a situation.
Another example is the following. Imagine a long rod in the horizontal x-y plane. The rod
points in the x direction, and a bead is free to slide frictionlessly along it. At t = 0, an external
machine is arranged to accelerate the rod in the negative y direction (that is, transverse to itself)
with acceleration −g. So y˙ = −gt. There is no internal potential energy in this system, so the
235
236
The Lagrangian method
Lagrangian is just the kinetic energy, L = m˙x2 /2 + m(gt)2 /2. Equation (6.52) therefore gives
E = m˙x2 /2 − m(gt)2 /2, which isn’t the energy. But Eq. (6.54) is still true, because
dE
∂L
=−
dt
∂t
⇐⇒
m˙xx¨ − mg 2 t = −mg 2 t
⇐⇒
x¨ = 0,
(6.57)
which is correct. However, this setup is exactly the same as projectile motion in the x-y plane,
where y is now the vertical axis, provided that we eliminate the rod and consider gravity
instead of the machine to be causing the acceleration in the y direction. But if we are thinking in
terms of gravity, then the normal thing to do is to say that the particle moves under the influence
of the potential V (y) = mgy. The Lagrangian for this closed system (bead plus earth) is
L = m(˙x2 + y˙ 2 )/2 − mgy, and so Eq. (6.52) gives E = m(˙x2 + y˙ 2 )/2 + mgy, which is indeed
the energy of the particle. But having said all this, most of the systems we’ll deal with are
closed, so you can usually ignore this remark and assume that the E in Eq. (6.52) gives the
energy. ♣
6.6
Noether’s theorem
We now present one of the most beautiful and useful theorems in physics. It deals
with two fundamental concepts, namely symmetries and conserved quantities.
The theorem (due to Emmy Noether) may be stated as follows.
Theorem 6.4 (Noether’s theorem) For each symmetry of the Lagrangian,
there is a conserved quantity.
By “symmetry,” we mean that if the coordinates are changed by some small
quantities, then the Lagrangian has no first-order change in these quantities. By
“conserved quantity,” we mean a quantity that does not change with time. The
result in Section 6.5.1 for cyclic coordinates is a special case of this theorem.
Proof: Let the Lagrangian be invariant, to first order in the small number ,
under the change of coordinates,
qi −→ qi + Ki (q).
(6.58)
Each Ki (q) may be a function of all the qi , which we collectively denote by the
shorthand, q.
Remark: As an example of what these Ki ’s might look like, consider the Lagrangian, L =
(m/2)(5˙x2 − 2˙xy˙ + 2˙y2 ) + C(2x − y). We’ve just pulled this out of a hat, although it happens to
be the type of L that arises in Atwood’s machine problems; see Problem 6.9 and Exercise 6.40.
This L is invariant under the transformation x → x + and y → y + 2 , because the derivative
terms are unaffected, and the difference 2x − y is unchanged. (It’s actually invariant to all
orders in , and not just first order. But this isn’t necessary for the theorem to hold.) Therefore,
Kx = 1 and Ky = 2, which happen to be independent of the coordinates. In the problems we’ll
be doing, the Ki ’s can generally be determined by simply looking at the potential term.
Of course, someone else might come along with Kx = 3 and Ky = 6, which is also a
symmetry. And indeed, any factor can be taken out of and put into the Ki ’s without changing
the quantity Ki (q) in Eq. (6.58). Any such modification will just bring an overall constant
factor (and hence not change the property of being conserved) into the conserved quantity in
Eq. (6.61) below. It is therefore irrelevant. ♣
6.6 Noether’s theorem
The fact that the Lagrangian does not change at first order in means that
0=
dL
=
d
i
∂L ∂qi
∂L ∂ q˙ i
+
∂qi ∂
∂ q˙ i ∂
i
∂L
∂L ˙
Ki .
Ki +
∂qi
∂ q˙ i
=
(6.59)
Using the E–L equation, Eq. (6.3), we can rewrite this as
0=
i
=
d
dt
d ∂L
dt ∂ q˙ i
i
Ki +
∂L ˙
Ki
∂ q˙ i
∂L
Ki .
∂ q˙ i
(6.60)
Therefore, the quantity
P(q, q˙ ) ≡
i
∂L
Ki (q)
∂ q˙ i
(6.61)
does not change with time. It is given the generic name of conserved momentum.
But it need not have the units of linear momentum.
As Noether most keenly observed
(And for which much acclaim is deserved),
It’s easy to see
That for each symmetry,
A quantity must be conserved.
Example 1: Consider the Lagrangian in the above remark, L = (m/2)(5˙x2 −2˙xy˙ +
2˙y2 ) + C(2x − y). We saw that Kx = 1 and Ky = 2. The conserved momentum is
therefore
P(x, y, x˙ , y˙ ) =
∂L
∂L
Kx +
Ky = m(5˙x − y˙ )(1) + m(−˙x + 2˙y)(2)
∂ x˙
∂ y˙
= m(3˙x + 3˙y).
(6.62)
The overall factor of 3m isn’t important.
Example 2: Consider a thrown ball. We have L = (m/2)(˙x2 + y˙ 2 + z˙ 2 )−mgz. This
is invariant under translations in x, that is, x → x + ; and also under translations in
y, that is, y → y + . (Both x and y are cyclic coordinates.) We need invariance only
to first order in for Noether’s theorem to hold, but this L is invariant to all orders.
237
238
The Lagrangian method
We therefore have two symmetries in our Lagrangian. The first has Kx = 1, Ky = 0,
and Kz = 0. The second has Kx = 0, Ky = 1, and Kz = 0. Of course, the nonzero
Ki ’s here can be chosen to be any constants, but we may as well pick them to be 1.
The two conserved momenta are
P1 (x, y, z, x˙ , y˙ , z˙ ) =
∂L
∂L
∂L
Kx +
Ky +
Kz = m˙x,
∂ x˙
∂ y˙
∂ z˙
P2 (x, y, z, x˙ , y˙ , z˙ ) =
∂L
∂L
∂L
Kx +
Ky +
Kz = m˙y.
∂ x˙
∂ y˙
∂ z˙
(6.63)
These are simply the x and y components of the linear momentum, as we saw in
the first example in Section 6.5.1. Note that any combination of these momenta, say
3P1 + 8P2 , is also conserved. (In other words, x → x + 3 , y → y + 8 , z → z is
a symmetry of the Lagrangian.) But the above P1 and P2 are the simplest momenta
to choose as a “basis” for the infinite number of conserved momenta (which is how
many you have, if there are two or more independent continuous symmetries).
Example 3: Consider a mass on a spring, with relaxed length zero, in the x-y plane.
The Lagrangian, L = (m/2)(˙x2 + y˙ 2 ) − (k/2)(x2 + y2 ), is invariant under the change
of coordinates, x → x + y and y → y − x, to first order in (as you can check).
So we have Kx = y and Ky = −x. The conserved momentum is therefore
P(x, y, x˙ , y˙ ) =
∂L
∂L
Kx +
Ky = m(˙xy − y˙ x).
∂ x˙
∂ y˙
(6.64)
This is the (negative of the) z component of the angular momentum. The angular
momentum is conserved here because the potential, V (x, y) ∝ x2 + y2 = r 2 , depends
only on the distance from the origin. We’ll discuss such potentials in Chapter 7.
In contrast with the first two examples above, the x → x + y, y → y − x transformation isn’t so obvious here. How did we get this? Well, unfortunately there doesn’t
seem to be any fail-safe method of determining the Ki ’s in general, so sometimes you
just have to guess around. But in many problems, the Ki ’s are simple constants which
are easy to see.
Remarks:
1. As we saw above, in some cases the Ki ’s are functions of the coordinates, and in some
cases they are not.
2. The cyclic-coordinate result in Eq. (6.48) is a special case of Noether’s theorem, for the
following reason. If L doesn’t depend on a certain coordinate qk , then qk −→ qk + is
certainly a symmetry. Hence Kk = 1 (with all the other Ki ’s equal to zero), and Eq. (6.60)
reduces to Eq. (6.48).
3. We use the word “symmetry” to describe the situation where the transformation in
Eq. (6.58) produces no first-order change in the Lagrangian. This is an appropriate choice
of word, because the Lagrangian describes the system, and if the system essentially
doesn’t change when the coordinates are changed, then we say that the system is symmetric. For example, if we have a setup that doesn’t depend on θ, then we say that the
setup is symmetric under rotations. Rotate the system however you want, and it looks
6.7 Small oscillations
4.
5.
6.
7.
6.7
the same. The two most common applications of Noether’s theorem are the conservation
of angular momentum, which arises from symmetry under rotations; and conservation of
linear momentum, which arises from symmetry under translations.
In simple systems, as in Example 2 above, it is clear why the resulting P is conserved. But
in more complicated systems, as in Example 1 above, the resulting P might not have an
obvious interpretation. But at least you know that it is conserved, and this will invariably
help in understanding a setup.
Although conserved quantities are extremely useful in studying a physical situation, it
should be stressed that there is no more information contained in them than there is in
the E–L equations. Conserved quantities are simply the result of integrating the E–L
equations. For example, if you write down the E–L equations for Example 1 above, and
then add the “x” equation (which is 5m¨x − m¨y = 2C) to twice the “y” equation (which
is −m¨x + 2m¨y = −C), then you arrive at 3m(¨x + y¨ ) = 0. In other words, 3m(˙x + y˙ ) is
constant, as we found from Noether’s theorem.
Of course, you might have to do some guesswork to find the proper combination of
the E–L equations that gives a zero on the right-hand side. But you’d have to do some
guesswork anyway, to find the symmetry for Noether’s theorem. At any rate, a conserved
quantity is useful because it is an integrated form of the E–L equations. It puts you one
step closer to solving the problem, compared with where you would be if you started with
the second-order E–L equations.
Does every system have a conserved momentum? Certainly not. The one-dimensional
problem of a falling ball (m¨z = −mg) doesn’t have one. And if you write down an
arbitrary potential in 3-D, odds are that there won’t be one. In a sense, things have to
contrive nicely for there to be a conserved momentum. In some problems, you can just
look at the physical system and see what the symmetry is, but in others (for example, in
the Atwood’s-machine problems for this chapter), the symmetry is not at all obvious.
By “conserved quantity,” we mean a quantity that depends on (at most) the coordinates
and their first derivatives (that is, not on their second derivatives). If we don’t make
this restriction, then it is trivial to construct quantities that are independent of time. For
example, in Example 1 above, the “x” E–L equation (which is 5m¨x − m¨y = 2C) tells
us that 5m¨x − m¨y has its time derivative equal to zero. Note that an equivalent way of
excluding these trivial cases is to say that the value of a conserved quantity depends on
the initial conditions (that is, the velocities and positions). The quantity 5m¨x − m¨y doesn’t
satisfy this criterion, because its value is always constrained to be 2C. ♣
Small oscillations
In many physical systems, a particle undergoes small oscillations around an
equilibrium point. In Section 5.2, we showed that the frequency of these small
oscillations is
ω=
V (x0 )
,
m
(6.65)
where V (x) is the potential energy, and x0 is the equilibrium point. However, this
result holds only for one-dimensional motion (we’ll see below why this is true).
In more complicated systems, such as the one described below, it is necessary
to use another procedure to obtain the frequency ω. This procedure is a fail-safe
one, applicable in all situations. It is, however, a bit more involved than simply
writing down Eq. (6.65). So in one-dimensional problems, Eq. (6.65) is still what
you want to use. We’ll demonstrate our fail-safe method through the following
problem.
239
240
The Lagrangian method
r
m
l-r
M
Fig. 6.5
θ
Problem: A mass m is free to slide on a frictionless table and is connected, via
a string that passes through a hole in the table, to a mass M that hangs below (see
Fig. 6.5). Assume that M moves in a vertical line only, and assume that the string
always remains taut.
(a) Find the equations of motion for the variables r and θ shown in the figure.
(b) Under what condition does m undergo circular motion?
(c) What is the frequency of small oscillations (in the variable r) about this circular
motion?
Solution:
(a) Let the string have length (this length won’t matter). Then the Lagrangian
(we’ll call it “L” here, to save “L” for the angular momentum, which arises
below) is
L=
1
1
M r˙ 2 + m(˙r 2 + r 2 θ˙ 2 ) + Mg( − r).
2
2
(6.66)
For the purposes of the potential energy, we’ve taken the table to be at height
zero, but any other value could be chosen. The E–L equations of motion
obtained from varying θ and r are
d
˙ = 0,
(mr 2 θ)
dt
(6.67)
(M + m)¨r = mr θ˙ 2 − Mg.
The first equation says that angular momentum is conserved. The second
equation says that the Mg gravitational force accounts for the acceleration of
the two masses along the direction of the string, plus the centripetal acceleration
of m.
(b) The first of Eqs. (6.67) says that mr 2 θ˙ = L, where L is some constant (the
angular momentum) which depends on the initial conditions. Plugging θ˙ =
L/mr 2 into the second of Eqs. (6.67) gives
(M + m)¨r =
L2
− Mg.
mr 3
(6.68)
Circular motion occurs when r˙ = r¨ = 0. Therefore, the radius of the circular
orbit is given by
r03 =
L2
.
Mmg
(6.69)
˙ Eq. (6.69) is equivalent to
Since L = mr 2 θ,
mr0 θ˙ 2 = Mg,
(6.70)
6.7 Small oscillations
which can also be obtained by letting r¨ = 0 in the second of Eqs. (6.67). In
other words, the gravitational force on M exactly accounts for the centripetal
acceleration of m if the motion is circular. Given r0 , Eq. (6.70) determines
what θ˙ must be in order to have circular motion, and vice versa.
(c) To find the frequency of small oscillations about the circular motion, we need
to look at what happens to r if we perturb it slightly from its equilibrium value
of r0 . Our fail-safe procedure is the following.
Let r(t) ≡ r0 + δ(t), where δ(t) is very small (more precisely, δ(t)
r0 ),
and expand Eq. (6.68) to first order in δ(t). Using
1
1
3δ
1
1
1
≡
≈ 3
= 3
≈ 3 1−
3
3
2
r0
r
(r0 + δ)
r0 + 3r0 δ
r0 (1 + 3δ/r0 )
r0
,
(6.71)
we obtain
(M + m)δ¨ ≈
L2
mr03
1−
3δ
r0
− Mg.
(6.72)
The terms not involving δ on the right-hand side cancel, by the definition of r0
given in Eq. (6.69). This cancellation always occurs in such a problem at this
stage, due to the definition of the equilibrium point. We are therefore left with
δ¨ +
3L2
(M + m)mr04
δ ≈ 0.
(6.73)
This is a good old simple-harmonic-oscillator equation in the variable δ.
Therefore, the frequency of small oscillations about a circle of radius r0 is
ω≈
3L2
(M + m)mr04
=
3M
M +m
g
,
r0
(6.74)
where we have used Eq. (6.69) to eliminate L in the second expression.
To sum up, the above frequency is the frequency of small oscillations in
the variable r. In other words, if you have nearly circular motion, and if you
plot r as a function of time (and ignore what θ is doing), then you will get a
nice sinusoidal graph whose frequency is given by Eq. (6.74). Note that this
frequency need not have anything to do with the other relevant frequency in this
√
√
problem, namely the frequency of the circular motion, which is M /m g/r0 ,
from Eq. (6.70).
√
Remarks: Let’s look at some limits. For a given r0 , if m
M , then ω ≈ 3Mg/mr0 ≈√0.
This makes sense, because everything is moving
√ very slowly. This frequency equals 3
times the frequency of circular motion,
namely Mg/mr0 , which isn’t at all obvious. For
√
M , then ω ≈ 3g/r0 , which isn’t so obvious, either.
a given r0 , if m
The frequency of small oscillations is equal to the frequency of circular motion if
M = 2m, which, once again, isn’t obvious. This condition is independent of r0 . ♣
241
242
The Lagrangian method
The above procedure for finding the frequency of small oscillations can be
summed up in three steps: (1) find the equations of motion; (2) find the equilibrium point; and (3) let x(t) ≡ x0 + δ(t), where x0 is the equilibrium point of the
relevant variable, and expand one of the equations of motion (or a combination
of them) to first order in δ, to obtain a simple-harmonic-oscillator equation for δ.
If the equilibrium point happens to be at x = 0 (which is often the case), then
everything is greatly simplified. There is no need to introduce δ, and the expansion in the third step above simply entails ignoring powers of x that are higher
than first order.
Remark: If you just use the potential energy for the above problem (which is Mgr, up to an
additive constant) in Eq. (6.65), then you will obtain a frequency of zero, which is incorrect.
You can use Eq. (6.65) to find the frequency, if you instead use the “effective potential” for
this problem, namely L2 /(2mr 2 ) + Mgr, and if you use the total mass, M + m, as the mass
in Eq. (6.65), as you can check. The reason why this works will become clear in Chapter 7
when we introduce the effective potential. In many problems, however, it isn’t obvious what
the appropriate modified potential is that should be used, or what mass goes in Eq. (6.65). So
it’s generally much safer to take a deep breath and go through an expansion similar to the one
in part (c) of the example above. ♣
The one-dimensional result in Eq. (6.65) is, of course, a special case of our
above expansion procedure. We can repeat the derivation of Section 5.2 in the
present language. In one dimension, the E–L equation of motion is m¨x = −V (x).
Let x0 be the equilibrium point, so V (x0 ) = 0. And let x(t) ≡ x0 + δ(t).
Expanding m¨x = −V (x) to first order in δ, we have mδ¨ = −V (x0 ) − V (x0 )δ,
plus higher-order terms. Since V (x0 ) = 0, we have mδ¨ ≈ −V (x0 )δ, as
desired.
6.8
Other applications
The formalism developed in Section 6.2 works for any function L(x, x˙ , t). If our
goal is to find the stationary points of S ≡ L, then Eq. (6.15) holds, no matter
what L is. There is no need for L to be equal to T − V , or indeed, to have anything
to do with physics. And t need not have anything to do with time. All that is
required is that the quantity x depend on the parameter t, and that L depend only
on x, x˙ , and t (and not, for example, on x¨ ; see Exercise 6.34). The formalism is
very general and powerful, as the following example demonstrates.
Fig. 6.6
y(x)
c1
a1
Fig. 6.7
c2
a2
Example (Minimal surface of revolution): A surface of revolution has two
given parallel rings as its boundary; see Fig. 6.6. What should the shape of the surface
be so that it has the minimum possible area? We will present three solutions. A fourth
is left for Problem 6.22.
First solution: Let the surface be generated by rotating the curve y = y(x) around
the x axis. The boundary conditions are y(a1 ) = c1 and y(a2 ) = c2 ; see Fig. 6.7.
6.8 Other applications
Slicing the surface up into vertical rings, we see that the area is given by
A=
a2
a1
2πy 1 + y 2 dx.
(6.75)
Our goal is to find the function y(x) that minimizes this integral. We therefore have
exactly the same situation as in Section 6.2, except that x is now the parameter (instead
of t), and y is now the function (instead of x). Our “Lagrangian” is thus L ∝ y 1 + y 2 .
To minimize the integral A, we “simply” have to write down the E–L equation,
d ∂L
dx ∂y
=
∂L
,
∂y
(6.76)
and calculate the derivatives. This calculation however, gets a bit tedious, so I’ve
relegated it to Lemma 6.5 at the end of this section. For now we’ll just use the result
in Eq. (6.86) which gives (with f (y) = y here)
1 + y 2 = By2 .
(6.77)
At this point we can cleverly guess (motivated by the fact that 1 + sinh2 z = cosh2 z)
that the solution is
y(x) =
1
cosh b(x + d),
b
(6.78)
√
is a constant of integration. Or we can separate variables to
where b = B, and d √
obtain (again with b = B)
dx =
dy
(by)2 − 1
,
(6.79)
and then use the fact that the integral of 1/ z 2 − 1 is cosh−1 z, to obtain the same
result. The answer to our problem, therefore, is that y(x) takes the form of Eq. (6.78),
with b and d determined by the boundary conditions,
c1 =
1
cosh b(a1 + d),
b
and
c2 =
1
cosh b(a2 + d).
b
(6.80)
In the symmetrical case where c1 = c2 , we know that the minimum occurs in the
middle, so we may choose d = 0 and a1 = −a2 .
Solutions for b and d exist only for certain ranges of the a’s and c’s. Basically, if
a2 − a1 is too large, then there is no solution. In this case, the minimal “surface” turns
out to be the two given circles, attached by a line (which isn’t a nice two-dimensional
surface). If you perform an experiment with soap bubbles (which want to minimize
their area), and if you pull the rings too far apart, then the surface will break and
disappear as it tries to form the two circles. Problem 6.23 deals with this issue.
Second solution: Consider the curve that we rotate around the x axis to be
described now by the function x(y). That is, let x be a function of y. The area is
then given by
A=
c2
c1
2πy 1 + x 2 dy,
(6.81)
243
244
The Lagrangian method
where x ≡ dx/dy. Note that the function x(y) may be double-valued, so it may not
really be a function. But it looks like a function locally, and all of our formalism deals
with local variations.
Our “Lagrangian” is now L ∝ y 1 + x 2 , and the E–L equation is
d ∂L
dy ∂x
∂L
∂x
=
=⇒
d
dy
yx
1+x2
= 0.
(6.82)
The nice thing about this solution is the “0” on the right-hand side, which arises
from the fact that L doesn’t depend on x (that is, x is a cyclic coordinate). Therefore,
yx / 1 + x 2 is constant. If we define this constant to be 1/b, then we can solve for
x and then separate variables to obtain
dx =
dy
(by)2 − 1
,
(6.83)
in agreement with Eq. (6.79). The solution proceeds as above.
Third solution: The “Lagrangian” in the first solution above, L ∝ y 1 + y 2 , is
independent of x. Therefore, in analogy with conservation of energy (which arises
from a Lagrangian that is independent of t), the quantity
E≡y
∂L
−L=
∂y
y 2y
1+y2
−y 1+y2 =
−y
1+y2
(6.84)
is constant (that is, independent of x). This statement is equivalent to Eq. (6.77), and
the solution proceeds as above. As demonstrated by the brevity of the second and
third solutions here, it is highly advantageous to make use of conserved quantities
whenever you can.
Let us now prove the following lemma, which we invoked in the first solution
above. This lemma is very useful, because it is common to encounter problems
where the quantity to be extremized depends on the arclength, 1 + y 2 , and takes
the form of f (y) 1 + y 2 dx. We will give two proofs. The first proof uses the
Euler–Lagrange equation. The calculation gets a bit messy, so it’s a good idea to
work through it once and for all and then just invoke the result whenever needed.
This derivation isn’t something you’d want to repeat too often. The second proof
makes use of a conserved quantity. And in contrast with the first proof, this
method is exceedingly clean and simple. It actually is something you’d want to
repeat quite often. But we’ll still do it once and for all.
Lemma 6.5 Let f (y) be a given function of y. Then the function y(x) that
extremizes the integral,
x2
x1
f (y) 1 + y 2 dx,
(6.85)
6.8 Other applications
satisfies the differential equation,
1 + y 2 = Bf (y)2 ,
(6.86)
where B is a constant of integration.11
First proof: Our goal is to find the function y(x) that extremizes the integral
in Eq. (6.85). We therefore have exactly the same situation as in Section 6.2,
except with x in place of t, and y in place of x. Our “Lagrangian” is thus L =
f (y) 1 + y 2 , and the Euler–Lagrange equation is
d ∂L
dx ∂y
=
∂L
∂y
d
f ·y ·
dx
=⇒
1
1+y2
=f
1 + y 2,
(6.87)
where f ≡ df /dy. We must now perform some straightforward (albeit tedious)
differentiations. Using the product rule on the three factors on the left-hand side,
and making copious use of the chain rule, we obtain
f y2
1+y2
+
fy
1+y2
−
fy 2 y
=f
(1 + y 2 )3/2
1 + y 2.
(6.88)
Multiplying through by (1 + y 2 )3/2 and simplifying gives
fy = f (1 + y 2 ).
(6.89)
We have completed the first step of the proof, namely producing the Euler–
Lagrange differential equation. We must now integrate it. Equation (6.89)
happens to be integrable for arbitrary functions f (y). If we multiply through
by y and rearrange, we obtain
yy
f y
=
.
1+y2
f
(6.90)
Taking the dx integral of both sides gives (1/2) ln(1 + y 2 ) = ln(f ) + C, where
C is a constant of integration. Exponentiation then gives (with B ≡ e2C )
1 + y 2 = Bf (y)2 ,
(6.91)
as we wanted to show. In an actual problem, we would solve this equation for y ,
and then separate variables and integrate. But we would need to be given a specific
function f (y) to be able to do this.
11
The constant B and also one other constant of integration (arising when Eq. (6.86) is integrated to
obtain y) are determined by the boundary conditions on y(x); see, for example, Eq. (6.80). This
situation, where the two constants are determined by the values of the function at two points, is
slightly different from the situation in the physics problems we’ve done where the two constants
are determined by the value (that is, the initial position) and the slope (that is, the speed) at just
one point. But either way, two given facts must be used.
245
246
The Lagrangian method
Second proof: Our “Lagrangian,” L = f (y) 1 + y 2 , is independent of x.
Therefore, in analogy with the conserved energy given in Eq. (6.52), the quantity
E≡y
∂L
−f (y)
−L=
∂y
1+y2
(6.92)
√
is independent of x. Call it 1/ B. Then we have easily reproduced Eq. (6.91).
For practice, you can also prove this lemma by considering x to be a function of
y, as we did in the second solution in the minimal-surface example above.
m
6.9
M
θ
Problems
Section 6.1: The Euler–Lagrange equations
6.1. Moving plane **
A block of mass m is held motionless on a frictionless plane of mass M
and angle of inclination θ (see Fig. 6.8). The plane rests on a frictionless
horizontal surface. The block is released. What is the horizontal acceleration of the plane? (This problem already showed up as Problem 3.8.
If you haven’t already done so, try solving it using F = ma. You will
then have a greater appreciation for the Lagrangian method.)
Fig. 6.8
ε
m
r
m
r
6.2. Two falling sticks **
Two massless sticks of length 2r, each with a mass m fixed at its middle,
are hinged at an end. One stands on top of the other, as shown in Fig. 6.9.
The bottom end of the lower stick is hinged at the ground. They are held
such that the lower stick is vertical, and the upper one is tilted at a
small angle with respect to the vertical. They are then released. At this
instant, what are the angular accelerations of the two sticks? Work in
the approximation where is very small.
Fig. 6.9
l
6.3. Pendulum with an oscillating support **
A pendulum consists of a mass m and a massless stick of length .
The pendulum support oscillates horizontally with a position given by
x(t) = A cos(ωt); see Fig. 6.10. What is the general solution for the
angle of the pendulum as a function of time?
m
Fig. 6.10
θ
m
Fig. 6.11
m
r
6.4. Two masses, one swinging ***
Two equal masses m, connected by a massless string, hang over two
pulleys (of negligible size), as shown in Fig. 6.11. The left one moves
in a vertical line, but the right one is free to swing back and forth in the
plane of the masses and pulleys. Find the equations of motion for r and
θ , as shown.
Assume that the left mass starts at rest, and the right mass undergoes
small oscillations with angular amplitude (with
1). What is the
6.9 Problems
247
initial average acceleration (averaged over a few periods) of the left
mass? In which direction does it move?
6.5. Inverted pendulum ****
A pendulum consists of a mass m at the end of a massless stick of length
. The other end of the stick is made to oscillate vertically with a position
given by y(t) = A cos(ωt), where A
. See Fig. 6.12. It turns out
that if ω is large enough, and if the pendulum is initially nearly upsidedown, then surprisingly it will not fall over as time goes by. Instead, it
will (sort of) oscillate back and forth around the vertical position. If you
want to do the experiment yourself, see the 28th demonstration of the
entertaining collection in Ehrlich (1994).
Find the equation of motion for the angle of the pendulum (measured
relative to its upside-down position). Explain why the pendulum doesn’t
fall over, and find the frequency of the back and forth motion.
Section 6.2: The principle of stationary action
6.6. Minimum or saddle **
(a) In Eq. (6.26), let t1 = 0 and t2 = T , for convenience. And let
ξ(t) be an easy-to-deal-with “triangular” function, of the form
ξ(t) =
t/T ,
(1 − t/T ),
0 ≤ t ≤ T /2,
T /2 ≤ t ≤ T .
(6.93)
Under what condition is the harmonic-oscillator S in Eq. (6.26)
negative?
(b) Answer the same question, but now with ξ(t) = sin(π t/T ).
Section 6.3: Forces of constraint
6.7. Normal force from a plane **
A mass m slides down a frictionless plane that is inclined at an angle θ.
Show, using the method in Section 6.3, that the normal force from the
plane is the familiar mg cos θ .
Section 6.5: Conservation laws
6.8. Bead on a stick *
A stick is pivoted at the origin and is arranged to swing around in a
horizontal plane at constant angular speed ω. A bead of mass m slides
frictionlessly along the stick. Let r be the radial position of the bead.
Find the conserved quantity E given in Eq. (6.52). Explain why this
quantity is not the energy of the bead.
m
l
Fig. 6.12
248
The Lagrangian method
Section 6.6: Noether’s theorem
x
y
4m
3m
m
Fig. 6.13
Section 6.7: Small oscillations
6.10. Hoop and pulley **
A mass M is attached to a massless hoop of radius R that lies in a vertical
plane. The hoop is free to rotate about its fixed center. M is tied to a
string which winds part way around the hoop, then rises vertically up
and over a massless pulley. A mass m hangs on the other end of the string
(see Fig. 6.14). Find the equation of motion for the angle of rotation of
the hoop. What is the frequency of small oscillations? Assume that m
moves only vertically, and assume M > m.
m
R
M
Fig. 6.14
ω
6.11. Bead on a rotating hoop **
A bead is free to slide along a frictionless hoop of radius R. The hoop
rotates with constant angular speed ω around a vertical diameter (see
Fig. 6.15). Find the equation of motion for the angle θ shown. What are
the equilibrium positions? What is the frequency of small oscillations
about the stable equilibrium? There is one value of ω that is rather
special; what is it, and why is it special?
R
θ
Fig. 6.15
θ
ω
R
(top view)
Fig. 6.16
M
l
m
Fig. 6.17
6.9. Atwood’s machine **
Consider the Atwood’s machine shown in Fig. 6.13. The masses are
4m, 3m, and m. Let x and y be the heights of the left and right masses,
relative to their initial positions. Find the conserved momentum.
r
6.12. Another bead on a rotating hoop **
A bead is free to slide along a frictionless hoop of radius r. The plane of
the hoop is horizontal, and the center of the hoop travels in a horizontal
circle of radius R, with constant angular speed ω, about a given point
(see Fig. 6.16). Find the equation of motion for the angle θ shown. Also,
find the frequency of small oscillations about the equilibrium point.
6.13. Mass on a wheel **
A mass m is fixed to a given point on the rim of a wheel of radius R that
rolls without slipping on the ground. The wheel is massless, except for
a mass M located at its center. Find the equation of motion for the angle
through which the wheel rolls. For the case where the wheel undergoes
small oscillations, find the frequency.
6.14. Pendulum with a free support **
A mass M is free to slide along a frictionless rail. A pendulum of length
and mass m hangs from M (see Fig. 6.17). Find the equations of motion.
For small oscillations, find the normal modes and their frequencies.
6.9 Problems
6.15. Pendulum support on an inclined plane **
A mass M is free to slide down a frictionless plane inclined at an
angle β. A pendulum of length
and mass m hangs from M ;
see Fig. 6.18 (assume that M extends a short distance beyond the side
of the plane, so the pendulum can hang down). Find the equations
of motion. For small oscillations, find the normal modes and their
frequencies.
6.16. Tilting plane ***
A mass M is fixed at the right-angled vertex where a massless rod of
length is attached to a very long massless rod (see Fig. 6.19). A mass m
is free to move frictionlessly along the long rod (assume that it can pass
through M ). The rod of length is hinged at a support, and the whole
system is free to rotate, in the plane of the rods, about the hinge. Let θ
be the angle of rotation of the system, and let x be the distance between
m and M . Find the equations of motion. Find the normal modes when θ
and x are both very small.
249
M
l
m
Fig. 6.18
θ l
x
m
M
Fig. 6.19
y
6.17. Rotating curve ***
The curve y(x) = b(x/a)λ is rotated around the y axis with constant
frequency ω (see Fig. 6.20). A bead moves frictionlessly along the curve.
Find the frequency of small oscillations about the equilibrium point.
Under what conditions do oscillations exist? (This problem gets a little
messy.)
6.18. Motion in a cone ***
A particle slides on the inside surface of a frictionless cone. The cone
is fixed with its tip on the ground and its axis vertical. The half-angle
at the tip is α (see Fig. 6.21). Let r be the distance from the particle to
the axis, and let θ be the angle around the cone. Find the equations of
motion.
If the particle moves in a circle of radius r0 , what is the frequency, ω,
of this motion? If the particle is then perturbed slightly from this circular
motion, what is the frequency, , of the oscillations about the radius r0 ?
Under what conditions does = ω?
ω
b
x
a
Fig. 6.20
r0
α
Fig. 6.21
6.19. Double pendulum ****
Consider a double pendulum made of two masses, m1 and m2 , and two
rods of lengths 1 and 2 (see Fig. 6.22). Find the equations of motion.
For small oscillations, find the normal modes and their frequencies
for the special case 1 = 2 (and consider the cases m1 = m2 , m1
m2 ,
and m1
m2 ). Do the same for the special case m1 = m2 (and consider
the cases 1 = 2 , 1
2 , and 1
2 ).
l1
m1
l2
m2
Fig. 6.22
250
The Lagrangian method
Section 6.8: Other applications
6.20. Shortest distance in a plane *
In the spirit of Section 6.8, show that the shortest path between two
points in a plane is a straight line.
6.21. Index of refraction **
Assume that the speed of light in a given slab of material is proportional
to the height above the base of the slab.12 Show that light moves in
circular arcs in this material; see Fig. 6.23. You may assume that light
takes the path of least time between two points (Fermat’s principle of
least time).
Fig. 6.23
x1
6.22. Minimal surface **
Derive the shape of the minimal surface discussed in Section 6.8, by
demanding that a cross-sectional “ring” (that is, the region between the
planes x = x1 and x = x2 ) is in equilibrium; see Fig. 6.24. Hint: The
tension must be constant throughout the surface (assuming that we’re
ignoring gravity, which we are).
x2
6.23. Existence of a minimal surface **
Consider the minimal surface from Section 6.8, and look at the special
case where the two rings have the same radius r (see Fig. 6.25). Let
2 be the distance between the rings. What is the largest value of /r
for which a minimal surface exists? You will need to solve something
numerically here.
Fig. 6.24
r
2l
r
6.24. The brachistochrone ***
A bead is released from rest at the origin and slides down a frictionless
wire that connects the origin to a given point, as shown in Fig. 6.26.
You wish to shape the wire so that the bead reaches the endpoint in the
shortest possible time. Let the desired curve be described by the function
y(x), with downward taken to be positive. Show that y(x) satisfies
Fig. 6.25
1+y2 =
x
B
,
y
(6.94)
where B is a constant. Then show that x and y may be written as
x = a(θ − sin θ),
y = a(1 − cos θ).
(6.95)
y
12
Fig. 6.26
If you want to make the equivalent statement in terms of the material’s “index of refraction,”
commonly denoted by n, then you can say: As a function of the height y, the index n is given by
n(y) = y0 /y, where y0 is some length that is larger than the height of the slab. This is equivalent
to the original statement because the speed of light in a material equals c/n.
6.10 Exercises
251
This is the parametrization of a cycloid, which is the path taken by a
point on the rim of a rolling wheel.
6.10
Exercises
Section 6.1: The Euler–Lagrange equations
6.25. Spring on a T **
A rigid T consists of a long rod glued perpendicular to another rod of
length that is pivoted at the origin. The T rotates around in a horizontal
plane with constant frequency ω. A mass m is free to slide along the
long rod and is connected to the intersection of the rods by a spring
with spring constant k and relaxed length zero (see Fig. 6.27). Find r(t),
where r is the position of the mass along the long rod. There is a special
value of ω; what is it, and why is it special?
6.26. Spring on a T, with gravity ***
Consider the setup in the previous exercise, but now let the T swing
around in a vertical plane with constant frequency ω. Find r(t). There is
a special value of ω; what is it, and why is it special? (You may assume
ω < k/m.)
6.27. Coffee cup and mass **
A coffee cup of mass M is connected to a mass m by a string. The coffee
cup hangs over a frictionless pulley of negligible size, and the mass m
is initially held with the string horizontal, as shown in Fig. 6.28. The
mass m is then released. Find the equations of motion for r (the length
of string between m and the pulley) and θ (the angle that the string to
m makes with the horizontal). Assume that m somehow doesn’t run into
the string holding the cup up.
The coffee cup will initially fall, but it turns out that it will reach a
lowest point and then rise back up. Write a program (see Section 1.4)
that numerically determines the ratio of the r at this lowest point to the r
at the start, for a given value of m/M . (To check your program, a value
of m/M = 1/10 yields a ratio of about 0.208.)
6.28. Three falling sticks ***
Three massless sticks of length 2r, each with a mass m fixed at its
middle, are hinged at their ends, as shown in Fig. 6.29. The bottom
end of the lower stick is hinged at the ground. They are held such that
the lower two sticks are vertical, and the upper one is tilted at a small
angle with respect to the vertical. They are then released. At this
instant, what are the angular accelerations of the three sticks? Work
(top view)
m
k
ω
l
Fig. 6.27
r
m
M
Fig. 6.28
ε
m
m
r
m
r
Fig. 6.29
252
The Lagrangian method
in the approximation where
Problem 6.2 first.)
α
Fig. 6.30
is very small. (You may want to look at
6.29. Cycloidal pendulum ****
√
The standard pendulum frequency of g/ holds only for small oscillations. The frequency becomes smaller as the amplitude grows. It turns
out that if you want to build a pendulum whose frequency is independent
of the amplitude, you should hang it from the cusp of a cycloid of a certain size, as shown in Fig. 6.30. As the string wraps partially around the
cycloid, the effect is to decrease the length of string in the air, which in
turn increases the frequency back up to a constant value. In more detail:
A cycloid is the path taken by a point on the rim of a rolling wheel.
The upside-down cycloid in Fig. 6.30 can be parametrized by (x, y) =
R(θ −sin θ , −1+cos θ), where θ = 0 corresponds to the cusp. Consider
a pendulum of length 4R hanging from the cusp, and let α be the angle
the string makes with the vertical, as shown.
(a) In terms of α, find the value of the parameter θ associated with
the point where the string leaves the cycloid.
(b) In terms of α, find the length of string touching the cycloid.
(c) In terms of α, find the Lagrangian.
(d) Show that the quantity sin α undergoes simple harmonic motion
√
with frequency g/4R, independent of the amplitude.
(e) In place of parts (c) and (d), solve the problem again by using
F = ma. This actually gives a much quicker solution.
Section 6.2: The principle of stationary action
6.30. Dropped ball *
Consider the action, from t = 0 to t = 1, of a ball dropped from rest.
From the E–L equation (or from F = ma), we know that y(t) = −gt 2 /2
yields a stationary value of the action. Show explicitly that the particular
function y(t) = −gt 2 /2+ t(t −1) yields an action that has no first-order
dependence on .
6.31. Explicit minimization *
For a ball thrown upward, guess a solution for y of the form y(t) =
a2 t 2 + a1 t + a0 . Assuming that y(0) = y(T ) = 0, this quickly becomes
y(t) = a2 (t 2 − Tt). Calculate the action between t = 0 and t = T , and
show that it is minimized when a2 = −g/2.
6.32. Always a minimum *
For a ball thrown up in the air, show that the stationary value of the
action is always a global minimum.
6.10 Exercises
253
6.33. Second-order change *
Let xa (t) ≡ x0 (t) + aβ(t). Equation (6.19) gives the first derivative of
the action with respect to a. Show that the second derivative is
d2
S[xa (t)] =
da2
t2
t1
∂ 2L 2
∂ 2L 2
∂ 2L
˙+
β
β˙ dt.
β
β
+
2
∂x2
∂x∂ x˙
∂ x˙ 2
(6.96)
6.34. x¨ dependence *
Assume that there is x¨ dependence (in addition to x, x˙ , t dependence) in
the Lagrangian in Theorem 6.1. There will then be the additional term
(∂L/∂ x¨ a )β¨ in Eq. (6.19). It is tempting to integrate this term by parts
twice, and then arrive at a modified form of Eq. (6.22):
∂L
d ∂L
−
∂x0
dt ∂ x˙ 0
+
d 2 ∂L
dt 2 ∂ x¨ 0
= 0.
(6.97)
Is this a valid result? If not, where is the error in the reasoning?
Section 6.3: Forces of constraint
6.35. Constraint on a circle *
A bead of mass m slides with speed v around a horizontal hoop of
radius R. What force does the hoop apply to the bead? (Ignore gravity.)
6.36. Atwood’s machine *
Consider the standard Atwood’s machine in Fig. 6.31, with masses m1
and m2 . Find the tension in the string.
6.37. Cartesian coordinates **
In Eq. (6.35), take two time derivatives of the
equation to obtain
R2 (x¨x + y¨y) + (x˙y − y˙x)2 = 0,
x2 + y2 − R = 0
(6.98)
and then combine this with the other two equations to solve for F in terms
of x, y, x˙ , y˙ . Convert the result to polar coordinates (with θ measured
from the vertical) and show that it agrees with Eq. (6.32).
6.38. Constraint on a curve ***
Let the horizontal plane be the x-y plane. A bead of mass m slides with
speed v along a curve described by the function y = f (x). What force
does the curve apply to the bead? (Ignore gravity.)
m1 m2
Fig. 6.31
254
The Lagrangian method
Section 6.5: Conservation laws
6.39. Bead on stick, using F = ma *
After doing Problem 6.8, show again that the quantity E is conserved,
but now use F = ma. Do this in two ways:
(a) Use the first of Eqs. (3.51). Hint: multiply through by r˙ .
(b) Use the second of Eqs. (3.51) to calculate the work done on the
bead, and use the work–energy theorem.
Section 6.6: Noether’s theorem
4m
x
y
5m 3m
Fig. 6.32
6.40. Atwood’s machine **
Consider the Atwood’s machine shown in Fig. 6.32. The masses are
4m, 5m, and 3m. Let x and y be the heights of the right two masses,
relative to their initial positions. Use Noether’s theorem to find the
conserved momentum. (The solution to Problem 6.9 gives some other
methods, too.)
Section 6.7: Small oscillations
k
M
6.42. Spring on a spoke **
A spring with spring constant k and relaxed length zero lies along a spoke
of a massless wheel of radius R. One end of the spring is attached to the
center, and the other end is attached to a mass m that is free to slide along
the spoke. When the system is in its equilibrium position with the spring
hanging vertically, how far (in terms of R) should the mass hang down
(you are free to adjust k) so that for small oscillations, the frequency of
the spring oscillations equals the frequency of the rocking motion of the
wheel? Assume that the wheel rolls without slipping.
Fig. 6.33
R
Fig. 6.34
6.41. Spring and a wheel *
The top of a wheel of mass M and radius R is connected to a spring
(at its equilibrium length) with spring constant k, as shown in Fig. 6.33.
Assume that all the mass of the wheel is at its center. If the wheel rolls
without slipping, what is the frequency of (small) oscillations?
θ θ
6.43. Oscillating hoop **
Two equal masses are glued to a massless hoop of radius R that is free to
rotate about its center in a vertical plane. The angle between the masses
is 2θ , as shown in Fig. 6.34. Find the frequency of small oscillations.
6.44. Oscillating hoop with a pendulum ***
A massless hoop of radius R is free to rotate about its center in a vertical
√
plane. A mass m is attached at one point, and a pendulum of length 2R
6.11 Solutions
255
(and also of mass m) is attached at another point 90◦ away, as shown in
Fig. 6.35. Let θ be the angle of the hoop relative to the position shown,
and let α be the angle of the pendulum with respect to the vertical. Find
the normal modes for small oscillations.
6.45. Mass sliding on a rim **
A mass m is free to slide frictionlessly along the rim of a wheel of radius R
that rolls without slipping on the ground. The wheel is massless, except
for a mass M located at its center. Find the normal modes for small
oscillations.
6.46. Mass sliding on a rim, with a spring ***
Consider the setup in the previous exercise, but now let the mass m be
attached to a spring with spring constant k and relaxed length zero,
the other end of which is attached to a point on the rim. Assume
that the spring is constrained to run along the rim, and assume that
the mass can pass freely over the point where the spring is attached
to the rim. To keep things from getting too messy here, you can
set M = m.
R
45 45
m
2R
m
Fig. 6.35
(a) Find the frequencies of the normal modes for small oscillations.
Check the g = 0 limit, and (if you’ve done the previous exercise)
the k = 0 limit.
(b) For the special case where√g/R = k/m, show that the frequencies
can be written as k/m( 5 ± 1)/2. This numerical factor is the
golden ratio (and its inverse). Describe what the normal modes
look like.
6.47. Vertically rotating hoop ***
A bead is free to slide along a frictionless hoop of radius r. The plane
of the hoop is vertical, and the center of the hoop travels in a vertical
circle of radius R with constant angular speed ω about a given point
(see Fig. 6.36). Find the equation of motion for the angle θ shown. For
large ω (which implies small θ), find the amplitude of the “particular”
solution with frequency ω. What happens if r = R?
θ
r
ω
R
(side view)
Fig. 6.36
m
6.11
Solutions
x1
6.1. Moving plane
Let x1 be the horizontal coordinate of the plane (with positive x1 to the left), and let x2
be the horizontal coordinate of the block (with positive x2 to the right); see Fig. 6.37.
The relative horizontal distance between the plane and the block is x1 + x2 , so the
height fallen by the block is (x1 + x2 ) tan θ. The Lagrangian is therefore
L=
1
1
M x˙ 12 + m x˙ 22 + (˙x1 + x˙ 2 )2 tan2 θ + mg(x1 + x2 ) tan θ.
2
2
(6.99)
x2
M
Fig. 6.37
θ
256
The Lagrangian method
The equations of motion obtained from varying x1 and x2 are
M x¨ 1 + m(¨x1 + x¨ 2 ) tan2 θ = mg tan θ ,
m¨x2 + m(¨x1 + x¨ 2 ) tan2 θ = mg tan θ .
(6.100)
Note that the difference of these two equations immediately yields conservation of
momentum, M x¨ 1 − m¨x2 = 0 =⇒ (d/dt)(M x˙ 1 − m˙x2 ) = 0. Equations (6.100) are two
linear equations in the two unknowns, x¨ 1 and x¨ 2 , so we can solve for x¨ 1 . After a little
simplification, we arrive at
x¨ 1 =
θ2
m
θ1 m
mg sin θ cos θ
M + m sin2 θ
.
(6.101)
For some limiting cases, see the remarks in the solution to Problem 3.8.
6.2. Two falling sticks
Let θ1 (t) and θ2 (t) be defined as in Fig. 6.38. Then the position of the bottom mass
in Cartesian coordinates is (r sin θ1 , r cos θ1 ), and the position of the top mass is
(2r sin θ1 − r sin θ2 , 2r cos θ1 + r cos θ2 ). So the potential energy of the system is
V (θ1 , θ2 ) = mgr(3 cos θ1 + cos θ2 ).
(6.102)
The kinetic energy is somewhat more complicated. The kinetic energy of the bottom
mass is simply mr 2 θ˙12 /2. Taking the derivative of the top mass’s position given above,
we find that the kinetic energy of the top mass is
Fig. 6.38
1 2
mr (2 cos θ1 θ˙1 − cos θ2 θ˙2 )2 + (−2 sin θ1 θ˙1 − sin θ2 θ˙2 )2 .
2
(6.103)
We can simplify this, using the small-angle approximations. The terms involving sin θ
are fourth order in the small θ’s, so we can neglect them. Also, we can approximate
cos θ by 1, because this entails dropping only terms of at least fourth order. So the
top mass’s kinetic energy becomes (1/2)mr 2 (2θ˙1 − θ˙2 )2 . In retrospect, it would have
been easier to obtain the kinetic energies of the masses by first applying the small-angle
approximations to the positions, and then taking the derivatives to obtain the velocities.
This strategy shows that both masses move essentially horizontally (initially). You will
probably want to use this strategy when solving Exercise 6.28.
Using the small-angle approximation cos θ ≈ 1 − θ 2 /2 to rewrite the potential
energy in Eq. (6.102), we have
L≈
1 2
3
1
mr 5θ˙12 − 4θ˙1 θ˙2 + θ˙22 − mgr 4 − θ12 − θ22 .
2
2
2
(6.104)
The equations of motion obtained from varying θ1 and θ2 are, respectively,
3g
θ1
r
g
−2θ¨1 + θ¨2 = θ2 .
r
5θ¨1 − 2θ¨2 =
x(t)
(6.105)
At the instant the sticks are released, we have θ1 = 0 and θ2 = . Solving Eq. (6.105)
for θ¨1 and θ¨2 gives
θ
θ¨1 =
l
m
Fig. 6.39
2g
,
r
and
θ¨2 =
5g
.
r
(6.106)
6.3. Pendulum with an oscillating support
Let θ be defined as in Fig. 6.39. With x(t) = A cos(ωt), the position of the mass m is
given by
(X , Y )m = (x + sin θ , − cos θ ).
(6.107)
6.11 Solutions
Taking the derivative to obtain the velocity, we find that the square of the speed is
Vm2 = X˙ 2 + Y˙ 2 =
θ + x˙ 2 + 2 x˙ θ˙ cos θ ,
2 ˙2
(6.108)
which also follows from applying the law of cosines to the horizontal x˙ and tangential
θ˙ parts of the velocity vector. The Lagrangian is therefore
1
m( 2 θ˙ 2 + x˙ 2 + 2 x˙ θ˙ cos θ ) + mg cos θ .
2
The equation of motion for θ is
L=
d
(m 2 θ˙ + m x˙ cos θ) = −m x˙ θ˙ sin θ − mg sin θ
dt
=⇒
θ¨ + x¨ cos θ = −g sin θ .
(6.109)
(6.110)
Plugging in the explicit form of x(t), we have
θ¨ − Aω2 cos(ωt) cos θ + g sin θ = 0.
(6.111)
In retrospect, this makes sense. Someone in the reference frame of the support, which
has horizontal acceleration x¨ = −Aω2 cos(ωt), may as well be living in a world
where the acceleration from gravity has a component g downward and a component
Aω2 cos(ωt) to the right. Equation (6.111) is just the F = ma equation in the tangential
direction in this accelerating world.
A small-angle approximation in Eq. (6.111) gives
θ¨ + ω02 θ = aω2 cos(ωt),
√
(6.112)
where ω0 ≡ g/ and a ≡ A/ . This equation is simply the equation for a driven
oscillator, which we solved in Chapter 4. The solution is
θ(t) =
aω2
cos(ωt) + C cos(ω0 t + φ),
− ω2
ω02
(6.113)
where C and φ are determined by the initial conditions.
If ω happens to equal ω0 , then it appears that the amplitude goes to infinity. However,
as soon as the amplitude becomes large, our small-angle approximation breaks down,
and Eqs. (6.112) and (6.113) are no longer valid.
6.4. Two masses, one swinging
The Lagrangian is
1 2 1
(6.114)
m˙r + m(˙r 2 + r 2 θ˙ 2 ) − mgr + mgr cos θ .
2
2
The last two terms are the (negatives of the) potentials of each mass, relative to where
they would be if the right mass were located at the right pulley. The equations of motion
obtained from varying r and θ are
L=
2¨r = r θ˙ 2 − g(1 − cos θ ),
(6.115)
d 2
˙ = −gr sin θ .
(r θ)
dt
The first equation deals with the forces and accelerations along the direction of the
string. The second equation equates the torque from gravity with the change in angular
momentum of the right mass. If we do a (coarse) small-angle approximation and keep
only terms up to first order in θ, we find that at t = 0 (using the initial condition,
r˙ = 0), Eqs. (6.115) become
r¨ = 0,
θ¨ +
g
θ = 0.
r
(6.116)
257
258
The Lagrangian method
These equations say that the left mass stays still, and the right mass behaves just like
a pendulum.
If we want to find the leading term in the initial acceleration of the left mass (that
is, the leading term in r¨ ), we need to be a little less coarse in our approximation. So
let’s keep terms in Eq. (6.115) up to second order in θ . We then have at t = 0 (using
the initial condition, r˙ = 0)
2¨r = r θ˙ 2 −
θ¨ +
1 2
gθ ,
2
g
θ = 0.
r
(6.117)
The second equation still says that the right mass undergoes harmonic motion. We are
told that the amplitude is , so we have
θ (t) = cos(ωt + φ),
√
where ω = g/r. Plugging this into the first equation gives
2¨r =
2
g sin2 (ωt + φ) −
1
cos2 (ωt + φ) .
2
(6.118)
(6.119)
If we average this over a few periods, both sin2 α and cos2 α average to 1/2, so we find
m
θ
r¨avg =
2g
8
.
(6.120)
This is a small second-order effect. It is positive, so the left mass slowly begins to
climb.
l
y(t)
6.5. Inverted pendulum
Let θ be defined as in Fig. 6.40. With y(t) = A cos(ωt), the position of the mass m is
given by
(X , Y ) = ( sin θ , y + cos θ ).
Fig. 6.40
(6.121)
Taking the derivative to obtain the velocity, we find that the square of the speed is
V 2 = X˙ 2 + Y˙ 2 =
θ + y˙ 2 − 2 y˙ θ˙ sin θ,
2 ˙2
(6.122)
which also follows from applying the law of cosines to the vertical y˙ and tangential θ˙
parts of the velocity vector. The Lagrangian is therefore
L=
1
m( 2 θ˙ 2 + y˙ 2 − 2 y˙ θ˙ sin θ ) − mg(y + cos θ ).
2
(6.123)
The equation of motion for θ is
d
dt
∂L
∂ θ˙
=
∂L
∂θ
=⇒
θ¨ − y¨ sin θ = g sin θ .
(6.124)
Plugging in the explicit form of y(t), we have
θ¨ + sin θ Aω2 cos(ωt) − g = 0.
(6.125)
In retrospect, this makes sense. Someone in the reference frame of the support, which
has vertical acceleration y¨ = −Aω2 cos(ωt), may as well be living in a world where
the acceleration from gravity is g − Aω2 cos(ωt) downward. Equation (6.125) is just
the F = ma equation in the tangential direction in this accelerating world.
Assuming θ is small, we may set sin θ ≈ θ , which gives
θ¨ + θ aω2 cos(ωt) − ω02 = 0,
(6.126)
6.11 Solutions
θ
theta
0.1
θ
theta
1.75
1.5
1.25
1
0.75
0.5
0.25
0.05
0.2 0.4 0.6 0.8
1
1.2 1.4
tt
0.05
0.2 0.4 0.6 0.8
1
tt
1.2 1.4
0.1
Fig. 6.41
θ
theta
0.1
θ
theta
0.02
0.04
0.06
0.08
0.1
tt
0.0995
0.05
2
4
6
8
tt
0.099
0.0985
0.05
0.098
0.1
Fig. 6.42
√
where ω0 ≡ g/ and a ≡ A/ . Equation (6.126) cannot be solved exactly, but we
can still get a good idea of how θ depends on time. We can do this both numerically
and (approximately) analytically.
Figure 6.41 shows how θ depends on time for parameters with values = 1 m,
A = 0.1 m, and g = 10 m/s2 . So a = 0.1, and ω02 = 10 s−2 . We produced these plots
numerically using Eq. (6.126), with the initial conditions of θ (0) = 0.1 and θ˙ (0) = 0.
In the first plot, ω = 10 s−1 . And in the second plot, ω = 100 s−1 . The stick falls over
in first case, but it undergoes oscillatory motion in the second case. Apparently, if ω is
large enough the stick won’t fall over.
Let’s now explain this phenomenon analytically. At first glance, it’s rather surprising
that the stick stays up. It seems like the average (over a few periods of the ω oscillations)
of the tangential acceleration in Eq. (6.126), namely −θ (aω2 cos(ωt) − ω02 ), equals
the positive quantity θω02 , because the cos(ωt) term averages to zero (or so it appears).
So you might think that there is a net force making θ increase, causing the stick to
fall over.
The error in this reasoning is that the average of the −aω2 θ cos(ωt) term is not zero,
because θ undergoes tiny oscillations with frequency ω, as seen in the second plot in
Fig. 6.42. Both of these plots have a = 0.005, ω02 = 10 s−2 , and ω = 1000 s−1 (we’ll
work with small a and large ω from now on; more on this below). The second plot is
a zoomed-in version of the first one near t = 0. The important point here is that the
tiny oscillations in θ shown in the second plot are correlated with cos(ωt). It turns out
that the θ value at the t where cos(ωt) = 1 is larger than the θ value at the t where
cos(ωt) = −1. So there is a net negative contribution to the −aω2 θ cos(ωt) part of
the acceleration. And it may indeed be large enough to keep the pendulum up, as we
will now show.
To get a handle on the −aω2 θ cos(ωt) term, let’s work in the approximation where
ω02 ,
ω is large and a ≡ A/ is small. More precisely, we will assume a
1 and aω2
for reasons we will explain below. Look at one of the little oscillations in the second
plot in Fig. 6.42. These oscillations have frequency ω, because they are due to the
259
260
The Lagrangian method
support moving up and down. When the support moves up, θ increases; and when the
support moves down, θ decreases. Since the average position of the pendulum doesn’t
change much over one of these small periods, we can look for an approximate solution
to Eq. (6.126) of the form
θ (t) ≈ C + b cos(ωt),
(6.127)
where b
C. C will change over time, but on the scale of 1/ω it is essentially
constant, if a ≡ A/ is small enough. Plugging this guess for θ into Eq. (6.126), and
ω02 , we find −bω2 cos(ωt) + Caω2 cos(ωt) = 0, to leading
using a
1 and aω2
13
order. So we must have b = aC. Our approximate solution for θ is therefore
θ ≈ C 1 + a cos(ωt) .
(6.128)
Let’s now determine how C gradually changes with time. From Eq. (6.126), the average
acceleration of θ , over a period T = 2π/ω, is
θ¨ = −θ aω2 cos(ωt) − ω02
≈ −C 1 + a cos(ωt)
aω2 cos(ωt) − ω02
= −C a2 ω2 cos2 (ωt) − ω02
= −C
a2 ω2
− ω02
2
≡ −C
2
,
(6.129)
where
=
g
a2 ω2
− .
2
(6.130)
¨ Equating
But if we take two derivatives of Eq. (6.127), we see that θ¨ simply equals C.
this value of θ¨ with the one in Eq. (6.129) gives
¨ +
C(t)
2
C(t) ≈ 0.
(6.131)
This equation describes nice simple-harmonic motion. Therefore, C oscillates sinusoidally with the frequency given in Eq. (6.130). This is the overall back and
√ forth
motion seen in the first plot in Fig. 6.42. Note that we must have aω > 2ω0 if
this frequency is to be real so that the pendulum stays up. Since we have assumed
ω02 , which is consistent with our
a
1, we see that a2 ω2 > 2ω02 implies aω2
initial assumption above.
√
If aω
ω0 , then Eq. (6.130) gives ≈ aω/ 2. This is the case if we change the
setup and just have the pendulum lie flat on a horizontal table where the acceleration
from gravity is zero. In this limit where g is irrelevant, dimensional analysis implies
that the frequency of the C oscillations must be a multiple of ω, because ω is the only
13
The reasons for the a
1 and aω2
ω02 qualifications are the following. If aω2
ω02 , then
the aω2 cos(ωt) term dominates the ω02 term in Eq. (6.126). The one exception to this is when
cos(ωt) ≈ 0, but this occurs for a negligibly small amount of time if aω2
ω02 . If a
1, then we
¨
can legally ignore the C term when Eq. (6.127) is substituted into Eq. (6.126). This is true because
we will find below in Eq. (6.129) that our assumptions lead to C¨ being roughly proportional to
Ca2 ω2 . Since the other terms in Eq. (6.126) are proportional to Caω2 , we need a
1 in order
for the C¨ term to be negligible. In short, a
1 is the condition under which C varies slowly on
the time scale of 1/ω.
6.11 Solutions
quantity
√ in the problem with units of frequency. It just so happens that the multiple
is a/ 2.
As a double check that we haven’t messed up somewhere, the value resulting
from the parameters in Fig. 6.42 (namely a = 0.005, ω02 = 10 s−2 , and ω = 1000 s−1 )
√
is = 25/2 − 10 = 1.58 s−1 . This corresponds to a period of 2π/ ≈ 3.97 s. And
indeed, from the first plot in the figure, the period looks to be about 4 s (or maybe a
hair less). For more on the inverted pendulum, see Butikov (2001).
6.6. Minimum or saddle
(a) For the given ξ(t), the integrand in Eq. (6.26) is symmetric around the midpoint,
so we obtain
T /2
2
m 2
k 2T
t 2
m
dt =
−
.
(6.132)
−k
S=
T
T
2T
24
0
√
This is negative if T > 12m/k ≡ 2 3/ω. Since the period of the oscillation
√
is τ ≡ 2π/ω, we see that T must be greater than ( 3/π)τ in order for S to
be negative, assuming that we are using the given triangular function for ξ .
(b) With ξ(t) = sin(π t/T ), the integrand in Eq. (6.26) becomes
S=
1
2
T
m
0
π
cos(π t/T )
T
2
−k
sin(πt/T )
2
dt.
m 2π 2
k 2T
−
,
(6.133)
4T
4
2
2
where we have used the fact that the average value of sin θ and cos θ over half
of a period is 1/2 (or you can just do the integrals). This result for S is negative
if T > π m/k ≡ π/ω = τ/2, where τ is the period.
=
Remark: It turns out that the ξ(t) ∝ sin(π t/T ) function gives the best chance
of making S negative. You can show this by invoking a theorem from Fourier
analysis that says that any function satisfying ξ(0) = ξ(T ) = 0 can be written
as the sum ξ(t) = ∞
1 cn sin(nπ t/T ), where the cn are numerical coefficients.
When this sum is plugged into Eq. (6.26), you can show that all the cross terms
(terms involving two different values of n) integrate to zero. Using the fact that
the average value of sin2 θ and cos2 θ is 1/2, the rest of the integral yields
S=
1
4
∞
cn2
1
mπ 2 n2
− kT .
T
(6.134)
In order to obtain the smallest value of T that can make this sum negative, we
want only the n = 1 term to exist. We then have ξ(t) = c1 sin(πt/T ), and
Eq. (6.134) reduces to Eq. (6.133), as it should.
As mentioned in Remark 4 in Section 6.2, it is always possible to make S
positive by picking a ξ(t) function that is small but wiggles very fast. Therefore,
we see that for a harmonic oscillator, if T > τ/2, then the stationary value of
S is a saddle point (some ξ ’s make S positive, and some make it negative),
but if T < τ/2, then the stationary value of S is a minimum (all ξ ’s make S
positive). In the latter case, the point is that T is small enough so that there is
no way for ξ to get large, without making ξ˙ large also. ♣
6.7. Normal force from a plane
First solution: The most convenient coordinates in this problem are w and z, where
w is the distance upward along the plane, and z is the distance perpendicularly away
from it. The Lagrangian is then
1
m(w˙ 2 + z˙ 2 ) − mg(w sin θ + z cos θ ) − V (z),
2
(6.135)
261
262
The Lagrangian method
where V (z) is the (very steep) constraining potential. The two equations of motion are
mw¨ = −mg sin θ ,
dV
.
(6.136)
dz
At this point we invoke the constraint z = 0. So z¨ = 0, and the second equation gives
m¨z = −mg cos θ −
y
Fc ≡ −V (0) = mg cos θ ,
(6.137)
as desired. We also obtain the usual result, w¨ = −g sin θ .
x
m
θ
Fig. 6.43
Second solution: We can also solve this problem by using the horizontal and vertical
coordinates, x and y. We’ll choose (x, y) = (0, 0) to be at the top of the plane; see
Fig. 6.43. The (very steep) constraining potential is V (z), where z ≡ x sin θ + y cos θ
is the distance from the mass to the plane (as you can verify). The Lagrangian is then
1
L = m(˙x2 + y˙ 2 ) − mgy − V (z).
(6.138)
2
Keeping in mind that z ≡ x sin θ + y cos θ, the two equations of motion are (using the
chain rule)
dV ∂z
m¨x = −
= −V (z) sin θ ,
dz ∂x
(6.139)
dV ∂z
m¨y = −mg −
= −mg − V (z) cos θ.
dz ∂y
At this point we invoke the constraint condition z = 0 =⇒ x = −y cot θ . This
condition, along with the two E–L equations, allows us to solve for the three unknowns,
x¨ , y¨ , and V (0). Using x¨ = −¨y cot θ in Eq. (6.139), we find
x¨ = g cos θ sin θ ,
y¨ = −g sin2 θ ,
Fc ≡ −V (0) = mg cos θ .
(6.140)
The first two results here are simply the horizontal and vertical components of the
acceleration along the plane, which is g sin θ .
6.8. Bead on a stick
There is no potential energy here, so the Lagrangian consists of just the kinetic energy,
T , which comes from the radial and tangential motions:
1
1
(6.141)
L = T = m˙r 2 + mr 2 ω2 .
2
2
Equation (6.52) therefore gives
1
1
E = m˙r 2 − mr 2 ω2 .
(6.142)
2
2
Claim 6.3 says that this quantity is conserved, because ∂L/∂t = 0. But it is not the
energy of the bead, due to the minus sign in the second term.
The point here is that in order to keep the stick rotating at a constant angular speed,
there must be an external force acting on it. This force in turn causes work to be done
on the bead, thereby increasing its kinetic energy. The kinetic energy T is therefore not
conserved. From Eqs. (6.141) and (6.142), we see that E = T − mr 2 ω2 is the quantity
that is constant in time. See Exercise 6.39 for some F = ma ways to show that the
quantity E in Eq. (6.142) is conserved.
6.9. Atwood’s machine
First solution: If the left mass goes up by x and the right mass goes up by y, then
conservation of string says that the middle mass must go down by x + y. Therefore,
the Lagrangian of the system is
1
1
1
L = (4m)˙x2 + (3m)(−˙x − y˙ )2 + m˙y2 − (4m)gx + (3m)g(−x − y) + mgy
2
2
2
7
(6.143)
= m˙x2 + 3m˙xy˙ + 2m˙y2 − mg(x − 2y).
2
6.11 Solutions
263
This is invariant under the transformation x → x + 2 and y → y + . Hence, we can
use Noether’s theorem, with Kx = 2 and Ky = 1. The conserved momentum is then
P=
∂L
∂L
Kx +
Ky = m(7˙x + 3˙y)(2) + m(3˙x + 4˙y)(1) = m(17˙x + 10˙y).
∂ x˙
∂ y˙
(6.144)
This P is constant. In particular, if the system starts at rest, then x˙ always equals
−(10/17)˙y.
Second solution:
The Euler–Lagrange equations are, from Eq. (6.143),
7m¨x + 3m¨y = −mg,
(6.145)
3m¨x + 4m¨y = 2mg.
Adding the second equation to twice the first gives
17m¨x + 10m¨y = 0
=⇒
d
17m˙x + 10m˙y = 0.
dt
(6.146)
Third solution: We can also solve this problem using F = ma. Since the tension T
is the same throughout the rope, we see that the three F = dP/dt equations are
2T − 4mg =
dP4m
,
dt
2T − 3mg =
dP3m
,
dt
2T − mg =
dPm
.
dt
(6.147)
The three forces depend on only two quantities (T and mg), so there must be some
combination of them that adds up to zero. If we set a(2T − 4mg) + b(2T − 3mg) +
c(2T − mg) = 0, then we have a + b + c = 0 and 4a + 3b + c = 0, which is satisfied
by a = 2, b = −3, and c = 1. Therefore,
d
(2P4m − 3P3m + Pm )
dt
d
2(4m)˙x − 3(3m)(−˙x − y˙ ) + m˙y
=
dt
d
= (17m˙x + 10m˙y).
dt
0=
(6.148)
6.10. Hoop and pulley
Let the radius to M make an angle θ with the vertical (see Fig. 6.44). Then the coordinates of M relative to the center of the hoop are R(sin θ, − cos θ ). The height of
m, relative to its position when M is at the bottom of the hoop, is y = −Rθ . The
Lagrangian is therefore (and yes, we’ve chosen a different y = 0 reference point for
each mass, but such a definition changes the potential only by a constant amount,
which is irrelevant)
L=
1
(M + m)R2 θ˙ 2 + MgR cos θ + mgRθ .
2
Fig. 6.44
(6.150)
This is just F = ma along the direction of the string (because Mg sin θ is the tangential
component of the gravitational force on M ).
Equilibrium occurs when θ˙ = θ¨ = 0. From Eq. (6.150), we see that this happens at
sin θ0 = m/M . Letting θ ≡ θ0 + δ, and expanding Eq. (6.150) to first order in δ, gives
δ¨ +
Mg cos θ0
δ = 0.
(M + m)R
θ
(6.149)
The equation of motion is then
(M + m)Rθ¨ = g(m − M sin θ ).
m
(6.151)
R
M
264
The Lagrangian method
The frequency of small oscillations is therefore
ω=
M cos θ0
M +m
where we have used cos θ0 =
g
=
R
M −m
M +m
1/4
g
,
R
(6.152)
1 − sin θ02 .
√
Remarks: If M
m, then θ0 ≈ 0, and ω ≈ g/R. This makes sense, because m
can be ignored, so M essentially oscillates around the bottom of the hoop, just like a
pendulum of length R.
If M is only slightly greater than m, then θ0 ≈ π/2, and ω ≈ 0. This also makes
sense, because if θ ≈ π/2, then the restoring force g(m − M sin θ ) doesn’t change
much as θ changes (the derivative of sin θ is zero at θ = π/2), so it’s as if we have a
pendulum in a very weak gravitational field.
We can actually derive the frequency in Eq. (6.152) without doing any calculations. Look at M at the equilibrium position. The tangential forces on it cancel, and
the radially inward force from the hoop must be Mg cos θ0 to balance the radial outward component of the gravitational force. Therefore, for all the mass M knows, it
is sitting at the bottom of a hoop of radius R in a world where gravity has strength
g = g cos θ0 . The general formula for the frequency of a pendulum (as you can
quickly show) is ω = F /M R, where F is the gravitational force (which is Mg
here), and M is the total mass being accelerated (which is M + m here). This gives the
ω in Eq. (6.152). (This reasoning is a little subtle; food for thought.) ♣
6.11. Bead on a rotating hoop
Breaking the velocity up into the component along the hoop plus the component
perpendicular to the hoop, we find
1
m(ω2 R2 sin2 θ + R2 θ˙ 2 ) + mgR cos θ.
2
The equation of motion is then
L=
Rθ¨ = sin θ (ω2 R cos θ − g).
(6.153)
(6.154)
The F = ma interpretation of this is that the component of gravity pulling downward
along the hoop accounts for the acceleration along the hoop plus the component of the
centripetal acceleration along the hoop.
Equilibrium occurs when θ˙ = θ¨ = 0. The right-hand side of Eq. (6.154) equals zero
when either sin θ = 0 (that is, θ = 0 or θ = π ) or cos θ = g/(ω2 R). Since cos θ must
be less than or equal to 1, this second condition is possible only if ω2 ≥ g/R. So we
have two cases:
• If ω2 < g/R, then θ = 0 and θ = π are the only equilibrium points.
The θ = π case is unstable. This is fairly intuitive, but it can also be seen
mathematically by letting θ ≡ π + δ, where δ is small. Equation (6.154) then
becomes
δ¨ − δ(ω2 + g/R) = 0.
(6.155)
The coefficient of δ is negative, so δ undergoes exponential instead of oscillatory
motion.
The θ = 0 case turns out to be stable. For small θ , Eq. (6.154) becomes
θ¨ + θ (g/R − ω2 ) = 0.
(6.156)
The coefficient of θ is positive, so we have sinusoidal solutions. The frequency
√
of small oscillations is g/R − ω2 . This goes to zero as ω → g/R.
2
2
• If ω ≥ g/R, then θ = 0, θ = π, and cos θ0 ≡ g/(ω R) are all equilibrium
points. The θ = π case is again unstable, by looking at Eq. (6.155). And the
6.11 Solutions
265
θ = 0 case is also unstable, because the coefficient of θ in Eq. (6.156) is now
negative (or zero, if ω2 = g/R).
Therefore, cos θ0 ≡ g/(ω2 R) is the only stable equilibrium. To find the frequency of small oscillations, let θ ≡ θ0 + δ in Eq. (6.154), and expand to first
order in δ. Using cos θ0 ≡ g/(ω2 R), we find
δ¨ + ω2 sin2 θ0 δ = 0.
(6.157)
ω2
− g 2 /ω2 R2 .
The frequency of small√oscillations is therefore ω sin θ0 =
The frequency ω = g/R is the critical frequency above which there is a stable
equilibrium at θ = 0, that is, above which the mass wants to move away from the
bottom of the hoop.
√
Remark: This frequency of small oscillations goes to zero as ω → g/R.
And it approximately equals ω as ω → ∞. This second limit can be viewed
in the following way. For very large ω, gravity isn’t important, and the bead
feels a centripetal force (the normal force from the hoop) essentially equal to
mω2 R as it moves near θ = π/2. So for all the bead knows, it is a pendulum of
length R in a world where “gravity” pulls sideways with a force mω2 R ≡ mg
(outward, so that it is approximately canceled by the inward-pointing normal
force, just as the downward gravitational force is approximately canceled by the
upward tension in a regular pendulum). The frequency of such a pendulum is
g /R = ω2 R/R = ω. ♣
6.12. Another bead on a rotating hoop
With the angles ωt and θ defined as in Fig. 6.45, the Cartesian coordinates for the bead
are
(x, y) = R cos ωt + r cos(ωt + θ), R sin ωt + r sin(ωt + θ ) .
θ
R
ωt
(6.158)
The velocity is then
Fig. 6.45
(x, y) = −ωR sin ωt − r(ω + θ˙ ) sin(ωt + θ ),
ωR cos ωt + r(ω + θ˙ ) cos(ωt + θ ) .
(6.159)
The square of the speed is therefore
˙ 2
v 2 = R2 ω2 + r 2 (ω + θ)
˙ sin ωt sin(ωt + θ ) + cos ωt cos(ωt + θ )
+ 2Rrω(ω + θ)
˙ 2 + 2Rrω(ω + θ)
˙ cos θ .
= R2 ω2 + r 2 (ω + θ)
(6.160)
This speed can also be obtained by using the law of cosines to add the velocity
of the center of the hoop to the velocity of the bead with respect to the center (as
you can show).
There is no potential energy, so the Lagrangian is simply L = mv 2 /2. The equation
of motion is then, as you can show,
r θ¨ + Rω2 sin θ = 0.
(6.161)
Equilibrium occurs when θ˙ = θ¨ = 0, so Eq. (6.161) tells us that the equilibrium
is located at θ = 0, which makes intuitive sense. (Another solution is θ = π , but
that’s an unstable equilibrium.) A small-angle approximation in Eq.
√ (6.161) gives
θ¨ + (R/r)ω2 θ = 0, so the frequency of small oscillations is = ω R/r.
Remarks: If R
r, then ≈ 0. This makes sense, because the frictionless hoop
is essentially not moving. If R = r, then
=√ω. If R
r, then is very large.
In this case, we can double-check the
= ω R/r result in the following way. In
m
(top view)
r
266
The Lagrangian method
the accelerating frame of the hoop, the bead feels a centrifugal force (discussed in
Chapter 10) of m(R + r)ω2 . For all the bead knows, it is in a gravitational field with
strength g ≡ (R+r)ω2 . So the bead (which acts like a pendulum of length r), oscillates
with a frequency equal to
g
=
r
R
m
θ
M
(R + r)ω2
R
≈ω
r
r
(for R
r).
(6.162)
Note that if we try to use this “effective gravity” argument as a double check for smaller
values of R, we√
get the wrong answer. For example, if√R = r, we obtain an oscillation
frequency of ω 2R/r, instead of the correct value ω R/r. This is because in reality
the centrifugal force fans out near the equilibrium point, while our “effective gravity”
argument assumes that the field lines are parallel (and so it gives a frequency that is
too large). ♣
6.13. Mass on a wheel
Let the angle θ be defined as in Fig. 6.46, with the convention that θ is positive if M is
to the right of m. Then the position of m in Cartesian coordinates, relative to the point
where m would be in contact with the ground, is
Fig. 6.46
(x, y)m = R(θ − sin θ , 1 − cos θ ).
(6.163)
We have used the nonslipping condition to say that the present contact point is a
distance Rθ to the right of where m would be in contact with the ground. Differentiating
2 = 2R2 θ˙ 2 (1 − cos θ ).
Eq. (6.163), we find that the square of m’s speed is vm
2 = R2 θ˙ 2 .
The position of M is (x, y)M = R(θ , 1), so the square of its speed is vM
The Lagrangian is therefore
L=
1
MR2 θ˙ 2 + mR2 θ˙ 2 (1 − cos θ ) + mgR cos θ ,
2
(6.164)
where we have measured both potential energies relative to the height of M . The
equation of motion is
MRθ¨ + 2mRθ¨ (1 − cos θ ) + mRθ˙ 2 sin θ + mg sin θ = 0.
(6.165)
In the case of small oscillations, we can use cos θ ≈
≈ θ . The second
and third terms in Eq. (6.165) are then third order in θ and can be neglected (basically,
the middle term in Eq. (6.164), which is the kinetic energy of m, is negligible), so we
find
mg
θ = 0.
(6.166)
θ¨ +
MR
1−θ 2 /2 and sin θ
The frequency of small oscillations is therefore
ω=
m
M
g
.
R
(6.167)
Remarks: If M
m, then ω → 0. This makes sense. If m
M , then ω → ∞.
This also makes sense, because the huge mg force makes the situation similar to one
where the wheel is bolted to the ground, in which case the wheel vibrates with a high
frequency.
Equation (6.167) can actually be derived in a much quicker way, using torque. For
small oscillations, the gravitational force on m produces a torque of −mgRθ around
the contact point on the ground. For small θ, m has essentially no moment of inertia
around the contact point, so the total moment of inertia is simply MR2 . Therefore,
τ = I α gives −mgRθ = MR2 θ¨ , from which the result follows. ♣
6.11 Solutions
267
6.14. Pendulum with a free support
Let x be the coordinate of M , and let θ be the angle of the pendulum (see Fig. 6.47).
Then the position of the mass m in Cartesian coordinates is (x + sin θ, − cos θ ).
Taking the derivative to find the velocity, and then squaring to find the speed, gives
2 = x˙ 2 + 2 θ˙ 2 + 2 x˙ θ˙ cos θ. The Lagrangian is therefore
vm
L=
1
1
M x˙ 2 + m(˙x2 +
2
2
2 ˙2
θ + 2 x˙ θ˙ cos θ ) + mg cos θ .
x
M
θ l
(6.168)
The equations of motion obtained from varying x and θ are
(M + m)¨x + m θ¨ cos θ − m θ˙2 sin θ = 0,
θ¨ + x¨ cos θ + g sin θ = 0.
m
(6.169)
If θ is small, we can use the small-angle approximations, cos θ ≈ 1 − θ 2 /2 and
sin θ ≈ θ. Keeping only the terms that are first-order in θ , we obtain
(M + m)¨x + m θ¨ = 0,
x¨ + θ¨ + gθ = 0.
(6.170)
The first equation expresses momentum conservation. Integrating it twice gives
x=−
m
θ + At + B.
M +m
(6.171)
The second equation is F = ma in the tangential direction. Eliminating x¨ from
Eq. (6.170) gives
θ¨ +
M +m g
θ = 0.
M
(6.172)
Therefore, θ (t) = C cos(ωt + φ), where
ω=
1+
m
M
g
.
(6.173)
The general solutions for θ and x are therefore
θ(t) = C cos(ωt + φ),
x(t) = −
Cm
cos(ωt + φ) + At + B.
M +m
(6.174)
The constant B is irrelevant, so we’ll ignore it. The two normal modes are:
• A = 0: In this case, x = −θ m /(M +m). Both masses oscillate with the frequency
ω given in Eq. (6.173), always moving in opposite directions. The center of mass
does not move (as you can verify).
• C = 0: In this case, θ = 0 and x = At. The pendulum hangs vertically, with both
masses moving horizontally at the same speed. The frequency of oscillations is
zero in this mode.
√
Remarks: If M
m, then ω = g/ , as expected, because the support essentially
stays still.
√
√
If m
M , then ω → m/M g/ → ∞. This makes sense, because the tension in the rod is very large. We can actually be quantitative about this limit. For
small oscillations and for m
M , the tension of mg in the rod produces a sideways force of mgθ on M . So the horizontal F = Ma equation for M is mgθ = M x¨ .
But x ≈ − θ in this limit, so we have mgθ = −M θ¨ , which gives the desired
frequency. ♣
Fig. 6.47
268
The Lagrangian method
6.15. Pendulum support on an inclined plane
Let z be the coordinate of M along the plane, and let θ be the angle of the pendulum
(see Fig. 6.48). In Cartesian coordinates, the positions of M and m are
M
z
θ l
m
(x, y)M = (z cos β, −z sin β),
(x, y)m = (z cos β + sin θ , −z sin β − cos θ ).
β
(6.175)
Differentiating these positions, we find that the squares of the speeds are
2
vM
= z˙ 2 ,
Fig. 6.48
2
vm
= z˙ 2 + 2 θ˙ 2 + 2 z˙ θ˙ (cos β cos θ − sin β sin θ ).
(6.176)
The Lagrangian is therefore
1
1
M z˙ 2 + m z˙ 2 +
2
2
θ + 2 z˙ θ˙ cos(θ + β) + Mgz sin β + mg(z sin β + cos θ ).
2 ˙2
(6.177)
The equations of motion obtained from varying z and θ are
(M + m)¨z + m
θ¨ cos(θ + β) − θ˙ 2 sin(θ + β) = (M + m)g sin β,
θ¨ + z¨ cos(θ + β) = −g sin θ.
(6.178)
Let us now consider small oscillations about the equilibrium point (where θ¨ = θ˙ = 0).
We must first determine where this point is. The first equation above gives z¨ = g sin β.
The second equation then gives g sin β cos(θ + β) = −g sin θ. By expanding the
cosine term, we find tan θ = − tan β, so θ = −β. (θ = π − β is also a solution, but
this is an unstable equilibrium.) The equilibrium position of the pendulum is therefore
where the string is perpendicular to the plane.14
To find the normal modes and frequencies for small oscillations, let θ ≡ −β + δ,
and expand Eq. (6.178) to first order in δ. Letting η¨ ≡ z¨ − g sin β for convenience, we
obtain
(M + m)η¨ + m δ¨ = 0,
(6.179)
η¨ + δ¨ + (g cos β)δ = 0.
Using the determinant method (or using the method in Problem 6.14; either way works),
we find the frequencies of the normal modes to be
ω1 = 0,
and
ω2 =
1+
m
M
g cos β
.
(6.180)
These are the same as the frequencies in the previous problem (where M moves horizontally), but with g cos β in place of g; compare Eq. (6.179) with Eq. (6.170).15
Looking at Eq. (6.174), and recalling the definition of η, we see that the general
solutions for θ and z are
g sin β 2
Cm
cos(ωt + φ) +
t + At + B.
θ(t) = −β + C cos(ωt + φ), z(t) = −
M +m
2
(6.181)
14
15
This makes sense. The tension in the string is perpendicular to the plane, so for all the pendulum
bob knows, it may as well be sliding down a plane parallel to the given one, a distance away.
Given the same initial speed, the two masses slide down their two “planes” with equal speeds at
all times.
This makes sense, because in a frame that accelerates down the plane at g sin β, the only external
force on the masses is a gravitational force of g cos β perpendicular to the plane. As far as M
and m are concerned, they live in a world where gravity pulls “downward” (perpendicular to the
plane) with strength g = g cos β.
6.11 Solutions
The constant B is irrelevant, so we’ll ignore it. The basic difference between these
normal modes and the ones in the previous problem is the acceleration down the plane.
If you go to a frame that accelerates down the plane at g sin β, and if you tilt your head
at an angle β and accept the fact that g = g cos β in your world, then the setup is
identical to the one in the previous problem.
6.16. Tilting plane
Relative to the support, the positions of the masses are
(x, y)M = ( sin θ, − cos θ),
(x, y)m = ( sin θ + x cos θ , − cos θ + x sin θ ).
(6.182)
Differentiating these positions, we find that the squares of the speeds are
2
vM
= 2 θ˙ 2 ,
2
vm
= ( θ˙ + x˙ )2 + x2 θ˙ 2 .
(6.183)
by noting that ( θ˙ + x˙ ) is the speed along the long rod, and xθ˙
You can also obtain
is the speed perpendicular to it. The Lagrangian is
2
vm
L=
1
M
2
2 ˙2
θ +
1
m ( θ˙ + x˙ )2 + x2 θ˙ 2 + Mg cos θ + mg( cos θ − x sin θ ).
2
(6.184)
The equations of motion obtained from varying x and θ are
θ¨ + x¨ = xθ˙ 2 − g sin θ ,
M
2¨
θ + m ( θ¨ + x¨ ) + mx2 θ¨ + 2mx˙xθ˙ = −(M + m)g sin θ − mgx cos θ.
(6.185)
Let us now consider the case where both x and θ are small (or more precisely, θ
and x/
1). Expanding Eq. (6.185) to first order in θ and x/ gives
( θ¨ + x¨ ) + gθ = 0,
M ( θ¨ + gθ) + m ( θ¨ + x¨ ) + mg θ + mgx = 0.
1
(6.186)
We can simplify these a bit. Using the first equation to substitute −gθ for ( θ¨ + x¨ ),
and also −¨x for ( θ¨ + gθ), in the second equation gives
θ¨ + x¨ + gθ = 0,
(6.187)
−M x¨ + mgx = 0.
The normal modes can be found using the determinant method, or we can find them just
by inspection.
√ The second equation says that either x(t) ≡ 0, or x(t) = A cosh(αt +β),
where α = mg/M . So we have two cases:
• If x(t) = 0, then the first equation in (6.187) says that the normal mode is
θ
x
=B
1
0
cos(ωt + φ),
(6.188)
√
where ω ≡ g/ . This mode is fairly clear. With the proper initial conditions,
m will stay right where M is. The normal force from the long rod will be exactly
what is needed in order for m to undergo the same oscillatory motion as M . The
two masses may as well be two pendulums of length swinging side by side.
• If x(t) = A cosh(αt + β), then the first equation in (6.187) can be solved (by
guessing a particular solution for θ of the same form) to give the normal mode,
θ
x
=C
−m
(M + m)
cosh(αt + β),
(6.189)
√
where α = mg/M . This mode is not as clear. And indeed, its range of validity
is rather limited. The exponential behavior will quickly make x and θ large, and
269
270
The Lagrangian method
thus outside the validity of our small-variable approximations. You can show that
in this mode the center of mass remains directly below the pivot. This can occur,
for example, by having m move down to the right as the rods rotate and swing M
up to the left. There is no oscillation in this mode; the positions keep growing.
The CM falls, to provide for the increasing kinetic energy.
6.17. Rotating curve
The speed along the curve is x˙ 1 + y 2 , and the speed perpendicular to the curve is
ωx. So the Lagrangian is
L=
1
m ω2 x2 + x˙ 2 (1 + y 2 ) − mgy,
2
(6.190)
where y(x) = b(x/a)λ . The equation of motion is then
d
dt
∂L
∂ x˙
=
∂L
∂x
=⇒
x¨ (1 + y 2 ) + x˙ 2 y y = ω2 x − gy .
(6.191)
Equilibrium occurs when x˙ = x¨ = 0, so Eq. (6.191) says that the equilibrium value of
x satisfies
x0 =
gy (x0 )
.
ω2
(6.192)
The F = ma explanation for this (writing y (x0 ) as tan θ , where θ is the angle of the
curve, and then multiplying through by ω2 cos θ ) is that the component of gravity along
the curve accounts for the component of the centripetal acceleration along the curve.
Using y(x) = b(x/a)λ , Eq. (6.192) yields
x0 = a
a2 ω2
λgb
1/(λ−2)
.
(6.193)
As λ → ∞, we see that x0 goes to a. This makes sense, because the curve essentially
equals zero up to a, and then it rises very steeply. You can check numerous other limits.
Letting x ≡ x0 + δ in Eq. (6.191), and expanding to first order in δ, gives
δ¨ 1 + y (x0 )2 = δ ω2 − gy (x0 ) .
(6.194)
The frequency of small oscillations is therefore given by
2
=
gy (x0 ) − ω2
.
1 + y (x0 )2
(6.195)
Using the explicit form of y, along with Eq. (6.193), we find
2
=
(λ − 2)ω2
1+
a2 ω 4
g2
a2 ω2
λgb
2/(λ−2)
.
(6.196)
We see that λ must be greater than 2 in order for there to be oscillatory motion around
the equilibrium point. For λ < 2, the equilibrium point is unstable, that is, to the left
the force is inward, and to the right the force is outward.
For the case λ = 2, we have y(x) = b(x/a)2 , so the equilibrium condition,
Eq. (6.192), gives x0 = (2gb/a2 ω2 )x0 . For this to be true for some x0 , we must
have ω2 = 2gb/a2 . But if this holds, then Eq. (6.192) is true for all x. So in the special
case of λ = 2, the bead happily sits anywhere on the curve if ω2 = 2gb/a2 . (In the
rotating frame of the curve, the tangential components of the centrifugal and gravitational forces exactly cancel at all points.) If λ = 2 and ω2 = 2gb/a2 , then the particle
feels a force either always inward or always outward.
Remarks: For ω → 0, Eqs. (6.193) and (6.196) give x0 → 0 and → 0. And for
→ 0. In both cases
→ 0, because in both
ω → ∞, they give x0 → ∞ and
6.11 Solutions
cases the equilibrium position is at a place where the curve is very flat (horizontally
or vertically, respectively), and the restoring force ends up being small.
For λ → ∞, we have x0 → a and → ∞. The frequency is large here because the
equilibrium position at a is where the curve has a sharp corner, so the restoring force
changes quickly with position. Or, you can think of it as a pendulum with a very small
length, if you approximate the “corner” by a tiny circle. ♣
6.18. Motion in a cone
If the particle’s distance from the axis is r, then its height is r/ tan α, and its distance
up along the cone is r/ sin α. Breaking the velocity into components up along the cone
and around the cone, we see that the square of the speed is v 2 = r˙ 2 / sin2 α + r 2 θ˙ 2 . The
Lagrangian is therefore
L=
mgr
r˙ 2
1
+ r 2 θ˙ 2 −
m
.
2
tan α
sin2 α
(6.197)
The equations of motion obtained from varying θ and r are
d
˙ =0
(mr 2 θ)
dt
(6.198)
r¨ = r θ˙ 2 sin2 α − g cos α sin α.
The first of these equations expresses conservation of angular momentum. The second
equation is more transparent if we divide through by sin α. With x ≡ r/ sin α being
the distance up along the cone, we have x¨ = (r θ˙ 2 ) sin α − g cos α. This is the F = ma
statement for the diagonal x direction.
Letting mr 2 θ˙ ≡ L, we can eliminate θ˙ from the second equation to obtain
L2 sin2 α
− g cos α sin α.
m2 r 3
We will now calculate the two desired frequencies.
r¨ =
(6.199)
• Frequency of circular oscillations, ω: For circular motion with r = r0 , we have
r˙ = r¨ = 0, so the second of Eqs. (6.198) gives
g
.
r0 tan α
ω ≡ θ˙ =
(6.200)
• Frequency of oscillations about a circle,
: If the orbit were actually the circle
r = r0 , then Eq. (6.199) would give (with r¨ = 0)
L2 sin2 α
= g cos α sin α.
m2 r03
(6.201)
˙
This is equivalent to Eq. (6.200), which can be seen by writing L as mr02 θ.
We will now use our standard procedure of letting r(t) = r0 + δ(t), where δ(t)
is very small, and then plugging this into Eq. (6.199) and expanding to first order
in δ. Using
3δ
1
1
1
1
≈ 3
= 3
≈ 3 1−
(r0 + δ)3
r0
r0 + 3r02 δ
r0 (1 + 3δ/r0 )
r0
,
(6.202)
we have
δ¨ =
L2 sin2 α
m2 r03
1−
3δ
r0
− g cos α sin α.
(6.203)
Recalling Eq. (6.201), the terms not involving δ cancel, and we are left with
δ¨ = −
3L2 sin2 α
m2 r04
δ.
(6.204)
271
272
The Lagrangian method
Using Eq. (6.201) again to eliminate L we have
δ¨ +
3g
sin α cos α δ = 0.
r0
(6.205)
3g
sin α cos α.
r0
(6.206)
Therefore,
=
Having found the two desired frequencies in Eqs. (6.200) and (6.206), we see that their
ratio is
√
= 3 sin α.
(6.207)
ω
√
This ratio /ω is independent of r0 . The two frequencies are equal if sin α = 1/ 3,
◦
˜ If α = α,
˜ then after one revolution around the cone, r
that is, if α ≈ 35.3 ≡ α.
returns to the value it had at the beginning of the revolution. So the particle undergoes
periodic motion.
θ1
Remarks: In the limit α → 0 (that is, the cone is very thin), Eq. (6.207) says that
/ω → 0. In fact, Eqs. (6.200) and (6.206) say that ω → ∞ and
→ 0. So the
particle spirals around many times during one complete r cycle. This seems intuitive.
In the limit α → π/2 (that is, the cone
√ is almost a flat plane), both ω and go to
zero, and Eq.√(6.207) says that /ω → 3. This result is not at all obvious.
If /ω = 3 sin α is a rational number, then the particle undergoes periodic motion.
so it takes two complete circles for r to go
For example, if α = 60◦ , then /ω = 3/2,√
through three cycles. Or, if α = arcsin(1/2 3) ≈ 16.8◦ , then /ω = 1/2, so it takes
two complete circles for r to go through one cycle. ♣
l1
m1
θ 2 l2
m2
Fig. 6.49
6.19. Double pendulum
Relative to the pivot point, the Cartesian coordinates of m1 and m2 are, respectively
(see Fig. 6.49),
(x, y)1 = ( 1 sin θ1 , − 1 cos θ1 ),
(x, y)2 = ( 1 sin θ1 + 2 sin θ2 , − 1 cos θ1 − 2 cos θ2 ).
(6.208)
Taking the derivative to find the velocities, and then squaring, gives
v12 = 21 θ˙12 ,
v22 = 21 θ˙12 + 22 θ˙22 + 2 1 2 θ˙1 θ˙2 (cos θ1 cos θ2 + sin θ1 sin θ2 ).
(6.209)
The Lagrangian is therefore
L=
1
1
m1 21 θ˙12 + m2 21 θ˙12 + 22 θ˙22 + 2 1 2 θ˙1 θ˙2 cos(θ1 − θ2 )
2
2
+ m1 g 1 cos θ1 + m2 g( 1 cos θ1 + 2 cos θ2 ).
(6.210)
The equations of motion obtained from varying θ1 and θ2 are
0 = (m1 + m2 ) 21 θ¨1 + m2 1 2 θ¨2 cos(θ1 − θ2 ) + m2 1 2 θ˙22 sin(θ1 − θ2 )
+ (m1 + m2 )g 1 sin θ1 ,
0 = m2 22 θ¨2 + m2 1 2 θ¨1 cos(θ1 − θ2 ) − m2 1 2 θ˙12 sin(θ1 − θ2 )
+ m2 g 2 sin θ2 .
(6.211)
6.11 Solutions
This is a bit of a mess, but it simplifies greatly if we consider small oscillations.
Using the small-angle approximations and keeping only the leading-order terms,
we obtain
0 = (m1 + m2 ) 1 θ¨1 + m2 2 θ¨2 + (m1 + m2 )gθ1 ,
0 = 2 θ¨2 + 1 θ¨1 + gθ2 .
(6.212)
Consider now the special case, 1 = 2 ≡ . We can find the frequencies of the
normal modes by using the determinant method, discussed in Section 4.5. You
can show that the result is
ω± =
m1 m2 + m22
m1 + m2 ±
m1
g
.
The normal modes turn out to be, after some simplification,
√
∓ m2
θ1 (t)
∝ √
cos(ω± t + φ± ).
θ2 (t) ±
m1 + m2
(6.213)
(6.214)
Some special cases are:
• m1 = m2 : The frequencies are
ω± =
2±
√
g
.
(6.215)
cos(ω± t + φ± ).
(6.216)
2
The normal modes are
θ1 (t)
θ2 (t)
• m1
±
∝
∓1
√
2
m2 : With m2 /m1 ≡ , the frequencies are (to leading nontrivial order
in )
ω± = (1 ±
The normal modes are
θ1 (t)
θ2 (t)
√
/2)
g
.
(6.217)
cos(ω± t + φ± ).
(6.218)
√
±
∝
∓
1
In both modes, the upper (heavy) mass essentially stands still, and the lower (light)
mass oscillates like a pendulum of length .
• m1
m2 : With m1 /m2 ≡ , the frequencies are (to leading order in )
2g
ω+ =
,
ω− =
g
.
2
(6.219)
The normal modes are
θ1 (t)
θ2 (t)
±
∝
∓1
1
cos(ω± t + φ± ).
(6.220)
In the first mode, the lower (heavy) mass essentially stands still (from the x2 in
Eq. (6.208)), and the upper (light) mass vibrates back and forth at a high frequency
(because there is a very large tension in the rods). In the second mode, the rods
form a straight line, and the system is essentially a pendulum of length 2 .
273
274
The Lagrangian method
Consider now the special case, m1 = m2 . Using the determinant method, you can
show that the frequencies of the normal modes are
ω± =
2+ 2
1
2
1+ 2±
√
g
.
(6.221)
1 2
The normal modes turn out to be, after some simplification,
θ1 (t)
∝
θ2 (t) ±
2
2+ 2
1
2
2− 1∓
cos(ω± t + φ± ).
(6.222)
Some special cases are:
• 1 = 2 : We already considered this case above. You can show that Eqs. (6.221)
•
and (6.222) agree with Eqs. (6.215) and (6.216), respectively.
1
2 : With 2 / 1 ≡ , the frequencies are (to leading order in )
2g
ω+ =
,
2
ω− =
g
.
(6.223)
1
The normal modes are
θ1 (t)
θ2 (t)
θ1 (t)
θ2 (t)
+
−
∝
−
2
∝
1
1
cos(ω+ t + φ+ ),
(6.224)
cos(ω− t + φ− ).
In the first mode, the masses essentially move equal distances in opposite directions, at a high frequency (assuming 2 is small). The factor of 2 in the frequency
arises because the angle of 2 is twice what it would be if m1 were bolted in
place; so m2 feels double the tangential force. In the second mode, the rods form a
straight line, and the masses move just like a mass of 2m. The system is essentially
a pendulum of length 1 .
• 1
2 : With 1 / 2 ≡ , the frequencies are (to leading order in )
2g
ω+ =
,
1
ω− =
g
.
(6.225)
2
The normal modes are
θ1 (t)
θ2 (t)
θ1 (t)
θ2 (t)
+
−
∝
1
−
∝
1
2
cos(ω+ t + φ+ ),
(6.226)
cos(ω− t + φ− ).
In the first mode, the bottom mass essentially stands still, and the top mass oscillates at a high frequency (assuming 1 is small). The factor of 2 in the frequency
arises because the top mass essentially lives in a world where the acceleration
from gravity is g = 2g (because of the extra mg force downward from the lower
mass). In the second mode, the system is essentially a pendulum of length 2 .
The factor of 2 in the angles is what is needed to make the tangential force on
the top mass roughly equal to zero (because otherwise it would oscillate at a high
frequency, since 1 is small).
6.11 Solutions
6.20. Shortest distance in a plane
Let the two given points be (x1 , y1 ) and (x2 , y2 ), and let the path be described
by the function y(x). (Yes, we’ll assume it can be written as a function. Locally,
we don’t have to worry about any double-valued issues.) Then the length of the
path is
x2
=
(6.227)
1 + y 2 , so the Euler–Lagrange equation is
The “Lagrangian” is L =
d ∂L
dx ∂y
1 + y 2 dx.
x1
=
∂L
∂y
d
dx
=⇒
y
1+y2
= 0.
(6.228)
We see that y / 1 + y 2 is constant. Therefore, y is also constant, so we have
a straight line, y(x) = Ax + B, where A and B are determined by the endpoint
conditions.
6.21. Index of refraction
Let the path be described by y(x). The speed at height y is v ∝ y. Therefore, the
time to go from (x1 , y1 ) to (x2 , y2 ) is
T =
x2 ds
x1
x2
∝
v
x1
1+y2
dx.
y
(6.229)
The “Lagrangian” is therefore
L∝
1+y2
.
y
(6.230)
At this point, we could apply the E–L equation to this L, but let’s just use
Lemma 6.5, with f (y) = 1/y. Equation (6.86) gives
1 + y 2 = Bf (y)2
=⇒
B
1+y2 = 2 .
y
(6.231)
We must now integrate this. Solving for y , and then separating variables and
integrating, gives
dx = ±
y dy
B − y2
=⇒
x + A = ∓ B − y2 .
(6.232)
Therefore, (x + A)2 + y2 = B, which is the equation for a circle. Note that the
circle is centered at a point with y = 0, that is, at a point on the bottom of the
slab. This is the point where the perpendicular bisector of the line joining the two
given points intersects the bottom of the slab.
6.22. Minimal surface
By “tension” in a surface, we mean the force per unit length in the surface. The
tension throughout the surface must be constant, because it is in equilibrium. If
the tension at one point were larger than at another, then some patch of the surface
between these points would move.
The ratio of the circumferences of the circular boundaries of the ring is
y2 /y1 . Therefore, the condition that the horizontal forces on the ring cancel is
275
276
The Lagrangian method
y1 cos θ1 = y2 cos θ2 , where the θ’s are the angles of the surface, as shown
in Fig. 6.50. In other words, y cos θ is constant throughout the surface. But
cos θ = 1/ 1 + y 2 , so we have
θ2
θ1
y1
y
y2
1+y2
= C.
(6.233)
This is equivalent to Eq. (6.77), and the solution proceeds as in Section 6.8.
6.23. Existence of a minimal surface
The general solution for y(x) is given in Eq. (6.78) as y(x) = (1/b) cosh b(x + d).
If we choose the origin to be midway between the rings, then d = 0. Both boundary
condition are thus
Fig. 6.50
r=
1
cosh b .
b
(6.234)
We will now determine the maximum value of /r for which the minimal surface
exists. If /r is too large, then we will see that there is no solution for b in
Eq. (6.234). If you perform an experiment with soap bubbles (which want to
minimize their area), and if you pull the rings too far apart, then the surface will
break and disappear as it tries to form the two boundary circles.
Define the dimensionless quantities,
η≡
w
w = cosh(ηz)
,
and
z ≡ br.
(6.235)
Then Eq. (6.234) becomes
w= z
z
Fig. 6.51
r
z = cosh ηz.
(6.236)
If we make a rough plot of the graphs of w = z and w = cosh ηz for a few values
of η (see Fig. 6.51), we see that there is no solution to Eq. (6.236) if η is too large.
The limiting value of η for which there exists a solution occurs when the curves
w = z and w = cosh ηz are tangent; that is, when the slopes are equal in addition
to the functions being equal. Let η0 be the limiting value of η, and let z0 be the
place where the tangency occurs. Then equality of the values and the slopes gives
z0 = cosh(η0 z0 ),
and
1 = η0 sinh(η0 z0 ).
(6.237)
Dividing the second of these equations by the first gives
1 = (η0 z0 ) tanh(η0 z0 ).
(6.238)
This must be solved numerically. The solution is
η0 z0 ≈ 1.200.
(6.239)
Plugging this into the second of Eqs. (6.237) gives
r max
≡ η0 ≈ 0.663.
(6.240)
(Note also that z0 = 1.200/η0 = 1.810.) We see that if /r is larger than
0.663, then there is no solution for y(x) that is consistent with the boundary
6.11 Solutions
2l
r
Fig. 6.52
conditions. Above this value of /r, the soap bubble minimizes its area by heading toward the shape of just two disks, but it will pop well before it reaches that
configuration.
To get a sense of the rough shape of the limiting minimal surface, note that the
ratio of the radius of the “middle” circle to the radius of the boundary rings is
y(0)
cosh(0)
1
1
=
=
=
≈ 0.55.
y( )
cosh(b )
cosh(η0 z0 )
z0
(6.241)
Remarks:
1. We glossed over one issue above, namely that there may be more than one solution
for the constant b in Eq. (6.234). In fact, Fig. 6.51 shows that for any η < 0.663,
there are two solutions for z in Eq. (6.236), and hence two solutions for b in
Eq. (6.234). This means that there are two possible surfaces that might solve
our problem. Which one do we want? It turns out that the surface corresponding
to the smaller value of b is the one that minimizes the area, while the surface
corresponding to the larger value of b is the one that (in some sense) maximizes
the area.
We say “in some sense” because the surface with the larger b is actually a
saddle point for the area. It can’t be a maximum, after all, because we can always
make the area larger by adding little wiggles to it. It’s a saddle point because there
does exist a class of variations for which it has the maximum area, namely ones
where the “dip” in the curve is continuously made larger ( just imagine lowering
the midpoint in a smooth manner). Such a set of variations is shown in Fig. 6.52.
If we start with a cylinder for a surface and then gradually pinch in the center, the
area decreases at first (the decrease in the cross-sectional area is the dominant
effect at the start). But then as the dip becomes very deep, the area increases
because the surface starts to look like the two disks, and these two disks have a
larger area than the original narrow cylinder. The surface eventually resembles
two nearly flat cones connected by a line. As these cones finally flatten out to
the two disks, the area decreases. Therefore, the area must have achieved a local
maximum (at least with respect to this class of variations) somewhere in between.
This local maximum (or rather, saddle point) arises because the Euler–Lagrange
technique simply sets the “derivative” equal to zero and doesn’t differentiate
between maxima, minima, and saddle points.
If η ≡ /r > 0.663 (so that the initial cylinder in now wide instead of narrow),
there exists at least one class of variations for which the area decreases monotonically from the area of the cylinder down to the area of the two disks. If you
draw a series of pictures (for a wide cylinder) analogous to those in Fig. 6.52, it
is quite believable that this is the case.
2. How does the area of the limiting surface (with η0 = 0.663) compare with the
area of the two disks? The area of the two disks is Ad = 2πr 2 . And the area of
277
278
The Lagrangian method
the limiting surface is
As =
2πy 1 + y 2 dx.
−
(6.242)
Using Eq. (6.234), this becomes
As =
−
2π
cosh2 bx dx =
b
−
π
(1 + cosh 2bx) dx
b
π sinh 2b
2π
.
+
=
b
b2
(6.243)
But from the definitions of η and z, we have = η0 r and b = z0 /r for the limiting
surface. Therefore, As can be written as
2η0
sinh 2η0 z0
+
z0
z02
As = π r 2
.
(6.244)
Plugging in the numerical values (η0 ≈ 0.663 and z0 ≈ 1.810) gives
Ad ≈ (6.28)r 2 ,
x
and
As ≈ (7.54)r 2 .
(6.245)
The ratio of As to Ad is approximately 1.2 (it’s actually η0 z0 , as you can show).
The limiting surface therefore has a larger area. This is expected, because for
/r > η0 the surface tries to run off to one with a smaller area, and there are no
other stable configurations besides the cosh solution we found. ♣
6.24. The brachistochrone
y
Fig. 6.53
(x0 , y0 )
First solution: In Fig. 6.53, the boundary conditions are y(0) = 0 and y(x0 ) =
y0 , with downward taken to be the positive
√ y direction. From conservation of
energy, the speed as a function of y is v = 2gy. The total time is therefore
T =
x0 ds
0
v
=
x0
0
1+y2
dx.
√
2gy
(6.246)
Our goal is to find the function y(x) that minimizes this integral, subject to the
boundary conditions above. We can therefore apply the results of the variational
technique, with a “Lagrangian” equal to
L∝
1+y2
.
√
y
(6.247)
At this point, we could apply the E–L equation to this L, but let’s just use Lemma
√
6.5, with f (y) = 1/ y. Equation (6.86) gives
1 + y 2 = Bf (y)2
=⇒
1+y2 =
B
,
y
(6.248)
6.11 Solutions
as desired. We must now integrate this. Solving for y and separating variables
gives
√
y dy
= ± dx.
(6.249)
√
B−y
A helpful change of variables to get rid of the square root in the denominator is
y ≡ B sin2 φ. Then dy = 2B sin φ cos φ dφ, and Eq. (6.249) simplifies to
2B sin2 φ dφ = ± dx.
(6.250)
We can now use sin2 φ = (1 − cos 2φ)/2 to integrate this. After multiplying
through by 2, the result is B(2φ − sin 2φ) = ± 2x − C, where C is a constant
of integration. Now note that we can rewrite our definition of φ (which was
y ≡ B sin2 φ) as 2y = B(1 − cos 2φ). If we then define θ ≡ 2φ, we have
x = ± a(θ − sin θ ) ± d,
y = a(1 − cos θ).
(6.251)
where a ≡ B/2, and d ≡ C/2. The particle starts at (x, y) = (0, 0). Therefore, θ
starts at θ = 0, since this corresponds to y = 0. The starting condition x = 0 then
implies that d = 0. Also, we are assuming that the wire heads down to the right,
so we choose the positive sign in the expression for x. Therefore, we finally have
x = a(θ − sin θ ),
y = a(1 − cos θ),
(6.252)
as desired. This is the parametrization of a cycloid, which is the path taken by a
point on the rim of a rolling wheel. The initial slope of the y(x) curve is infinite,
as you can check.
Remark: The above method derived the parametric form in (6.252) from scratch.
But since Eq. (6.252) was given in the statement of the problem, another route is to
simply verify that this parametrization satisfies Eq. (6.248). To this end, assume that
x = a(θ − sin θ) and y = a(1 − cos θ), which gives
y ≡
dy
dy/dθ
sin θ
=
=
.
dx
dx/dθ
1 − cos θ
(6.253)
Therefore,
1+y2 =1+
sin2 θ
2
2a
=
=
,
(1 − cos θ)2
1 − cos θ
y
which agrees with Eq. (6.248), with B ≡ 2a.
(6.254)
♣
Second solution: Let’s use a variational argument again, but now with y as the
independent variable. That is, let the chain be described by the function x(y). The
arclength is now given by ds = 1 + x 2 dy. Therefore, instead of the Lagrangian
in Eq. (6.247), we have
L∝
1+x2
.
√
y
(6.255)
The Euler–Lagrange equation is
d ∂L
dy ∂x
=
∂L
∂x
=⇒
d
1
√
dy
y
x
1+x2
= 0.
(6.256)
279
280
The Lagrangian method
The zero on the right-hand side makes things nice and easy because it means√
that
the quantity in parentheses is a constant. If we define this constant to be 1/ B,
then we can solve for x and then separate variables to obtain
√
y dy
= ± dx.
(6.257)
√
B−y
in agreement with Eq. (6.249). The solution proceeds as above.
Third solution:
in Eq. (6.247) as
The “Lagrangian” in the first solution above, which is given
L∝
1+y2
,
√
y
(6.258)
is independent of x. Therefore, in analogy with conservation of energy (which
arises from a Lagrangian that is independent of t), the quantity
E≡y
y2
∂L
−L= √
−
∂y
y 1+y2
1+y2
−1
= √
√
y
y 1+y2
(6.259)
is constant (that is, independent of x). This statement is equivalent to Eq. (6.248),
and the solution proceeds as above.
Chapter 7
Central forces
A central force is by definition a force that points radially and whose magnitude
depends only on the distance from the source (that is, not on the angle around the
source).1 Equivalently, we may say that a central force is one whose potential
depends only on the distance from the source. That is, if the source is located at
the origin, then the potential energy is of the form V (r) = V (r). Such a potential
does indeed yield a central force, because
F(r) = −∇V (r) = −
dV
rˆ ,
dr
(7.1)
which points radially and depends only on r. Gravitational and electrostatic forces
are central forces, with V (r) ∝ 1/r. The spring force is also central, with V (r) ∝
(r − )2 , where is the equilibrium length.
There are two important facts concerning central forces: (1) they are ubiquitous
in nature, so we had better learn how to deal with them, and (2) dealing with them
is much easier than you might think, because crucial simplifications occur in the
equations of motion when V is a function of r only. These simplifications will
become evident in the following two sections.
7.1
Conservation of angular momentum
Angular momentum plays a key role in dealing with central forces because, as
we will show, it is constant over time. For a point mass, we define the angular
momentum L by
L = r × p,
(7.2)
where the “cross product” is defined inAppendix B. L depends on r, so it therefore
depends on where you pick the origin of your coordinate system. Note that L is
a vector, and that it is orthogonal to both r and p, by nature of the cross product.
You might wonder why we care enough about r × p to give it a name. Why not
look at r 3 p5 r × (r × p), or something else? The answer is that L has some very
nice properties, one of which is the following.
1
Taken literally, the term “central force” would imply only the radial nature of the force. But a
physicist’s definition also includes the dependence solely on the distance from the source.
281
282
Central forces
Theorem 7.1 If a particle is subject to a central force only, then its angular
momentum is conserved.2 That is,
If V (r) = V (r),
then
dL
= 0.
dt
(7.3)
Proof: We have
dL d
= (r × p)
dt
dt
dr
dp
=
×p+r×
dt
dt
= v × (mv) + r × F
= 0,
(7.4)
because F ∝ r, and the cross product of two parallel vectors is zero.
We’ll prove this theorem again in the next section, using the Lagrangian
method. Let’s now prove another theorem which is probably obvious, but good
to show anyway.
Theorem 7.2 If a particle is subject to a central force only, then its motion
takes place in a plane.
Proof: At a given instant t0 , consider the plane P containing the position vector r0 (with the source of the potential taken to be the origin) and the velocity
vector v0 . We claim that r lies in P at all times.3 This is true because P is
defined as the plane orthogonal to the vector n0 ≡ r0 × v0 . But in the proof of
Theorem 7.1, we showed that the vector r × v ≡ (r × p)/m does not change with
time. Therefore, r × v = n0 for all t. Since r is certainly orthogonal to r × v, we
see that r is orthogonal to n0 for all t. Hence, r must always lie in P.
An intuitive look at this theorem is the following. Since the position, velocity,
and acceleration (which is proportional to F, which in turn is proportional to the
position vector r) vectors initially all lie in P, there is a symmetry between the
two sides of P. Therefore, there is no reason for the particle to head out of P on
one side rather than the other. The particle therefore remains in P. We can then
use this same reasoning again a short time later, and so on.
This theorem shows that we need only two coordinates, instead of the usual
three, to describe the motion. But since we’re on a roll, why stop there? We’ll
show below that we really need only one coordinate. Not bad, three coordinates
reduced down to one.
2
3
This is a special case of the fact that torque equals the rate of change of angular momentum. We’ll
talk about this in great detail in Chapter 8.
The plane P is not well defined if v0 = 0, or if r0 = 0, or if v0 is parallel to r0 . But in these cases,
you can quickly show that the motion is always radial, which is even more restrictive than planar.
7.2 The effective potential
7.2
The effective potential
The effective potential provides a sneaky and useful method for simplifying a
three-dimensional central-force problem down to a one-dimensional problem.
Here’s how it works. Consider a particle of mass m subject to a central force
only, described by the potential V (r). Let r and θ be the polar coordinates in
the plane of the motion. In these polar coordinates, the Lagrangian (which we’ll
label as “L”, to save “L” for the angular momentum) is
L=
1
m(˙r 2 + r 2 θ˙ 2 ) − V (r).
2
(7.5)
The equations of motion obtained from varying r and θ are
m¨r = mr θ˙ 2 − V (r),
d
(mr 2 θ˙ ) = 0.
dt
(7.6)
Since −V (r) equals the force F(r), the first of these equations is the radial
F = ma equation, complete with the centripetal acceleration, in agreement with
the first of Eqs. (3.51). The second equation is the statement of conservation of
angular momentum, because mr 2 θ˙ = r(mr θ˙ ) = rpθ (where pθ is the momentum
in the angular direction) is the magnitude of L = r × p, from Eq. (B.9). We
therefore see that the magnitude of L is constant. And since the direction of L is
always perpendicular to the fixed plane of the motion, the vector L is constant in
time. We have therefore just given a second proof of Theorem 7.1. In the present
Lagrangian language, the conservation of L follows from the fact that θ is a
cyclic coordinate, as we saw in Example 2 in Section 6.5.1. Since mr 2 θ˙ does not
change with time, let us denote its constant value by
˙
L ≡ mr 2 θ.
(7.7)
L is determined by the initial conditions. It can be specified, for example, by
giving the initial values of r and θ˙ . Using θ˙ = L/(mr 2 ), we can eliminate θ˙ from
the first of Eqs. (7.6). The result is
m¨r =
L2
− V (r).
mr 3
(7.8)
Multiplying by r˙ and integrating with respect to time yields
1 2
m˙r +
2
L2
+ V (r) = E,
2mr 2
(7.9)
where E is a constant of integration. E is simply the energy, which can be seen
by noting that this equation could also have been obtained by using Eq. (7.7) to
eliminate θ˙ in the energy equation, (m/2)(˙r 2 + r 2 θ˙ 2 ) + V (r) = E.
283
284
Central forces
Equation (7.9) is rather interesting. It involves only the variable r. And it
looks a lot like the equation for a particle moving in one dimension (labeled by
the coordinate r) under the influence of the potential,
Veff (r) ≡
L2
+ V (r).
2mr 2
(7.10)
The subscript “eff ” here stands for “effective.” Veff (r) is called the effective
potential. The “effective force” is easily read off from Eq. (7.8) to be
Feff (r) =
L2
− V (r),
mr 3
(7.11)
which agrees with Feff = −Veff (r), as it should. This “effective” potential
concept is a marvelous result and should be duly appreciated. It says that if
we want to solve a two-dimensional problem (which could have come from a
three-dimensional problem) involving a central force, then we can recast the
problem into a simple one-dimensional problem with a slightly modified potential. We can forget that we ever had the variable θ, and we can solve this
one-dimensional problem (as we’ll demonstrate below) to obtain r(t). Hav˙
ing found r(t), we can use θ(t)
= L/mr 2 to solve for θ(t) (in theory, at
least). This whole procedure works only because there is a quantity involving
r and θ (or rather, θ˙ ) that is independent of time. The variables r and θ are
therefore not independent, so the problem is really one-dimensional instead of
two-dimensional.
To get a general idea of how r behaves with time, all we have to do is draw the
graph of Veff (r). Consider the example where V (r) = Ar 2 . This is the potential
for a spring with relaxed length zero. Then
Veff (r) =
Veff (r) =
L2
______
2mr 2
+ Ar 2
E
r1
Fig. 7.1
r2
r
L2
+ Ar 2 .
2mr 2
(7.12)
To plot Veff (r), we must be given L (determined by the initial conditions), along
with A and m (determined by the system we’re dealing with). But the general
shape looks like the curve in Fig. 7.1. The energy E (determined by the initial
conditions), which must be given too, is also drawn. The coordinate r bounces
back and forth between the turning points, r1 and r2 , which satisfy Veff (r1,2 ) =
E. 4 This is true because it is impossible for the particle to be located at an r
for which E < Veff , since Eq. (7.9) would then imply an imaginary value for r˙ .
If E equals the minimum of Veff (r), then r1 = r2 , so r is stuck at this one value,
which means that the motion is a circle.
Remark: The L2 /2mr 2 term in the effective potential is sometimes called the angular momentum barrier. It has the effect of keeping the particle from getting too close to the origin. Basically,
the point is that L ≡ mr 2 θ˙ is constant, so as r gets smaller, θ˙ gets bigger. But θ˙ increases at a
4
It turns out that for our Ar 2 spring potential, the motion in space is an ellipse, with semi-axis lengths
r1 and r2 (see Problem 7.5). But for a general potential, the motion isn’t so nice.
7.3 Solving the equations of motion
˙ So eventually we end up
greater rate than r decreases, due to the square of the r in L = mr 2 θ.
with a tangential kinetic energy, mr 2 θ˙ 2 /2, that is greater than what is allowed by conservation
of energy.5
As he walked past the beautiful belle,
The attraction was easy to tell.
But despite his persistence,
He was kept at a distance
By that darn conservation of L! ♣
Note that it is by no means necessary to introduce the concept of the effective
potential. You can simply solve the equations of motion in Eq. (7.6) as they are.
But introducing Veff makes it much easier to see what’s going on in a central-force
problem.
When using potentials, effective,
Remember the one main objective:
The goal is to shun
All dimensions but one,
And then view things with 1-D perspective.
7.3
Solving the equations of motion
If we want to be quantitative, we must solve the equations of motion in Eq. (7.6).
Equivalently, we must solve their integrated forms, Eqs. (7.7) and (7.9), which
are the conservation of L and E statements,
mr 2 θ˙ = L,
1 2
L2
m˙r +
+ V (r) = E.
2
2mr 2
(7.13)
The word “solve” is a little ambiguous here, because we should specify what
quantities we want to solve for in terms of what other quantities. There are
essentially two things we can do. We can solve for r and θ in terms of t. Or
we can solve for r in terms of θ . The former has the advantage of immediately yielding velocities and, of course, the information of where the particle
is at time t. The latter has the advantage of explicitly showing what the trajectory looks like in space, even though we don’t know how quickly it is being
traversed. We’ll deal mainly with this latter case, particularly when we discuss the gravitational force and Kepler’s laws below. But let’s look at both
cases now.
5
This argument doesn’t hold if V (r) goes to −∞ faster than −1/r 2 . You can see this by drawing the graph of Veff (r), which heads to −∞ instead of +∞ as r → 0. V (r) decreases
fast enough to allow for the increase in kinetic energy. But such potentials don’t come up that
often.
285
286
Central forces
7.3.1
Finding r (t ) and θ(t )
The value of r˙ at any point is found from Eq. (7.13) to be
L2
dr
2
E−
− V (r) .
=±
dt
m
2mr 2
(7.14)
To get an actual r(t) out of this, we must be supplied with E and L (determined
˙ and also the function V (r). To solve this
by the initial values of r, r˙ , and θ),
differential equation, we “simply” have to separate variables and then (in theory)
integrate:
dr
E−
L2
2mr 2
− V (r)
=±
2
2
dt = ±
(t − t0 ).
m
m
(7.15)
We need to evaluate this (rather unpleasant) integral on the left-hand side, to
obtain t as a function of r. Having found t(r), we can then (in theory) invert the
result to obtain r(t). Finally, substituting this r(t) into the relation θ˙ = L/mr 2
from Eq. (7.13) gives θ˙ as a function of t, which we can (in theory) integrate to
obtain θ (t).
As you might have guessed, this procedure has the potential to yield some
stress. For most V (r)’s, the integral in Eq. (7.15) is not calculable in closed form.
There are only a few “nice” potentials V (r) for which we can evaluate it. And
even in those cases, the remaining tasks are still a pain.6 But the good news
is that these “nice” potentials are precisely the ones we are most interested in.
In particular, the gravitational potential, which goes like 1/r and which we will
concentrate on during the remainder of this chapter, leads to a calculable integral
(the spring potential ∼ r 2 does also). However, having said all this, we’re not
going to apply this procedure to gravity. It’s nice to know that it exists, but we
won’t be doing anything else with it. Instead, we’ll use the following strategy to
solve for r as a function of θ.
7.3.2
Finding r (θ)
We can eliminate the dt from Eqs. (7.13) by getting the r˙ 2 term alone on the left
side of the second equation, and then dividing by the square of the first equation.
The dt 2 factors cancel, and we obtain
1 dr
r 2 dθ
2
=
2mE
1
2mV (r)
− 2−
.
2
L
r
L2
(7.16)
We can now (in theory) take a square root, separate variables, and integrate to
obtain θ as a function of r. We can then (in theory) invert to obtain r as a function
6
Of course, if you run out of patience or hit a brick wall, you always have the option of doing things
numerically. See Section 1.4 for a discussion of this.
7.4 Gravity, Kepler’s laws
of θ . To do this, we must be given the function V (r). So let’s now finally give
ourselves a V (r) and do a problem all the way through. We’ll study the most
important potential of all, or perhaps the second most important one, gravity.7
7.4
7.4.1
Gravity, Kepler’s laws
Calculation of r (θ)
Our goal in this subsection is to obtain r as a function of θ, for a gravitational
potential. Let’s assume that we’re dealing with the earth and the sun, with masses
M and m, respectively. The gravitational potential energy of the earth–sun
system is
V (r) = −
α
,
r
where α ≡ GM m.
(7.17)
In the present treatment, we’ll consider the sun to be bolted down at the origin of
our coordinate system. Since M
m, this is approximately true for the earth–
sun system. (If we want to do the problem exactly, we must use the reduced mass,
which is the topic of Section 7.4.5.) Equation (7.16) becomes
1 dr
r 2 dθ
2
=
2mE
1
2mα
− 2+
.
L2
r
rL2
(7.18)
As stated above, we could take a square root, separate variables, integrate to
find θ (r), and then invert to find r(θ ). This method, although straightforward, is
rather messy. So let’s solve for r(θ ) in a slick way.
With all the 1/r terms floating around, it might be easier to solve for 1/r instead
of r. Using d(1/r)/dθ = −(dr/dθ )/r 2 , and letting y ≡ 1/r for convenience,
Eq. (7.18) becomes
dy
dθ
2
= −y2 +
2mα
2mE
y+ 2 .
2
L
L
(7.19)
At this point, we could also use the separation-of-variables technique, but let’s
continue to be slick. Completing the square on the right-hand side gives
dy
dθ
2
=− y−
mα
L2
2
+
2mE
+
L2
mα
L2
2
.
(7.20)
Defining z ≡ y − mα/L2 for convenience yields
dz
dθ
7
2
= −z 2 +
mα
L2
2
1+
2EL2
mα 2
≡ −z 2 + B2 ,
(7.21)
The two most important potentials in physics are certainly the gravitational and harmonic-oscillator
ones. They both lead to doable integrals, and interestingly both lead to elliptical orbits.
287
288
Central forces
where
B≡
mα
L2
1+
2EL2
.
mα 2
(7.22)
At this point, in the spirit of being slick, we can just look at Eq. (7.21) and observe
that
z = B cos(θ − θ0 )
(7.23)
is the solution, because cos2 x + sin2 x = 1. But lest we feel guilty about not
doing separation-of-variables at least once in this problem, let’s solve Eq. (7.21)
that way, too. The integral is nice and doable, and we have
√
dz
B2
− z2
=
=⇒
dθ
cos−1
z
= θ − θ0 ,
B
(7.24)
which gives z = B cos(θ − θ0 ). It is customary to pick the axes so that θ0 = 0,
so we’ll drop the θ0 from here on. Recalling our definition z ≡ 1/r − mα/L2
and also the definition of B from Eq. (7.22), Eq. (7.23) becomes
mα
1
= 2 (1 + cos θ),
r
L
(7.25)
where
1+
≡
2EL2
mα 2
(7.26)
is the eccentricity of the particle’s motion. We will see shortly exactly what
signifies.
This completes the derivation of r(θ ) for the gravitational potential,
V (r) ∝ 1/r. It was a little messy, but not unbearably painful. At any rate,
we just discovered the basic motion of objects under the influence of gravity,
which takes care of virtually all of the gazillion tons of stuff in the universe. Not
bad for one page of work.
Newton said as he gazed off afar,
“From here to the most distant star,
These wond’rous ellipses
And solar eclipses
All come from a 1 over r.”
What are the limits on r in Eq. (7.25)? The minimum value of r is obtained
when the right-hand side reaches its maximum value, which is (mα/L2 )(1 + ).
Therefore,
rmin =
L2
.
mα(1 + )
(7.27)
7.4 Gravity, Kepler’s laws
289
What is the maximum value of r? The answer depends on whether is greater
than or less than 1. If < 1 (which corresponds to circular or elliptical orbits, as
we’ll see below), then the minimum value of the right-hand side of Eq. (7.25) is
(mα/L2 )(1 − ). Therefore,
rmax =
L2
mα(1 − )
(if
< 1).
(7.28)
If ≥ 1 (which corresponds to parabolic or hyperbolic orbits, as we’ll see below),
then the right-hand side of Eq. (7.25) can become zero (when cos θ = −1/ ).
Therefore,
rmax = ∞ (if
7.4.2
≥ 1).
(7.29)
The orbits
Let’s examine in detail the various cases for .
• Circle ( = 0)
If = 0, then Eq. (7.26) says that E = −mα 2 /2L2 . The negative E means that the potential energy is more negative than the kinetic energy is positive, so the particle is trapped
in the potential well. Equations (7.27) and (7.28) give rmin = rmax = L2 /mα. Therefore, the particle moves in a circular orbit with radius L2 /mα. Equivalently, Eq. (7.25)
says that r is independent of θ .
Note that it isn’t necessary to do all the work of Section 7.4.1 if we just want to look at
circular motion. For a given L, the energy −mα 2 /2L2 is the minimum value that the E
given by Eq. (7.13) can take. This is true because to achieve the minimum, we certainly
want r˙ = 0. And you can show that minimizing the effective potential, L2 /2mr 2 − α/r,
yields this value for E. If we plot Veff (r), we have the situation shown in Fig. 7.2.
The particle is trapped at the bottom of the potential well, so it has no motion in the r
direction.
• Ellipse (0 < < 1)
If 0 < < 1, then Eq. (7.26) says that −mα 2 /2L2 < E < 0. Equations (7.27) and
(7.28) give rmin and rmax . It isn’t obvious that the resulting motion is an ellipse. We’ll
demonstrate this below.
If we plot Veff (r), we have the situation shown in Fig. 7.3. The particle oscillates
between rmin and rmax . The energy is negative, so the particle is trapped in the potential
well.
• Parabola ( = 1)
If = 1, then Eq. (7.26) says that E = 0. This value of E implies that the particle
barely makes it out to infinity (its speed approaches zero as r → ∞). Equation (7.27)
gives rmin = L2 /2mα, and Eq. (7.29) gives rmax = ∞. Again, it isn’t obvious that
the resulting motion is a parabola. We’ll demonstrate this below.
L2 - __
α
Veff (r) = ______
2mr 2 r
rmin = rmax
r
E
Fig. 7.2
Veff (r)
rmin
rmax
r
E
Fig. 7.3
290
Central forces
Veff (r)
rmin
E
Fig. 7.4
Veff (r)
E
If we plot Veff (r), we have the situation shown in Fig. 7.4. The particle does not
oscillate back and forth in the r direction. It moves inward (or possibly not, if it was
initially moving outward), turns around at rmin = L2 /2mα, and then heads out to
infinity forever.
• Hyperbola ( > 1)
If > 1, then Eq. (7.26) says that E > 0. This value of E implies that the particle
makes it out to infinity with energy to spare. The potential goes to zero as r → ∞, so
√
the particle’s speed approaches the nonzero value 2E/m as r → ∞. Equation (7.27)
gives rmin , and Eq. (7.29) gives rmax = ∞. Again, it isn’t obvious that the resulting
motion is a hyperbola. We’ll demonstrate this below.
If we plot Veff (r), we have the situation shown in Fig. 7.5. As in the parabola case,
the particle does not oscillate back and forth in the r direction. It moves inward (or
possibly not, if it was initially moving outward), turns around at rmin , and then heads
out to infinity forever.
r
7.4.3
rmin
Proof of conic orbits
Let’s now prove that Eq. (7.25) does indeed describe the conic sections stated
above. We’ll also show that the origin (the source of the potential) is a focus
of the conic section. These proofs are straightforward, although the ellipse and
hyperbola cases get a bit messy. In what follows, we’ll find it easier to work with
Cartesian coordinates. For convenience, let
Fig. 7.5
L2
.
mα
Multiplying Eq. (7.25) through by kr, and using cos θ = x/r, gives
k≡
(7.30)
k = r + x.
(7.31)
Solving for r and squaring yields
x2 + y2 = k 2 − 2k x +
2 2
x .
(7.32)
Let’s look at the various cases for . We will invoke without proof various facts
about conic sections (focal lengths, etc.).
y
• Circle ( = 0)
k
x
In this case, Eq. (7.32) becomes x2 + y2 = k 2 . So we have a circle with radius k =
L2 /mα, with its center at the origin (see Fig. 7.6).
• Ellipse (0 < < 1)
In this case, Eq. (7.32) can be written (after completing the square for the x terms, and
expending some effort) as
x+ k
Fig. 7.6
1−
a2
2
2
y2
+ 2 = 1,
b
where a =
k
,
1− 2
and b =
k
1− 2
.
(7.33)
7.4 Gravity, Kepler’s laws
291
y
This is the equation for an ellipse with its center located at (−k /(1 − 2 ), 0). The
semi-major and semi-minor axes are a and b, respectively. And the focal length is
c = a2 − b2 = k /(1 − 2 ). Therefore, one focus is located at the origin (see
Fig. 7.7). Note that c/a equals the eccentricity, .
• Parabola ( = 1)
In this case, Eq. (7.32) becomes y2 = k 2 −2kx, which can be written as y2 = −2k(x− k2 ).
This is the equation for a parabola with vertex at (k/2, 0) and focal length k/2. (The
focal length of a parabola written in the form y2 = 4ax is a.) So we have a parabola
with its focus located at the origin (see Fig. 7.8).
• Hyperbola ( > 1)
a
b
a
c
x
Fig. 7.7
y
In this case, Eq. (7.32) can be written (after completing the square for the x terms)
x − 2k
−1
a2
2
y2
− 2 = 1,
b
k
where a = 2
,
−1
and b =
k
2−1
k/2
.
This is the equation for a hyperbola with its center (defined to be the intersection of the
asymptotes) located at (k /( 2 −1), 0). The focal length is c = a2 + b2 = k /( 2 −1).
Therefore, the focus is located at the origin (see Fig. 7.9). Note that c/a equals the
eccentricity, .
The impact parameter (usually denoted by the letter b) of a trajectory is defined to be
the closest distance to the origin the particle would achieve if it moved in the straight line
determined by its initial velocity far from the origin (that is, along the dotted line in the
Fig. 7.9). You might think that choosing the letter b here would cause a problem, because
we already defined b in Eq. (7.34). However, it turns out that these two definitions are
identical (see Exercise 7.14), so all is well.
Equation (7.34) actually describes an entire hyperbola, that is, it also describes a
branch that opens up to the right with its focus located at (2k /( 2 − 1), 0). However,
this right branch was introduced in the squaring operation that produced Eq. (7.32). It
isn’t a solution to the original equation we wanted to solve, Eq. (7.31). It turns out that
the right-opening branch (or its reflection across the y axis, depending on your sign
convention for B and ) is relevant for a repulsive, instead of attractive, 1/r potential;
see Exercise 7.21.
7.4.4
x
(7.34)
Kepler’s laws
We can now, with minimal extra work, write down Kepler’s laws. Kepler (1571–
1630) lived before Newton (1642–1727), so he didn’t have Newton’s laws at his
disposal. Kepler arrived at his laws via observational data, which was a rather
impressive feat. It was known since the time of Copernicus (1473–1543) that
the planets move around the sun, but it was Kepler who first gave a quantitative
description of the orbits. Kepler’s laws assume that the sun is massive enough so
that its position is essentially fixed in space. This is a very good approximation,
Fig. 7.8
y
b
c
c
x
a
Fig. 7.9
292
Central forces
but the following subsection on the reduced mass shows how to modify the laws
and solve things exactly.
rdθ
dθ
Fig. 7.10
r
• First law: The planets move in elliptical orbits with the sun at one focus.
We proved this in Eq. (7.33).8 There are undoubtedly also objects flying past the sun in
hyperbolic orbits, but we don’t call these things planets, because we never see the same
one twice.
• Second law: The radius vector to a planet sweeps out area at a rate that is independent
of its position in the orbit.
This law is nothing other than a fancy way of stating conservation of angular
momentum. The area swept out by the radius vector during a short period of time
is dA = r(r dθ )/2, because r dθ is the base of the thin triangle in Fig. 7.10. Therefore,
˙
we have (using L = mr 2 θ)
dA
r 2 θ˙
L
=
=
,
dt
2
2m
(7.35)
which is constant, because L is constant for a central force. This quick proof is
independent of all the work we did in Sections 7.4.1–7.4.3.
• Third law: The square of the period of an orbit, T , is proportional to the cube of the
semi-major-axis length, a. More precisely,
T2 =
4π 2 a3
,
GM
(7.36)
where M is the mass of the sun. Note that the planet’s mass m doesn’t appear in this
equation.
Proof: Integrating Eq. (7.35) over the time of a whole orbit gives
A=
LT
.
2m
(7.37)
But the area of an ellipse is A = π ab, where a and b are the semi-major and semi-minor
axes, respectively. Squaring Eq. (7.37) and using Eq. (7.33) to write b = a 1 − 2
gives
π 2 a4 =
L2
m(1 − 2 )
T2
.
4m
(7.38)
We have grouped the right-hand side in this way because we can now use the L2 ≡ mαk
relation from Eq. (7.30) to transform the term in parentheses into αk/(1 − 2 ) ≡ αa,
where a is given in Eq. (7.33). But αa ≡ (GM m)a, so we obtain
π 2 a4 =
(GM ma)T 2
,
4m
(7.39)
which gives Eq. (7.36), as desired.
8
For an alternate geometrical proof due to Feynman (and also Maxwell), see Goodstein and
Goodstein (1996).
7.4 Gravity, Kepler’s laws
These three laws describe the motion of all the planets (and asteroids, comets,
and such) in the solar system. But our solar system is only the tip of the iceberg.
There’s a lot of other stuff out there, and it’s all governed by gravity (although
Newton’s inverse square law must be supplanted by Einstein’s General Relativity
theory of gravitation). There’s a whole universe around us, and as time passes on
we see and understand more and more of it, both experimentally and theoretically.
In recent years, we’ve even begun to look for friends we might have out there.
Why? Because we can. There’s nothing wrong with looking under the lamppost
now and then. It just happens to be a very big one in this case.
As we grow up, we open an ear,
Exploring the cosmic frontier.
In this coming of age,
We turn in our cage,
All alone on a tiny blue sphere.
7.4.5
Reduced mass
We assumed in Section 7.4.1 that the sun is large enough so that it is only
negligibly affected by the presence of the planets. That is, it is essentially
fixed at the origin. But how do we solve a problem in which the masses of
the two interacting bodies are comparable in size? Equivalently, how do we
solve the earth–sun problem exactly? It turns out that the only modification
required is a replacement of the earth’s mass with the reduced mass, defined
below. The following discussion actually holds for any central force, not just
gravity.
The Lagrangian of a general central-force system consisting of the interacting
masses m1 and m2 is
L=
1
1
m1 r˙ 12 + m2 r˙ 22 − V (|r1 − r2 |).
2
2
(7.40)
We have written the potential in this form, dependent only on the distance |r1 −r2 |,
because we are assuming a central force. Let us define
R≡
m1 r1 + m2 r2
,
m1 + m 2
and
r ≡ r1 − r2 .
(7.41)
R and r are the position of the center of mass and the vector between the masses,
respectively. Invert these equations to obtain
r1 = R +
m2
r,
M
and
r2 = R −
m1
r,
M
(7.42)
293
294
Central forces
where M ≡ m1 + m2 is the total mass of the system. In terms of R and r, the
Lagrangian becomes
1
˙ + m2 r˙ 2 + 1 m2 R
˙ − m1 r˙
L = m1 R
2
M
2
M
m1 m2
1 ˙2 1
= MR
r˙ 2 − V (r)
+
2
2 m1 + m 2
2
− V (|r|)
1 ˙2 1 2
= MR
+ µ˙r − V (r),
2
2
(7.43)
where the reduced mass, µ, is defined by
1
1
1
+
.
≡
µ
m1
m2
(7.44)
˙ but not on R.
We now note that the Lagrangian in Eq. (7.43) depends on R,
˙
Therefore, the Euler–Lagrange equations say that R is constant. That is, the CM
moves at constant velocity (this is just the statement that there are no external
forces). The CM motion is therefore trivial, so let’s ignore it. Our Lagrangian
therefore becomes
L→
1 2
µ˙r − V (r).
2
(7.45)
But this is simply the Lagrangian for a particle of mass µ that moves around a
fixed origin under the influence of the potential V (r). For gravity, we have
L=
1 2 α
µ˙r +
2
r
(where α ≡ GM m).
(7.46)
To solve the earth–sun system exactly, we therefore just need to replace (in the
calculation in Section 7.4.1) the earth’s mass, m, with the reduced mass, µ,
given by
1
1
1
.
≡ +
µ
m M
(7.47)
The resulting value of r in Eq. (7.25) is the distance between the earth and sun.
The earth and sun are therefore distances of (M /M )r and (m/M )r, respectively,
away from the CM, from Eq. (7.42). These distances are just scaled-down versions
of the distance r, which represents an ellipse, so we see that the earth and sun
move in elliptical orbits (whose sizes are in the ratio M /m) with the CM as a
focus. Note that the m’s that are buried in L and in Eq. (7.25) must be changed
to µ’s. But α is still defined to be GM m, so the m in this definition does not get
replaced with µ.
For the earth–sun system, the µ in Eq. (7.47) is essentially equal to m, because
M is so large. Using m = 5.97 · 1024 kg, and M = 1.99 · 1030 kg, we find that
7.4 Gravity, Kepler’s laws
µ is smaller than m by only one part in 3 · 105 . Our fixed-sun approximation is
therefore a very good one. You can show that the CM is about 5 · 105 m from
the center of the sun, which is well inside the sun (about a thousandth of the
radius).
How are Kepler’s laws modified when we solve for the orbits exactly using
the reduced mass?
• First law: The elliptical statement in the first law is still true, but with the CM
(not the sun) located at a focus. The sun also travels in an ellipse with the CM at
a focus.9 Whatever is true for the earth must also be true for the sun, because they
come into Eq. (7.43) symmetrically. The only difference is in the size of various
quantities.
• Second law: In the second law, we need to consider the position vector from the CM
(not the sun) to the earth. This vector sweeps out equal areas in equal times, because the
angular momentum of the earth (and the sun, too) relative to the CM is constant. This
is true because the gravitational force always points through the CM, so the force is a
central force with the CM chosen as the origin.
• Third law: The period of the earth’s orbit (and the sun’s, too) is the same as the period of
the orbit of our hypothetical particle of mass µ orbiting around a fixed origin under the
influence of the potential −α/r ≡ −GM m/r. This is true because the radius vectors in
all three of these systems are always in the same ratio. To find the period of the particle’s
orbit, we can repeat the derivation leading up to Eq. (7.39). But in that equation, the m
on the bottom is replaced by µ, while the m on the top remains m, because this is the m
that appears in α. Therefore, we obtain10
T2 =
4π 2 a3µ µ
GM m
=
4π 2 a3µ
GM
,
(7.48)
where we have used µ ≡ M m/(M + m) ≡ M m/M . This result is symmetric in
M and m, as it must be because if we interchange the labels of M and m, we still
have the same system. And it also correctly reduces to Eq. (7.36) when M
m.
If you want to write Eq. (7.48) in terms of the semi-major axis of the earth’s elliptical
orbit, which is aE = (M /M )aµ , then just plug in aµ = (M /M )aE to obtain
T2 =
4π 2 (M /M )3 a3E
=
GM
M2
M2
4π 2 a3E
.
GM
(7.49)
Let’s perform a check on this formula by considering the special case of equal masses
m orbiting around the same circular path of radius r (at diametrically opposite points)
9
10
Well, this statement is true only if there is just one planet. With many planets, the tiny motion of
the sun is very complicated. This is perhaps the best reason to work in the approximation where
it is essentially bolted down.
We have put the subscript µ on the length a to remind us that it is the semi-major axis of our
hypothetical particle’s orbit, and not the semi-major axis of the earth’s orbit.
295
296
Central forces
with their CM at the center of the circle. For this simple system, we can solve for the
period from scratch using F = ma:
Gm2
mv 2
=
r
(2r)2
=⇒
Gm2
m(2πr/T )2
=
r
(2r)2
=⇒
T2 =
16π 2 r 3
,
Gm
(7.50)
which agrees with Eq. (7.49), in different notation, when M = 2M .
Remark: There’s actually a fairly quick way to see where the M 2 /M 2 factor in Eq. (7.49)
comes from. Imagine a new system where the earth’s orbit has the same dimensions but
where the (slightly) moving sun is replaced by a stationary mass bolted down at the location
of the earth–sun CM. This new mass is now a fraction M /M as far away from the earth
as the sun was. Therefore, since the gravitational force is proportional to 1/r 2 , if we make
our new mass equal to (M /M )2 M , then it will exert the same force on the earth that the
sun exerted. So if mother earth has her eyes closed, she’ll never know the difference. The
periods of these two systems must therefore be the same. But from Eq. (7.36), the period of
the second system, which has a bolted-down mass, is
T2 =
in agreement with Eq. (7.49) .
7.5
4π 2 a3E
,
G · (M /M )2 M
(7.51)
♣
Problems
Section 7.2: The effective potential
7.1. Exponential spiral *
Given L, find the V (r) that leads to a spiral path of the form r = r0 eaθ .
Choose E to be zero. Hint: Obtain an expression for r˙ that contains no
θ ’s, and then use Eq. (7.9).
7.2. Cross section **
A particle moves in a potential, V (r) = −C/(3r 3 ).
(a) Given L, find the maximum value of the effective potential.
(b) Let the particle come in from infinity with speed v0 and impact
parameter b. In terms of C, m, and v0 , what is the largest value of
b (call it bmax ) for which the particle is captured by the potential?
In other words, what is the “cross section” for capture, π b2max , for
this potential?
7.3. Maximum L ***
2 2
A particle moves in a potential, V (r) = −V0 e−λ r .
(a) Given L, find the radius of the stable circular orbit. An implicit
equation is fine.
(b) It turns out that if L is too large, then no circular orbit exists. What
is the largest value of L for which a circular orbit does in fact exist?
7.5 Problems
If r0 is the radius of the circle in this cutoff case, what is the value
of Veff (r0 )?
Section 7.4: Gravity, Kepler’s laws
7.4. r k potential ***
A particle of mass m moves in a potential given by V (r) = βr k . Let the
angular momentum be L.
(a) Find the radius, r0 , of the circular orbit.
(b) If the particle is given a tiny kick so that the radius oscillates
around r0 , find the frequency, ωr , of these small oscillations in r.
(c) What is the ratio of the frequency ωr to the frequency of the
˙ Give a few values of k for
(nearly) circular motion, ωθ ≡ θ?
which the ratio is rational, that is, for which the path of the nearly
circular motion closes back on itself.
7.5. Spring ellipse ***
Aparticle moves in a V (r) = βr 2 potential. Following the general strategy
in Sections 7.4.1 and 7.4.3, show that the particle’s path is an ellipse.
7.6. β/r 2 potential ***
A particle is subject to a V (r) = β/r 2 potential. Following the general
strategy in Section 7.4.1, find the shape of the particle’s path. You will
need to consider various cases for β.
7.7. Rutherford scattering ***
A particle of mass m travels in a hyperbolic orbit past a mass M , whose
position is assumed to be fixed. The speed at infinity is v0 , and the impact
parameter is b (see Exercise 7.14).
(a) Show that the angle through which the particle is deflected is
φ = π − 2 tan−1(γ b)
=⇒
b=
1
φ
cot
γ
2
,
(7.52)
where γ ≡ v02 /GM .
(b) Let dσ be the cross-sectional area (measured when the particle
is initially at infinity) that gets deflected into a solid angle of size
d at angle φ.11 Show that
dσ
1
.
=
2
d
4γ sin4 (φ/2)
11
(7.53)
The solid angle of a patch on a sphere is the area of the patch divided by the square of the sphere’s
radius. So a whole sphere subtends a solid angle of 4π steradians (the name for one unit of solid
angle).
297
298
Central forces
This quantity is called the differential cross section. Note: the
label of this problem, Rutherford scattering, actually refers to
the scattering of charged particles. But since the electrostatic and
gravitational forces are both inverse-square laws, the scattering
formulas look the same, except for a few constants.
7.6
Exercises
Section 7.1: Conservation of angular momentum
7.8. Wrapping around a pole *
A puck of mass m sliding on frictionless ice is attached by a horizontal
string of length to a thin vertical pole of radius R. The puck initially
travels in (essentially) a circle around the pole at speed v0 . The string
wraps around the pole, and the puck gets drawn in and eventually hits
the pole. What quantity is conserved during this motion? What is the
puck’s speed right before it hits the pole?
7.9. String through a hole *
Ablock of mass m sliding on a frictionless table is attached to a horizontal
string that passes through a tiny hole in the table. The block initially travels in a circle of radius around the hole at speed v0 . If you slowly pull
the string down through the hole, what quantity is conserved during this
motion? What is the block’s speed when it is a distance r from the hole?
Section 7.2: The effective potential
7.10. Power-law spiral **
Given L, find the V (r) that leads to a spiral path of the form r = r0 θ k .
Choose E to be zero. Hint: Obtain an expression for r˙ that contains
no θ’s, and then use Eq. (7.9).
Section 7.4: Gravity, Kepler’s laws
7.11. Circular orbit *
For a circular orbit, derive Kepler’s third law from scratch, using
F = ma.
7.12. Falling into the sun *
Imagine that the earth is suddenly (and tragically) stopped in its orbit,
and then allowed to fall radially into the sun. How long will this take?
Use data from Appendix J, and assume that the initial orbit is essentially
circular. Hint: Consider the radial path to be part of a very thin ellipse.
7.6 Exercises
7.13. Intersecting orbits **
Two masses, m and 2m, orbit around their CM. If the orbits are circular,
they don’t intersect. But if they are very elliptical, they do. What is the
smallest value of the eccentricity for which they intersect?
7.14. Impact parameter **
Show that the distance b defined in Eq. (7.34) and Fig. 7.9 is equal to
the impact parameter. Do this:
(a) Geometrically, by showing that b is the distance from the origin
to the dotted line in Fig. 7.9.
(b) Analytically, by letting the particle come in from infinity at speed
v0 and impact parameter b , and then showing that the b in
Eq. (7.34) equals b .
7.15. Closest approach **
A particle with speed v0 and impact parameter b starts far away from a
planet of mass M .
(a) Starting from scratch (that is, without using any of the results from
Section 7.4), find the distance of closest approach to the planet.
(b) Use the results of the hyperbola discussion in Section 7.4.3
to show that the distance of closest approach to the planet is
k/( + 1), and then show that this agrees with your answer to
part (a).
7.16. Skimming a planet **
A particle travels in a parabolic orbit in a planet’s gravitational field and
skims the surface at its closest approach. The planet has mass density ρ.
Relative to the center of the planet, what is the angular velocity of the
particle as it skims the surface?
7.17. Parabola L **
A mass m orbits around a planet of mass M in a parabolic orbit of the
form y = x2 /(4 ), which has focal length . Find the angular momentum
in three different ways:
(a) Find the speed at closest approach.
(b) Use Eq. (7.30).
(c) Consider the point (x, x2 /4 ), where x is very large. Find approximate expressions for the speed and impact parameter at this
point.
7.18. Circle to parabola **
A spaceship travels in a circular orbit around a planet. It applies a sudden
thrust and increases its speed by a factor f . If the goal is to change the
orbit from a circle to a parabola, what should f be if the thrust points in
299
300
Central forces
the tangential direction? Is your answer any different if the thrust points
in some other direction? What is the distance of closest approach if the
thrust points in the radial direction?
7.19. Zero potential **
A particle is subject to a constant potential, which we will take to be zero
(equivalently, consider the α ≡ GMm = 0 limit). Following the general
strategy in Section 7.4, show that the particle’s path is a straight line.
7.20. Ellipse axes **
Taking it as given that Eq. (7.25) describes an ellipse for 0 < < 1,
calculate the lengths of the semi-major and semi-minor axes, and show
that the results agree with Eq. (7.33).
7.21. Repulsive potential **
Consider an “anti-gravitational” potential (or more mundanely, the electrostatic potential between two like charges), V (r) = α/r, where α > 0.
What is the basic change in the analysis of Section 7.4? Show that
circular, elliptical, and parabolic orbits do not exist. Draw the figure
analogous to Fig. 7.9 for the hyperbolic orbit.
7.7
Solutions
7.1. Exponential spiral
The given information r = r0 eaθ yields (using θ˙ = L/mr 2 )
r˙ = a(r0 eaθ )θ˙ = ar
L
mr 2
=
aL
.
mr
(7.54)
Plugging this into Eq. (7.9) gives
2
m aL
2 mr
+
L2
+ V (r) = E = 0.
2mr 2
(7.55)
Therefore,
V (r) = −
(1 + a2 )L2
.
2mr 2
(7.56)
7.2. Cross section
(a) The effective potential is
Veff (r)
Veff (r) =
E
E
Fig. 7.11
(7.57)
Setting the derivative equal to zero gives r = mC/L2 . Plugging this into
Veff (r) gives
max
Veff
rmin
L2
C
− 3.
2mr 2
3r
r
max
Veff
=
L6
.
6m3 C 2
(7.58)
max , then the particle will reach a
(b) If the energy E of the particle is less than Veff
minimum value of r, and then head back out to infinity (see Fig. 7.11). If E is
7.7 Solutions
301
max , then the particle will head all the way in to r = 0, never to
greater than Veff
max < E. Using L = mv b and
return. The condition for capture is therefore Veff
0
2
E = E∞ = mv0 /2, this condition becomes
mv02
(mv0 b)6
<
3
2
6m C
2
=⇒
b<
1/6
3C 2
m2 v04
≡ bmax .
(7.59)
The cross section for capture is therefore
σ = π b2max = π
1/3
3C 2
m2 v04
.
(7.60)
It makes sense that this should increase with C and decrease with m and v0 .
7.3. Maximum L
(a) The effective potential is
Veff (r) =
L2
2 2
− V0 e−λ r .
2mr 2
(7.61)
A circular orbit exists at the value(s) of r for which Veff (r) = 0. Setting the
derivative equal to zero and solving for L2 gives
L2 = (2mV0 λ2 )r 4 e−λ
2 r2
.
(7.62)
This implicitly determines r. As long as L isn’t too large, Veff (r) looks something
like the graph in Fig. 7.12, although it doesn’t necessarily dip down to negative
values; see the remark below. You can arrive at this picture by noting that for any
L, Veff (r) behaves like 1/r 2 for both r → 0 and r → ∞; and for sufficiently
small L, Veff (r) reaches negative values somewhere in between, due to the
−V0 term. The curve must therefore look like the one shown, which has two
locations where Veff (r) = 0. The smaller solution is the one with the stable orbit.
However, if L is too large, then there are no solutions to Veff (r) = 0, because
Veff (r) decreases monotonically to zero (because L2 /2mr 2 does so). We’ll be
quantitative about this in part (b).
2 2
(b) The function r 4 e−λ r on the right-hand side of Eq. (7.62) has a maximum
value, because it goes to zero for both r → 0 and r → ∞. Therefore, there
is a maximum value of L for which a solution for r exists. The maximum of
2 2
r 4 e−λ r occurs where
0=
d(r 4 e−λ
dr
2 r2
)
= e−λ
2 r2
4r 3 + r 4 (−2λ2 r)
Veff (r)
=⇒
r2 =
r
Fig. 7.12
2
≡ r02 .
λ2
(7.63)
Plugging r0 into Eq. (7.62) gives
L2max =
Plugging r0 and
L2max
8mV0
.
λ2 e2
(7.64)
Veff (r)
into Eq. (7.61) gives
V0
(for L = Lmax ).
(7.65)
e2
Note that this is greater than zero. For the L = Lmax case, the graph of Veff is
shown in Fig. 7.13. This is the cutoff case between having a dip in the graph,
and decreasing monotonically to zero.
Veff (r0 ) =
Remark: A common error in this problem is to say that the condition for a
circular orbit to exist is that Veff (r) < 0 at the point where Veff (r) is minimum.
r
Fig. 7.13
302
Central forces
The logic here is that since the goal is to have a well in which the particle can be
trapped, it seems like we just need Veff to achieve a value less than the value at
r = ∞, namely 0. However, this gives the wrong answer (L2max = 2mV0 /λ2 e,
as you can show), because Veff (r) can look like the graph in Fig. 7.14. This has
a local minimum with Veff (r) > 0. ♣
Veff (r)
r
Fig. 7.14
7.4. r k potential
(a) A circular orbit exists at the value of r for which the derivative of the effective
potential (which is the negative of the effective force) is zero. This is simply the
statement that the right-hand side of Eq. (7.8) equals zero, so that r¨ = 0. Since
V (r) = βkr k−1 , Eq. (7.8) gives
L2
− βkr k−1 = 0
mr 3
=⇒
r0 =
L2
mβk
1/(k+2)
.
(7.66)
If k is negative, then β must also be negative if there is to be a real solution
for r0 .
(b) The long method of finding the frequency is to set r(t) ≡ r0 + (t), where
represents the small deviation from the circular orbit, and to then plug this
expression for r into Eq. (7.8). The result (after making some approximations) is
a harmonic-oscillator equation of the form ¨ = −ωr2 . This general procedure,
which was described in detail in Section 6.7, will work fine here (as you are
encouraged to show), but let’s use an easier method.
By introducing the effective potential, we have reduced the problem to a onedimensional problem in the variable r. Therefore, we can make use of the result
in Section 5.2, where we found in Eq. (5.20) that to find the frequency of small
oscillations, we just need to calculate the second derivative of the potential. For
the problem at hand, we must use the effective potential, because that is what
determines the motion of the variable r. We therefore have
ωr =
Veff (r0 )
.
m
(7.67)
If you work through the r ≡ r0 + method described above, you will find
that you are basically calculating the second derivative of Veff , but in a rather
cumbersome way.
Using the form of the effective potential, we have
Veff (r0 ) =
3L2
1
+ βk(k − 1)r0k−2 = 4
mr04
r0
3L2
+ βk(k − 1)r0k+2 .
m
(7.68)
Using the r0 from Eq. (7.66), this simplifies to
Veff (r0 ) =
L2 (k + 2)
mr04
=⇒
ωr =
√
Veff (r0 )
L k +2
.
=
m
mr02
(7.69)
We could get rid of the r0 here by using Eq. (7.66), but this form of ωr will be
more useful in part (c).
Note that we must have k > −2 for ωr to be real. If k < −2, then Veff (r0 ) < 0,
which means that we have a local maximum of Veff , instead of a local minimum.
In other words, the circular orbit is unstable. Small perturbations grow, instead
of oscillating around zero.
(c) Since L = mr02 θ˙ for the circular orbit, we have ωθ ≡ θ˙ = L/(mr02 ). Combining
this with Eq. (7.69), we find
√
ωr
= k + 2.
(7.70)
ωθ
7.7 Solutions
k = -1
k=7
k=2
k = -7/4
Fig. 7.15
A few values of k that yield rational values for this ratio are (the plots of the
orbits are shown in Fig. 7.15):
• k = −1 =⇒ ωr /ωθ = 1: This is the gravitational potential. The variable r
makes one oscillation for each complete revolution of the (nearly) circular
orbit.
• k = 2 =⇒ ωr /ωθ = 2: This is the spring potential. The variable r makes
two oscillations for each complete revolution.
• k = 7 =⇒ ωr /ωθ = 3: The variable r makes three oscillations for each
complete revolution.
• k = −7/4 =⇒ ωr /ωθ = 1/2: The variable r makes half of an oscillation
for each complete revolution. So we need to have two revolutions to get
back to the same value of r.
There is an infinite number of k values that yield closed orbits. But note that
this statement applies only to orbits that are nearly circular. Also, the “closed”
nature of the orbits is only approximate, because it is based on Eq. (7.67) which
is an approximate result based on small oscillations. The only k values that
lead to exactly closed orbits for any initial conditions are k = −1 (gravity) and
k = 2 (spring), and in both cases the orbits are ellipses. This result is known as
Bertrand’s theorem; see Brown (1978).
7.5. Spring ellipse
With V (r) = βr 2 , Eq. (7.16) becomes
1 dr
r 2 dθ
2
=
2mE
1
2mβr 2
− 2 −
.
2
L
r
L2
(7.71)
As stated in Section 7.4.1, we could take a square root, separate variables, integrate to
find θ(r), and then invert to find r(θ ). But let’s solve for r(θ) in a slick way, as we
did for the gravitational case, where we made the change of variables, y ≡ 1/r. Since
there are lots of r 2 terms floating around in Eq. (7.71), it’s reasonable to try the change
of variables, y ≡ r 2 or y ≡ 1/r 2 . The latter turns out to be the better choice. So, using
y ≡ 1/r 2 and dy/dθ = −2(dr/dθ)/r 3 , and multiplying Eq. (7.71) through by 1/r 2 ,
we obtain
1 dy
2 dθ
2
=
2mEy
2mβ
− y2 − 2 .
L2
L
=− y−
2
mE
L2
−
2mβ
+
L2
mE
L2
2
.
(7.72)
Defining z ≡ y − mE/L2 for convenience, we have
dz
dθ
2
= −4z 2 + 4
mE
L2
≡ −4z 2 + 4B2 .
2
1−
2βL2
mE 2
(7.73)
303
304
Central forces
As in Section 7.4.1, we can just look at this equation and observe that
z = B cos 2(θ − θ0 )
(7.74)
is the solution. We can rotate the axes so that θ0 = 0, so we’ll drop the θ0 from
here on. Recalling our definition z ≡ 1/r 2 − mE/L2 and also the definition of B from
Eq. (7.73), Eq. (7.74) becomes
mE
1
= 2 (1 + cos 2θ ),
r2
L
(7.75)
where
≡
1−
2βL2
.
mE 2
(7.76)
It turns out, as we’ll see below, that is not the eccentricity of the ellipse, as it was in
the gravitational case.
We will now use the procedure in Section 7.4.3 to show that Eq. (7.76) represents
an ellipse. For convenience, let
L2
.
mE
k≡
(7.77)
Multiplying Eq. (7.75) through by kr 2 , and using
cos 2θ = cos2 θ − sin2 θ =
x2
y2
− 2,
r2
r
(7.78)
and also r 2 = x2 + y2 , we obtain k = (x2 + y2 ) + (x2 − y2 ). This can be written as
x2
y2
+ 2 = 1,
a2
b
y
b
b
c
a
Fig. 7.16
x
a=
where
k
1+
,
k
1−
and b =
.
(7.79)
This is the equation for an ellipse with its center located at the origin (as opposed to a
focus located at the origin, as in the gravitational case). In Fig. 7.16, the√semi-major
and semi-minor axes are b and a, respectively, and the focal length is c = b2 − a2 =
√
2k /(1 − 2 ). The eccentricity is c/b = 2 /(1 + ).
Remark: If = 0, then a = b, which means that the ellipse is actually a circle. Let’s
see if this makes sense. Looking at Eq. (7.76), we see that we want to show that circular
motion implies 2βL2 = mE 2 . For circular motion, the radial F = ma equation is
mv 2 /r = 2βr =⇒ v 2 = 2βr 2 /m. The energy is therefore E = mv 2 /2 + βr 2 = 2βr 2 .
Also, the square of the angular momentum is L2 = m2 v 2 r 2 = 2mβr 4 . Therefore,
2βL2 = 2β(2mβr 4 ) = m(2βr 2 )2 = mE 2 , as we wanted to show. ♣
7.6. β/r 2 potential
With V (r) = β/r 2 , Eq. (7.16) becomes
1 dr
r 2 dθ
2
=
2mE
1
2mβ
− 2 − 2 2
L2
r
r L
=
2mE
1
− 2
L2
r
1+
2mβ
L2
.
(7.80)
Letting y ≡ 1/r, and using dy/dθ = −(1/r 2 )(dr/dθ ), this becomes
dy
dθ
2
+ a2 y 2 =
2mE
,
L2
where a2 ≡ 1 +
2mβ
.
L2
(7.81)
7.7 Solutions
We must now consider various possibilities for a2 . These possibilities depend on how
β compares to L2 , which depends on the initial conditions of the motion. In what
follows, note that the effective potential equals
Veff (r) =
L2
β
a 2 L2
+ 2 =
.
2
2mr
r
2mr 2
2mE
sin aθ .
L2
Veff (r) ~ 1/r2
(7.82)
E
Case 1: a2 > 0, or equivalently β > −L2 /2m. In this case, the effective potential
looks like the graph in Fig. 7.17. The solution for y in Eq. (7.81) is a trig function,
which we will take to be a “sin” by appropriately rotating the axes. Using y ≡ 1/r, we
obtain
1
1
=
r
a
305
(7.83)
θ = 0 and θ = π/a make the right-hand side equal to zero, so they correspond to
r = ∞. And θ = π/2a makes the right-hand side maximum, so it corresponds to
the minimum value of r, which is rmin = a L2 /2mE. This minimum r can also be
obtained in a much quicker manner by finding where Veff (r) = E.
If the particle comes in from infinity at θ = 0, it eventually heads back out to infinity
at θ = π/a. The angle that the outgoing path makes with the incoming path is therefore
π/a. So if a is large (that is, if β is large and positive, or if L is small), then the particle
bounces nearly straight backward. If a is small (that is, if β is negative and if L2 is
only slightly larger than −2mβ), then the particle spirals around many times before it
pops back out to infinity.
A few special cases are: (1) β = 0 =⇒ a = 1, which means that the total angle
is π , that is, there is no net deflection. In fact, the particle’s path is a straight line,
because the potential is zero; see Exercise 7.19. (2) L2 = −8mβ/3 =⇒ a = 1/2,
which means that the total angle is 2π , that is, the final line of the particle’s motion
is (anti)parallel to the initial line. These lines are shifted sideways from one another,
with the separation depending on the initial conditions.
r
rmin
Fig. 7.17
E
Veff (r) = 0
r
Fig. 7.18
E
−L2 /2m.
Case 2: a = 0, or equivalently β =
In this case, the effective potential is
identically zero, as shown in Fig. 7.18. Equation (7.81) becomes
dy
dθ
2
=
2mE
.
L2
(7.84)
The solution to this is y = θ 2mE/L2 + C, which gives
r=
1
θ
L2
,
2mE
Veff (r) ~ -1/r2
(7.85)
Fig. 7.19
where we have set the integration constant C equal to zero by choosing θ = 0 to be
2
the angle that
√ corresponds to r = ∞. Note that we can use β = −L /2m to write r as
r = (1/θ) −β/E.
Since the effective potential is flat, the rate of change of r is constant. Therefore, if
the particle has r˙ < 0, it will reach the origin in finite time, even though Eq. (7.85)
says that it will spiral around the origin an infinite number of times (because θ → ∞
as r → 0).
Case 3: a2 < 0, or equivalently β < −L2 /2m. In this case, we have the situation
shown in either Fig. 7.19 or Fig. 7.20, depending on the sign of E. For convenience,
let b be the positive real number such that b2 = −a2 . Then Eq. (7.81) becomes
dy
dθ
2
− b2 y 2 =
2mE
.
L2
r
(7.86)
rmax
r
E
Veff (r) ~ -1/r2
Fig. 7.20
306
Central forces
The solution to this equation is a hyperbolic trig function. But we must consider two
cases:
• E > 0: Using the identity cosh2 z − sinh2 z = 1, and recalling y ≡ 1/r, we see
that the solution to Eq. (7.86) is12
1
1
=
r
b
2mE
sinh bθ.
L2
(7.87)
Unlike the a2 > 0 case above, the sinh function has no maximum value. Therefore,
the right-hand side can head to infinity, which means that r can head to zero, if
the initial r˙ is negative. It does so in a finite time, because r˙ only becomes more
negative as time goes by, from Fig. 7.19. For large z, we have sinh z ≈ ez /2, so
r heads to zero like e−bθ . The particle will therefore spiral around the origin an
infinite number of times (because θ → ∞ as r → 0).
• E < 0: In this case, Eq. (7.86) can be rewritten as
b2 y2 −
dy
dθ
2
=
2m|E|
.
L2
(7.88)
The solution to this equation is13
1
1 2m|E|
=
cosh bθ .
(7.89)
r
b
L2
As in the sinh case, the cosh function has no maximum value. Therefore, the
right-hand side can head to infinity, which means that r can head to zero. But in
the present cosh case, the right-hand side does achieve a nonzero minimum value,
when θ = 0. So r achieves a maximum value (if the initial r˙ is positive) equal to
rmax = b L2 /2m|E|. This is clear from Fig. 7.20. This maximum r can also be
obtained by simply finding where Veff (r) = E. After reaching rmax , the particle
heads back down to the origin with behavior similar (for large θ ) to the sinh case.
y
b
7.7. Rutherford scattering
a
(a) From Exercise 7.14, we know that the impact parameter b equals the distance b
shown in Fig. 7.9. Therefore, Fig. 7.21 tells us that the angle of deflection (the
angle between the initial and final velocity vectors) is
x
φ
φ = π − 2 tan−1
Fig. 7.21
b
.
a
(7.90)
But from Eqs. (7.34) and (7.26), we have
b
=
a
2
−1=
2EL2
=
mα 2
2(mv02 /2)(mv0 b)2
v2 b
= 0 .
2
m(GMm)
GM
(7.91)
Substituting this into Eq. (7.90), with γ ≡ v02 /(GM ), gives the first expression
in Eq. (7.52). Dividing by 2 and taking the cotangent of both sides then gives
the second expression,
b=
φ
1
cot
γ
2
.
(7.92)
It actually isn’t necessary to go through all the work of Section 7.4.3 to obtain
this result via a and b. We can just use Eq. (7.25), which says that r → ∞ when
12
13
More generally, we should write sinh(θ −θ0 ) here. But we can eliminate the need for θ0 by picking
θ = 0 to be the angle that corresponds to r = ∞.
Again, we should write cosh(θ − θ0 ) here. But we can eliminate the need for θ0 by picking θ = 0
to be the angle that corresponds to the maximum value of r.
7.7 Solutions
cos θ →√−1/ . This then
√ implies that the dotted lines in Fig. 7.21 have slope
2 − 1, which reproduces Eq. (7.91).
tan θ = sec2 θ − 1 =
(b) Imagine a wide beam of particles moving in the positive x direction toward
the mass M . Consider a thin cross-sectional ring in this beam, with radius
b and thickness db. Now consider a very large sphere centered at M . Any
particle that passes through the cross-sectional ring of radius b will hit this
sphere in a ring located at an angle φ relative to the x axis, with an angular
spread of dφ. The relation between db and dφ is found from Eq. (7.92). Using
d(cot β)/dβ = −1/ sin2 β, we have
1
db
.
=
dφ
2γ sin2 (φ/2)
(7.93)
The area of the incident cross-sectional ring is dσ = 2πb |db|. What is the solid
angle subtended by a ring at angle φ with thickness dφ? Taking the radius of
the large sphere to be R (which will cancel out), the radius of the ring is R sin φ,
and the width is R |dφ|. The area of the ring is therefore 2π(R sin φ)(R |dφ|),
and so the solid angle subtended by the ring is d = 2π sin φ |dφ| steradians.
Therefore, the differential cross section is
2π b |db|
dσ
=
=
d
2π sin φ |dφ|
=
=
b
sin φ
db
dφ
(1/γ ) cot(φ/2)
2 sin(φ/2) cos(φ/2)
1
4γ 2 sin4 (φ/2)
1
2γ sin2 (φ/2)
.
(7.94)
Remarks: What does this “differential cross section” result tell us? It tells us
that if we want to find out how much cross-sectional area gets mapped into
the solid angle d at the angle φ, then we can use Eq. (7.94) to say (recalling
γ ≡ v02 /GM ),
dσ =
G2 M 2
4v04 sin4 (φ/2)
d
=⇒
dσ =
G 2 M 2 m2
16E 2 sin4 (φ/2)
d ,
(7.95)
where we have used E = mv02 /2 to obtain the second expression. Let’s look at
some special cases. If φ ≈ 180◦ (that is, backward scattering), then the amount
of area that gets scattered into a nearly backward solid angle of d equals
dσ = (G 2 M 2 /4v04 ) d . If v0 is small, then we see that dσ is large, that is, a
large area gets deflected nearly straight backward. This makes sense, because
with v0 ≈ 0, the orbit is essentially parabolic, which means that the initial and
final velocities at infinity are (anti)parallel. (If you release a particle from rest far
away from a gravitational source, it will come back to you. Assuming it doesn’t
bump into the source, of course.) If v0 is large, then we see that dσ is small, that
is, only a small area gets deflected backward. This makes sense, because the
particle is more likely to fly past M without much deflection if it is moving fast,
because the force has hardly any time to act. The particle needs to start with a
very small b value (which corresponds to a very small area) in order to get close
enough to M to allow there to be a large enough force to swing it around.
Another special case is φ ≈ 0, that is, negligible deflection. In this case,
Eq. (7.95) tells us that the amount of area that gets scattered into a nearly forward
solid angle of d is dσ ≈ ∞. This makes sense, because if the impact parameter
b is large (and there is an infinite cross-sectional area for which this is true), then
307
308
Central forces
the particle will hardly feel the mass M , so it will continue to move essentially
in a straight line.14
What if we consider the electrostatic force, instead of the gravitational force?
What is the differential cross section in this case? To answer this, note that we
can rewrite γ as
γ =
v02
GM
=
2(mv02 /2)
2E
≡
.
GMm
α
(7.96)
In the case of electrostatics, the force takes the form, Fe = kq1 q2 /r 2 . This
looks like the gravitational force, Fg = Gm1 m2 /r 2 , except that the constant
α is now kq1 q2 , instead of Gm1 m2 . Therefore, the γ in Eq. (7.96) becomes
γe = 2E/(kq1 q2 ). Substituting this into Eq. (7.94), or equivalently replacing
GMm by kq1 q2 in Eq. (7.95), gives the differential cross section for electrostatic
scattering,
k 2 q12 q22
dσ
.
=
2
d
16E sin4 (φ/2)
(7.97)
This is the Rutherford-scattering differential-cross-section formula. Around
1910, Rutherford and his students bombarded metal foils with alpha particles.
Their results for the distribution of scattering angles were consistent with the
above formula. In particular, they observed backward scattering of the alpha
particles. Since the above formula is based on the assumption of a point source
for the potential, this led Rutherford to his theory that atoms contained a dense
positively charged nucleus, as opposed to being made of a spread-out “plum
pudding” distribution of charge, which (as a special case of not yielding the
correct distribution of scattering angles in general) doesn’t yield backward
scattering. ♣
14
Remember, all we care about is the angle here. So when you’re picturing the large sphere of radius
R centered at M , don’t say, “If b is large (for example, R/2), then a straight-line trajectory will hit
the sphere at a large angle up above the x axis (for example, 30◦ ).” This is incorrect. If you want
to think in terms of a physical sphere of radius R, it is understood that R is infinitely large. Or
more precisely, R
b, for any impact parameter b you might choose. So even if b is “large,” it is
still small compared with R, so a straight-line trajectory will hit the sphere of radius R at an angle
that is essentially zero. Alternatively, you can just think in terms of angles, and not visualize an
actual large sphere centered at M .
Chapter 8
Angular momentum, Part I
ˆ)
(Constant L
The angular momentum of a point mass, relative to a given origin, is defined by
L = r × p.
(8.1)
For a collection of particles, the total L is simply the sum of the L’s of
all the particles. The vector r × p is a useful thing to study because it has
many nice properties. One of these is the conservation law presented in Theorem 7.1, which allowed us to introduce the “effective potential” in Section
7.2. And later in this chapter we will introduce the concept of torque, τ,
which appears in the bread-and-butter statement, τ = dL/dt (analogous to
Newton’s F = dp/dt law).
There are two basic types of angular momentum problems in the world. Since
the solution to any rotational problem invariably comes down to using τ = dL/dt,
as we will see, we must determine how L changes in time. And since L is a vector,
it can change because (1) its length changes, or (2) its direction changes (or
ˆ
through some combination of these effects). In other words, if we write L = LL,
ˆ is the unit vector in the L direction, then L can change because L changes,
where L
ˆ changes, or both.
or because L
ˆ is the easily understood one.
The first of these cases, that of constant L,
Consider a spinning record, with the center chosen as the origin. The vector
L=
r × p is perpendicular to the record, because every term in the sum has
this property. If we give the record a tangential force in the proper direction, then
it will speed up (in a precise way which we will soon determine). There is nothing
mysterious going on here. If we push on the record, it goes faster. L points in the
same direction as before, but now simply with a larger magnitude. In fact, in this
type of problem, we can completely forget that L is a vector. We can deal just
with its magnitude L, and everything will be fine. This first case is the subject of
the present chapter.
The second case, however, where L changes direction, can get rather confusing. This is the subject of the following chapter, where we will talk about
twirling tops and other kinds of spinning objects that have a tendency to make
one’s head spin too. In these situations, the entire point is that L is actually a
309
ˆ
Angular momentum, Part I (Constant L)
310
ˆ case, we really have to visualize things in
vector. And unlike in the constant-L
three dimensions to see what’s going on.1
The angular momentum of a point mass is given by the simple expression in
Eq. (8.1). But in order to deal with setups in the real world, which invariably
consist of many particles, we must learn how to calculate the angular momentum
of an extended object. This is the task of the Section 8.1. In this chapter, we’ll
deal only with rotations around the z axis, or an axis parallel to the z axis. We’ll
save the general 3-D motion for Chapter 9.
y
ω
8.1
CM
V
x
Pancake object in x -y plane
Consider a flat rigid body undergoing arbitrary motion (both translating and
spinning) in the x-y plane; see Fig. 8.1. What is the angular momentum of this
body, relative to the origin of the coordinate system?2 If we imagine the body to
consist of particles of mass mi , then the angular momentum of the entire body is
the sum of the angular momenta of each mi , which is Li = ri × pi . So the total
angular momentum is
Fig. 8.1
L=
ri × p i .
(8.2)
i
For a continuous distribution of mass, we would have an integral instead of a
sum. L depends on the locations and momenta of the masses. The momenta in
turn depend on how fast the body is translating and spinning. Our goal here is
to find the dependence of L on the distribution and motion of the constituent
masses. The result will involve the geometry of the body in a specific way, as we
will show.
In this section, we will deal only with pancake-like objects that move in the
x-y plane. We will calculate L relative to the origin, and we will also derive
an expression for the kinetic energy. We will deal with non-pancake objects in
Section 8.2. Note that since both the r and p of all the masses in our pancake-like
objects always lie in the x-y plane, the vector L =
r × p always points in
the zˆ direction. As mentioned above, this fact is what makes these pancake cases
easy to deal with. L changes only because its length changes, not its direction.
So when we eventually get to the τ = dL/dt equation, it will take on a simple
form. Let’s first look at a special case, and then we’ll look at general motion in
the x-y plane.
1
2
The difference between these two cases is essentially the same as the difference between the two
basic F = dp/dt cases. The vector p can change because its magnitude changes, in which case
we have F = ma (assuming that m is constant). Or, p can change because its direction changes,
in which case we have the centripetal-acceleration statement, F = mv 2 /r. (Or there could be a
combination of these effects.) The former case seems a bit more intuitive than the latter.
Remember, L is defined relative to a chosen origin, because it has the vector r in it. So it makes
no sense to ask what L is, without specifying what origin we’ve chosen.
8.1 Pancake object in x -y plane
8.1.1
y
Rotation about the z axis
The pancake in Fig. 8.2 is pivoted at the origin and rotates with angular speed ω
around the z axis, in the counterclockwise direction (as viewed from above).
Consider a little piece of the body, with mass dm and position (x, y). This
little piece travels in a circle around the origin with speed v = ωr, where
r = x2 + y2 . Therefore, the angular momentum of this piece (relative to the
origin) equals L = r × p = r(v dm)ˆz = dm r 2 ωˆz. The zˆ direction arises from
the cross product of the (orthogonal) vectors r and p. The angular momentum of
the entire body is therefore
L=
r 2 ωˆz dm =
(x2 + y2 )ωˆz dm,
(8.3)
where the integration runs over the area of the body. If the density of the object
is constant, as is usually the case, then we have dm = ρ dx dy. If we define the
moment of inertia around the z axis to be
Iz ≡
r 2 dm =
(x2 + y2 ) dm,
(8.4)
then the z component of L is
Lz = Iz ω,
(8.5)
and both Lx and Ly are zero. In the case where the rigid body is made up of a
collection of point masses mi in the x-y plane, the moment of inertia in Eq. (8.4)
takes the discretized form,
Iz ≡
mi ri2 .
(8.6)
i
Given any rigid body in the x-y plane, we can calculate Iz . And given ω, we
can then multiply it by Iz to find Lz . In Section 8.3.1, we’ll get some practice
calculating various moments of inertia.
What is the kinetic energy of our object? We need to add up the energies of
all the little pieces. A little piece has energy dm v 2 /2 = dm(rω)2 /2. So the total
kinetic energy is
T =
r 2 ω2
dm.
2
(8.7)
With our definition of Iz in Eq. (8.4), this becomes
T =
311
Iz ω 2
.
2
(8.8)
This is easy to remember, because it looks a lot like the kinetic energy of a point
mass, mv 2 /2.
ω
x
Fig. 8.2
ˆ
Angular momentum, Part I (Constant L)
312
y
8.1.2
ω
CM
V
x
Fig. 8.3
y
r'
CM
R
r
x
General motion in x -y plane
How do we deal with general motion in the x-y plane? For the motion in Fig. 8.3,
where the object is both translating and spinning, the various pieces of mass don’t
travel in circles around the origin, so we can’t write v = ωr as we did above.
It turns out to be highly advantageous to write the angular momentum, L, and
the kinetic energy, T , in terms of the center-of-mass (CM) coordinates and the
coordinates relative to the CM. The expressions for L and T take on very nice
forms when written this way, as we now show.
Let the position of the CM relative to a fixed origin be R = (X , Y ). And let
the position of a given point relative to the CM be r = (x , y ). Then the position
of the given point relative to the fixed origin is r = R + r (see Fig. 8.4). Let
the velocity of the CM be V, and let the velocity relative to the CM be v . Then
v = V + v . Let the body rotate with angular speed ω around the CM (around
an instantaneous axis parallel to the z axis, so that the pancake remains in the x-y
plane at all times).3 Then v = ω r .
Let’s look at L first. Let M be the total mass of the pancake. The angular
momentum relative to the origin is
L=
Fig. 8.4
r × v dm
=
(R + r ) × (V + v ) dm
=
R × V dm +
= MR × V +
r × v dm
(cross terms vanish; see below)
r 2 ω dm zˆ
≡ R × P + IzCM ω zˆ .
(8.9)
In going from the second to third line above, the cross terms, r × V dm and
R × v dm, vanish by definition of the CM, which says that r dm = 0.
(See Eq. (5.58); basically, the position of the CM in the CM frame is zero.) This
implies that v dm = d( r dm)/dt also equals zero. And since we can pull the
constant vectors V and R out of the above integrals, we are therefore left with
zero. The quantity IzCM in the final result is the moment of inertia around an axis
through the CM, parallel to the z axis. Equation (8.9) is a very nice result, and it
is important enough to be called a theorem. In words, it says:
Theorem 8.1 The angular momentum (relative to the origin) of a body can be
found by treating the body like a point mass located at the CM and finding the
3
What we mean here is the following. Consider a coordinate system whose origin is the CM and
whose axes are parallel to the fixed x and y axes. Then the pancake rotates with angular speed ω
with respect to this system.
8.1 Pancake object in x -y plane
angular momentum of this point mass relative to the origin, and by then adding
on the angular momentum of the body relative to the CM. 4
Note that if we have the special case where the CM travels around the origin
in a circle with angular speed
(so that V = R), then Eq. (8.9) becomes
L = MR2 + IzCM ω zˆ .
Now let’s look at T . The kinetic energy is
T =
=
=
=
≡
1 2
v dm
2
1
|V + v |2 dm
2
1
1
V 2 dm +
v 2 dm (cross term vanishes; see below)
2
2
1
1
MV 2 +
r 2 ω 2 dm
2
2
1
1
MV 2 + IzCM ω 2 .
(8.10)
2
2
In going from the second to third line above, the cross term V · v dm =
V · v dm vanishes by definition of the CM, as in the above calculation of L.
Again, Eq. (8.10) is a very nice result. In words, it says:
Theorem 8.2 The kinetic energy of a body can be found by treating the body
like a point mass located at the CM, and by then adding on the kinetic energy of
the body due to the motion relative to the CM. 5
To calculate E, my dear class,
Just add up two things, and you’ll pass.
Take the CM point’s E,
And then add on with glee,
The E ’round the center of mass.
Example (Cylinder on a ramp): A cylinder of mass m, radius r, and moment of
inertia I = (1/2)mr 2 (this is the I for a solid cylinder around its center, as we’ll see
in Section 8.3.1) rolls without slipping down a plane inclined at an angle θ. What is
the acceleration of the center of the cylinder?
4
5
This theorem works only if we use the CM as the location of the imagined point mass. True, in
the above analysis we could have chosen a point P other than the CM, and then written things
in terms of the coordinates of P and the coordinates relative to P (which could also be described
by a rotation). But then the cross terms in Eq. (8.9) wouldn’t vanish, and we’d end up with an
unenlightening mess.
We already knew this from Section 5.6.2. It’s just that now we know that the kinetic energy in the
CM frame takes the form of IzCM ω 2 /2.
313
ˆ
Angular momentum, Part I (Constant L)
314
Solution: We’ll use conservation of energy to determine the speed v of the center
of the cylinder after it has moved a distance d down the plane, and then
√ we’ll read
off a from the standard constant-acceleration kinematic relation, v = 2ad.
The loss in potential energy of the cylinder is mgd sin θ. This shows up as kinetic
energy, which equals mv 2 /2+I ω2 /2 from Theorem 8.2. But the nonslipping condition
is v = ωr. Therefore, ω = v/r, and conservation of energy gives
1 2
mv +
2
1
= mv 2 +
2
mgd sin θ =
=
1 2
Iω
2
1 1 2
mr
2 2
v 2
r
3 2
mv .
4
So the speed as a function of distance is v = (4/3)gd sin θ. Hence, v =
gives a = (2/3)g sin θ, which is independent of r.
(8.11)
√
2ad
Remarks: Our answer is 2/3 of the g sin θ result for a block sliding down a frictionless
plane. It is smaller because there is kinetic energy “wasted” in the rotational motion.
Alternatively, it is smaller because there is a friction force pointing up the plane (to provide
the torque necessary to get the cylinder rotating, but we’ll talk about torque in Section 8.4),
so this decreases the net force down the plane.
If we let I take the general form I = βmr 2 , where β is a numerical factor, then you can
show that the acceleration becomes a = g sin θ/(1 + β). So if β = 0 (all the mass is at
the center), then we simply have a = g sin θ, so the cylinder behaves just like a sliding
block. If β = 1 (all the mass is on the rim), then a = (1/2)g sin θ. If β → ∞,6 then
a → 0. After reading Section 8.4, you can think about these special cases alternatively in
terms of the forces and torques involved.
Although this problem can also be solved by using force and torque (the task of Exercise 8.37), the conservation-of-energy method is generally quicker in more complicated
problems of this type, as you can see by doing, for example, Exercises 8.28 and 8.46. ♣
8.1.3
ω
stick
glu
e
y
CM
R
x
The parallel-axis theorem
Consider the special case where the CM rotates around the origin at the same
rate as the body rotates around the CM. This can be achieved, for example, by
gluing a stick across the pancake and pivoting one end of the stick at the origin;
see Fig. 8.5. In this special case, we have the simplified situation where all points
in the pancake travel in circles around the origin. Let their angular speed be ω.
Then the speed of the CM is V = ωR, so Eq. (8.9) gives the angular momentum
around the origin as
Lz = (MR2 + IzCM )ω.
(8.12)
Fig. 8.5
6
This can be obtained by adding on long spokes protruding from the cylinder and having them pass
through a deep groove in the plane, or by having a spool with a very small inner radius roll down
a thin plane with only its inner “axle” rolling on the plane.
8.1 Pancake object in x -y plane
In other words, the moment of inertia around the origin is
Iz = MR2 + IzCM .
(8.13)
This is the parallel-axis theorem. It says that once you’ve calculated the moment
of inertia of an object around the axis passing through the CM (namely IzCM ),
then if you want to calculate the moment of inertia around a parallel axis, you
simply have to add on MR2 , where R is the distance between the two axes, and
M is the mass of the object. Note that since the parallel-axis theorem is a special
case of the result in Eq. (8.9), it is valid only with the CM, and not with any other
point. The parallel-axis theorem actually holds for arbitrary nonplanar objects
too, as we’ll see in Section 8.2. And we’ll also derive a more general form of the
theorem in Chapter 9.
We can also look at the kinetic energy in this special case where the CM
rotates around the origin at the same rate as the body rotates around the CM.
Using V = ωR in Eq. (8.10), we find
T =
1
1
MR2 + IzCM ω2 = Iz ω2 .
2
2
(8.14)
Example (A stick): Let’s verify the parallel-axis theorem for a stick of mass m
and length , in the case where we want to compare the moment of inertia around an
axis through an end (perpendicular to the stick) with the moment of inertia around an
axis through the CM (perpendicular to the stick).
For convenience, let ρ = m/ be the density. The moment of inertia around an axis
through an end is
I end =
0
x2 dm =
0
x2 ρ dx =
1 3
1
1
ρ = (ρ ) 2 = m 2 .
3
3
3
(8.15)
The moment of inertia around an axis through the CM is
I CM =
/2
− /2
x2 dm =
/2
− /2
x2 ρ dx =
1 3
1
1
ρ =
(ρ ) 2 =
m 2.
12
12
12
(8.16)
This is consistent with the parallel-axis theorem, Eq. (8.13), because
I end = m
2
2
+ I CM .
(8.17)
Remember that this works only with the CM. If we instead want to compare I end
with the I around a point, say, /6 from that end, then we cannot say that they differ
by m( /6)2 . But we can compare each of them to I CM and say that they differ by
( /2)2 − ( /3)2 = 5 2 /36.
315
ˆ
Angular momentum, Part I (Constant L)
316
y
8.1.4
pancake
The perpendicular-axis theorem
This theorem is valid only for pancake objects. Consider a pancake object in the
x-y plane (see Fig. 8.6). Then the perpendicular-axis theorem says that
Iz = Ix + Iy ,
(8.18)
x
Fig. 8.6
where Ix and Iy are defined analogously to the Iz in Eq. (8.4). That is, to find Ix ,
imagine spinning the object around the x axis at angular speed ω, and then define
Ix ≡ Lx /ω. (Only the distance from the x axis matters in calculating the speed of
a given point. So the fact that the object has extent along the x direction and is
therefore not a pancake in the y-z plane is irrelevant. The following section has
further discussion of this.) Likewise for Iy . In other words,
Ix ≡
(y2 + z 2 ) dm,
Iy ≡
(z 2 + x2 ) dm,
Iz ≡
(x2 + y2 ) dm. (8.19)
To prove the perpendicular-axis theorem, we simply use the fact that z = 0
for our pancake object. Equation (8.19) then gives Iz = Ix + Iy . In the limited
number of situations where this theorem is applicable, it can save you some
trouble. A few examples are given in Section 8.3.1.
8.2
Nonplanar objects
In Section 8.1, we restricted the discussion to pancake objects in the x-y plane.
However, nearly all the results we derived carry over to nonplanar objects, provided that the axis of rotation is parallel to the z axis, and provided that we are
concerned only with Lz , and not Lx or Ly . So let’s drop the pancake assumption
and run through the results we obtained above.
First, consider an object rotating around the z axis. Let the object have extension in the z direction. If we imagine slicing the object into pancakes parallel to
the x-y plane, then Eqs. (8.4) and (8.5) correctly give the Lz for each pancake.
And since the Lz of the whole object is the sum of the Lz ’s of all the pancakes,
we see that the Iz of the whole object is the sum of the Iz ’s of all the pancakes.
The difference in the z values of the pancakes is irrelevant. Therefore, for any
object rotating around the z axis, we have
Iz =
(x2 + y2 ) dm,
and
Lz = Iz ω,
(8.20)
where the integration runs over the entire volume of the body. We’ll calculate the
Iz for many nonplanar objects in Section 8.3.1. Note that even though Eq. (8.20)
gives the Lz for an arbitrary object, the analysis in this chapter is still not completely general because (1) we are restricting the axis of rotation to be the (fixed)
z axis, and (2) even with this restriction, an object outside the x-y plane might
8.2 Nonplanar objects
317
have nonzero x and y components of L, but we found only the z component in
Eq. (8.20). This second fact is strange but true. We’ll deal with it in great detail
in Section 9.2.
As far as the kinetic energy goes, the T for a nonplanar object rotating around
the z axis is still given by Eq. (8.8), because we can obtain the total T by adding
up the T ’s of all the pancake slices.
Also, Eqs. (8.9) and (8.10) continue to hold for a nonplanar object in the
case where the CM is translating while the object is spinning around it (or more
precisely, spinning around an axis parallel to the z axis and passing through
the CM). The velocity V of the CM can actually point in any direction, and
these two equations are still valid. But we’ll assume in this chapter that all
velocities are in the x-y plane.
Lastly, the parallel-axis theorem still holds for a nonplanar object (the derivation using Eq. (8.9) is the same). But as mentioned in Section 8.1.4, the
perpendicular-axis theorem does not hold. This is the one instance where we
need the planar assumption.
Finding the CM
The center of mass has come up repeatedly in this chapter. For example, when we
used the parallel-axis theorem, we needed to know where the CM was. In some
cases, such as with a stick or a disk, the location is obvious. But in other cases,
it isn’t so clear. So let’s get a little practice calculating the location of the CM.
Depending on whether the mass distribution is discrete or continuous, the position
of the CM is defined by (see Eq. (5.58))
RCM =
ri mi
,
M
or
RCM =
r dm
,
M
(8.21)
where M is the total mass. We’ll do an example with a continuous mass distribution here. As is often the case with problems involving an integral, the main step
in the solution is deciding how you want to slice up the object to do the integral.
Example (Hemispherical shell): Find the location of the CM of a hollow
hemispherical shell, with uniform mass density and radius R.
Rsinθ
Solution: By symmetry, the CM is located on the line above the center of the base.
So our task reduces to finding the height, yCM . Let the mass density be σ . We’ll slice
the hemisphere up into horizontal rings, described by the angle θ above the horizontal,
as shown in Fig. 8.7. If the angular thickness of a ring is dθ, then its mass is
dm = σ dA = σ (length)(width) = σ (2πR cos θ)(R dθ).
(8.22)
θ
Rcosθ
Fig. 8.7
ˆ
Angular momentum, Part I (Constant L)
318
All points on the ring have a y value of R sin θ. Therefore,
yCM =
1
M
y dm =
π/2
1
(R sin θ)(2πR2 σ cos θ dθ)
(2πR2 )σ 0
=R
π/2
0
sin θ cos θ dθ
=
R sin2 θ π/2
2
0
=
R
.
2
(8.23)
The simple factor of 1/2 here is nice, but it’s not all that obvious. It comes from the
fact that each value of y is represented equally. If you solved the problem by doing a
dy integral instead of a dθ one, you would find that there is the same area (and hence
the same mass) in each ring of vertical height dy. In short, as y increases, the larger
tilt of the surface cancels out the smaller radius of the rings, yielding the same area.
You are encouraged to work this out.
The calculation of a CM is very similar to the calculation of a moment of
inertia. Both involve an integration over the mass of an object, but the former
has one power of a length in the integrand, whereas the latter has two powers.
8.3
Calculating moments of inertia
8.3.1
Lots of examples
Let’s now calculate the moments of inertia of various objects, around specified
axes. We’ll use ρ to denote the mass density (per unit length, area, or volume, as
appropriate), and we’ll assume that this density is uniform throughout the object.
For the more complicated objects in the list below, it is generally a good idea to
slice them up into pieces for which I is already known. The problem then reduces
to integrating over these known I ’s. There is usually more than one way to do this
slicing. For example, a sphere can be looked at as a series of concentric shells
or a collection of disks stacked on top of each other. In the examples below, you
may want to play around with slicings other than the ones given. Consider at least
a few of these examples to be problems and try to work them out for yourself.
R
1. A ring of mass M and radius R (axis through center, perpendicular to plane; Fig. 8.8):
R
Fig. 8.8
I=
r 2 dm =
2π
0
R2 ρR dθ = (2π Rρ)R2 = MR2 ,
as expected, because all of the mass is a distance R from the axis.
(8.24)
8.3 Calculating moments of inertia
319
2. A ring of mass M and radius R (axis through center, in plane; Fig. 8.8): The distance
from the axis is (the absolute value of ) R sin θ . Therefore,
r 2 dm =
I=
2π
0
(R sin θ )2 ρR dθ =
1
(2πRρ)R2 = 12 MR2 ,
2
(8.25)
where we have used sin2 θ = (1 − cos 2θ )/2. You can also find I by using the
perpendicular-axis theorem. In the notation of Section 8.1.4, we have Ix = Iy , by
symmetry. Therefore, Iz = 2Ix . Using Iz = MR2 from Example 1 then gives Ix =
MR2 /2.
3. A disk of mass M and radius R (axis through center, perpendicular to plane; Fig. 8.9):
I=
r 2 dm =
2π
0
R
0
r 2 ρr dr dθ = (R4 /4)2πρ =
1
(ρπ R2 )R2 = 12 MR2 .
2
(8.26)
You can save the (trivial) step of integrating over θ by considering the disk to be
made up of many concentric rings, and invoking Example 1. The mass of each ring is
ρ2πr dr. Integrating over the rings gives I = 0R (ρ2πr dr)r 2 = πR4 ρ/2 = MR2 /2,
as above. Slicing up the disk is fairly inconsequential in this example, but it will save
you some trouble in others.
4. A disk of mass M and radius R (axis through center, in plane; Fig. 8.9): Slice the disk
up into rings, and use Example 2.
I=
R
0
(1/2)(ρ2πr dr)r 2 = (R2 /4)ρπ =
1
(ρπR2 )R2 = 14 MR2 .
4
x2 dm =
L/2
−L/2
x2 ρ dx =
1
1 ML2 .
(ρL)L2 = 12
12
R
(8.27)
Or just use Example 3 and the perpendicular-axis theorem.
5. A thin uniform rod of mass M and length L (axis through center, perpendicular to
rod; Fig. 8.10): We already found this I and the next one in Section 8.1.3, but we’ll
include them here for completeness.
I=
R
Fig. 8.9
L
(8.28)
L
6. A thin uniform rod of mass M and length L (axis through end, perpendicular to rod;
Fig. 8.10):
I=
x2 dm =
L
0
Fig. 8.10
1
x2 ρ dx = (ρL)L2 = 13 ML2 .
3
(8.29)
7. A spherical shell of mass M and radius R (any axis through center; Fig. 8.11): Let’s
slice the sphere into horizontal ring-like strips. In spherical coordinates, the radius
of a ring is given by r = R sin θ , where θ is the angle down from the north pole.
ˆ
Angular momentum, Part I (Constant L)
320
sin3 θ =
The area of a strip is then 2π(R sin θ)R dθ. Using
− cos θ + cos3 θ/3, we have
π
r 2 dm =
I=
0
(R sin θ)2 2πρ(R sin θ)R dθ = 2πρR4
= 2πρR4 (4/3) =
R
Fig. 8.11
L
2β
L
Fig. 8.12
0
sin3 θ
2
(4πR2 ρ)R2 = 23 MR2 .
3
(8.30)
(x2 + y2 + z 2 ) dm = 2
r 2 dm.
(8.31)
If we apply this to a spherical shell, where r always equals the radius R, then the
right-hand side is 2MR2 . And since the I ’s are all equal by symmetry, they must all
be 2MR2/3.
8. A solid sphere of mass M and radius R (any axis through center; Fig. 8.11): A sphere
is made up of concentric spherical shells. The volume of a shell is 4πr 2 dr. Using
Example 7, we have
R
I=
L
π
An alternate and slick way of deriving this is to use the following result, similar in
spirit to the perpendicular-axis theorem: Adding up the three moments of inertia in
Eq. (8.19) gives
Ix + Iy + Iz = 2
R
sin θ(1 − cos2 θ) =
0
(2/3)(4πρr 2 dr)r 2 = (R5 /5)(8πρ/3) =
2
(4πR3 ρ/3)R2 = 25 MR2 .
5
(8.32)
The task of Exercise 8.33 is to derive this result by slicing the sphere into horizontal
disks.
9. An infinitesimally thin triangle of mass M and length L (axis through tip, perpendicular
to plane; Fig. 8.12): Let the base have length a, where a is infinitesimally small. Then
a tiny vertical slice a distance x from the tip has length a(x/L). If the slice has thickness
dx, then it is essentially a point mass of mass dm = ρax dx/L. Therefore,
I=
x2 dm =
L
0
x2 ρax/L dx =
1
(ρaL/2)L2 = 12 ML2 ,
2
(8.33)
because aL/2 is the area of the triangle. This has the same form as the disk in
Example 3, because a disk is made up of many of these triangles.
10. An isosceles triangle of mass M , vertex angle 2β, and common-side length L (axis
through tip, perpendicular to plane; Fig. 8.12): Let h be the altitude of the triangle
(so h = L cos β). Slice the triangle into thin strips parallel to the base. Let x be the
distance from a strip to the vertex. Then the length of a strip is = 2x tan β, and its
mass is dm = ρ(2x tan β dx). Using Example 5 above, along with the parallel-axis
theorem, we have
I=
2
dm
= 2ρ tan β
12
0
h
+ x2
=
h
tan2 β
1+
0
3
(ρ2x tan β dx)
(2x tan β)2
+ x2
12
x3 dx = 2ρ tan β 1 +
tan2 β
3
h4
.
4
(8.34)
8.3 Calculating moments of inertia
321
But the area of the entire triangle is h2 tan β, so we have I = (Mh2 /2) 1 +
(1/3) tan2 β . In terms of L = h/ cos β, this is
I = (ML2/2)(cos2 β + sin2 β/3) = 12 ML2 1 − 23 sin2 β
.
(8.35)
11. A regular N -gon of mass M and “radius” R (axis through center, perpendicular to
plane; Fig. 8.13): The N -gon is made up of N isosceles triangles, so we can use
Example 10, with β = π/N . The masses of the triangles simply add, so if M is the
mass of the whole N -gon, we have
π
I = 12 MR2 1 − 23 sin2 N
.
8.3.2
N=6
(8.36)
We can list the values of I for a few N . With the shorthand notation (N , I /MR2 ),
5 ), (∞, 1 ). These values of I form a nice arithmetic
Eq. (8.36) gives (3, 14 ), (4, 13 ), (6, 12
2
progression.
12. A rectangle of mass M and sides of length a and b (axis through center, perpendicular
to plane; Fig. 8.13): Let the z axis be perpendicular to the plane. We know that
Ix = Mb2/12 and Iy = Ma2/12 (because the extent of an object along an axis doesn’t
affect the moment around that axis, when written in terms of the total M ). So the
perpendicular-axis theorem gives
1 M (a2 + b2 ) .
Iz = Ix + Iy = 12
R
(8.37)
A neat trick
For some objects with certain symmetries, it’s possible to calculate the moment
of inertia without doing any integrals. The only things we need are a scaling
argument and the parallel-axis theorem. We’ll illustrate this technique by finding
the I for a stick around its center (Example 5 above). You’ll find other applications
in the problems for this chapter.
In the present example, the basic trick is to compare the I for a stick of length L
with the I for a stick of length 2L (and same density ρ). A quick scaling argument
shows that the latter is eight times the former. This is true because the integral
x2 dm = x2 ρ dx has three powers of x in it (yes, the dx counts). So a change
of variables, x = 2y, brings in a factor of 23 = 8. Equivalently, if we imagine
expanding the smaller stick into the larger one, then a corresponding piece in the
larger stick will be twice as far from the axis, and also twice as massive. The
integral x2 dm therefore increases by a factor of 22 · 2 = 8.
The technique is most easily illustrated with pictures. If we denote the moment
of inertia of an object by a picture of the object, with a dot signifying the axis,
b
a
Fig. 8.13
322
ˆ
Angular momentum, Part I (Constant L)
then we have:
L
L
= 8
L
= 2
( )2
__
L
+ M 2
=
The first line comes from the scaling argument, the second line comes from the
fact that moments of inertia simply add (the left-hand side is two copies of the
right-hand side, attached at the pivot), and the third line comes from the parallelaxis theorem. Equating the right-hand sides of the first two equations gives
= 4
Plugging this expression for
result,
into the third equation gives the desired
1 ML2
= __
12
Note that sooner or later we must use real live numbers, which enter here through
the parallel-axis theorem. Using only scaling arguments isn’t sufficient, because
they provide only linear equations homogeneous in the I ’s, and therefore give
no means of picking up the proper dimensions. (For an interesting account of
Galileo’s discovery of scaling laws, see Peterson (2002).)
Once you’ve mastered this trick and applied it to the fractal objects in Problem 8.8, you can impress your friends by saying that you know how to “use
scaling arguments, along with the parallel-axis theorem, to calculate moments of
inertia of objects with fractal dimension.” And you never know when that might
come in handy!
8.4
Torque
We will now show that under certain conditions (stated below), the rate of change
of angular momentum is equal to a certain quantity, τ, which we call the torque.
That is, τ = dL/dt. This is the rotational analog of our old friend F = dp/dt
involving linear momentum. The basic idea here is straightforward, but there are
two subtle issues. One deals with internal forces within a collection of particles.
The other deals with the possible acceleration of the origin (the point relative to
which the torque and angular momentum are calculated). To keep things straight,
we’ll prove the general result by dealing with three increasingly complicated
situations.
8.4 Torque
323
Our derivation of τ = dL/dt here holds for completely general motion, so
we can take the result and use it in the following chapter, too. If you wish, you
can construct a more specific proof of τ = dL/dt for the special case where the
axis of rotation is parallel to the z axis. But since the general proof is no more
difficult, we’ll present it here in this chapter and do it once and for all.
8.4.1
y
Point mass, ﬁxed origin
Consider a point mass at position r relative to a fixed origin (see Fig. 8.14). The
time derivative of the angular momentum, L = r × p, is
r
d
dL
= (r × p)
dt
dt
dp
dr
×p+r×
=
dt
dt
x
Fig. 8.14
= v × (mv) + r × F
= 0 + r × F,
(8.38)
where F is the force acting on the particle. This is the same calculation as in
Theorem 7.1, except that here we are considering an arbitrary force instead of a
central one. If we define the torque on the particle as
τ ≡ r × F,
(8.39)
then Eq. (8.38) becomes
τ=
dL
.
dt
(8.40)
It is understood that the r in the torque is measured with respect to the same
origin as the r in the angular momentum.
8.4.2
Extended mass, ﬁxed origin
In an extended object, there are internal forces acting on the various pieces of the
object, in addition to whatever external forces exist. For example, the external
force on a given molecule in a body might come from gravity, while the internal
forces come from the adjacent molecules. How do we deal with these different
types of forces?
In what follows, we will deal only with internal forces that are central forces,
so that the force between two objects is directed along the line between them.
This is a valid assumption for the pushing and pulling forces between molecules
in a solid. (It isn’t valid, for example, when dealing with magnetic forces. But
we won’t be interested in such things here.) We will invoke Newton’s third law,
ˆ
Angular momentum, Part I (Constant L)
324
y
which says that the force that particle 1 applies to particle 2 is equal and opposite
to the force that particle 2 applies to particle 1.
For concreteness, let’s assume that we have a collection of N discrete particles
labeled by the index i (see Fig. 8.15). In the continuous case, we would need to
replace the following sums with integrals. The total angular momentum of the
system is
N=4
r2
r1
r3
x
N
r4
L=
ri × p i .
(8.41)
i=1
Fig. 8.15
int
The force acting on each particle is Fext
i + Fi = dpi /dt. Therefore,
d
dL
=
dt
dt
=
i
=
ri × p i
i
dri
× pi +
dt
ri ×
i
vi × (mvi ) +
i
dpi
dt
int
ri × (Fext
i + Fi )
i
=0+
ri × Fext
i
i
≡
τ ext
i .
(8.42)
i
The second-to-last line follows because vi × vi = 0, and also because i ri ×
Fint
i = 0, as you can show in Problem 8.9. In other words, the internal forces
provide no net torque. This is quite reasonable. It basically says that a rigid object
with no external forces won’t spontaneously start rotating.
Note that the right-hand side involves the total external torque acting on the
body, which may come from forces acting at many different points. Note also
that nowhere did we assume that the particles are rigidly connected to each other.
Equation (8.42) still holds even if there is relative motion among the particles.
But in that case, it’s usually hard to get a handle on L, because it doesn’t take the
nice I ω form.
y
r1
r1 - r0
r0
r2 - r0
r2
Fig. 8.16
8.4.3
x
Extended mass, nonﬁxed origin
Let the position of the origin be r0 (see Fig. 8.16), and let the positions of the
particles be ri . The vectors r0 and ri are measured with respect to a given fixed
coordinate system. The total angular momentum of the system, relative to the
8.4 Torque
(possibly accelerating) origin r0 , is7
L=
(ri − r0 ) × mi (˙ri − r˙ 0 ).
(8.43)
i
Therefore,
dL
d
=
dt
dt
=
(ri − r0 ) × mi (˙ri − r˙ 0 )
i
(˙ri − r˙ 0 ) × mi (˙ri − r˙ 0 ) +
i
(ri − r0 ) × mi (¨ri − r¨ 0 )
i
=0+
int
(ri − r0 ) × (Fext
¨ 0 ),
i + Fi − mi r
(8.44)
i
int
because mi r¨ i is the net force (namely Fext
i + Fi ) acting on the ith particle. But
a quick corollary to Problem 8.9 is that the term involving Fint
i vanishes (as you
should check). And since
mi ri = M R (where M =
mi is the total mass,
and R is the position of the center of mass), we have
dL
=
dt
− M (R − r0 ) × r¨ 0 .
(ri − r0 ) × Fext
i
(8.45)
i
The first term here is the external torque, measured relative to the origin r0 . The
second term is something we wish would go away. And indeed, it usually does.
It vanishes if any of the following three conditions is satisfied.
1. R = r0 , that is, the origin is the CM.
2. r¨ 0 = 0, that is, the origin is not accelerating.
3. (R − r0 ) is parallel to r¨ 0 . This condition is rarely invoked.
If any of these conditions is satisfied, then we are free to write
dL
=
dt
(ri − r0 ) × Fext
i ≡
i
τ ext
i .
(8.46)
i
In other words, we can equate the total external torque with the rate of change of
the total angular momentum. An immediate corollary of this result is:
Corollary 8.3 If the total external torque on a system is zero, then its angular
momentum is conserved. In particular, the angular momentum of an isolated
system (one that is subject to no external forces) is conserved.
7
More precisely, we are calculating the angular momentum relative to a coordinate system whose
origin is r0 and whose axes remain parallel to the fixed axes. If we allowed for a rotation of
the axes, then we would have to deal with all the fictitious-force issues that are the subject
of Chapter 10. As it is, we will still end up dealing with one fictitious force (see the remark
below).
325
ˆ
Angular momentum, Part I (Constant L)
326
Everything up to this point is valid for arbitrary motion. The particles can
be moving relative to each other, and the various Li ’s can point in different
directions, etc. But let’s now restrict the motion. In the present chapter, we are
dealing only with cases where Lˆ is constant (taken to point in the z direction).
ˆ
ˆ If in addition we consider only rigid
Therefore, dL/dt = d(LL)/dt
= (dL/dt)L.
objects (where the relative distances among the particles are fixed) that undergo
pure rotation around a given point, then L = I ω, which gives dL/dt = I ω˙ ≡ I α.
Taking the magnitude of both sides of Eq. (8.46) then gives
τext = I α.
(8.47)
Invariably, we will calculate angular momentum and torque around either the
CM or a fixed point (or a point that moves with constant velocity, but this doesn’t
come up often). These are the “safe” origins, in the sense that Eq. (8.46) holds. As
long as you always use one of these safe origins, you can simply apply Eq. (8.46)
and not worry much about its derivation.
Remark: You’ll probably never end up invoking the third condition above, but it’s interesting
to note that there’s a simple way of understanding it in terms of accelerating reference frames.
This is the topic of Chapter 10, so we’re getting a little ahead of ourselves here, but the reasoning
is as follows. Let r0 be the origin of a reference frame that is accelerating with acceleration
r¨ 0 . Then all objects in this accelerating frame feel a mysterious fictitious force of −m¨r0 . For
example, on a train accelerating to the right with acceleration a, you feel a strange force of
ma pointing to the left. If you don’t counter this with another force (by grabbing a handle,
for example), then you will fall over. This fictitious force acts just like a gravitational force,
because it is proportional to the mass. Therefore, it effectively acts at the CM, producing a
torque of (R − r0 ) × (−M r¨ 0 ). This is the second term in Eq. (8.45). This term vanishes if the
CM is directly “above” or “below” (as far as the fictitious gravitational force is concerned) the
origin, in other words, if (R − r0 ) is parallel to r¨ 0 . See Problem 10.8 for further discussion of
this in terms of fictitious forces.
There is one common situation where the third condition can be invoked. Consider a wheel
rolling without slipping on the ground. Mark a dot on the rim. At the instant this dot is in
contact with the ground, it is a valid choice for the origin. This is true because (R − r0 ) points
vertically. And r¨ 0 also points vertically, because a dot on a rolling wheel traces out a cycloid.
Right before the dot hits the ground, it is moving straight downward. And right after it hits the
ground, it is moving straight upward. But having said this, there’s usually nothing to be gained
by picking such an origin. So the safe thing to do is to always pick your origin to be either the
CM or a fixed point, even if the third condition holds.
For conditions that number but three,
We say, “Torque is dL by dt.”
But though they’re all true,
I’ll stick to just two;
It’s CM’s and fixed points for me! ♣
M
m
θ
Fig. 8.17
Example: A string wraps around a uniform cylinder of mass M , which rests on a
fixed plane. The string passes up over a massless pulley and is connected to a mass m,
as shown in Fig. 8.17. Assume that the cylinder rolls without slipping on the plane,
and that the string is parallel to the plane. What is the acceleration of the mass m?
8.4 Torque
327
What is the condition on the ratio M /m for which the cylinder accelerates down the
plane?
First solution: The friction, tension, and gravitational forces are shown in
Fig. 8.18. Define positive a1 , a2 , and α as shown. These three accelerations, along
with T and F, are five unknowns. We therefore need to produce five equations.
They are:
1.
2.
3.
4.
5.
F = ma on m =⇒ T − mg = ma2 .
F = ma on M =⇒ Mg sin θ − T − F = Ma1 .
τ = I α on M (around the CM) =⇒ FR − TR = (MR2 /2)α.
Nonslipping condition =⇒ α = a1 /R.
Conservation of string =⇒ a2 = 2a1 .
α
Mgsinθ
a1
A few comments on these equations: The normal force and the gravitational force
perpendicular to the plane cancel, so we can ignore them. We have picked positive F
to point up the plane, but if it happens to point down the plane and thereby turn out
to be negative, that’s fine (but it won’t); we don’t need to worry about which way it
really points. In (3), we are using the CM of the cylinder as our origin, but we can
also use a fixed point; see the second solution below. In (5), we have used the fact that
the top of a rolling wheel moves twice as fast as the center. This is true because the
top has the same speed relative to the center as the center has relative to the ground.
We can go about solving these five equations in various ways. Three of the equations
involve only two variables, so it’s not so bad. (3) and (4) give F −T = Ma1 /2. Adding
this to (2) gives Mg sin θ − 2T = 3Ma1 /2. Using (1) to eliminate T , and using (5) to
write a1 in terms of a2 , then gives
Mg sin θ − 2(mg + ma2 ) =
3Ma2
4
=⇒
a2 =
(M sin θ − 2m)g
.
(3/4)M + 2m
(8.48)
And a1 = a2 /2. We see that a1 is positive (that is, the cylinder rolls down the plane)
if M /m > 2/ sin θ. If θ → 0, then this gives M /m → ∞, which makes sense.
If θ → π/2, then M /m → 2. The basic reason for this is that the friction force F, in
addition to the tension T , is holding the cylinder up (the coefficient of friction must
be very large in this case if no slipping is to occur).
Second solution: In using τ = dL/dt, we can also pick a fixed point as our origin,
instead of the CM. The most sensible point is one that is located somewhere along
the plane. The Mg sin θ force now provides a torque, but the friction does not. And
the lever arm for the tension is now 2R. The angular momentum of the cylinder with
respect to a point on the plane is L = I ω + M v1 R, where the second term comes from
the angular momentum due to the object being treated like a point mass at the CM.
So dL/dt = I α + Ma1 R, and τ = dL/dt gives
(Mg sin θ )R − T (2R) = (MR2 /2)α + Ma1 R.
(8.49)
This turns out to be the sum of the third equation plus R times the second equation in
the first solution. We therefore obtain the same result.
T
T
F
a2
θ
mg
Fig. 8.18
328
ˆ
Angular momentum, Part I (Constant L)
8.5
Collisions
In Section 5.7, we looked at collisions involving point particles or otherwise
nonrotating objects. The fundamental ingredients there that enabled us to solve
problems were conservation of linear momentum and conservation of energy
(if the collision was elastic). With conservation of angular momentum now at
our disposal, we can extend our study of collisions to ones with rotating objects.
The additional fact of conservation of L is compensated for by the new degree of
freedom of the rotation. Therefore, provided that the problem is set up properly,
we will still have the same number of equations as unknowns.
Conservation of energy can be used in a collision only if it is elastic (by definition). But conservation of angular momentum is similar to conservation of linear
momentum, in that it can always be used (see the remark below), assuming that
the system is isolated. However, conservation of L is a little different from conservation of p, because we have to pick an origin before we can proceed. In view
of the three conditions that are necessary for Corollary 8.3 to hold, we must pick
our origin to be either a fixed point or the CM of the system (we’ll ignore the third
condition, since it’s rarely used). If we unwisely choose an accelerating point,
then τ = dL/dt does not hold, so we have no right to claim that dL/dt equals
zero just because the torque is zero (as it is for an isolated system). In collision
problems, it is easy to fall into the trap of picking an accelerating point as your
origin. For example, you might choose the center of a stick as your origin. But if
another object collides with the stick, then the center will accelerate, making it
an invalid choice for the origin.
Remark: As far as conservation goes, the way that E differs from p and L is that energy
can be hidden in the microscopic motion of molecules in the body, in the form of heat. This
motion consists of little vibrations with small amplitudes but high speeds. The energy of these
vibrations can be large enough to be on the same order of magnitude as the overall energy
of the system. But because the vibrations are too small to see, it appears that energy is lost.
However, note that even though they’re too small to see, you can still feel them with your hand,
as heat.
Linear momentum, however, can’t be hidden. If an object (not necessarily rigid) has nonzero
momentum, then it has to be moving as a whole, and there’s no way around that. In short, since
P = MVCM , we see that if P is nonzero, then VCM is also. So the motion must be on a
macroscopic scale; there’s no way for it to be hidden on a microscopic scale.
With angular momentum, things are a little trickier. If the object is rigid, then it can’t
have hidden angular momentum, for reasons similar to those in the linear momentum case;
since L = I ω, we see that if L is nonzero, then ω is also. However, if the object isn’t
rigid (consider, say, a gas of particles), then it turns out that it can theoretically have hidden angular momentum in microscopic motion, although in practice it ends up being too small
to notice. This hidden angular momentum can arise from little swirling regions throughout
the system. In contrast with linear momentum, it is possible to have angular momentum without any overall motion. So in this respect, this microscopic swirling motion is similar to the
microscopic vibrational motion that yields hidden energy. There are, however, three main
differences.
First, if we’re assuming that the swirling motion takes place on a microscopic scale, then the
r in L = mr v is very small, and this leads to a negligible L. This argument doesn’t hold for E,
because the energy of the vibrations doesn’t involve r. Instead, it involves only v , in the form of
8.5 Collisions
329
mv 2 /2, so it can end up being large. Second, there’s no easy way to start up the circular motion
of many little swirls by means of a collision, in contrast with the easily started random linear
motion that makes up the energy of heat; you just smash two things together. And third, in the
case where the object is rigid, the molecules can easily vibrate, but they can’t rotate indefinitely
because this would involve ripping apart the bonds between adjacent molecules. The issue is
that in vibrational motion all coordinates remain small, whereas in rotational motion they don’t,
because θ eventually becomes large.
There is, however, one very common phenomenon, namely magnetism, that (in a sense) is
an exception to all three of the above points. Although magnetism isn’t an angular momentum,
it does come (roughly speaking) from the “circular” motion of electrons around the nuclei in
atoms. (In general, it actually comes more from the “spin” of the electrons than their orbital
motion around the nuclei, but let’s not worry about that here. To get everything right, we’d
have to think in terms of quantum mechanics, anyway. Let’s just work in a rough classical
approximation.) Electrons throughout a magnetic material move in tiny little correlated loops.
We can escape the above three points of reasoning because first, the magnetic field involves the
electric charge e, and this is large enough (on the scale of things) to cancel out the smallness of
the r factor. (The actual angular momentum of the electrons in a magnetic material is negligible
because their mass is so small. There isn’t a large quantity like e to save the day.) Second, it
is quite easy to get the electrons moving in correlated circles by means of magnetic forces;
there’s no need to smash things together. And third, the electrons are free to move around in
little circles (in a classical sense) in atoms without ripping things apart. ♣
It is important to remember that you are free to choose your origin from the
legal possibilities of fixed points or the CM. Since it is generally the case that one
choice is better than others (in that it makes the calculations easier), you should
take advantage of this freedom. Let’s do two examples. First an elastic collision,
and then an inelastic one.
Example (Elastic collision): A mass m travels perpendicular to a stick of mass
m and length , which is initially at rest. At what location should the mass collide
elastically with the stick, so that the mass and the center of the stick move with equal
speeds after the collision?
Solution: Let the initial speed of the mass be v0 . We have three unknowns in the
problem (see Fig. 8.19), namely the desired distance from the middle of the stick,
h; the final (equal) speeds of the stick and the mass, v; and the final angular speed
of the stick, ω. We can solve for these three unknowns by using our three available
conservation laws:
• Conservation of p:
v0
.
(8.50)
mv0 = mv + mv =⇒ v =
2
• Conservation of E: Remembering that the energy of the stick equals the energy
of the rotational motion around the center, plus the energy of the effective point
mass at the center, we have
√
mv02
m v0 2 1 m 2 2
6v0
m v0 2
+
+
=⇒ ω =
=
ω
.
2
2 2
2 2
2 12
(8.51)
ω
v0
m
v
h
l
m
Fig. 8.19
v
ˆ
Angular momentum, Part I (Constant L)
330
• Conservation of L: Let’s pick our origin to be the fixed point in space that
coincides with the initial location of the center of the stick. Then conservation
of L gives
mv0 h = m
v0
h+
2
m 2
ω+0 .
12
(8.52)
The zero here comes from the fact that the CM of the stick moves directly away
from the origin, so there is no contribution to L from the first of the two parts in
Theorem 8.1. Plugging the ω from Eq. (8.51) into Eq. (8.52) gives
√
1
m 2
6v0
mv0 h =
=⇒ h = √ .
(8.53)
2
12
6
You are encouraged to solve this problem again with a different choice of origin,
for example, the fixed point that coincides with the spot where the mass hits the
stick, or the CM of the entire system.
Remark: After the stick makes half of a revolution, it will hit the backside of m. The
resulting motion will have the stick sitting at rest (both translationally and rotationally)
and the mass moving with its initial speed v0 . You can show this by working though the
second collision, using the quantities we found above. Or you can just use the fact that this
scenario certainly satisfies conservation of p, E, and L with the initial conditions, so it must
be what happens (because the quadratic conservation statements have only two solutions,
and the other one corresponds to the intermediate√motion). Note that the time it takes the
stick to make
is π/ω = π / 6v0 . So the stick travels a distance of
√ half of a revolution
√
(v0 /2)(π / 6v0 ) = (π /2 6) before it ends up at rest. This distance is independent of
v0 (which follows from dimensional analysis). ♣
Let’s now look at an inelastic collision, where one object sticks to another. We
won’t be able to use conservation of E now. But conservation of p and L will be
sufficient since there is one fewer degree of freedom in the final motion, because
the objects don’t move independently.
v0
ω
m
CM
l
m
Fig. 8.20
Example (Inelastic collision): A mass m travels at speed v0 perpendicular to a
stick of mass m and length , which is initially at rest. The mass collides completely
inelastically with the stick at one of its ends and sticks to it. What is the resulting
angular velocity of the system?
Solution: The first thing to note is that the CM of the system is /4 from the end,
as shown in Fig. 8.20. After the collision, the system rotates about the CM as the CM
moves in a straight line. Conservation of momentum quickly tells us that the speed
of the CM is v0 /2. Also, using the parallel-axis theorem, the moment of inertia of the
system around the CM is
stick + I mass =
ICM = ICM
CM
2
2
m 2
5
+m
m 2.
=
+m
12
4
4
24
(8.54)
8.6 Angular impulse
There are now many ways to proceed, depending on what point we choose as our
origin.
First method: Choose the origin to be the fixed point that coincides with the location
of the CM right when the collision happens (that is, the point /4 from the end of the
stick). Conservation of L says that the initial L of the ball must equal the final L of
the system. This gives
mv0
4
=
5
m 2 ω+0
24
=⇒
ω=
6v0
.
5
(8.55)
The zero here comes from the fact that the CM of the stick moves directly away from
the origin, so there is no contribution to L from the first of the two parts in Theorem
8.1. Note that we didn’t need to use conservation of p in this method.
Second method: Choose the origin to be the fixed point that coincides with the
initial center of the stick. Then conservation of L gives
mv0
2
=
5
v0
m 2 ω + (2m)
24
2
=⇒
4
ω=
6v0
.
5
(8.56)
The right-hand side is the angular momentum of the system relative to the CM, plus
the angular momentum (relative to the origin) of a point mass of mass 2m located
at the CM.
Third method: Choose the origin to be the CM of the system. This point moves to
the right with speed v0 /2, along the line a distance /4 below the top of the stick.
Relative to the CM, the mass m moves to the right, and the stick moves to the left,
both with speed v0 /2. Conservation of L gives
m
v0
2
4
+ 0+m
v0
2
4
=
5
m 2 ω
24
=⇒
ω=
6v0
.
5
(8.57)
The zero here comes from the fact that the stick initially has no L around its center.
A fourth reasonable choice for the origin is the fixed point that coincides with the
initial location of the top of the stick. You can work this one out for practice.
8.6
Angular impulse
In Section 5.5.1, we defined the impulse, which we’ll label as I, to be the time
integral of the force applied to an object. From Newton’s second law, F = dp/dt,
the impulse is therefore the net change in linear momentum. That is,
I≡
t2
F(t) dt =
p.
(8.58)
t1
We now define the angular impulse, Iθ , to be the time integral of the torque
applied to an object. From τ = dL/dt, the angular impulse is therefore the net
331
332
ˆ
Angular momentum, Part I (Constant L)
change in angular momentum. That is,
Iθ ≡
t2
τ(t) dt =
L.
(8.59)
t1
These are just definitions, devoid of any content. The place where the physics
comes in is the following. Consider a situation where F(t) is always applied
at the same position relative to the origin around which τ(t) is calculated (this
origin must be a legal one, of course). Let this position be R. Then we have
τ(t) = R × F(t). Plugging this into Eq. (8.59), and taking the constant R outside
the integral, gives Iθ = R × I. In other words,
L = R × ( p)
for F(t) applied at one position .
(8.60)
This is a very useful result. It gives the relation between the net changes in L
and p, as opposed to the individual values of each. Even if F(t) is changing in
some arbitrary manner as time goes by, so that we have no idea what L and
p themselves are, we still know that they are related by Eq. (8.60). In many
cases, we don’t have to worry about the cross product in Eq. (8.60), because the
lever arm R is perpendicular to the change in momentum p. In such cases,
we have
| L| = R| p|.
(8.61)
Also, in many cases the object starts at rest, so we don’t have to bother with
the ’s. The following example is a classic application of angular impulse
and Eq. (8.61).
Example (Striking a stick): A stick of mass m and length , initially at rest, is
struck with a hammer. The blow is made perpendicular to the stick at one end. Let
the blow occur quickly, so that the stick doesn’t have time to move much while the
hammer is in contact. If the CM of the stick ends up moving at speed v, what are the
velocities of the ends right after the blow?
Solution: Although we have no hope of knowing exactly what F(t) looks like, or
the length of time it is applied, we still know from Eq. (8.61) that L = ( /2) p,
where we have chosen our origin to be the CM, which gives a lever arm of
/2. Therefore, (m 2 /12)ω = ( /2)mv, so the final v and ω are related by
ω = 6v/ .
The velocities of the ends right after the blow are obtained by adding (or subtracting)
the rotational motion to the CM’s translational motion. The rotational velocities of the
ends relative to the CM are ±ω( /2) = ±(6v/ )( /2) = ±3v. Therefore, the end
that is hit moves with velocity v + 3v = 4v, and the other end moves with velocity
v − 3v = −2v (that is, backward).
8.7 Problems
333
What L was, he just couldn’t tell.
And p? He was clueless as well.
But despite his distress,
He wrote down the right guess
For their quotient: the lever arm’s .
Impulse is also useful for “collisions” that occur over extended times. See, for
example, Problem 8.24.
8.7
Problems
Section 8.1: Pancake object in x-y plane
8.1. Massive pulley *
Consider the Atwood’s machine shown in Fig. 8.21. The masses are m
and 2m, and the pulley is a uniform disk of mass m and radius r. The
string is massless and does not slip with respect to the pulley. Find the
acceleration of the masses. Use conservation of energy.
8.2. Leaving the sphere **
A ball with moment of inertia βmr 2 rests on top of a fixed sphere.
There is friction between the ball and the sphere. The ball is given an
infinitesimal kick, and it rolls down without slipping. Assuming that
r is much smaller than the radius of the sphere, at what point does
the ball lose contact with the sphere? How does your answer change
if the size of the ball is comparable to, or larger than, the size of the
sphere? You may want to solve Problem 5.3 first, if you haven’t already
done so.
8.3. Sliding ladder ***
A ladder of length and uniform mass density stands on a frictionless
floor and leans against a frictionless wall. It is initially held motionless,
with its bottom end an infinitesimal distance from the wall. It is then
released, whereupon the bottom end slides away from the wall, and the
top end slides down the wall (see Fig. 8.22). When it loses contact with
the wall, what is the horizontal component of the velocity of the center
of mass?
8.4. Leaning rectangle ***
A rectangle of height 2a and width 2b rests on top of a fixed cylinder of
radius R (see Fig. 8.23). The moment of inertia of the rectangle around
its center is I . The rectangle is given an infinitesimal kick and then
“rolls” on the cylinder without slipping. Find the equation of motion for
the tilt angle of the rectangle. Under what conditions will the rectangle
m
m
2m
Fig. 8.21
l
Fig. 8.22
2a
b
b
R
Fig. 8.23
ˆ
Angular momentum, Part I (Constant L)
334
fall off the cylinder, and under what conditions will it oscillate back and
forth? Find the frequency of these small oscillations.
m
8.5. Mass in a tube ***
A tube of mass M and length is free to swing around a pivot at one end.
A mass m is positioned inside the (frictionless) tube at this end. The tube
is held horizontal and then released (see Fig. 8.24). Let θ be the angle
of the tube with respect to the horizontal, and let x be the distance the
mass has traveled along the tube. Find the Euler–Lagrange equations for
θ and x, and then write them in terms of θ and η ≡ x/ (the fraction of
the distance along the tube).
These equations can only be solved numerically, and you must pick
a numerical value for the ratio r ≡ m/M in order to do this. Write a
program (see Section 1.4) that produces the value of η when the tube is
vertical (θ = π/2). Give this value of η for a few values of r.
M
l
Fig. 8.24
z =1
Section 8.3: Calculating moments of inertia
z =0
8.6. Minimum I *
A moldable blob of matter of mass M is to be situated between the planes
z = 0 and z = 1 (see Fig. 8.25) so that the moment of inertia around the
z axis is as small as possible. What shape should the blob take?
Fig. 8.25
8.7. Slick calculations of I **
In the spirit of Section 8.3.2, find the moments of inertia of the following
objects (see Fig. 8.26).
l
(a) A uniform square of mass m and side (axis through center,
perpendicular to plane).
(b) A uniform equilateral triangle of mass m and side (axis through
center, perpendicular to plane).
l
8.8. Slick calculations of I for fractal objects ***
In the spirit of Section 8.3.2, find the moments of inertia of the following
fractal objects. Be careful how the mass scales.
Fig. 8.26
(a) Take a stick of length , and remove the middle third. Then remove
the middle third from each of the remaining two pieces. Then
remove the middle third from each of the remaining four pieces,
and so on, forever. Let the final object have mass m, and let the axis
be through the center, perpendicular to the stick; see Fig. 8.27.8
l
Fig. 8.27
8
This object is the Cantor set, for those who like such things. It has no length, so the density of
the remaining mass is infinite. If you suddenly develop an aversion to point masses with infinite
density, simply imagine the above iteration being carried out only, say, a million times.
8.7 Problems
(b) Take an equilateral triangle of side , and remove the “middle”
triangle (1/4 of the area). Then remove the “middle” triangle from
each of the remaining three triangles, and so on, forever. Let the
final object have mass m, and let the axis be through the center,
perpendicular to the plane; Fig. 8.28.
(c) Take a square of side , and remove the “middle” square (1/9
of the area). Then remove the “middle” square from each of the
remaining eight squares, and so on, forever. Let the final object
have mass m, and let the axis be through the center, perpendicular
to the plane; see Fig. 8.29.
335
l
Fig. 8.28
Section 8.4: Torque
8.9. Zero torque from internal forces **
Given a collection of particles with positions ri , let the force on the ith
particle due to all the others be Fint
i . Assuming that the force between
any two particles is directed along the line between them, use Newton’s
third law to show that i ri × Fint
i = 0.
l
Fig. 8.29
8.10. Removing a support *
(a) A uniform rod of length and mass m rests on supports at its ends.
The right support is quickly removed (see Fig. 8.30). What is the
force from the left support immediately thereafter?
(b) A rod of length 2r and moment of inertia βmr 2 rests on top of
two supports, each of which is a distance d away from the center.
The right support is quickly removed (see Fig. 8.30). What is the
force from the left support immediately thereafter?
8.11. Falling stick *
A massless stick of length b has one end pivoted on a support and the
other end glued perpendicular to the middle of a stick of mass m and
length .
l
r
r
d d
Fig. 8.30
b
(a) If the two sticks are held in a horizontal plane (see Fig. 8.31) and
then released, what is the initial acceleration of the CM?
(b) If the two sticks are held in a vertical plane (see Fig. 8.31) and
then released, what is the initial acceleration of the CM?
8.12. Pulling a cylinder **
In Exercise 8.50 below, the cylinder moves directly to the right. The fact
that it doesn’t have any transverse motion follows from the fact that the
two segments of the string pull only to the right and therefore cannot
supply a transverse force. Demonstrate this result again by explicitly
l
b
g
l
g
Fig. 8.31
ˆ
Angular momentum, Part I (Constant L)
336
R
8.13. Oscillating ball **
A small ball with radius r and uniform density rolls without slipping
near the bottom of a fixed cylinder of radius R (see Fig. 8.32). What is
the frequency of small oscillations? Assume r
R.
Fig. 8.32
θ
l
h
r
(a) Assuming that θ is very small, how does r depend on ?
(b) Assuming that θ is very close to π/2, how does h depend on ?
R
Fig. 8.34
Fig. 8.35
8.14. Oscillating cylinders **
A hollow cylinder of mass M1 and radius R1 rolls without slipping on
the inside surface of another hollow cylinder of mass M2 and radius
R2 . Assume R1
R2 . Both axes are horizontal, and the larger cylinder is free to rotate about its axis. What is the frequency of small
oscillations?
8.15. Lengthening the string **
A mass hangs from a massless string and swings around in a horizontal circle, as shown in Fig. 8.33. The length of the string is then
very slowly increased (or decreased). Let θ, , r, and h be defined as
shown.
Fig. 8.33
R
integrating the string’s force on the cylinder over the semicircle of contact. (The N = T dθ result from the “Rope wrapped around a pole”
example in Section 2.1 will come in handy.)
R
8.16. A triangle of cylinders ***
Three identical cylinders with moments of inertia I = βmR2 are situated
in a triangle as shown in Fig. 8.34. Find the initial downward acceleration
of the top cylinder for the following two cases. Which case has a larger
acceleration?
(a) There is friction between the bottom two cylinders and the ground
(so they roll without slipping), but there is no friction between
any of the cylinders.
(b) There is no friction between the bottom two cylinders and the
ground, but there is friction between the cylinders (so they don’t
slip with respect to each other).
8.17. Falling chimney ****
A chimney initially stands upright. It is given a tiny kick, and it topples
over. At what point along its length is it most likely to break? In doing this
problem, work with the following two-dimensional simplified model of
a chimney. Assume that the chimney consists of boards stacked on top
of each other, and that each board is attached to the two adjacent ones
with tiny rods at each end (see Fig. 8.35). The goal is to determine
which rod in the chimney achieves the maximum tension. Work in the
8.7 Problems
337
approximation where the width of the chimney is very small compared
with the height.
Section 8.5: Collisions
8.18. Ball hitting stick **
A ball of mass M collides with a stick with moment of inertia I = βm 2
(relative to its center, which is its CM). The ball is initially traveling
at speed V0 perpendicular to the stick. The ball strikes the stick at a
distance d from the center (see Fig. 8.36). The collision is elastic. Find
the resulting translational and rotational speeds of the stick, and also the
resulting speed of the ball.
8.19. A ball and stick theorem **
Consider the setup in Problem 8.18. Show that the relative speed
of the ball and the point of contact on the stick is the same before
and immediately after the collision. (This result is analogous to
the “relative speed” result for a 1-D collision, Theorem 5.3 in
Section 5.7.1)
Section 8.6: Angular impulse
8.20. The superball **
A ball with radius R and I = (2/5)mR2 is thrown through the air. It
spins around the axis perpendicular to the (vertical) plane of the motion.
Call this the x-y plane. The ball bounces off a floor without slipping
during the time of contact. Assume that the collision is elastic, and that
the magnitude of the vertical vy is the same before and after the bounce.
Show that vx and ω after the bounce are related to vx and ω before the
bounce by
vx
Rω
=
1
7
3
−4
−10 −3
vx
Rω
,
(8.62)
where positive vx is to the right, and positive ω is counterclockwise.
8.21. Many bounces *
Using the result of Problem 8.20, describe what happens over the course
of many superball bounces.
8.22. Rolling over a bump **
A ball with radius R and moment of inertia I = (2/5)mR2 rolls with
speed V0 without slipping on the ground. It encounters a step of height
h and rolls up over it. Assume that the ball sticks to the corner of
m
2
M
V0
Fig. 8.36
l
d
ˆ
Angular momentum, Part I (Constant L)
338
the step briefly (until the center of the ball is directly above the corner). Show that if the ball is to climb over the step, then V0 must
satisfy
V0 ≥
5h
10gh
1−
7
7R
−1
.
(8.63)
8.23. Falling toast **
After buttering your toast (assumed to be a uniform rigid square of
side length , for the sake of having a doable problem) one morning and
holding it in a horizontal position with buttered side up, you accidentally
drop it from a height H above a counter, which itself is a height h above
the ground. The toast is oriented “parallel” to the counter, and as it falls,
one edge barely clips the counter (elastically), causing the toast to spin.
What value of H , in terms of h and , yields the unfortunate scenario
where the toast makes half of a revolution, landing buttered side down
on the floor? There is a special value of in terms of h. What is it, and
why is it special?
V0
R
(a) Without knowing anything about the nature of the friction force,
find the speed of the ball when it begins to roll without slipping.
Also, find the kinetic energy lost while sliding.
(b) Now consider the special case where the coefficient of kinetic
friction is µ, independent of position. At what time, and at what
distance, does the ball begin to roll without slipping? Verify that
the work done by friction equals the energy loss calculated in part
(a). (Be careful on this.)
Fig. 8.37
m1
m2
2r
m3
m4
....
Fig. 8.38
8.24. Sliding to rolling **
Aball initially slides without rotating on a horizontal surface with friction
(see Fig. 8.37). The initial speed is V0 , and the moment of inertia around
the center is I = βmR2 .
8.25. Lots of sticks ***
Consider a collection of rigid sticks of length 2r, masses mi , and
moments of inertia βmi r 2 , with m1
m2
m3
· · · . The CM
of each stick is located at its center. The sticks are placed on a horizontal frictionless surface, as shown in Fig. 8.38. The ends overlap a tiny
distance in the y direction and are a tiny distance apart in the x direction. The first (heaviest) stick is given an instantaneous blow (as shown)
which causes it to translate and rotate. The first stick will strike the
second stick, which will then strike the third stick, and so on. Assume
that all the collisions are elastic. Depending on the size of β, the speed
of the nth stick will either (1) approach zero, (2) approach infinity, or
8.8 Exercises
339
(3) be independent of n, as n → ∞. Show that the special value of β
corresponding to the third of these three scenarios is β = 1/3, which
happens to correspond to a uniform stick.
8.8
Exercises
L
Section 8.1: Pancake object in x-y plane
8.26. Swinging stick **
A uniform stick of length L is pivoted at its bottom end and is initially
held vertical. It is given an infinitesimal kick, and it swings down around
the pivot. After three-quarters of a turn (in the horizontal position shown
in Fig. 8.39), the pivot is somehow vaporized, and the stick flies freely
up in the air. What is the maximum height of the center of the stick in
the resulting motion? At what angle is the stick tilted when the center
reaches this maximum height?
8.27. Atwood’s with a cylinder **
A massless string of negligible thickness is wrapped around a uniform
cylinder of mass m and radius r. The string passes up over a massless
pulley and is tied to a block of mass m at its other end, as shown in
Fig. 8.40. The system is released from rest. What are the accelerations
of the block and the cylinder? Assume that the string does not slip with
respect to the cylinder. Use conservation of energy (after applying a
quick F = ma argument to show that the two objects move downward
with the same acceleration).
8.28. Board and cylinders **
A board lies on top of two uniform cylinders that lie on a fixed plane
inclined at an angle θ , as shown in Fig. 8.41. The board has mass m, and
each of the cylinders has mass m/2. The system is released from rest. If
there is no slipping between any of the surfaces, what is the acceleration
of the board? Use conservation of energy.
Fig. 8.39
m
m
Fig. 8.40
m/2
m
m/2
θ
Fig. 8.41
8.29. Moving plane ***
A ball of mass m and moment of inertia I = βmr 2 is held motionless on a
plane of mass M and angle of inclination θ (see Fig. 8.42). The plane rests
on a frictionless horizontal surface. The ball is released. Assuming that
it rolls without slipping on the plane, what is the horizontal acceleration
of the plane? Hint: You might want to do Problem 3.8 first. But as
with all the exercises in this section, use conservation of energy instead
of force and torque; this problem gets extremely messy with the latter
strategy.
m
M
θ
Fig. 8.42
ˆ
Angular momentum, Part I (Constant L)
340
Section 8.2: Nonplanar objects
8.30. Semicircle CM *
A wire is bent into a semicircle of radius R. Find the CM.
8.31. Hemisphere CM *
Find the CM of a solid hemisphere.
Section 8.3: Calculating moments of inertia
8.32. A cone *
Find the moment of inertia of a solid cone (mass M , base radius R)
around its symmetry axis.
8.33. A sphere *
Find the moment of inertia of a solid sphere (mass M , radius R) around
a diameter. Do this by slicing the sphere into disks.
l
8.34. A triangle, the slick way **
In the spirit of Section 8.3.2, find the moment of inertia of a uniform
equilateral triangle of mass m and side , around a line joining a vertex
and the opposite side; see Fig. 8.43.
Fig. 8.43
l
8.35. Fractal triangle **
Take an equilateral triangle of side , and remove the “middle” triangle (1/4 of the area). Then remove the “middle” triangle from each
of the remaining three triangles, and so on, forever. Let the final
fractal object have mass m. In the spirit of Section 8.3.2, find the
moment of inertia around a line joining a vertex and the opposite side;
see Fig. 8.44.
Section 8.4: Torque
Fig. 8.44
8.36. Swinging your arms *
You are standing on the edge of a step on some stairs, facing up the
stairs. You feel yourself starting to fall backward, so you start swinging
your arms around in vertical circles, like a windmill. This is what people
tend to do in such a situation, but does it actually help you not to fall, or
does it simply make you look silly? Explain your reasoning.
8.37. Rolling down the plane *
A ball with moment of inertia βmr 2 rolls without slipping down a plane
inclined at an angle θ . What is its linear acceleration?
8.8 Exercises
8.38. Coin on a plane *
A uniform coin rolls down a plane inclined at an angle θ . If the coefficient
of static friction between the coin and the plane is µ, what is the largest
angle θ for which the coin doesn’t slip?
8.39. Accelerating plane *
A ball with I = (2/5)MR2 is placed on a plane inclined at an angle θ. The
plane is accelerated upwards (along its direction) with acceleration a;
see Fig. 8.45. For what value of a does the CM of the ball not move?
Assume that there is sufficient friction so that the ball doesn’t slip with
respect to the plane.
341
a
θ
Fig. 8.45
8.40. Bowling ball on paper *
A bowling ball sits on a piece of paper on the floor. You grab the paper
and pull it horizontally along the floor, with acceleration a0 . What is the
acceleration of the center of the ball? Assume that the ball does not slip
with respect to the paper.
8.41. Spring and cylinder *
The axle of a solid cylinder of mass m and radius r is connected to a
spring with spring constant k, as shown in Fig. 8.46. If the cylinder rolls
without slipping, what is the frequency of the oscillations?
8.42. Falling quickly *
A massless stick of length L is pivoted at one end and has a mass m
attached to its other end. It is held in a horizontal position, as shown in
Fig. 8.47. Where should a second mass m be attached to the stick, so
that the stick falls as fast as possible when dropped?
8.43. Maximum frequency *
A pendulum is made of a uniform stick of length L. It is allowed to swing
in a vertical plane. Where should the pivot be placed on the stick so that
the frequency of (small) oscillations is maximum?
8.44. Massive pulley *
Solve Problem 8.1 again, but now use force and torque instead of
conservation of energy.
8.45. Atwood’s with a cylinder **
Solve Exercise 8.27 again, but now use force and torque instead of
conservation of energy.
8.46. Board and cylinders **
Solve Exercise 8.28 again, but now use force and torque instead of
conservation of energy.
k
m
Fig. 8.46
pivot
m
x
Fig. 8.47
L
m
ˆ
Angular momentum, Part I (Constant L)
342
R
m
r
8.47. The spool **
A spool of mass m and moment of inertia I is free to roll without slipping
on a table. It has an inner radius r, and an outer radius R. If you pull
on the string with tension T at an angle θ (see Fig. 8.48), what is the
acceleration of the spool? Which way does it move?
T
θ
8.48. Stopping the coin **
A coin stands vertically on a table. It is projected forward (in the plane
of itself) with speed v and angular speed ω, as shown in Fig. 8.49. The
coefficient of kinetic friction between the coin and the table is µ. What
should v and ω be so that the coin comes to rest (both translationally
and rotationally) a distance d from where it started?
Fig. 8.48
ω
R
v
8.49. Measuring g **
(a) Consider an extended pendulum whose CM is a distance from
the pivot, and whose moment of inertia around the pivot is I . Show
√
that the frequency of small oscillations is ω = mg /I , which
√
gives T = 2π/ω = 2π I /mg , and hence g = 4π 2 I /(m T 2 ).
Therefore, by measuring I , m, , and T , you can determine g.
However, if the pendulum has an odd shape, it may be difficult
to determine I . Consider, then, the following alternative method
of measuring g.
(b) For simplicity, assume that the pendulum is planar. Pick an arbitrary point as the pivot and measure the period, T , of small
oscillations. Then with the pendulum at rest, draw a vertical line
through this pivot. By trial and error, find another pivot point
on this line on the same side of the CM 9 (you may need to
extend the line with a massless extension) that yields the same
period T . Let L be the sum of the lengths from these two points
to the CM.10 Show that g is given by g = 4π 2 L/T 2 , which is
independent of m and I .
µ
Fig. 8.49
8.50. Pulling a cylinder **
Asolid cylinder of mass m and radius r lies flat on a frictionless horizontal
table, with a massless string running halfway around it, as shown in
9
10
The CM can be found by hanging the pendulum from a point not on the drawn line, and then
drawing a vertical line through this point. The intersection of the two lines is the CM.
There are also two points on the other side of the CM that yield the same period (except in the
special case where the two points coincide and produce the minimum period, as in Exercise 8.43,
but don’t worry about this). These two other points are the same distances from the CM as the
original two points, as you can show. So if you happen to find all four points, you can obtain L by
instead measuring the distance from the “inside” point on one side to the “outside” point on the
other. This method doesn’t require knowing the location of the CM.
8.8 Exercises
Fig. 8.50. A mass also of mass m is attached to one end of the string,
and you pull on the other end with a force T . The circumference of the
cylinder is sufficiently rough so that the string does not slip with respect
to it. What is the acceleration of the mass m attached to the end of the
string?
8.51. Coin and plank **
A coin of mass M and radius R stands vertically on the right end of
a horizontal plank of mass M and length L, as shown in Fig. 8.51.
The system starts at rest. The plank is then pulled to the right with a
constant force F. Assume that the coin does not slip with respect to
the plank. What are the accelerations of the plank and coin? How far
to the right does the coin move by the time the left end of the plank
reaches it?
8.52. Cylinder, board, and spring **
A board of mass m, which is free to slide on a frictionless floor, is
connected by a spring (with spring constant k) to a wall. A cylinder,
also of mass m (and I = mR2 /2), rests on top of the board, as shown in
Fig. 8.52, and is free to roll without slipping on the board. If the board
and the cylinder are pulled some distance to the left and then released
from rest, what is the frequency of the resulting oscillatory motion?
8.53. Swirling around a cone **
A fixed hollow frictionless cone is positioned with its tip pointing down.
A particle is released from rest on the inside surface. After it has slid
halfway down to the tip, it bounces elastically off a platform. The platform is positioned at a 45◦ angle along the surface of the cone, so the
particle ends up being deflected horizontally along the surface (in other
words, into the page in Fig. 8.53). The particle then swirls up and
around the cone before coming down. Measured from the tip of the
cone, show that the ratio of the
√ particle’s maximum swirling height to
the height of the platform is ( 5 + 1)/2, which just happens to be the
golden ratio.
343
(top view)
T
m
m
Fig. 8.50
m
L
m
F
Fig. 8.51
m
k
m
Fig. 8.52
platform
Fig. 8.53
8.54. Raising a hoop **
A bead of mass m is positioned at the top of a frictionless hoop of mass
M and radius R, which stands vertically on the ground. A wall touches
the hoop on its left, and a short wall of height R touches the hoop on
its right, as shown in Fig. 8.54. All surfaces are frictionless. The bead
is given a tiny kick, and it slides down the hoop, as shown. What is the
largest value of m/M for which the hoop never rises up off the ground?
Note: It’s possible to solve this problem by using only force, but solve
it here by using torque.
m
M
Fig. 8.54
R
ˆ
Angular momentum, Part I (Constant L)
344
µ=1
m
m
θ
Fig. 8.55
8.56. Falling and sliding stick ***
One end of a uniform stick is attached to a pivot, and the pivot is free
to slide along a frictionless horizontal rail. The stick is held at an initial
angle θ0 with respect to the (upward) vertical direction and then released;
see Fig. 8.56. Assume that the stick can somehow swing down below
the horizontal position without running into the rail (perhaps by having
the pivot attached to the side of the rail, so that the stick is shifted
horizontally a small distance from the rail).
θ0
(a) Show that when the stick is horizontal, the normal force N from
the rail equals mg/4, independent of θ0 .
(b) If θ0 = 0 (a tiny kick is allowed), show that N = 13mg when the
stick is at the bottom of its motion (at θ = π ).
(c) If θ0 = 0, show that the minimum N occurs at θ ≈ 61.5◦ and that
the value is Nmin ≈ (0.165)mg. You will obtain a cubic; feel free
to solve it numerically.
...
...
Fig. 8.56
a
Fig. 8.57
8.55. Block and cylinder **
A block and a cylinder (with I = βmR2 ), both of mass m, lie on a plane
(inclined at an angle θ ), touching each other as shown in Fig. 8.55.
There is sufficient friction between the cylinder and the plane so that
it rolls without slipping. But the bottom of the block is coated with
grease, so there is no friction between it and the plane. However, there
is a coefficient of kinetic friction µ = 1 between the block’s right face
and the cylinder. What is the acceleration of the block? How does your
answer compare with the result for a lone cylinder rolling down a plane?
Assume that θ is small enough so that the block’s bottom face remains in
contact with the plane at all times; what is the condition on θ for which
this is true?
8.57. Tower of cylinders ****
Consider the infinitely tall system of massless planks and identical massive cylinders shown in Fig. 8.57. The moment of inertia of the cylinders
is I = MR2/2. There are two cylinders at each level, and the number of
levels is infinite. The cylinders do not slip with respect to the planks,
but the bottom plank is free to slide on a table. If you pull on the bottom
plank so that it accelerates horizontally with acceleration a, what is the
horizontal acceleration of the bottom row of cylinders?
Section 8.5: Collisions
8.58. Pendulum collision *
A stick of mass m and length is pivoted at an end. It is held horizontal
and then released. It swings down, and when it is vertical the free end
8.8 Exercises
345
elastically collides with a ball, as shown in Fig. 8.58. (Assume that
the ball is initially held at rest and then released a split second before
the stick strikes it.) If the stick loses half of its angular velocity during
the collision, what is the mass of the ball? What is the speed of the ball
right after the collision?
8.59. No final rotation *
A stick of mass m and length spins around on a frictionless horizontal
table, with its CM at rest (but not fixed by a pivot). A ball of mass M is
placed on the table, and one end of the stick collides elastically with it,
as shown in Fig. 8.59. What should M be so that after the collision the
stick has translational motion, but no rotational motion?
8.60. Same final speeds *
A stick slides perpendicular to itself (without rotating) across a frictionless horizontal table and collides elastically at one of its ends with a
stationary ball. Both the stick and the ball have mass m. The mass of the
stick is distributed in such a way that the moment of inertia around the
CM (which is at the center of the stick) is I = Am 2 , where A is some
number. What should A be so that the ball moves at the same speed as
the center of the stick after the collision?
8.61. Perpendicular deflection **
A mass M moves at speed V0 perpendicular to a dumbbell at rest on
a frictionless horizontal table, as shown in Fig. 8.60. The dumbbell
consists of two masses m at the ends of a massless rod of length .
The mass M collides elastically with one of the masses (not head-on),
and afterwards it is observed that M moves perpendicular to its original
direction, with speed u. What is u in terms of V0 , m, and M ? What is the
smallest value of m (in terms of M ) for which this scenario is possible?
8.62. Glancing off a stick **
A frictionless stick of mass m and length lies at rest on a frictionless
horizontal table. A mass km (where k is some number) moves with speed
v0 at a 45◦ angle to the stick and collides elastically with it very close
to an end; see Fig. 8.61. What should k be so that the mass ends up
moving in the y-direction, as shown? Hint: Remember that the stick is
frictionless.
8.63. Sticking to a dumbbell *
A mass m moves at speed v perpendicular to a dumbbell at rest on
a frictionless horizontal table, as shown in Fig. 8.62. The dumbbell
consists of two masses also of mass m at the ends of a massless rod
of length . The moving mass collides and sticks to one of the masses.
start
m
l
Fig. 8.58
M
(top view)
l
m
Fig. 8.59
M V0
m
l
(top view)
m
Fig. 8.60
final velocity
of mass
v0
km
(top view)
Fig. 8.61
45
m
ˆ
Angular momentum, Part I (Constant L)
346
m v
m
What is the resulting ω of the system? What is the velocity of the end of
the rod that has the two masses on it, after the rod has made one half of
a revolution?
l
(top view)
8.64. Colliding sticks **
On a frictionless horizontal table, a stick of mass m and length spins
around a pivot at one of its ends with angular frequency ω. It collides and
sticks to an identical stick, with an overlap length equal to x, as shown
in Fig. 8.63. Immediately before the collision, the pivot is removed.
What should x be so that after the collision the double-stick system has
translational but no rotational motion?
m
Fig. 8.62
no pivot
8.65. Lollipop **
A hockey puck of mass m and radius R slides across frictionless ice, as
shown in Fig. 8.64 (the view is from above). It has translational speed
v to the right and rotational speed ω clockwise. It grazes the “top” end
of a rod of mass m and length 2R which is initially at rest on the ice.
It sticks to the rod, forming a rigid object that looks like a lollipop.
l
x
ω
l
ω
(top view)
(a) In the special case of v = Rω, what is the resulting angular speed
of the lollipop?
(b) How much energy is lost during the collision? How do you explain
the fact that energy is lost, given that the v = Rω condition implies
that the contact point on the puck touches the rod with zero relative
speed (and thus doesn’t crash into it, as is commonly the case with
inelastic collisions)?
(c) Given ω, show that v should equal 6Rω/5 if you want the
minimum amount of energy to be lost.
Fig. 8.63
ω
v
R
2R
(top view)
Fig. 8.64
8.66. Pencil on a plane ****
This exercise deals with the terminal velocity of a “pencil” rolling down
a plane. To simplify things, we’ll assume that the pencil has all its mass
on its axis. And to avoid messy complications, we’ll assume that the
cross section of the pencil looks like a wheel with six equally spaced
massless spokes and no rim (see Fig. 8.65).11 Let the length of the spokes
be r, and let the plane be inclined at an angle θ . Assume that there is
sufficient friction to prevent the spokes from slipping on the plane, and
assume that the pencil does not bounce when a spoke hits the plane.
θ
Fig. 8.65
(a) Explain qualitatively why the pencil reaches a terminal (average)
velocity, assuming that it remains in contact with the plane at all
times.
11
If the pencil instead looks like a hexagon with flat sides, then it is impossible to say how it
behaves, because if the sides bow outward an infinitesimal amount then the system conserves
energy, whereas if they bow inward an infinitesimal amount then it does not (for reasons you will
figure out).
8.8 Exercises
347
(b) Assume that conditions have been set up so that the pencil
eventually reaches a nonzero terminal (average) velocity, while
remaining in contact with the plane at all times. Describe this
terminal velocity. You may do this by stating the maximum speed
of the axis in the limiting steady state.
(c) What is the minimum value of θ for which a nonzero terminal
velocity exists? An initial kick to the pencil is allowed.
(d) What is the maximum value of θ for which the pencil remains in
contact with the plane at all times? As a check on your answer,
the difference between the answers in parts (c) and (d) is about
5.09◦ .
Section 8.6: Angular impulse
8.67. Striking a pool ball *
At what height should you horizontally strike a pool ball so that it
immediately rolls without slipping?
8.68. Center of percussion *
You hold one end of a uniform stick of length L. The stick is struck
with a hammer. Where should this blow occur so that the end you are
holding doesn’t move (immediately after the blow)? In other words,
where should the blow occur so that you don’t feel a “sting” in your
hand? This point is called the center of percussion.
8.69. Another center of percussion *
You hold the top vertex of a solid equilateral triangle of side length L. The
plane of the triangle is vertical. It is struck with a hammer, somewhere
along the vertical axis. Where should this blow occur so that the point
you are holding doesn’t move (immediately after the blow)? Use the
fact that the moment of inertia about any axis through the CM of an
equilateral triangle is mL2/24.
8.70. Not hitting the pole **
A (possibly non-uniform) stick of mass m and length lies on frictionless
ice. Its midpoint (which is also its CM) touches a thin pole sticking out
of the ice. One end of the stick is struck with a quick blow perpendicular
to the stick, as shown in Fig. 8.66, so that the CM moves away from the
pole. What is the minimum value of the stick’s moment of inertia that
allows the stick not to hit the pole?
8.71. Pulling the paper **
A ball sits at rest on a piece of paper on a table. You pull the paper
in a straight line out from underneath the ball. You are free to pull the
(top view)
m
pole
l
strike
Fig. 8.66
ˆ
Angular momentum, Part I (Constant L)
348
paper in an arbitrary (straight line) manner, frontward or backward. You
may even give it abrupt, jerky motions, so that the ball slips with respect
to it. After the ball comes off the paper, it will eventually roll without
slipping on the table. Show (and you are encouraged to experimentally
verify this) that the ball in fact ends up at rest. (The generalization of
this fact is given in Problem 9.29.) Is it possible to pull the paper in such
a way that the ball ends up exactly where it started?
8.72. Up, down, and twisting **
A uniform stick is held horizontally and then released. At the same
instant, one end is struck with a quick upwards blow. If the stick ends
up horizontal when it returns to its original height, what are the possible
values for the maximum height to which the center rises?
8.73. Doing work **
(a) A pencil of mass m and length lies at rest on a frictionless table.
You push on it at its midpoint (perpendicular to it), with a constant
force F for a time t. Find the final speed and the distance traveled.
Verify that the work you do equals the final kinetic energy.
(b) Assume that you apply the same force F for the same time t
as above, but that you now apply it at one of the pencil’s ends
(perpendicular to the pencil). Assume that t is small, so that the
pencil doesn’t have much time to rotate (this means that you can
assume that your force is always essentially perpendicular to the
pencil, as far as the torque is concerned). Find the final CM speed,
the final angular speed, and the distance your hand moves. Verify
that the work you do equals the final kinetic energy.
(top view)
L
θ
Fig. 8.67
l
θ
8.74. Bouncing between bricks ***
A stick of length slides on frictionless ice. It bounces elastically
between two parallel fixed bricks, a distance L apart, in such a way
that the same end touches both bricks, and the stick hits the bricks at
an angle θ each time. See Fig. 8.67. What is θ in terms of L and
(an implicit equation is fine)? Draw a reasonably accurate picture of
what the situation looks like in the limit L
.
What should θ be in terms of L and (again, an implicit equation
is fine), if the stick makes an additional n half revolutions between the
bricks? What is the minimum value of L/ for which n half revolutions
are possible?
8.75. Repetitive bouncing *
Using the result of Problem 8.20, what must the relation between vx and
Rω be so that a superball continually bounces back and forth between
the same two points of contact on the ground?
8.9 Solutions
8.76. Bouncing under a table **
You throw a superball so that it bounces off the floor, then off the underside of a table, then off the floor again. What must the initial relation
between vx and Rω be so that the ball returns to your hand, with the
outward and return paths the same? Use the result of Problem 8.20, and
modifications thereof.12
8.77. Bouncing under a table again ****
Consider the setup in the previous exercise, where we assumed that the
outward and return paths were the same. Is this trajectory (where the
path retraces itself ) the only possible one for which the ball returns to
your hand? Show13 that the answer is “yes” unless t1 = 7t2 , where t1
is the time the ball spends between your hand and the floor (which is
the same out and back, because the magnitude of vy isn’t changed by
a bounce), and t2 is the time the ball spends between the floor and the
table (again, the same out and back). For the special case where t1 = 7t2 ,
you will find that the ball returns to your hand for any initial relation
between vx and Rω. 14
For a ball with a general moment of inertia I = βmr 2 , show that
the answer is always “yes,” without exception, if β ≤ 1/3 (which
corresponds to a wheel with massive spokes and a massless rim). Also,
show that if β = 1 (a hoop), then the t1 = 7t2 condition becomes t1 = t2 .
In other words, if you throw a “super-hoop” from the same height as
the table, then no matter how you throw it (as long as the throw is
downward and in the plane of the hoop), it will return to your hand. This
is a little more believable if you look at the remark in the solution to
Problem 8.20.
8.9
Solutions
8.1. Massive pulley
The two masses have equal speeds at all times. Let v be their common speed after they
have each moved a distance d. If 2m falls a distance d (so that m rises a distance d),
then the change in potential energy is −2mgd + 2mgd = −mgd. The total kinetic
energy is
K=
12
13
14
1
1 2 1
mv + (2m)v 2 + I ω2
2
2
2
You are strongly encouraged to bounce a ball in such a manner and have it magically come back
to your hand. It turns out that the required value of ω is small, so a natural throw with ω ≈ 0
essentially gets the job done.
You will want to use Mathematica or some other aid to keep track of the matrix multiplications,
especially in the second part of this exercise, which would be completely intractable otherwise.
I learned of this extremely bizarre fact from Howard Georgi.
349
350
ˆ
Angular momentum, Part I (Constant L)
=
1 2 1
1
mv + (2m)v 2 +
2
2
2
1 2
mr
2
v 2
r
7
= mv 2 ,
4
(8.64)
where we have used the nonslipping condition, v = rω. Conservation of energy
therefore gives
7 2
4
mv − mgd =⇒ v =
gd .
4
7
√
The usual kinematic result v = 2ad holds here, so we obtain a = 2g/7.
0=
(8.65)
8.2. Leaving the sphere
In this setup, as in Problem 5.3, the ball leaves the sphere when the normal force
becomes zero, that is, when
mv 2
= mg cos θ .
R
(8.66)
The only change from the solution to Problem 5.3 comes in the calculation of v . The
ball now has rotational energy, so conservation of energy gives mgR(1 − cos θ ) =
mv 2 /2 + I ω2 /2 = mv 2 /2 + βmr 2 ω2 /2. But the nonslipping condition is v = rω, so
we have
1
(1 + β)mv 2 = mgR(1 − cos θ )
2
=⇒
v=
2gR(1 − cos θ )
.
1+β
(8.67)
Plugging this into Eq. (8.66), we see that the ball leaves the sphere when
cos θ =
2
.
3+β
(8.68)
Remarks: For β = 0, this equals 2/3, as in Problem 5.3. For a uniform ball with
β = 2/5, we have cos θ = 10/17, so θ ≈ 54◦ . For β → ∞ (for example, a spool with
a very thin axle rolling down the rim of a circle), we have cos θ → 0, so θ ≈ 90◦ .
This makes sense because v is always very small, since most of the energy takes the
form of rotational energy. The coefficient of friction would have to be very large in
this case, of course, to keep the spool from slipping near θ ≈ 90◦ . ♣
If the size of the ball is comparable to, or larger than, the size of the sphere, then we
must take into account the fact that the CM of the ball does not move along a circle of
radius R. Instead, it moves along a circle of radius R + r, so Eq. (8.66) becomes
mv 2
= mg cos θ.
R+r
(8.69)
Also, the conservation of energy equation takes the form, mg(R + r)(1 − cos θ ) =
mv 2 /2 + βmr 2 ω2 /2. But v still equals rω (because the ball can be considered to be
instantaneously rotating around the contact point with angular speed ω), so the kinetic
energy still equals (1 + β)mv 2 /2. The conservation of energy statement is thus
1
(1 + β)mv 2 = mg(R + r)(1 − cos θ ).
2
(8.70)
We therefore have the same equations as above, except that R is replaced everywhere by
R + r. But R didn’t appear in the result for θ in Eq. (8.68), so the answer is unchanged.
Remark: The method of the second solution to Problem 5.3 does not work in this
problem, because there is a force available to make vx decrease, namely the friction
8.9 Solutions
351
force. And indeed, vx does decrease before the rolling ball
√ leaves the sphere. At a given
value of θ, the v in the present problem is simply 1/ 1 + β times the v in Problem
5.3, so the maximum vx is achieved here at cos θ = 2/3, just as in Problem 5.3. But
the angle in Eq. (8.68) is larger than this, so vx decreases while the ball is between
these two angles. (However, see the following problem for a setup involving rotations
where the max vx is relevant.) ♣
8.3. Sliding ladder
The important point to realize in this problem is that the ladder loses contact with the
wall before it hits the ground. So we need to find where this loss of contact occurs. Let
r = /2, for convenience. While the ladder is in contact with the wall, its CM moves
in a circle of radius r. This follows from the fact that the median to the hypotenuse of
a right triangle has half the length of the hypotenuse. Let θ be the angle between the
wall and the radius from the corner to the CM (see Fig. 8.68). This is also the angle
between the ladder and the wall.
We’ll solve this problem by assuming that the CM always moves in a circle, and
then determining the position at which the horizontal CM speed starts to decrease, that
is, the point at which the normal force from the wall would have to become negative.
Since the normal force of course can’t be negative, this is the point where the ladder
loses contact with the wall.
By conservation of energy, the kinetic energy of the ladder equals the loss in potential
energy, which is mgr(1 − cos θ). This kinetic energy can be broken up into the CM
translational energy plus the rotation energy. The CM translational energy is mr 2 θ˙ 2 /2,
because the CM travels in a circle of radius r. The rotational energy is I θ˙ 2 /2. The
same θ˙ applies here as in the CM translational motion, because θ is the angle between
the ladder and the vertical, and is thus the angle of rotation of the ladder. Letting
I ≡ βmr 2 to be general (β = 1/3 for our ladder), the conservation of energy statement
is (1+β)mr 2 θ˙ 2 /2 = mgr(1−cos θ). Therefore, the speed of the CM, which is v = r θ˙ ,
equals
v=
2gr(1 − cos θ )
.
1+β
(8.71)
The horizontal component of this is
vx =
2gr
1+β
(1 − cos θ ) cos θ.
(8.72)
√
Taking the derivative of (1 − cos θ) cos θ , we see that the horizontal speed is
maximum when cos θ = 2/3. Therefore the ladder loses contact with the wall when
2
(8.73)
=⇒ θ ≈ 48.2◦ ,
3
which is independent of β. This means that, for example, a dumbbell (two masses at
the ends of a massless rod, with β = 1) loses contact with the wall at the same angle.
Plugging the value of θ from Eq. (8.73) into Eq. (8.72), and using β = 1/3, we obtain
a final horizontal speed of
√
√
2gr
g
vx =
≡
.
(8.74)
3
3
√
Note that this is 1/3 of the 2gr horizontal speed that the ladder would have if it
were arranged (perhaps by having the top end slide down a curve) to eventually slide
horizontally along the ground. You are encouraged to compare various aspects of this
problem with those in Problem 8.2 and Problem 5.3.
cos θ =
Remark: The normal force from the wall is zero at the start and zero at the finish, so
it must reach a maximum at some intermediate value of θ. Let’s find this θ. Taking the
θ r
θ r
Fig. 8.68
r
ˆ
Angular momentum, Part I (Constant L)
352
derivative
√ of vx in Eq. (8.72) to find the CM’s horizontal acceleration ax , and then using
θ˙ ∝ 1 − cos θ from Eq. (8.71), we see that the force from the wall is proportional to
ax ∝
sin θ (3 cos θ − 2)θ˙
∝ sin θ (3 cos θ − 2).
√
1 − cos θ
(8.75)
Taking the derivative
of this, we find that the force from the wall is maximum when
√
cos θ = (1 + 19)/6 =⇒ θ ≈ 26.7◦ ♣
2b
a
Rθ
R
θ
2a
8.4. Leaning rectangle
We must first find the position of the rectangle’s CM when it has rotated through an
angle θ. Using Fig. 8.69, we can obtain this position (relative to the center of the
cylinder) by adding up the distances along the three shaded triangles. Because there
is no slipping, the contact point has moved a distance Rθ along the rectangle. We find
that the position of the CM is
(x, y) = R(sin θ , cos θ ) + Rθ (− cos θ, sin θ ) + a(sin θ, cos θ ).
(8.76)
We’ll now use the Lagrangian method to find the equation of motion and the frequency
of small oscillations. Using Eq. (8.76), you can show that the square of the speed of
the CM is
Fig. 8.69
v 2 = x˙ 2 + y˙ 2 = (a2 + R2 θ 2 )θ˙ 2 .
(8.77)
The simplicity of this result suggests that there is a quicker way to obtain it. And
indeed, the CM instantaneously rotates around the contact
√ point with angular speed
˙ and from Fig. 8.69, the distance to the contact point is a2 + R2 θ 2 . Therefore, the
θ,
√
speed of the CM is ωr = θ˙ a2 + R2 θ 2 .
The Lagrangian is
1
1
m(a2 + R2 θ 2 )θ˙ 2 + I θ˙ 2 − mg (R + a) cos θ + Rθ sin θ . (8.78)
2
2
The equation of motion is, as you can check,
L=T −V =
(ma2 + mR2 θ 2 + I )θ¨ + mR2 θ θ˙ 2 = mga sin θ − mgRθ cos θ .
(8.79)
Let us now consider small oscillations. Using the small-angle approximations,
sin θ ≈ θ and cos θ ≈ 1 − θ 2 /2, and keeping terms up to first order in θ , we obtain
(ma2 + I )θ¨ + mg(R − a)θ = 0.
(8.80)
The coefficient of θ is positive if a < R. Therefore, oscillatory motion occurs if a < R.
Note that this condition is independent of b. The frequency of small oscillations is
ω=
mg(R − a)
.
ma2 + I
(8.81)
If a ≥ R, then the rectangle falls off the cylinder.
Remarks: Let’s look at some special cases. If I = 0 (that is, all of the rectangle’s
mass is located at the CM), we have ω = g(R − a)/a2 . If in addition a
R, then
ω ≈ gR/a2 . You can also derive these results by considering the CM to be a point
mass sliding in a parabolic potential. If the rectangle is instead a uniform horizontal
stick, so that a
R, a
b, and I ≈ mb2 /3, then we have ω ≈ 3gR/b2 . If the
rectangle is a vertical stick (satisfying a < R), so that b
a and I ≈ ma2 /3, then we
2
have ω ≈ 3g(R − a)/4a . If in addition a
R, then ω ≈ 3gR/4a2 .
Without doing much work, there are two other ways we can determine the condition
under which there is oscillatory motion. The first is to look at the height of the CM
(although this is essentially what we ended up doing in the solution above). Using
small-angle approximations in Eq. (8.76), the height of the CM is y ≈ (R + a) +
8.9 Solutions
(R − a)θ 2 /2. Therefore, if a < R, the potential energy increases with θ , so the
rectangle wants to decrease its θ and fall back down to the middle. But if a > R, the
potential energy decreases with θ , so the rectangle wants to increase its θ and fall off
the cylinder.
The second way is to look at the horizontal positions of the CM and the contact
point. Small-angle approximations in Eq. (8.76) show that the former equals aθ and
the latter equals Rθ. Therefore, if a < R then the CM is to the left of the contact point,
so the torque from gravity (relative to the contact point) makes θ decrease, and the
motion is stable. But if a > R then the torque from gravity makes θ increase, and the
motion is unstable. ♣
8.5. Mass in a tube
The Lagrangian is
L=
1
2
1
M
3
2
θ˙ 2 +
1 2 2 1 2
mx θ˙ + m˙x + mgx sin θ + Mg
2
2
2
sin θ . (8.82)
The Euler–Lagrange equations are then
d
dt
∂L
∂ x˙
=
∂L
=⇒ m¨x = mxθ˙ 2 + mg sin θ,
∂x
d
dt
∂L
∂ θ˙
=
∂L
d
=⇒
∂θ
dt
=⇒
1
M
3
1
M
3
2
2˙
θ + mx2 θ˙ = mgx +
Mg
2
+ mx2 θ¨ + 2mx˙xθ˙ = mgx +
cos θ
Mg
2
(8.83)
cos θ .
In terms of η ≡ x/ , these equations become
η¨ = ηθ˙ 2 + g˜ sin θ ,
(1 + 3rη2 )θ¨ = 3r gη
˜ +
3g˜
2
˙
cos θ − 6rηη˙ θ,
(8.84)
where r ≡ m/M and g˜ ≡ g/ . Below is a Maple program that numerically finds the
value of η when θ equals π/2, in the case where r = 1. As mentioned in Problem
1.2, this value of η does not depend on g or , and hence not g.
˜ In the program, we’ll
denote g˜ by g, which we’ll give the arbitrary value of 10. We’ll use q for θ, and n for
η. Also, we’ll denote θ˙ by q1 and θ¨ by q2, etc. Even if you don’t know Maple, this
program should still be understandable. See Section 1.4 for more discussion of solving
differential equations numerically.
n:=0:
n1:=0:
q:=0:
q1:=0:
e:=.0001:
g:=10:
r:=1:
while q<1.57079 do
n2:=n*q1ˆ2+g*sin(q):
q2:=((3*r*g*n+3*g/2)*cos(q)
-6*r*n*n1*q1)/(1+3*r*nˆ2):
n:=n+e*n1:
n1:=n1+e*n2:
q:=q+e*q1:
q1:=q1+e*q2:
end do:
n;
#
#
#
#
#
#
#
#
#
#
initial n value
initial n speed
initial angle
initial angular speed
small time interval
value of g/l
value of m/M
do this process until
the angle is pi/2
the first E–L equation
#
#
#
#
#
#
#
the second E–L equation
how n changes
how n1 changes
how q changes
how q1 changes
stop the process
print the value of n
353
ˆ
Angular momentum, Part I (Constant L)
354
The resulting value of η is 0.378. If you actually run this program on Maple with
different values of g, you will find that the result for n doesn’t depend on g, as stated
above. A few results for η for various values of r are, in (r, η) notation: (0, 0.349),
(1, 0.378), (2, 0.410), (10, 0.872), (20, 3.290). It turns out that r ≈ 11.25 yields η ≈ 1.
That is, the mass m gets to the end of the tube right when the tube becomes vertical.
For η values larger than 1, we could imagine attaching a massless tubular extension
on the end of the given tube. It turns out that η → ∞ as r → ∞. In this case, the
mass m drops nearly straight down, causing the tube to quickly swing down to a nearly
vertical position. But m ends up slightly to one side and then takes a very long time to
move over to become directly below the pivot.
z
z=1
r1
r2
P1
P2
x
Fig. 8.70
8.6. Minimum I
The shape should be a cylinder with the z axis as its symmetry axis. A quick proof
(by contradiction) is as follows. Assume that the optimal blob is not a cylinder, and
consider the surface of the blob. If the blob is not a cylinder, then there exist two
points on the surface, P1 and P2 , that are located at different distances, r1 and r2 ,
from the z axis. Assume r1 < r2 (see Fig. 8.70). If we move a small piece of the
blob from P2 to P1 , then we decrease the moment of inertia, r 2 dm. Therefore, the
proposed noncylindrical blob cannot be the one with the smallest I . In order to avoid
this contradiction, all points on the surface must be equidistant from the z axis. The
only blob with this property is a cylinder.
8.7. Slick calculations of I
(a) We claim that the I for a square of side 2 is 16 times the I for a square of
side , assuming that the axes pass through any two corresponding points.
This is true because dm is proportional to the area, which is proportional to
length squared, so the corresponding dm’s differ by a factor of 4. And then
there are the two powers of r in the integrand. Therefore, when changing variables from one square to the other, there are four powers of 2 in the integral
r 2 dm = r 2 ρ dx dy.
As in Section 8.3.2, we can express the relevant relations in terms of pictures:
2l
l
= 16
= 4
=
( )
___
l
+ m
2
2
The first line comes from the scaling argument, the second comes from the fact
that moments of inertia simply add, and the third comes from the parallel-axis
theorem. Equating the right-hand sides of the first two, and then using the third
to eliminate
gives
l
1 2
= _6 ml
This agrees with the result of Example 12 in Section 8.3.1, with a = b = .
8.9 Solutions
(b) This is again a two-dimensional object, so the I for a triangle of side 2 is 16
times the I for a triangle of side , assuming that the axes pass through any two
corresponding points. With pictures, we have:
2l
l
= 16
(
=
+ 3
=
___
l
+ m
3
( )
)
2
The first line comes from the scaling argument, the second comes from the fact
that moments of inertia simply add, and the third comes from the parallel-axis
theorem. Equating the right-hand sides of the first two, and then using the third
to eliminate
gives
1
__
l
= 12 ml
2
This agrees with the result of Example 11√in Section 8.3.1, with N = 3. The
“radius” R used in that example equals / 3 in the present notation.
8.8. Slick calculations of I for fractal objects
(a) The scaling argument here is a little trickier than the one in Section 8.3.2. Our
object is self-similar to an object 3 times as big, so let’s increase the length by a
factor of 3 and see what happens to I . In the integral x2 dm, the x’s pick up a
factor of 3, so this gives a factor of 9. But what happens to the dm? Well, tripling
the size of our object increases its mass by a factor of 2, because the new object
is made up of two of the smaller ones, plus some empty space in the middle.
So the dm picks up a factor of 2. Therefore, the I for an object of length 3 is
18 times the I for an object of length , assuming that the axes pass through
any two corresponding points. With pictures, we have (the following symbols
denote our fractal object):
3l
= 18
= 2(
=
l
l /2
)
2
+ ml
The first line comes from the scaling argument, the second comes from the
fact that moments of inertia simply add, and the third comes from the parallelaxis theorem. Equating the right-hand sides of the first two, and then using the
355
356
ˆ
Angular momentum, Part I (Constant L)
third to eliminate
gives
l
1 2
= _8 ml
This is larger than the I for a uniform stick, namely m 2 /12, because the mass
here is generally farther away from the center.
Remark: When we increase the length of our object by a factor of 3, the factor
of 2 in the dm is larger than the factor of 1 relevant to a zero-dimensional object,
but smaller than the factor of 3 relevant to a one-dimensional object. So in some
sense our object has a dimension between 0 and 1. It is reasonable to define the
fractal dimension, d, of an object as the number for which r d is the increase in
“volume” when the dimensions are increased by a factor of r. In this problem,
we have 3d = 2, so d = log3 2 ≈ 0.63. ♣
(b) Again, the mass scales in a strange way. Let’s increase the dimensions of our
object by a factor of 2 and see what happens to I . In the integral x2 dm, the x’s
pick up a factor of 2, so this gives a factor of 4. But what happens to the dm?
Doubling the size of our object increases its mass by a factor of 3, because the
new object is made up of three of the smaller ones, plus an empty triangle in the
middle. So the dm picks up a factor of 3. Therefore, the I for an object of side
2 is 12 times the I for an object of side , assuming that the axes pass through
any two corresponding points. With pictures, we have:
2l
l
= 12
= 3
)
(
( )
l
+ m ___
3
=
2
The first line comes from the scaling argument, the second comes from the fact
that moments of inertia simply add, and the third comes from the parallel-axis
theorem. Equating the right-hand sides of the first two, and then using the third
to eliminate
gives
_1
l
= 9 ml
2
This is larger than the I for the uniform triangle in Problem 8.7, namely m 2 /12,
because the mass here is generally farther away from the center. Increasing the
size of our object by a factor of 2 increases the “volume” by a factor of 3, so the
fractal dimension is given by 2d = 3 =⇒ d = log2 3 ≈ 1.58.
(c) Again, the mass scales in a strange way. Let’s increase the dimensions of our
object by a factor of 3 and see what happens to I . In the integral x2 dm, the x’s
pick up a factor of 3, so this gives a factor of 9. But what happens to the dm?
8.9 Solutions
357
Tripling the size of our object increases its mass by a factor of 8, because the
new object is made up of eight of the smaller ones, plus an empty square in the
middle. So the dm picks up a factor of 8. Therefore, the I for an object of side
3 is 72 times the I for an object of side , assuming that the axes pass through
any two corresponding points. With pictures, we have:
3l
l
= 72
= 4
( ) + 4( )
=
+ ml 2
=
+ m ( 2 l)
2
The first line comes from the scaling argument, the second comes from the fact
that moments of inertia simply add, and the third and fourth come from the
parallel-axis theorem. Equating the right-hand sides of the first two, and then
using the third and fourth to eliminate
l
and
gives
3 ml 2
__
= 16
This is larger than the I for the uniform square in Problem 8.7, namely m 2 /6,
because the mass here is generally farther away from the center. Increasing the
size of our object by a factor of 3 increases the “volume” by a factor of 8, so the
fractal dimension is given by 3d = 8 =⇒ d = log3 8 ≈ 1.89.
Fijint
ri
8.9. Zero torque from internal forces
Let Fint
ij be the force that the ith particle feels due to the jth particle (see Fig. 8.71).
Then
Fint
i =
Fint
ij .
ri - rj
Fjiint
(8.85)
rj
j
The total internal torque on all the particles, relative to the chosen origin, is therefore
τ
int
ri × Fint
i
≡
i
ri × Fint
ij .
=
i
(8.86)
j
But if we interchange the indices (which were labeled arbitrarily), we have
τ int =
rj × Fint
ji = −
j
i
rj × Fint
ij ,
j
i
(8.87)
Fig. 8.71
ˆ
Angular momentum, Part I (Constant L)
358
int
where we have used Newton’s third law, Fint
ij = −Fji . Adding the two previous
equations gives
2τ int =
(ri − rj ) × Fint
ij .
i
(8.88)
j
But Fint
ij is parallel to (ri − rj ), by assumption. Therefore, each cross product in the
sum equals zero.
The above sums might make this solution look a bit involved. But the idea is simply
that the torques cancel in pairs. This is clear from Fig. 8.71, because the two forces
shown are equal and opposite, and they have the same lever arm relative to the origin.
l
8.10. Removing a support
mg
F
(a) First solution Let the desired force from the left support be F, and let the
downward acceleration of the stick’s CM be a. Then the F = ma and τ = I α
(relative to the fixed support; see Fig. 8.72) equations, along with the circularmotion relation between a and α, are, respectively,
Fig. 8.72
mg − F = ma,
mg
2
=
m 2
3
α,
a=
2
α.
(8.89)
The second equation gives α = 3g/2 . The third equation then gives a = 3g/4.
And the first equation then gives F = mg/4. Note that the right end of the stick
accelerates at 2a = 3g/2, which is larger than g.
Second solution Looking at torques around the CM, and also torques around
the fixed support, we have, respectively,
F
r
d
F
Fig. 8.73
mg
2
=
m 2
12
α,
and
mg
2
=
m 2
3
α.
(8.90)
Dividing the first of these equations by the second gives F = mg/4.
(b) First solution As in the first solution above, we have (using the parallel-axis
theorem; see Fig. 8.73)
mg − F = ma,
mgd = (βmr 2 + md 2 )α,
a = αd.
(8.91)
Solving for F gives F = mg/(1 + d 2 /βr 2 ). For d = r and β = 1/3, we obtain
the answer in part (a).
Second solution As in the second solution above, looking at torques around
the CM, and also torques around the fixed support, we have, respectively,
Fd = (βmr 2 )α,
and
mgd = (βmr 2 + md 2 )α.
(8.92)
Dividing the first of these equations by the second gives F = mg/(1 + d 2 /βr 2 ).
Remarks: For the special case d = r, we have the following: If β = 0 (point
mass in the middle) then F = 0; if β = 1 (dumbbell with masses at the ends)
then F = mg/2; and if β = ∞ (masses at the ends of long massless extensions
of the stick) then F = mg. These all make intuitive sense. In the limit d = 0, we
have F = mg. And in the limit d = ∞ (using √
a massless extension),
we have
√
β r and d
β r here. ♣
F = 0. Technically, we should be writing d
8.11. Falling stick
(a) Let’s calculate τ and L relative to the pivot point. The torque is due to gravity,
which effectively acts on the CM and has magnitude mgb. The moment of inertia
of the stick around the horizontal axis through the pivot (and perpendicular to
the massless stick) is simply mb2 . So when the stick starts to fall, the τ = dL/dt
8.9 Solutions
359
equation is mgb = (mb2 )α. Therefore, the initial acceleration of the CM, namely
bα, is
bα = g,
(8.93)
which is independent of and b. This answer makes sense. The stick initially
falls straight down, and the pivot provides no force because it doesn’t know right
away that the stick is moving.
(b) The only change from part (a) is the moment of inertia of the stick around the
horizontal axis through the pivot (and perpendicular to the massless stick). From
the parallel-axis theorem, this moment is mb2 +m 2 /12. So when the stick starts
to fall, the τ = dL/dt equation is mgb = (mb2 +m 2 /12)α. Therefore, the initial
acceleration of the CM is
g
bα =
.
(8.94)
1 + ( 2 /12b2 )
For
b, this goes to g, as it should. And for
b, it goes to zero, as it
should. In this case, a tiny movement of the CM corresponds to a very large
movement of the points far out along the stick. Therefore, by conservation of
energy, the CM must be moving very slowly.
8.12. Pulling a cylinder
Consider the net force in the y direction (the transverse direction) on an infinitesimal
arclength of the cylinder. Measure θ clockwise from the bottom, as shown in Fig. 8.74.
The tension increases as θ increases (assuming T1 > T2 ), and the friction force Ff on
the cylinder equals the difference dT in the tension from one end of the small arc to the
other (using Newton’s third law, and noting that there can be no net force on the string
because it is massless). The friction force on the cylinder therefore produces a force
component dT sin θ in the y direction. From the example in Section 2.1, the normal
force on the cylinder is N = T dθ, which yields a force component T dθ cos θ in the
y direction. The net y-force on the cylinder at the little arc is therefore
dFy = dT sin θ + T dθ cos θ = d(T sin θ ).
T1
dθ
θ
Ff
N
T2
Fig. 8.74
(8.95)
The total Fy on the cylinder is thus
Fy =
dFy =
d(T sin θ) =
(T sin θ ).
(8.96)
But θ runs from 0 to π, which means that sin θ starts and ends at zero. Therefore, the
total change in T sin θ equals zero, and so Fy = 0, as desired. Note that nowhere in
this solution did we assume anything about T . The string might be slipping, and the
cylinder might be rough in some places and smooth in others, and it doesn’t matter.
The total Fy is still zero (as we know it must be, from the simpler reasoning that T1
and T2 pull only in the x direction). Physically, what happens is that N is larger in the
top half of the semicircle than in the bottom half, and the resulting net downward force
cancels the upward force from the friction.
The above Fy = (T sin θ) result holds in general, and not just when the string
wraps around a semicircle. Because of our convention for θ , the signs work out so that
(T sin θ) is simply the sum of the y components of the tensions T1 and T2 , which is
the answer we expect.
θ
8.13. Oscillating ball
Let the angle from the bottom of the cylinder to the ball be θ (see Fig. 8.75), and let
Ff be the friction force. Then the tangential F = ma equation is
Ff − mg sin θ = ma,
R
(8.97)
where we have chosen rightward to be the positive direction for a and Ff . Also, the
τ = I α equation (relative to the CM) is
2
(8.98)
−rFf = mr 2 α,
5
Ff
Fig. 8.75
ˆ
Angular momentum, Part I (Constant L)
360
where we have chosen clockwise to be the positive direction for α. Using the nonslipping condition rα = a, the torque equation becomes Ff = −(2/5)ma. Plugging
this into Eq. (8.97), and using sin θ ≈ θ, we obtain mgθ + (7/5)ma = 0. Under the
assumption r
R, the center of the ball moves along a circle with radius essentially
equal to R, so we have a ≈ Rθ¨ , We therefore arrive at
5g
θ = 0.
7R
θ¨ +
(8.99)
This is the equation for simple harmonic motion with frequency
5g
.
7R
ω=
(8.100)
In general, if the ball has a moment of inertia equal to βmr 2 , you can show that
the frequency of small oscillations is g/(1 + β)R. Note that we needed to use two
different expressions for a in this solution, namely rα and Rθ¨ .
√
Remarks: The answer in Eq. (8.100) is slightly smaller than the g/R answer for
the case where the ball slides. In terms of forces, the reason for this is that the friction
force causes there to be a smaller net tangential force. In terms of energy, the reason
is that energy is “lost” in the rotational motion, so the ball ends up moving slower.
If we omit the r
R assumption, then the rα = a relation still holds, because
we can consider the ball to be instantaneously rotating around the contact point. But
¨ because the center of the ball
the a = Rθ¨ relation is replaced by a = (R − r)θ,
moves along a circle of radius R − r. Therefore, the exact result for the frequency is
ω = 5g/7(R − r). This goes to infinity as r → R. ♣
8.14. Oscillating cylinders
The moments of inertia of the cylinders are I1 = M1 R21 and I2 = M2 R22 . Let F be the
force between the two cylinders, defined with rightward on the small cylinder being
positive. Let θ1 and θ2 be the angles of rotation of the cylinders, with counterclockwise
positive, relative to the position where the small cylinder is at the bottom of the big
cylinder. Then the torque equations are
θ
Fig. 8.76
R2
FR1 = M1 R21 θ¨1 ,
and
FR2 = −M2 R22 θ¨2 .
(8.101)
We are not so much concerned with θ1 and θ2 as we are with the angular position that
M1 makes with the vertical. Let this angle be θ (see Fig. 8.76). In the approximation
R1
R2 , the nonslipping condition says that R2 θ ≈ R2 θ2 − R1 θ1 , since both sides
of this equation are expressions for the arclength away from the bottom of the big
cylinder. Adding Eqs. (8.101) after dividing through by the masses then gives
1
1
+
M1
M2
= −R2 θ¨ .
(8.102)
F − M1 g sin θ = M1 (R2 θ¨ ).
(8.103)
F
The tangential force equation on M1 is
Substituting the F from (8.102) into this equation gives, with sin θ ≈ θ ,
M1 +
1
1
M1
+
1
M2
θ¨ +
M1 g
R2
θ = 0.
(8.104)
After simplifying, the frequency of small oscillations is
ω=
g
R2
M1 + M2
.
M1 + 2M2
(8.105)
8.9 Solutions
361
√
Remarks: In the limit M2
M1 , we obtain ω ≈ g/R2 . In this case, there is
essentially no friction force between the cylinders, because otherwise the “massless”
M2 would have infinite angular acceleration. So there is only a normal force, and the
M2 , we
small cylinder
√ essentially acts like a pendulum of length R2 . In the limit M1
obtain ω ≈ g/2R2 . In this case, the big cylinder is essentially fixed, so we simply
have the setup mentioned in the solution to Problem 8.13, with β = 1. ♣
8.15. Lengthening the string
Consider the angular momentum relative to the support point P. The forces on the mass
are the tension in the string and gravity. The former provides no torque around P, and
the latter provides no torque in the z direction. Therefore, Lz is constant. The motion
is always approximately circular because the length of the string changes very slowly,
so if we let ω be the frequency of the circular motion when the string has length ,
then we can say that
Lz = mr 2 ω
(8.106)
is constant. The frequency ω can be obtained by using F = ma for the circular motion.
The tension in the string is essentially mg/ cos θ (to make the forces in the y direction
cancel), so the horizontal radial force is mg tan θ . Therefore,
g
g
=
.
(8.107)
mg tan θ = mrω2 = m( sin θ)ω2 =⇒ ω =
cos θ
h
Plugging this into Eq. (8.106), we see that the constant value of Lz is
g
Lz = mr 2
.
(8.108)
h
√
This holds at all times, so the quantity r 2 / h is constant. Let’s look at the two cases.
√
(a) For θ ≈ 0, we have h ≈ , so Eq. (8.108) says that r 2 / is constant. Therefore,
r ∝ 1/4 ,
(8.109)
which means that r grows very slowly as you let the string out when
θ ≈ 0.
√
(b) For θ ≈ π/2, we have r ≈ , so Eq. (8.108) says that 2 / h is constant.
Therefore,
h ∝ 4,
(8.110)
which means that h grows very quickly as you let the string out when θ ≈ π/2.
Note that Eq. (8.108) says that h ∝ r 4 for any value of θ . So no matter what
θ is, if you slowly lengthen the string so that r doubles, then h increases by a
factor of 16. Equivalently, if you draw the string in, the envelope of the motion
of the mass is a surface of revolution generated by a curve of the form y ∝ −x4 .
8.16. A triangle of cylinders
(a) Let N be the normal force between the cylinders, and let F be the friction force
from the ground (see Fig. 8.77). Let ax be the initial horizontal acceleration of
the right bottom cylinder (so α = ax /R is its angular acceleration), and let ay be
the initial vertical acceleration of the top cylinder, with downward taken to be
positive.
If we consider the torque around the center of one of the bottom cylinders,
then the only relevant force is F, because N , gravity, and the normal force from
the ground all point through the center. The equations for Fx = max on the
bottom right cylinder, Fy = may on the top cylinder, and τ = I α on the bottom
right cylinder are, respectively,
N cos 60◦ − F = max ,
mg − 2N sin 60◦ = may ,
(8.111)
FR = (βmR )(ax /R).
2
N
N
F
Fig. 8.77
N
N
F
ˆ
Angular momentum, Part I (Constant L)
362
We have four unknowns, N , F, ax , and ay , so we need one more equation.
Fortunately, ax and ay are related. The contact surface between the top and
bottom cylinders lies (initially) at an angle of 30◦ with the horizontal. Therefore,
if the bottom cylinders move a distance d to the side, then the top cylinder moves
a distance d tan 30◦ downward. Hence,
√
ax = 3ay .
(8.112)
F
N
Fig. 8.78
N N
F
F
We now have four equations and four unknowns. Solving for ay by your method
of choice gives
g
ay =
.
(8.113)
7 + 6β
F
N
(b) Let N be the normal force between the cylinders, and let F be the friction
force between the cylinders, with positive directions shown in Fig. 8.78. Let
ax be the initial horizontal acceleration of the right bottom cylinder, and let ay
be the initial vertical acceleration of the top cylinder, with downward taken to
be positive. Let α be the angular acceleration of the right bottom cylinder, with
counterclockwise taken to be positive. Note that α is not equal to ax /R, because
the bottom cylinders slip on the ground.
If we consider the torque around the center of one of the bottom cylinders,
then the only
√relevant force is F. And from the same reasoning as in part (a), we
have ax = 3ay . Therefore, the four equations analogous to Eqs. (8.111) and
(8.112) are
N cos 60◦ − F sin 60◦ = max ,
mg − 2N sin 60◦ − 2F cos 60◦ = may ,
FR = (βmR2 )α,
√
ax = 3ay .
(8.114)
We have five unknowns, N , F, ax , ay , and α, so we need one more equation.
The tricky part is relating α to ax . One way to do this is to ignore the y motion
of the top cylinder and imagine the bottom right cylinder to be rotating up and
around the top cylinder, which is held fixed. In this rotational motion, the center
of the bottom cylinder moves at an angle of 30◦ with respect to the horizontal
(at the start). So if it moves an infinitesimal distance d to the right, then its center
moves a distance d/ cos 30◦ up and to the right. So the√angle through which the
bottom cylinder rotates is θ = (d/ cos 30◦ )/R = (2/ 3)(d/R). Bringing back
in the vertical motion of the top cylinder doesn’t change this result. Therefore,
taking two derivatives of this relation gives
2 ax
.
α= √
3 R
(8.115)
We now have five equations and five unknowns. Solving for ay by your method
of choice gives
g
.
(8.116)
ay =
7 + 8β
Remarks: If β = 0, that is, if all the mass is at the center of the cylinders, then
the results in both parts (a) and (b) reduce to g/7. This β = 0 case is equivalent
to the case of frictionless cylinders (of any mass distribution), because then
nothing rotates. If β = 0, then the result in part (b) is smaller than that in
part (a). This isn’t so obvious, but the basic reason is that the bottom cylinders in
part (b) take
√ up more energy because they have to rotate slightly faster, because
α = (2/ 3)(ax /R) instead of α = ax /R. ♣
8.9 Solutions
8.17. Falling chimney
Let θ be the angle through which the chimney has fallen. Before we start dealing with
the forces in the rods, let’s first determine θ¨ as a function of θ . Let be the height
of the chimney. Then the moment of inertia around the pivot point on the ground is
m 2 /3 (if we ignore the width). And the torque (around the pivot point) due to gravity
is τ = mg( /2) sin θ . Therefore, τ = dL/dt gives mg( /2) sin θ = (1/3)m 2 θ¨ ,
and so
3g sin θ
θ¨ =
.
(8.117)
2
Let’s now determine the forces in the rods. Our strategy will be to imagine that the
chimney consists of a chimney of height h, with another chimney of height −h placed
on top of it. We’ll find the forces in the rods connecting these two “sub-chimneys,”
and then we’ll maximize one of these forces (T2 , defined below) as a function of h.
The forces on the top piece are gravity and also the forces from the two rods at each
end of the bottom board. Let’s break these latter forces up into transverse and longitudinal forces along the chimney. Let T1 and T2 be the two longitudinal components,
and let F be the sum of the transverse components, as shown in Fig. 8.79. We have
picked the positive directions for T1 and T2 so that positive T1 corresponds to compression in the left rod, and positive T2 corresponds to tension in the right rod (which
is what the forces will turn out to be, as we’ll see). It turns out that if the width (which
F (as we’ll see below), so the
we’ll call 2r) is much less than the height, then T2
tension in the right rod is essentially equal to T2 . We will therefore be concerned with
maximizing T2 .
In writing down the force and torque equations for the top piece, we have three
equations (the radial and tangential F = ma equations, and τ = dL/dt around the
CM), and three unknowns (F, T1 , and T2 ). If we define the fraction f ≡ h/ , then
the top piece has length (1 − f ) and mass (1 − f )m, and its CM travels in a circle of
radius (1 + f ) /2, Therefore, our three force and torque equations are, respectively,
T2 − T1 + (1 − f )mg cos θ = (1 − f )m
(1 + f )
2
θ˙ 2 ,
F + (1 − f )mg sin θ = (1 − f )m
(1 + f )
2
θ¨ ,
(T1 + T2 )r − F
(1 − f )2
(1 − f )
= (1 − f )m
2
12
2
(8.118)
θ¨ .
At this point, we could plow forward and solve this system of three equations in three
unknowns. But things simplify greatly in the limit r
. The third equation says that
T1 + T2 is of order 1/r, and the first equation says that T2 − T1 is of order 1. These
imply that T1 ≈ T2 , to leading order in 1/r. Therefore, we can set T1 + T2 ≈ 2T2 in the
third equation. Using this approximation, along with the value of θ¨ from Eq. (8.117),
the second and third equations become
3
(1 − f 2 )mg sin θ ,
4
(1 − f )
1
2rT2 − F
= (1 − f )3 mg sin θ.
2
8
This first of these equations gives
F + (1 − f )mg sin θ =
F=
(8.119)
mg sin θ
(−1 + 4f − 3f 2 ),
4
(8.120)
mg sin θ
f (1 − f )2 .
8r
(8.121)
and then the second gives
T2 ≈
363
l-h
T1
r
r
F
h
θ
Fig. 8.79
T2
364
ˆ
Angular momentum, Part I (Constant L)
As stated above, this is much greater than F (because /r
1), so the tension in the
right rod is essentially equal to T2 . Taking the derivative of T2 with respect to f , we
see that it is maximum at
f ≡
h
1
.
3
=
(8.122)
Therefore, the chimney is most likely to break at a point one-third of the way up
(assuming that the width is much less than the height). Interestingly, f = 1/3 makes
the force F in Eq. (8.120) exactly equal to zero. For more on the falling chimney, see
Madsen (1977) and Varieschi and Kamiya (2003).
8.18. Ball hitting stick
Let V , v , and ω be the speed of the ball, the speed of the stick’s CM, and the angular
speed of the stick, respectively, after the collision. Then conservation of momentum,
angular momentum (around the fixed point that coincides with the initial center of the
stick), and energy give
MV0 = MV + mv ,
MV0 d = MVd + βm 2 ω,
MV02
(8.123)
= MV + mv + βm ω .
2
2
2 2
We must solve these three equations for V , v , and ω. The first two equations quickly
give v d = β 2 ω. Solving for V in the first equation and plugging the result into the
third, and then eliminating ω through v d = β 2 ω gives
v=
2V0
1+
m
M
+
d2
β 2
=⇒
ω = V0
2 βd 2 V0
1+
m
M
+
d2
β 2
.
(8.124)
Having found v , the first equation above gives V as
V = V0
1−
m
M
+
1+
m
M
+
d2
β 2
d2
β 2
.
(8.125)
You are encouraged to check various limits of these answers. Another solution to
Eq. (8.123) is of course V = V0 , v = 0, and ω = 0. The initial conditions certainly
satisfy conservation of p, L, and E with the initial conditions (a fine tautology, indeed).
Nowhere in Eq. (8.123) does it say that the ball actually hits the stick.
8.19. A ball and stick theorem
As in the solution to Problem 8.18, we have
MV0 = MV + mv ,
MV0 d = MVd + I ω,
MV02
(8.126)
= MV + mv + I ω .
2
2
2
The speed of the contact point on the stick right after the collision equals the speed
of the CM plus the rotational speed relative to the CM. In other words, it equals v +ωd.
The desired relative speed is therefore (v + ωd) − V . We can determine the value of
this relative speed by solving the above three equations for V , v , and ω. Equivalently,
we can just use the results of Problem 8.18. There is, however, a much more appealing
method, which is the following.
The first two equations quickly give mv d = I ω. The last equation may then be
written as, using I ω2 = (I ω)ω = (mv d)ω,
M (V0 − V )(V0 + V ) = mv (v + ωd).
(8.127)
8.9 Solutions
If we now write the first equation as
M (V0 − V ) = mv ,
(8.128)
we can divide Eq. (8.127) by Eq. (8.128) to obtain V0 + V = v + ωd, or
V0 = (v + ωd) − V ,
(8.129)
as we wanted to show. In terms of velocities, the correct statement is that the final
relative velocity is the negative of the initial relative velocity. In other words, V0 − 0 =
− V − (v + ωd) .
8.20. The superball
Since we are told that |vy | is unchanged by the bounce, we can ignore it when applying
conservation of energy. And since the vertical impulse from the floor provides no
torque around the ball’s CM, we can completely ignore the y motion in this problem.
The horizontal impulse from the floor is responsible for changing both vx and ω.
With positive directions defined as in the statement of the problem, Eq. (8.61) gives
L=R p
=⇒
I (ω − ω) = Rm(vx − vx ).
(8.130)
And conservation of energy gives
1
1
1
1
2
mv + I ω 2 = mvx2 + I ω2
2 x
2
2
2
=⇒
2
I (ω 2 − ω2 ) = m(vx2 − vx ).
Dividing this equation by Eq. (8.130)
(8.131)
gives15
R(ω + ω) = −(vx + vx ).
(8.132)
We can now combine this equation with Eq. (8.130) which can be rewritten as, using
I = (2/5)mR2 ,
2
(8.133)
R(ω − ω) = vx − vx .
5
Given vx and ω, the previous two equations are two linear equations in the two
unknowns, vx and ω . Solving for vx and ω , and writing the result in matrix notation,
gives
vx
Rω
=
1
7
3
−10
−4
−3
vx
Rω
,
(8.134)
as desired. Note that Eq. (8.132), when written in the form of vx + Rω = −(vx + Rω ),
says that the relative velocity of the ball’s contact point and the ground simply changes
sign during the bounce.
Remark: For a ball with a general moment of inertia I = βmR2 , you can use the
above procedure to show that the matrix in Eq. (8.134) takes the general form,
1
1+β
15
1−β
−2
−2β
−(1 − β)
.
(8.135)
We have divided out the trivial ω = ω and vx = vx solution, which corresponds to slipping
motion on a frictionless plane. The nontrivial solution we will find shortly is the nonslipping one.
Basically, to conserve energy, there must be no work done by friction. And since work is force
times distance, this means that either (1) the plane is frictionless, so that the force is zero, or (2)
there is no relative motion between the ball’s contact point and the plane, so that the distance is
zero. The latter case is the one we are concerned with here.
365
366
ˆ
Angular momentum, Part I (Constant L)
For β = 1 (a hoop), this becomes
−1
0
0
−1
,
(8.136)
which means that the bounce simply interchanges and negates the values of vx and Rω.
In particular, if you throw a “super-hoop” sideways with no spin (that is, Rω = 0),
then it will bounce straight up in the air (that is, vx = 0) while spinning. ♣
8.21. Many bounces
Equation (8.62) gives the result after one bounce, so the result after two bounces is
vx
Rω
=
3/7
−10/7
−4/7
−3/7
=
3/7
−10/7
−4/7
−3/7
=
1
0
=
Rω
vx
vx
Rω
2
vx
Rω
vx
0
1
Rω
.
(8.137)
The square of the matrix turns out to be the identity. Therefore, after two bounces,
both vx and ω return to their original values. The ball then repeats the motion of
the previous two bounces (and so on, after every two bounces). The only difference
between successive pairs of bounces is that the ball may shift horizontally. You are
strongly encouraged to experimentally verify this interesting periodic behavior.
8.22. Rolling over a bump
We will use the fact that the angular momentum of the ball with respect to the corner
of the step (call this point P) is unchanged by the collision. This is true because any
forces exerted at point P provide zero torque around P. (The torque from gravity
will be relevant during the subsequent rising-up motion. But during the instantaneous
collision, L does not change.) This fact will allow us to find the energy of the ball right
after the collision, which we will then require to be greater than mgh.
Breaking the initial L into the contribution relative to the CM, plus the contribution
from the ball treated like a point mass located at the CM, we see that the initial angular
momentum is L = (2/5)mR2 ω0 + mV0 (R − h), where ω0 is the initial angular speed.
But the nonslipping condition tells us that ω0 = V0 /R, so we can write L as
L=
2
mRV0 + mV0 (R − h) = mV0
5
7R
−h .
5
(8.138)
Let ω be the angular speed of the ball around point P immediately after the collision. The parallel-axis theorem says that the moment of inertia around P is equal to
(2/5)mR2 + mR2 = (7/5)mR2 . Conservation of L around P during the collision then
gives
mV0
7R
7
− h = mR2 ω
5
5
=⇒
ω =
V0
R
1−
5h
.
7R
(8.139)
The energy of the ball right after the collision is therefore
E=
1
2
5h
7 2
7
mR ω 2 =
mV 2 1 −
5
10 0
7R
2
.
(8.140)
The ball will climb up over the step if E ≥ mgh, which gives
V0 ≥
5h
10gh
1−
7
7R
−1
.
(8.141)
8.9 Solutions
Remarks: It is possible for the ball to rise up over the step even if h > R, provided
that the ball sticks to the corner, without slipping. (If h > R, the step would have to be
“hollowed out” so that the ball doesn’t collide with the side of the step.) But note that
V0 → ∞ as h → 7R/5. For h ≥ 7R/5, it is impossible for the ball to make it up over
the step, no matter how large V0 is. The ball will get pushed down into the ground,
instead of rising up, if h > 7R/5.
For an object with a general moment of inertia I = βmR2 (so β = 2/5 in our
problem), you can show that the minimum initial speed is
V0 ≥
2gh
1+β
1−
h
(1 + β)R
−1
.
(8.142)
This decreases as β increases. It is smallest when the “ball” is a wheel with all the
mass on its rim (so that β = 1), in which case it is possible for the wheel to climb up
over the step even if h is close to 2R. ♣
8.23. Falling toast
the speeds of the CM right before and right after the collision (so we
Let v0 and v be √
know that v0 = 2gH ). The I for a square is the same as the I for a stick, (1/12)m 2 ,
so L = ( /2) p gives16
m 2
ω = − (mv − mv0 )
12
2
=⇒
v0 − v =
ω
.
6
(8.143)
Conservation of energy during the collision gives
1
1 m 2 2
1 2
mv0 = mv 2 +
ω
2
2
2 12
=⇒
(v0 + v )(v0 − v ) =
2 ω2
12
.
(8.144)
Dividing this by Eq. (8.143) gives v0 + v = ω/2. We now have two linear equations
for v and ω. Solving gives v = v0 /2 and ω = 3v0 / .
After the collision, the time to hit the ground is given by v t + gt 2 /2 = h. Solving
for t, setting ωt = π for half a rotation, and using the v and ω we just found, yields
⎛
⎞
v02
3v0 1 ⎝ v0
·
− +
+ 2gh ⎠ = π .
(8.145)
g
2
4
Factoring out a v02 and then using v02 = 2gH gives
3H
1+
4h
− 1 = π.
H
(8.146)
Isolating the square root, and then squaring and solving for H gives
H=
π2 2
.
6(6h − π )
(8.147)
The special value of is = (6/π )h (which would be a huge piece of toast), in which
case H = ∞. If is larger than (6/π )h, then there isn’t enough time to make half a
rotation before hitting the floor. The intuitive reason for this is that since ω = 6v /
(from above), there is no way to increase ω without also increasing v . The toast has the
best chance of making half a rotation if v is very large, because then gravity doesn’t
have any time to increase v . In this limit, the toast makes half a rotation upon hitting
the floor if π/ω = h/v . Plugging in ω = 6v / gives 6h = π , as desired.
16
The minus sign on the right-hand side comes from the fact that we’re defining the v’s to be positive
downward. Equivalently, the force from the counter increases the angular speed but decreases the
linear speed.
367
ˆ
Angular momentum, Part I (Constant L)
368
For the reasonable values of h = 1 m and = 10 cm, we obtain H ≈ 3 mm, which
is quite small. You should try this with a pencil to convince yourself that even this
small distance can yield the desired (or rather, undesired) half rotation.
ω
8.24. Sliding to rolling
R
Fig. 8.80
V
(a) Define all linear quantities to be positive to the right, and all angular quantities
to be positive clockwise, as shown in Fig. 8.80. Then, for example, the friction
force Ff is negative. The friction force slows down the translational motion and
speeds up the rotational motion, according to
Ff = ma,
and
− Ff R = I α.
(8.148)
Eliminating Ff , and using I = βmR2 , gives a = −βRα. Integrating this over
time, up to the time when the ball stops slipping, gives
V = −βR ω.
(8.149)
Note that we could have obtained this by simply using the impulse equation,
Eq. (8.61). Using V = Vf −V0 , and ω = ωf −ω0 = ωf , and also ωf = Vf /R
(the nonslipping condition), Eq. (8.149) gives
Vf =
V0
,
1+β
(8.150)
independent of the nature of Ff . Ff can depend on position, time, speed, or
anything else. The relation a = −βRα, and hence also Eq. (8.149), will still be
true at all times.
Remark: We can also calculate τ and L relative to a dot painted on the ground
that is the contact point at a given instant. There is zero torque relative to this
point. To find L, we must add the L of the CM and the L relative to the CM.
Therefore, τ = dL/dt gives 0 = (d/dt)(mv R + βmR2 ω), and so a = −βRα, as
above. ♣
Using Eq. (8.150), and also the relation ωf = Vf /R, the loss in kinetic energy is
K=
1
mV 2 −
2 0
1
1
mV 2 + I ωf2
2 f
2
=
1
β
1
mV 2 1 −
−
2 0
(1 + β)2
(1 + β)2
=
1
mV 2
2 0
β
1+β
.
(8.151)
For β → 0, no energy is lost, which makes sense. And for β → ∞ (a spool
sliding on its axle), all the energy is lost, which also makes sense, because we
essentially have a sliding block which can’t rotate.
(b) Let’s first find t. The friction force is Ff = −µmg, so F = ma gives −µg = a.
Therefore, V = at = −µgt. But Eq. (8.150) says that V ≡ Vf − V0 =
−V0 β/(1 + β). Therefore,
t=
V0
β
·
.
(1 + β) µg
(8.152)
For β → 0, we have t → 0, which makes sense. And for β → ∞, we have
t → V0 /(µg), which equals the time a sliding block would take to stop.
8.9 Solutions
Let’s now find d. We have d = V0 t +(1/2)at 2 . Using a = −µg, and plugging
in the t from Eq. (8.152), we obtain
d=
β(2 + β) V02
·
.
(1 + β)2 2µg
(8.153)
The two extreme cases for β check here.
To calculate the work done by friction, we might be tempted to write down the
product Ff d, with Ff = −µmg and d given in Eq. (8.153). But the result doesn’t
equal the loss in kinetic energy calculated in Eq. (8.151). What’s wrong with
this reasoning? The error is that the friction force does not act over a distance d.
To find the distance over which Ff acts, we must find how far the surface of
the ball moves relative to the ground. The speed of a dot on the ball that is
instantaneously the contact point is Vrel (t) = V (t) − Rω(t) = (V0 + at) − Rαt.
Using α = −a/βR and a = −µg, this becomes
1+β
µgt.
β
Vrel (t) = V0 −
(8.154)
Integrating this from t = 0 to the t given in Eq. (8.152) gives
drel =
Vrel (t) dt =
V2
β
· 0 .
1 + β 2µg
(8.155)
The work done by friction is Ff drel = −µmgdrel , which does indeed give the
loss in kinetic energy given in Eq. (8.151).
8.25. Lots of sticks
Consider the collision between two sticks. Let V be the speed of the contact point on
the heavy one. Since this stick is essentially infinitely heavy, we may consider it to
be an infinitely heavy ball, moving at speed V . The rotational degree of freedom of
the heavy stick is irrelevant, as far as the light stick is concerned. We can therefore
invoke the result of Problem 8.19 to say that the relative speed of the contact points is
the same before and after the collision. This implies that the contact point on the light
stick picks up a speed of 2V , because the heavy stick is essentially unaffected by the
collision and keeps moving at speed V .
Let us now find the speed of the other end of the light stick. This stick receives an
impulse from the heavy stick, so we can apply Eq. (8.61) to the light stick to obtain
L=r p
=⇒
βmr 2 ω = r(mvCM )
=⇒
rω =
vCM
β
.
(8.156)
The speed of the struck (top) end is vtop = rω + vCM , because the CM speed adds to the
rotational speed. The speed of the other (bottom) end is vbot = rω − vCM , because the
CM speed subtracts from the rotational speed.17 The ratio of these speeds is
vbot
=
vtop
vCM
β
− vCM
vCM
β
+ vCM
=
1−β
.
1+β
(8.157)
This is a general result whenever you strike the end of a stick with any force. In the
present problem, we have vtop = 2V . Therefore,
vbot = V
17
2(1 − β)
.
1+β
(8.158)
Since β ≤ 1 for any real stick, we have rω = vCM /β ≥ vCM . Therefore, rω − vCM is greater
than or equal to zero.
369
370
ˆ
Angular momentum, Part I (Constant L)
The same analysis holds for all the other collisions. Therefore, the bottom ends of the
sticks move with speeds that form a geometric progression with ratio 2(1 − β)/(1 +
β). If this ratio is less than 1 (that is, if β > 1/3), then the speeds go to zero as
n → ∞. If it is greater than 1 (that is, if β < 1/3), then the speeds go to infinity
as n → ∞. If it equals 1 (that is, if β = 1/3), then the speeds remain equal to V and
are thus independent of n, as we wanted to show. A uniform stick has β = 1/3 relative
to its center (which is usually written in the form I = m 2 /12, where = 2r).
Chapter 9
Angular momentum, Part II
ˆ)
(General L
In Chapter 8, we discussed situations where the direction of the vector L remained
constant, and only its magnitude changed. In this chapter, we will look at more
general situations where the direction of L is allowed to change. The vector
nature of L will prove to be vital here, and we will arrive at all sorts of strange
results for spinning tops and such things. This chapter is rather long, alas, but the
general outline is that the first three sections cover general theory, then Section 9.4
introduces some actual physical setups, and then Section 9.6 begins the discussion
of tops.
9.1
9.1.1
Preliminaries concerning rotations
The form of general motion
Before getting started, we should make sure we’re all on the same page concerning
a few important things about rotations. Because rotations generally involve three
dimensions, they can often be hard to visualize. A rough drawing on a piece of
paper might not do the trick. For this reason, this chapter is one of the more
difficult ones in this book. But to ease into it, the next few pages consist of some
definitions and helpful theorems. This first theorem describes the general form
of any motion. You might consider it obvious, but it’s a little tricky to prove.
z
Theorem 9.1 (Chasles’theorem) Consider a rigid body undergoing arbitrary
motion. Pick any point P in the body. Then at any instant (see Fig. 9.1), the
motion of the body can be written as the sum of the translational motion of P,
plus a rotation around some axis (which may change with time) through P.1
P
V
ω
y
Proof: The motion of the body can be written as the sum of the translational
motion of P, plus some other motion relative to P (this is true because relative
coordinates are additive quantities). We must show that this latter motion is a
rotation. This seems quite plausible, and it holds because the body is rigid; that is,
1
x
Fig. 9.1
In other words, a person at rest with respect to a frame whose origin is P, and whose axes are
parallel to the fixed-frame axes, sees the body undergoing a rotation around some axis through P.
371
ˆ
Angular momentum, Part II (General L)
372
all points keep the same distances relative to each other. If the body weren’t rigid,
then this theorem wouldn’t be true.
To be rigorous, consider a spherical shell fixed in the body, centered at P. The
motion of the body is completely determined by the motion of the points on this
sphere, so we need only examine what happens to the sphere. Because distances
are preserved in the rigid body, the points on the sphere must always remain the
same radial distance from P. And because we are looking at motion relative to P,
we have therefore reduced the problem to the following: In what manner can a
rigid sphere transform into itself? We claim that any such transformation has the
property that there exist two points that end up where they started.2 These two
points must then be diametrically opposite points (assuming that the whole sphere
doesn’t end up back where it started, in which case every point ends up where it
started), because distances are preserved; given one point that ends up where it
started, the diametrically opposite point must also end up where it started, to
maintain the distance of a diameter.
If this claim is true, then we are done, because for an infinitesimal transformation, a given point moves in only one direction, because there is no time to do
any turning. So a point that ends up where it started must have remained fixed
for the whole (infinitesimal) time. Therefore, all the points on the diameter joining the two fixed points must also have remained fixed the whole time, because
distances are preserved. So we are left with a rotation around this axis.
This “two points ending up where they started” claim is quite believable,
but nevertheless tricky to prove. Claims with these properties are always fun to
think about, so I’ve left this one as a problem (Problem 9.2). Try to solve it on
your own.
We will invoke this theorem repeatedly in this chapter (often without bothering
to say so). Note that we are assuming that P is a point in the body, because we
used the fact that P keeps the same distances from other points in the body.
old ω
ck
sti
P
new ω
Remark: A situation where this theorem isn’t so obvious is the following (this setup contains
only rotation, with no translation of the point P). Consider an object rotating around a fixed
axis, the stick shown in Fig. 9.2. But now imagine grabbing the stick and rotating it around
some other axis (the dotted line shown). It isn’t immediately obvious that the resulting motion
is (instantaneously) a rotation around some new axis through the point P (which remains fixed).
But indeed it is. We’ll be quantitative about this in the “Rotating sphere” example later in this
section. ♣
9.1.2
The angular velocity vector
Fig. 9.2
It is extremely useful to introduce the angular velocity vector, ω, which is defined
as the vector that points along the axis of rotation, and whose magnitude equals
2
This claim is actually true for any transformation of a rigid sphere into itself, but for the present
purposes we are concerned only with infinitesimal transformations, because we are looking only
at what happens at a given instant in time.
9.1 Preliminaries concerning rotations
373
the angular speed. The choice of the two possible directions along the axis is given
by the right-hand rule: if you curl your right-hand fingers in the direction of the
spin, then your thumb points in the direction of ω. For example, a spinning record
has ω perpendicular to the record, through the center (as shown in Fig. 9.3),3 with
its magnitude equal to the angular speed, ω. The points on the axis of rotation
are the ones that (instantaneously) do not move. Of course, the direction of ω
may change over time, so the points that were formerly on the axis may now be
moving.
ω
ω
Fig. 9.3
Remarks:
1. If you want, you can break the mold and use the left-hand rule to determine ω, as long as
you use it consistently. The direction of ω will be the opposite, but that doesn’t matter,
because ω isn’t really physical. Any physical result (for example, the velocity of a particle,
given below in Theorem 9.2) will come out the same, independent of which hand you
(consistently) use.
When studying vectors in school,
You’ll use your right hand as a tool.
But look in a mirror,
And then you’ll see clearer,
It’s just like the left-handed rule.
2. The fact that we can specify a rotation by specifying a vector ω is a peculiarity to three
dimensions. If we lived in one dimension, then there would be no such thing as a rotation.
If we lived in two dimensions, then all rotations would take place in that plane, so we
could label a rotation by simply giving its speed, ω. In three dimensions, rotations take
place in 32 = 3 independent planes. And we choose to label these, for convenience, by
the directions orthogonal to these planes, and by the angular speed in each plane. If we
lived in four dimensions, then rotations could take place in 42 = 6 planes, so we would
have to label a rotation by giving 6 planes and 6 angular speeds. Note that a vector, which
has four components in four dimensions, would not do the trick. ♣
In addition to specifying the points that are instantaneously motionless, ω
also easily produces the velocity of any point in the rotating object. Consider
the situation where the axis of rotation passes through the origin, which we’ll
generally assume to be the case in this chapter, unless otherwise stated. Then we
have the following theorem.
Theorem 9.2 Given an object rotating with angular velocity ω, the velocity
of a point at position r is given by
v = ω × r.
(9.1)
Proof: Drop a perpendicular from the point in question (call it P) to the axis ω.
Let Q be the foot of the perpendicular, and let r be the vector from Q to P
3
It’s actually meaningless to say that ω passes through the center of the record, because you can
draw the vector anywhere, and it’s still the same vector, as long as it has the correct magnitude and
direction. Nevertheless, it’s customary to draw ω along the axis of rotation and to say things like,
“An object rotates around ω . . .”
ˆ
Angular momentum, Part II (General L)
374
ω
Q
r'
P
θ
r
(see Fig. 9.4). From the properties of the cross product (see Appendix B), v =
ω × r is orthogonal to ω, r, and also r because r is a linear combination of ω
and r. Therefore, the direction of v is correct; it is always orthogonal to ω and
r , so it describes circular motion around the axis ω. Also, by the right-hand rule
in the cross product (or the left-hand rule, if you had chosen to be different and
defined ω that way), v has the proper orientation around ω, namely into the page
at the instant shown. And since
|v| = |ω||r| sin θ = ωr ,
(9.2)
Fig. 9.4
we see that v has the correct magnitude, because ωr is the speed of the circular
motion around ω. So v is indeed the correct velocity vector. (If we have the
special case where P lies along ω, then r is parallel to ω, so the cross product
gives a zero result for v, as it should.)
We’ll make good use of Eq. (9.1) and apply it repeatedly throughout this
chapter. Even if it’s hard to visualize what’s going on in a given rotation, all you
have to do to find the speed of any point is calculate the cross product ω × r.
Conversely, if the speed of every point in a body is given by v = ω × r, then the
body must be undergoing a rotation with angular velocity ω, because all points
on the axis ω are motionless, and all other points move with the proper speed for
this rotation.
A very nice thing about angular velocities is that they simply add. Stated more
precisely:
Theorem 9.3 Let coordinate systems S1 , S2 , and S3 have a common origin.
Let S1 rotate with angular velocity ω1,2 with respect to S2 , and let S2 rotate with
angular velocity ω2,3 with respect to S3 . Then S1 rotates (instantaneously) with
angular velocity
ω1,3 = ω1,2 + ω2,3
(9.3)
with respect to S3 .
Proof: If ω1,2 and ω2,3 point in the same direction, then the theorem is clear;
the angular speeds just add. If, however, they don’t point in the same direction,
then things are a bit harder to visualize. But we can prove the theorem by making
abundant use of the definition of ω.
Pick a point P1 at rest in S1 . Let r be the vector from the origin to P1 . The
velocity of P1 (relative to a very close point P2 at rest in S2 ) due to the rotation of
S1 around ω1,2 is VP1 P2 = ω1,2 × r. The velocity of P2 (relative to a very close
point P3 at rest in S3 ) due to the rotation of S2 around ω2,3 is VP2 P3 = ω2,3 × r,
because P2 is also located essentially at position r. Therefore, the velocity of P1
relative to P3 is VP1 P2 + VP2 P3 = (ω1,2 + ω2,3 ) × r. This holds for any point
P1 at rest in S1 , so the frame S1 rotates with angular velocity (ω1,2 + ω2,3 ) with
9.1 Preliminaries concerning rotations
375
respect to S3 . We see that the proof basically comes down to the facts that (1) the
linear velocities just add, as usual, and (2) the angular velocities differ from the
linear velocities by a cross product with r.
If ω1,2 is constant in S2 , then the vector ω1,3 = ω1,2 + ω2,3 will change with
respect to S3 as time goes by, because ω1,2 , which is fixed in S2 , is changing with
respect to S3 (assuming that ω1,2 and ω2,3 aren’t parallel). But at any instant,
ω1,3 may be obtained by adding the present values of ω1,2 and ω2,3 . Consider the
following example.
Example (Rotating sphere): A sphere rotates with angular speed ω3 around a
stick that initially points in the zˆ direction. You grab the stick and rotate it around the
yˆ axis with angular speed ω2 . What is the angular velocity of the sphere, with respect
to the lab frame, as time goes by?
Solution: In the language of Theorem 9.3, the sphere defines the S1 frame; the stick
and the yˆ axis define the S2 frame; and the lab frame is the S3 frame. The instant after
you grab the stick, we are given that ω1,2 = ω3 zˆ , and ω2,3 = ω2 yˆ . Therefore, the
angular velocity of the sphere with respect to the lab frame is ω1,3 = ω1,2 + ω2,3 =
ω3 zˆ + ω2 yˆ , as shown in Fig. 9.5. Convince yourself that the combination of these
two rotations yields zero motion for the points along the line of ω1,3 . As time goes
by, the stick (and hence ω1,2 ) rotates around the y axis, so ω1,3 = ω1,2 + ω2,3 traces
out a cone around the y axis, as shown.
Remark: Note the different behavior of ω1,3 for a slightly different statement of the
problem: Let the sphere initially rotate with angular velocity ω2 yˆ around a stick, and then
grab the stick and rotate it with angular velocity ω3 zˆ . For this situation, ω1,3 initially
points in the same direction as in the original statement of the problem (it initially equals
ω2 yˆ + ω3 zˆ ). But as time goes by, it is now the horizontal component (defined by the stick)
of ω1,3 that changes, so ω1,3 = ω1,2 + ω2,3 traces out a cone around the z axis, as shown
in Fig. 9.6. ♣
z
stick
ω1,3
ω1,2
y
ω2,3
Fig. 9.5
z
ω2,3
ω1,3
ω1,2
stick
An important point concerning rotations is that they are defined with respect
to a coordinate system. It makes no sense to ask how fast an object is rotating with
respect to a certain point, or even a certain axis. Consider, for example, an object
rotating with angular velocity ω = ω3 zˆ with respect to the lab frame. Saying only,
“The object has angular velocity ω = ω3 zˆ ,” is not sufficient, because someone
standing in the frame of the object would measure ω = 0, and would therefore be
very confused by your statement. Throughout this chapter, we’ll try to remember
to state the coordinate system with respect to which ω is measured. But if we
forget, the default frame is the lab frame.
This section was definitely a bit abstract, so don’t worry too much about it
at the moment. The best strategy is perhaps to read on, and then come back
for a second pass after digesting a few more sections. At any rate, we’ll be
Fig. 9.6
y
ˆ
Angular momentum, Part II (General L)
376
discussing many other aspects (probably more than you’d ever want to know)
of ω in Section 9.7.2, so you’re assured of getting a lot more practice with it.
For now, if you want to strain some brain cells thinking about ω vectors, you are
encouraged to solve Problem 9.3, and also to look at the three given solutions.
9.2
Given an object undergoing general motion, the inertia tensor is what relates the
angular momentum, L, to the angular velocity, ω. This tensor (which is just a
fancy name for “matrix” in this context) depends on the geometry of the object,
as we’ll see. In finding the L due to general motion, we’ll follow the strategy of
Section 8.1. We’ll first look at the special case of rotation around an axis through
the origin, then we’ll look at the most general possible motion.
z
ω
9.2.1
y
x
Fig. 9.7
The inertia tensor
Rotation around an axis through the origin
The three-dimensional object in Fig. 9.7 rotates with angular velocity ω. Consider
a little piece of the body, with mass dm and position r. The velocity of this piece is
v = ω×r, so its angular momentum (relative to the origin) is r×p = (dm)r×v =
(dm)r × (ω × r). The angular momentum of the entire body is therefore
L=
r × (ω × r) dm,
(9.4)
where the integration runs over the volume of the body. In the case where the rigid
body is made up of a collection of point masses mi , the angular momentum is
L=
mi ri × (ω × ri ).
(9.5)
i
The double cross product in Eqs. (9.4) and (9.5) looks a bit intimidating, but it’s
actually not so bad. First, we have
ω×r =
xˆ
ω1
x
yˆ
ω2
y
zˆ
ω3
z
= (ω2 z − ω3 y)ˆx + (ω3 x − ω1 z)ˆy + (ω1 y − ω2 x)ˆz.
(9.6)
We’re using the notation ω1 instead of ωx , etc., because there are already enough
x, y, z’s floating around here. The double cross product is then
r × (ω × r) =
=
xˆ
x
(ω2 z − ω3 y)
yˆ
y
(ω3 x − ω1 z)
zˆ
z
(ω1 y − ω2 x)
ω1 (y2 + z 2 ) − ω2 xy − ω3 zx xˆ
9.2 The inertia tensor
377
+ ω2 (z 2 + x2 ) − ω3 yz − ω1 xy yˆ
+ ω3 (x2 + y2 ) − ω1 zx − ω2 yz zˆ .
(9.7)
The angular momentum in Eq. (9.4) may therefore be written in the concise
matrix form,
⎛
⎞ ⎛
⎞
⎞⎛
L1
(y2 + z 2 )
− xy
− zx
ω1
⎝ L2 ⎠ = ⎝ − xy
(z 2 + x2 )
− yz ⎠ ⎝ ω2 ⎠
− zx
− yz
(x2 + y2 )
L3
ω3
⎛
⎞⎛
⎞
Ixx Ixy Ixz
ω1
≡ ⎝ Iyx Iyy Iyz ⎠ ⎝ ω2 ⎠
Izx Izy Izz
ω3
≡ Iω.
(9.8)
For the sake of clarity, we have not bothered to write the dm part of each integral
(and we’ll continue to drop it for most of the remainder of this section). The
matrix I is called the inertia tensor. If the word “tensor” scares you, just ignore
it. I is simply a matrix. It acts on a vector (the angular velocity) and produces
another vector (the angular momentum).
z
L
Example (Cube with origin at corner): Calculate the inertia tensor for a solid
cube of mass M and side length L, with the coordinate axes parallel to the edges of
the cube, and the origin at a corner (see Fig. 9.8).
Solution: Due to the symmetry of the cube, there are only two integrals we need
to calculate in Eq. (9.8). The diagonal entries are all equal to (y2 + z 2 ) dm, and the
off-diagonal entries are all equal to − xy dm. With dm = ρ dx dy dz, and ρ = M /L3 ,
these two integrals are
L
0
L
L
0
−
0
L
0
L
( y2 + z 2 )ρ dx dy dz = ρL2
L
0
L
0
0
xyρ dx dy dz = −ρL
y2 dy + ρL2
L
L
x dx
0
0
L
0
z 2 dz =
y dy = −
ML2
.
4
2
ML2 ,
3
(9.9)
Therefore,
⎛
2/3
⎜
I = ML2 ⎝ −1/4
−1/4
−1/4
2/3
−1/4
⎞
−1/4
⎟
−1/4 ⎠.
2/3
(9.10)
Having found I, we can calculate the angular momentum associated with any given
angular velocity. If, for example, the cube is rotating around the z axis with angular speed ω, then we can apply the matrix I to the vector (0, 0, ω) to find that the
angular momentum is L = ML2 ω(−1/4, −1/4, 2/3). Note the somewhat odd fact
L
x
Fig. 9.8
L
y
ˆ
Angular momentum, Part II (General L)
378
that Lx and Ly are nonzero, even though the rotation is only around the z axis. We’ll
discuss this issue after the following remarks.
Remarks:
1. The inertia tensor in Eq. (9.8) is a rather formidable-looking object. You will therefore
be very pleased to hear that you rarely have to use it. It’s nice to know that it’s there if
you need it, but the concept of principal axes (discussed in Section 9.3) provides a way
to avoid using the inertia tensor (or more precisely, to greatly simplify it) and is therefore
much more useful in solving problems.
2. I is a symmetric matrix, which is a fact that will be important in Section 9.3. There are
therefore only six independent entries, instead of nine.
3. In the case where the rigid body is made up of a collection of point masses mi , the entries
in the matrix are just sums. For example, the upper left entry is mi (yi2 + zi2 ).
4. I depends only on the geometry of the object, and not on ω.
5. To construct an I, you not only need to specify the origin, you also need to specify the x, y,
z axes of your coordinate system. And the basis vectors must be orthogonal, because the
cross product calculation above is valid only for an orthonormal basis. If someone else
comes along and chooses a different orthonormal basis (but the same origin), then her I
will have different entries, as will her ω, as will her L. But her ω and L will be exactly the
same vectors as your ω and L. They will appear different only because they are written
in a different coordinate system. A vector is what it is, independent of how you choose to
look at it. If you each point your arm in the direction of what you calculate L to be, then
you will both be pointing in the same direction.
6. For the case of a pancake object rotating in the x-y plane, we have z = 0 for all points in
the object. And ω = ω3 zˆ , so ω1 = ω2 = 0. The only nonzero term in the L in Eq. (9.8)
is therefore L3 = (x2 + y2 ) dm ω3 , which is simply the Lz = Iz ω result we found in
Eq. (8.5). ♣
This is all perfectly fine. Given any rigid body, we can calculate I (relative
to a given origin, using a given set of axes). And given ω, we can then apply I
to it to find L. But what do these entries in I really mean? How do we interpret
them? Note, for example, that ω3 appears not only in L3 in Eq. (9.8), but also in
L1 and L2 . But ω3 is relevant to rotations around the z axis, so what in the world
is it doing in L1 and L2 ? Consider the following examples.
Example 1 (Point mass in the x-y plane): Consider a point mass m traveling
in a circle of radius r (centered at the origin) in the x-y plane, with frequency ω3 , as
shown in Fig. 9.9. Using ω = (0, 0, ω3 ), x2 + y2 = r 2 , and z = 0 in Eq. (9.8) (with
a discrete sum of only one object, instead of the integrals), the angular momentum
with respect to the origin is
z
ω
r
x
Fig. 9.9
y
L = (0, 0, mr 2 ω3 ).
(9.11)
The z component is mr(rω3 ) = mrv, as it should be. And the x and y components
are zero, as they should be. This case where ω1 = ω2 = 0 and z = 0 is simply
the case we studied in Chapter 8, as mentioned in Remark 6 above.
9.2 The inertia tensor
379
Example 2 (Point mass in space): Consider a point mass m traveling in a circle
of radius r, with frequency ω3 . But now let the circle be centered at the point (0, 0, z0 ),
with the plane of the circle parallel to the x-y plane, as shown in Fig. 9.10. Using
ω = (0, 0, ω3 ), x2 + y2 = r 2 , and z = z0 in Eq. (9.8), the angular momentum with
respect to the origin is
z0
L = mω3 (−xz0 , −yz0 , r 2 ).
ω
z
(9.12)
The z component is mrv, as it should be. But surprisingly, we have nonzero L1 and
L2 , even though the mass is just rotating around the z axis. L does not point along ω
here. What’s going on?
Consider an instant when the mass is in the y-z plane, as shown in Fig. 9.10. The
velocity of the mass is then in the −ˆx direction. Therefore, the particle most certainly
has angular momentum around the y axis, as well as the z axis. Someone looking at a
split-second movie of the mass at this point can’t tell whether it’s rotating around the
y axis, the z axis, or undergoing some complicated motion. But the past and future
motion is irrelevant; at any instant in time, as far as the angular momentum goes, we
are concerned only with what is happening at this instant.
At this instant, the angular momentum around the y axis is L2 = −mz0 v, because
z0 is the distance from the y axis, and the minus sign comes from the right-hand rule.
Using v = ω3 r = ω3 y, we have L2 = −mz0 ω3 y, in agreement with Eq. (9.12). Also,
at this instant, L1 is zero, because the velocity is parallel to the x axis. This agrees
with Eq. (9.12), since x = 0. As an exercise, you can check that Eq. (9.12) is also
correct when the mass is at a general point (x, y, z0 ).
We see that, for example, the Iyz ≡ − yz entry in I tells us how much the ω3
component of the angular velocity contributes to the L2 component of the angular
momentum. And due to the symmetry of I, the Iyz = Izy entry in I also tells us how
much the ω2 component of the angular velocity contributes to the L3 component of
the angular momentum. In the former case, if we group the product of the various
quantities as − (ω3 y)z, we see that this is simply the appropriate component of the
velocity times the distance from the y axis. In the latter case with − (ω2 z)y, it is
the opposite grouping. But in both cases there is one factor of y and one factor of z,
hence the symmetry in I.
y
x
Fig. 9.10
z
Remark: For a point mass, L is actually more easily obtained by just calculating L =
r × p. The result for the instant shown in Fig. 9.10 is drawn in Fig. 9.11, where it is clear
that L has both y and z components, and thus also clear that L doesn’t point along ω. For a
more complicated object, the tensor I is generally used, because it is necessary to perform
the integral of the L = r × p contributions over the entire object, and the tensor has this
integral built into it. At any rate, whatever method you use, you will find that except in
special circumstances (see Section 9.3), L doesn’t point along ω.
Consider the vector of L,
And that of ω as well.
The erroneous claim
That they must aim the same
Is a view that you’ve got to dispel! ♣
r
z0
ω
x
Fig. 9.11
L
r
p
(into
page)
r
y
ˆ
Angular momentum, Part II (General L)
380
Example 3 (Two point masses): Let’s now add another point mass m to the
previous example. Let it travel in the same circle, at the diametrically opposite point,
as shown in Fig. 9.12. Using ω = (0, 0, ω3 ), x2 + y2 = r 2 , and z = z0 in Eq. (9.8),
you can show that the angular momentum with respect to the origin is
z
z0
r
ω
L = 2mω3 (0, 0, r 2 ).
y
x
Fig. 9.12
(9.13)
Since v = ω3 r, the z component is 2mrv, as it should be. And L1 and L2 are zero,
unlike in the previous example, because these components of the L’s of the two
particles cancel. This occurs because of the symmetry of the masses around the z
axis, which causes the Izx and Izy entries in the inertia tensor to vanish; they are each
the sum of two terms, with opposite x values, or opposite y values. Alternatively, you
can just note that adding on the mirror-image L vector in Fig. 9.10 produces canceling
x and y components.
Let’s now look at the kinetic energy of our object, which is rotating around an
axis passing through the origin. To find this, we must add up the kinetic energies
of all the little pieces. A little piece has energy (dm) v 2 /2 = dm |ω × r|2 /2.
Therefore, using Eq. (9.6), the total kinetic energy is
T =
1
2
(ω2 z − ω3 y)2 + (ω3 x − ω1 z)2 + (ω1 y − ω2 x)2 dm.
(9.14)
Multiplying this out, we see (after a little work) that we can write T as
⎞
⎞⎛
⎛
− xy
− zx
ω1
(y2 + z 2 )
1
T = (ω1 , ω2 , ω3 ) · ⎝ − xy
(z 2 + x2 )
− yz ⎠ ⎝ ω2 ⎠
2
ω3
− zx
− yz
(x2 + y2 )
=
1
1
ω · Iω = ω · L.
2
2
(9.15)
If ω = ω3 zˆ , then this reduces to T = Izz ω32 /2, which agrees with the result in
Eq. (8.8), with a slight change in notation.
z
V
9.2.2
ω
y
x
Fig. 9.13
General motion
How do we deal with general motion in space? That is, what if an object is both
translating and rotating? For the motion in Fig. 9.13, the various pieces of mass
aren’t traveling in circles around the origin, so we can’t write v = ω × r, as we
did prior to Eq. (9.4).
To determine L (relative to the origin), and also the kinetic energy T , we will
use Theorem 9.1 to write the motion as the sum of a translation plus a rotation.
In applying the theorem, we may choose any point in the body to be the point
P in the theorem. However, only in the case where P is the object’s CM can we
extract anything useful, as we’ll see. The theorem then says that the motion of
9.2 The inertia tensor
the body is the sum of the motion of the CM plus a rotation around the CM. So
let the CM move with velocity V, and let the body instantaneously rotate with
angular velocity ω around the CM (that is, with respect to the frame whose origin
is the CM, and whose axes are parallel to the fixed-frame axes).4
Let the position of the CM relative to the origin be R = (X, Y, Z), and let
the position of a given piece of mass relative to the CM be r = (x , y , z ). Then
r = R + r is the position of a piece of mass relative to the origin (see Fig. 9.14).
Let the velocity of a piece of mass relative to the CM be v (so v = ω × r ).
Then v = V + v is the velocity relative to the origin.
Let’s look at L first. The angular momentum is
381
z
CM
R
r × v dm =
=
(R + r ) × V + (ω × r ) dm
(R × V) dm +
r × (ω × r ) dm
= M (R × V) + LCM ,
(9.16)
where the cross terms vanish because the integrands are linear in r . More precisely, the integrals involve r dm, which is zero by definition of the CM
(because r dm/M is the position of the CM relative to the CM, which is zero).
LCM is the angular momentum relative to the CM.5
We see that as in the pancake case in Section 8.1.2, the angular momentum
(relative to the origin) of a body can be found by treating the body like a point
mass located at the CM and finding the angular momentum of this point mass
relative to the origin, and by then adding on the angular momentum of the body
relative to the CM. Note that these two parts of the angular momentum need not
point in the same direction, as they did in the case of the pancake moving in the
x-y plane.
Now let’s look at T . The kinetic energy is
T =
4
5
1 2
v dm =
2
1
|V + v |2 dm
2
1 2
1 2
=
V dm +
v dm
2
2
1
1
= MV 2 +
|ω × r |2 dm
2
2
1
1
≡ MV 2 + ω · LCM ,
2
2
(9.17)
It’s not necessary to put the prime on the ω here, because the angular velocity vector in the CM
frame is the same as in the lab frame. But we’ll use the prime just because we’ll have primes on
the other CM quantities below.
By this, we mean the angular momentum as measured in the coordinate system whose origin is the
CM, and whose axes are parallel to the fixed-frame axes.
r
y
x
Fig. 9.14
L =
r'
ˆ
Angular momentum, Part II (General L)
382
z
ω'
where the last line follows from the steps leading to Eq. (9.15). The cross term
V · v dm = V · (ω × r ) dm vanishes because the integrand is linear in r
and thus yields a zero integral, by definition of the CM. As in the pancake case
in Section 8.1.2, the kinetic energy of a body can be found by treating the body
like a point mass located at the CM, and by then adding on the kinetic energy of
the body due to the rotation around the CM.
ω'
CM
9.2.3
y
x
Fig. 9.15
The parallel-axis theorem
Consider the special case where the CM rotates around the origin with the same
angular velocity at which the body rotates around the CM (see Fig. 9.15), that is,
V = ω × R. This can be achieved, for example, by piercing the body with the
base of a rigid “T” and then rotating the T and the body around the (fixed) line of
the “upper” part of the T (the origin must pass through this line). We then have the
nice situation where all points in the body travel in fixed circles around the axis
of rotation. Mathematically, this follows from v = V + v = ω × R + ω × r =
ω × r. Dropping the prime on ω, Eq. (9.16) becomes
L = M R × (ω × R) +
r × (ω × r ) dm
(9.18)
Expanding the double cross products as in the steps leading to Eq. (9.8), we can
write this as
⎞
⎛ 2
⎞⎛
⎞
⎛
−XY
−ZX
Y + Z2
ω1
L1
⎝ L2 ⎠ = M ⎝ −XY
Z2 + X 2
−YZ ⎠ ⎝ ω2 ⎠
2
−ZX
−YZ
X + Y2
L3
ω3
⎛
⎞
⎞⎛
(y 2 + z 2 )
− xy
− zx
ω1
⎠ ⎝ ω2 ⎠
+⎝ − xy
(z 2 + x 2 )
− yz
2
2
(x + y )
− zx
− yz
ω3
≡ (IR + ICM )ω.
(9.19)
This is the generalized parallel-axis theorem. It says that once you’ve calculated
ICM relative to the CM, then if you want to calculate I relative to another point,
you simply have to add on the IR matrix, obtained by treating the object like a
point mass at the CM. So you have to compute six extra numbers (there are six,
instead of nine, because the IR matrix is symmetric) instead of just the one MR2
in the parallel-axis theorem in Chapter 8, given in Eq. (8.13). Problem 9.4 gives
another derivation of the parallel-axis theorem, without mentioning the angular
velocity.
Remark: The name “parallel-axis” theorem is actually a misnomer here. The inertia tensor
isn’t associated with one particular axis, as the moment of inertia in Chapter 8 was. The moment
of inertia is just one of the diagonal entries (associated with a given axis) in the inertia tensor.
The inertia tensor depends on the entire coordinate system. So in that sense we should call this
9.3 Principal axes
the “parallel-axes” theorem, because the coordinate axes in the CM frame are assumed to be
parallel to the ones in the fixed frame. At any rate, the point is that the parallel-axis theorem
in Chapter 8 dealt with shifting the axis, whereas the present theorem deals with shifting the
origin (and hence all three axes in general). ♣
As far as the kinetic energy goes, if ω and ω are equal, so that V = ω × R,
then Eq. (9.17) gives (dropping the prime on ω)
T =
1
M |ω × R|2 +
2
1
|ω × r |2 dm.
2
(9.20)
Performing the steps leading to Eq. (9.15), this becomes
T =
9.3
1
1
ω · (IR + ICM )ω = ω · L.
2
2
(9.21)
Principal axes
The cumbersome expressions in the previous section may seem a bit unsettling,
but it turns out that usually we can get by without them. The strategy for avoiding
all of the above mess is to use the principal axes of a body, which we will define
below.
In general, the inertia tensor I in Eq. (9.8) has nine nonzero entries, of which
six are independent due to the symmetry of I. In addition to depending on the
origin chosen, the inertia tensor depends on the set of orthonormal basis vectors
chosen for the coordinate system; the x, y, z variables in the integrals in I depend,
of course, on the coordinate system they’re measured with respect to. Given a
blob of material, and given an arbitrary origin,6 any orthonormal set of basis
vectors is usable, but there is one special set that makes all our calculations
very nice. These special basis vectors are called the principal axes. They can be
defined in various equivalent ways:
• The principal axes are the orthonormal basis vectors for which I is diagonal, that is,
for which7
⎛
⎞
I1 0 0
⎜
⎟
I = ⎝ 0 I2 0 ⎠ .
(9.22)
0 0 I3
I1 , I2 , and I3 are called the principal moments. For many objects, it is quite obvious
what the principal axes are. For example, consider a uniform rectangle in the x-y plane.
Pick the origin to be the CM, and let the x and y axes be parallel to the sides. Then the
6
7
The CM is often chosen to be the origin, but it need not be. There are principal axes associated
with any origin.
Technically, we should be writing I11 or Ixx instead of I1 , etc., in this matrix, because the oneindex object I1 looks like the component of a vector, not a matrix. But the two-index notation gets
cumbersome, so we’ll be sloppy and just use I1 , etc.
383
ˆ
Angular momentum, Part II (General L)
384
principal axes are clearly the x, y, and z axes, because all the off-diagonal elements
in the inertia tensor in Eq. (9.8) vanish, by symmetry. For example, Ixy ≡ − xy dm
equals zero, because for every point (x, y) in the rectangle, there is a corresponding point
(−x, y), so the contributions to xy dm cancel in pairs. Also, the integrals involving z
are identically zero, because z = 0.
ˆ That is, a principal axis is a special
• A principal axis is an axis ωˆ for which Iωˆ = I ω.
direction with the property that if ω points along it, then so does L. The principal axes
of an object are then the orthonormal set of three vectors ωˆ 1 , ωˆ 2 , ωˆ 3 with the property
that
Iωˆ 1 = I1 ωˆ 1 ,
ω2
x
Fig. 9.16
Iωˆ 3 = I3 ωˆ 3 .
(9.23)
The three statements in Eq. (9.23) are equivalent to Eq. (9.22), because the vectors ωˆ 1 ,
ωˆ 2 , and ωˆ 3 are simply (1, 0, 0), (0, 1, 0), and (0, 0, 1) in the frame in which they are the
basis vectors.
• Consider an object rotating around a fixed axis with constant angular speed. Then this
axis is a principal axis if there is no need for any torque. So in some sense, the object is
“happy” to spin around a principal axis. A set of three orthonormal axes, each of which
has this property, is by definition what we call a set of principal axes.
This definition of a principal axis is equivalent to the previous definition for the
following reason. Assume that the object rotates around a fixed axis ωˆ 1 for which
L = Iωˆ 1 = I1 ωˆ 1 , as in Eq. (9.23). Then since ωˆ 1 is assumed to be fixed, we see that L
is also fixed. Therefore, τ = dL/dt = 0.
Conversely, if the object is rotating around a fixed axis ωˆ1 , and if τ = dL/dt = 0,
then we claim that L points along ωˆ 1 (that is, L = I1 ωˆ 1 ). This is true because if L does
not point along ωˆ 1 , then imagine painting a dot on the object somewhere along the line
of L. A little while later, the dot will have rotated around the fixed vector ωˆ 1 . But the
line of L must always pass through the dot, because we could have rotated our axes
around ωˆ 1 and started the process at a slightly later time (this argument relies on ωˆ 1
being fixed). Therefore, we see that L has changed, in contradiction to the assumption
that dL/dt = 0. Hence, L must in fact point along ωˆ 1 .
ˆ the lack of need for any torque means that
For a rotation around a principal axis ω,
if the object is pivoted at the origin, and if the origin is the only place where any
force is applied (which implies that there is zero torque around it), then the object can
undergo rotation with constant angular velocity ω. If you try to set up this scenario with
a nonprincipal axis, it won’t work.
y
ω1
Iωˆ 2 = I2 ωˆ 2 ,
Example (Square with origin at corner): Consider the uniform square in
Fig. 9.16. In Appendix E, we show that the principal axes are the dotted lines drawn
(and also the z axis perpendicular to the page). But there is no need to use the techniques in the appendix to see this, because in this new basis it is clear by symmetry
that the integral x1 x2 is zero; for every x1 in the integral, there is a −x1 . And x3 ≡ z
9.3 Principal axes
is identically zero, which makes all the other off-diagonal terms in I also equal to
zero. Therefore, since I is diagonal in this new basis, these basis vectors are the
principal axes.
Furthermore, it is intuitively clear that the square will be happy to rotate around
any one of these axes indefinitely. During such a rotation, the pivot will certainly be
applying a force (if the axis is ωˆ 1 or zˆ , but not if it is ωˆ 2 ), to produce the centripetal
acceleration of the CM in its circular motion. But it won’t be applying a torque relative
to the origin (because the r in r × F is 0). This is good, because for a rotation around
one of these principal axes, dL/dt = 0, so there is no need for any torque.
In contrast with the off-diagonal zeros in the new basis, the integral xy in the
old basis is not zero, because every point gives a positive contribution. So the inertia
tensor is not diagonal in the old basis, which means that xˆ and yˆ are not principal axes.
Consistent with this, it is reasonably clear that it is impossible to make the square
rotate around, say, the x axis, assuming that its only contact with the outside world
is through a pivot (for example, a ball and socket) at the origin. The square simply
doesn’t want to remain in this circular motion. Mathematically, L (relative to the
origin) doesn’t point along the x axis, so it therefore precesses around the x axis along
with the square, tracing out the surface of a cone. This means that L is changing. But
there is no torque available (relative to the origin) to provide for this change in L.
Hence, such a rotation cannot exist.
At the moment, it is not at all obvious that an orthonormal set of principal axes
exists for an arbitrary object. But this is indeed the case, as stated in Theorem 9.4
below. Assuming for now that principal axes do exist, then in this basis the L and
T in Eqs. (9.8) and (9.15) take on the particularly nice forms,
L = (I1 ω1 , I2 ω2 , I3 ω3 ),
T =
1
I1 ω12 + I2 ω22 + I3 ω32 .
2
(9.24)
The quantities ω1 , ω2 , and ω3 here are the components of a general vector ω
written in the principal-axis basis; that is, ω = ω1 ωˆ 1 + ω2 ωˆ 2 + ω3 ωˆ 3 . Equation
(9.24) is a vast simplification over the general formulas in Eqs. (9.8) and (9.15).
We will therefore invariably work with principal axes in the remainder of this
chapter.
Note that the directions of the principal axes (relative to the body) depend only
on the geometry of the body. They may therefore be considered to be painted on
(or in) it. Hence, they will generally move around in space as the body rotates.
For example, if the object is rotating around a principal axis, then that axis
stays fixed while the other two principal axes rotate around it. In relations like
ω = (ω1 , ω2 , ω3 ) and L = (I1 ω1 , I2 ω2 , I3 ω3 ), the components ωi and Ii ωi are
measured along the instantaneous principal axes ωˆ i . Since these axes change
with time, it is quite possible that the components ωi and Ii ωi change with time,
as we’ll see in Section 9.5 (and onward).
385
386
ˆ
Angular momentum, Part II (General L)
Let’s now state the theorem that implies that a set of principal axes does indeed
exist for any body and any origin. The proof of this theorem involves a useful
but rather slick technique, but it’s slightly off the main line of thought, so we’ll
relegate it to Appendix D. Take a look at the proof if you wish, but if you want
to just accept the fact that principal axes exist, that’s fine.
Theorem 9.4 Given a real symmetric 3 × 3 matrix, I, there exist three
orthonormal real vectors, ωˆ k , and three real numbers, Ik , with the property
that
Iωˆ k = Ik ωˆ k .
(9.25)
Proof: See Appendix D.
Since the inertia tensor in Eq. (9.8) is indeed symmetric for any body and any
origin, this theorem says that we can always find three orthogonal basis vectors
that satisfy Eq. (9.23). Or equivalently, we can always find three orthogonal basis
vectors for which I is a diagonal matrix, as in Eq. (9.22). In other words, principal
axes always exist. Problem 9.7 gives another way to demonstrate the existence
of principal axes in the special case of a pancake object.
Invariably, it is best to work in a coordinate system that has principal axes as
its basis, due to the simplicity of Eq. (9.24). And as mentioned in Footnote 6, the
origin is generally chosen to be the CM, because from Section 8.4.3 the CM is
one of the origins for which τ = dL/dt is a valid statement. But this choice is
not necessary; there are principal axes associated with any origin.
For an object with a fair amount of symmetry, the principal axes are usually
the obvious choices and can be written down by simply looking at the object
(examples are given below). If, however, you are given an unsymmetrical body,
then the only way to determine the principal axes is to pick an arbitrary basis,
then find I in this basis, and then go through a diagonalization procedure. This
diagonalization procedure basically consists of the steps at the beginning of the
proof of Theorem 9.4 (given in Appendix D), with the addition of one more step
to get the actual vectors, so we’ll relegate it to Appendix E. There’s no need
to worry much about this method. Virtually every system you encounter will
involve an object with sufficient symmetry to enable you to just write down the
principal axes.
Let’s now prove two very useful (and very similar) theorems.
Theorem 9.5 If two principal moments are equal (I1 = I2 ≡ I ), then any axis
(through the chosen origin) in the plane of the corresponding principal axes is a
principal axis, and its moment is also I . Similarly, if all three principal moments
are equal (I1 = I2 = I3 ≡ I ), then any axis (through the chosen origin) in space
is a principal axis, and its moment is also I .
Proof: The first part was already proved at the end of the proof in Appendix D,
but we’ll do it again here. Since I1 = I2 ≡ I , we have Iu1 = I u1 , and Iu2 = I u2 ,
9.3 Principal axes
where the u’s are the principal axes. Hence, I(au1 + bu2 ) = I (au1 + bu2 ), for
any a and b. Therefore, any linear combination of u1 and u2 (that is, any vector
in the plane spanned by u1 and u2 ) is a solution to Iu = I u and is thus a principal
axis, by definition.
The proof of the second part proceeds in a similar manner. Since I1 = I2 =
I3 ≡ I , we have Iu1 = I u1 , Iu2 = I u2 , and Iu3 = I u3 . Hence, I(au1 + bu2 +
cu3 ) = I (au1 + bu2 + cu3 ). Therefore, any linear combination of u1 , u2 , and u3
(that is, any vector in space) is a solution to Iu = I u and is thus a principal axis,
by definition.
In short, if I1 = I2 ≡ I , then I is the identity matrix (up to a multiple) in the
space spanned by u1 and u2 . And if I1 = I2 = I3 ≡ I , then I is the identity matrix
(up to a multiple) in the entire space. Note that it isn’t required that the various
ui vectors be orthogonal. All we need is that they span the relevant space.
If two or three moments are equal, so that there is freedom in choosing the
principal axes, then it is possible to pick a nonorthogonal group of them. We
will, however, always choose ones that are orthogonal. So when we say “a set of
principal axes,” we mean an orthonormal set.
Theorem 9.6 If a pancake object is symmetric under a rotation through an
angle θ = 180◦ in the x-y plane (such as a hexagon), then every axis in the
x-y plane (with the origin chosen to be the center of the symmetry rotation) is a
principal axis with the same moment.
Proof: Let ωˆ 0 be a principal axis in the plane, and let ωˆ θ be the axis obtained
by rotating ωˆ 0 through the angle θ . Then ωˆ θ is also a principal axis with the same
principal moment, due to the symmetry of the object. Therefore, Iωˆ 0 = I ωˆ 0 , and
Iωˆ θ = I ωˆ θ .
Now, any vector ω in the x-y plane can be written as a linear combination of
ωˆ 0 and ωˆ θ , provided that θ = 180◦ (or zero, of course). That is, ωˆ 0 and ωˆ θ span
the x-y plane. Therefore, any vector ω can be written as ω = aωˆ 0 + bωˆ θ , and so
Iω = I(aωˆ 0 + bωˆ θ ) = aI ωˆ 0 + bI ωˆ θ = I ω.
(9.26)
Hence, ω is also a principal axis. Problem 9.8 gives another proof of this
theorem.
The theorem actually holds even without the “pancake” restriction. That is,
it holds for any object with a rotational symmetry around the z axis (excluding
θ = 180◦ ). This can be seen as follows. The z axis is a principal axis, because
if ω points along zˆ , then L must also point along zˆ , by symmetry. There are
therefore (at least) two principal axes in the x-y plane. Label one of these as ωˆ 0
and proceed as above.
Let’s now do some quick examples. We’ll state the principal axes for the
objects listed below (relative to the origin). Your task is to show that they
387
ˆ
Angular momentum, Part II (General L)
388
are correct. Usually, a quick symmetry argument shows that
⎛
I≡⎝
(y2 + z 2 )
− xy
− zx
− xy
(z 2 + x2 )
− yz
⎞
− zx
− yz ⎠
(x2 + y2 )
(9.27)
is diagonal. In all of these examples (see Fig. 9.17), the origin for the principal
axes is understood to be the origin of the given coordinate system (which is not
necessarily the CM). In describing the axes, they therefore all pass through the
origin, in addition to having the other properties stated.
Example 1: Point mass at the origin.
principal axes: any axes.
y
Example 2: Point mass at the point (x0 , y0 , z0 ).
principal axes: axis through point, any axes perpendicular to this.
x
Example 3: Rectangle centered at the origin, as shown.
principal axes: the x, y, and z axes.
z
Example 4: Cylinder with axis as z axis.
principal axes: z axis, any axes in x-y plane.
x
Example 5: Square with one corner at origin, as shown.
principal axes: z axis, axis through CM, axis perpendicular to this.
y
x
9.4
Fig. 9.17
Two basic types of problems
The previous three sections introduced a variety of abstract concepts. We will
now finally look at some actual physical systems. The concept of principal axes
gives us the ability to solve many kinds of problems. Two kinds, however, come
up again and again. There are variations on these, of course, but they may be
generally stated as follows.
• Strike a rigid object with an impulsive (that is, quick) blow. What is the motion of the
object immediately after the blow?
• An object rotates around a fixed axis. A given torque is applied. What is the frequency
of the rotation? Or conversely, given the frequency, what is the required torque?
We’ll work through an example for each of these problems. In both cases, the
solution involves a few standard steps, so we’ll write them out explicitly.
9.4 Two basic types of problems
9.4.1
389
Motion after an impulsive blow
C 2m
Problem: Consider the rigid object in Fig. 9.18. Three masses are connected
by three massless rods, in the shape of an isosceles right triangle with hypotenuse
length 4a. The mass at the right angle is 2m, and the other two masses are m.
Label them A, B, C, as shown. Assume that the object is floating freely in outer
space. Mass B is struck with a quick blow, directed into the page. Let the imparted
impulse have magnitude F dt = P. (See Section 8.6 for a discussion of impulse
and angular impulse.) What are the velocities of the three masses immediately
after the blow?
Solution: Our strategy will be to find the angular momentum of the system
(relative to the CM) using the angular impulse, and then calculate the principal
moments and find the angular velocity vector (which will give the velocities
relative to the CM), and then finally add on the CM motion.
The altitude from the right angle to the hypotenuse has length 2a, and the CM
is easily seen to be located at its midpoint (see Fig. 9.19). Picking the CM as our
origin, and letting the plane of the paper be the x-y plane, the positions of the
three masses are rA = (−2a, −a, 0), rB = (2a, −a, 0), and rC = (0, a, 0). There
are now five standard steps that we must perform.
• Find L: The positive z axis is directed out of the page, so the impulse vector is
P ≡ F dt = (0, 0, −P). Therefore, the angular momentum of the system (relative to
the CM) is
L=
τ dt =
(rB × F) dt = rB ×
F dt
= (2a, −a, 0) × (0, 0, −P) = aP(1, 2, 0),
(9.28)
as shown in Fig. 9.19. We have used the fact that rB is essentially constant during the
blow (because the blow is assumed to happen very quickly) in taking rB outside the
integral.
• Calculate the principal moments: The principal axes are the x, y, and z axes, because
the symmetry of the triangle makes I diagonal in this basis, as you can quickly check.
The moments (relative to the CM) are
Ix = ma2 + ma2 + (2m)a2 = 4ma2 ,
Iy = m(2a)2 + m(2a)2 + (2m)02 = 8ma2 ,
(9.29)
Iz = Ix + Iy = 12ma2 .
We have used the perpendicular-axis theorem to obtain Iz , although it won’t be needed
to solve the problem.
• Find ω: We now have two expressions for the angular momentum of the system. One
expression is in terms of the given impulse, Eq. (9.28). The other is in terms of the
B
A
m
m
4a
F
Fig. 9.18
y
L
ω
2m C
a
A
m
a
2a
Fig. 9.19
x
CM
2a
B
m
390
ˆ
Angular momentum, Part II (General L)
moments and the angular velocity components, Eq. (9.24). Equating these gives
(Ix ωx , Iy ωy , Iz ωz ) = aP(1, 2, 0)
=⇒
(4ma2 ωx , 8ma2 ωy , 12ma2 ωz ) = aP(1, 2, 0)
=⇒
(ωx , ωy , ωz ) =
P
(1, 1, 0),
4ma
(9.30)
as shown in Fig. 9.19.
• Calculate the velocities relative to the CM: Right after the blow, the object rotates
around the CM with the angular velocity found in Eq. (9.30). The velocities relative to
the CM are then ui = ω × ri . Thus,
P
(1, 1, 0) × (−2a, −a, 0) = (0, 0, P/4m),
4ma
P
(1, 1, 0) × (2a, −a, 0) = (0, 0, −3P/4m),
uB = ω × rB =
4ma
P
(1, 1, 0) × (0, a, 0) = (0, 0, P/4m).
uC = ω × rC =
4ma
uA = ω × rA =
(9.31)
As a check, it makes sense that uB is three times as large as uA and uC , because B is
three times as far from the axis of rotation as A and C are, as you can verify by doing a
little geometry in Fig. 9.19.
• Add on the velocity of the CM: The impulse (that is, the change in linear momentum)
supplied to the whole system is P = (0, 0, −P). The total mass of the system is M = 4m.
Therefore, the velocity of the CM is
VCM =
P
= (0, 0, −P/4m).
M
(9.32)
The total velocities of the masses are therefore
vA = uA + VCM = (0, 0, 0),
vB = uB + VCM = (0, 0, −P/m),
(9.33)
vC = uC + VCM = (0, 0, 0).
Remarks:
1. We see that masses A and C are instantaneously at rest immediately after the blow, and
mass B acquires all of the imparted impulse. In retrospect, this is clear. Basically, it is
possible for both A and C to remain at rest while B moves a tiny bit, so this is what
happens. If B moves into the page by a small distance , then A and C won’t know that B
has moved, because their distances to B will change (assuming hypothetically that they
don’t move) by a distance of order only 2 . If we changed the problem and added a mass
D at, say, the midpoint of the hypotenuse, then it would not be possible for A, C, and D
to remain at rest while B moved a tiny bit. So there would have to be some other motion
in addition to B’s. This setup is the topic of Exercise 9.38.
2. As time goes on, the system undergoes a rather complicated motion. What happens
is that the CM moves with constant velocity while the masses rotate around it in
9.4 Two basic types of problems
391
a messy manner. Since there are no torques acting on the system (after the initial
blow), we know that L forever remains constant. It turns out that ω moves around
L while the masses rotate around this changing ω. These matters are the subject of
Section 9.6, although in that discussion we restrict ourselves to symmetric tops, that
is, ones with two equal moments. But these issues aside, it’s good to know that
we can, without too much difficulty, determine what’s going on immediately after
the blow.
3. The object in this problem was assumed to be floating freely in space. If we instead have
an object that is pivoted at a given fixed point, then we should use this pivot as our origin.
There is then no need to perform the last step of adding on the velocity of the origin (which
was the CM, above), because this velocity is now zero. Equivalently, just consider the
pivot to be an infinite mass, which is therefore the location of the (motionless) CM. ♣
9.4.2
ω=?
Frequency of motion due to a torque
Problem: Consider a stick of length , mass m, and uniform mass density. The
stick is pivoted at its top end and swings around the vertical axis. Assume that
conditions have been set up so that the stick always makes an angle θ with the
vertical, as shown in Fig. 9.20. What is the frequency, ω, of this motion?
Solution: Our strategy will be to find the principal moments and then the
angular momentum of the system (in terms of ω), and then find the rate of
change of L, and then calculate the torque and equate it with dL/dt. We will
choose the pivot to be the origin.8 Again, there are five standard steps that we
must perform.
θ
l
m
Fig. 9.20
y
• Calculate the principal moments: The principal axes are the axis along the stick,
along with any two orthogonal axes perpendicular to the stick. So let the x and y axes
be as shown in Fig. 9.21. The positive z axis then points out of the page. The moments
(relative to the pivot) are Ix = m 2 /3, Iy = 0, and Iz = m 2 /3 (which won’t be needed).
• Find L: The angular velocity vector points vertically (however, see the third remark
following this solution), so in the basis of the principal axes, the angular velocity vector
is ω = (ω sin θ , ω cos θ, 0), where ω is yet to be determined. The angular momentum
of the system (relative to the pivot) is therefore
L = (Ix ωx , Iy ωy , Iz ωz ) = (1/3)m 2 ω sin θ, 0, 0 .
• Find dL/dt: The vector L in Eq. (9.34) points up to the right, along the x axis (at
the instant shown in Fig. 9.21), with magnitude L = (1/3)m 2 ω sin θ. As the stick
rotates around the vertical axis, L traces out the surface of a cone. That is, the tip of L
traces out a horizontal circle. The radius of this circle is the horizontal component of
L, which is L cos θ . The speed of the tip (which is the magnitude of dL/dt) is therefore
8
x
L
θ l
Fig. 9.21
(9.34)
This is a better choice than the CM because this way we won’t have to worry about any messy
forces acting at the pivot when computing the torque. The task of Exercise 9.41 is to work through
the more complicated solution which has the CM as the origin.
ω
ˆ
Angular momentum, Part II (General L)
392
(L cos θ )ω, because L rotates around the vertical axis with the same frequency as the
stick. So dL/dt has magnitude
1
dL
= (L cos θ)ω = m 2 ω2 sin θ cos θ,
dt
3
(9.35)
and it points into the page.
Remark: With more complicated objects where Iy = 0, L won’t point nicely along a
principal axis, so the length of its horizontal component (the radius of the circle that L
traces out) won’t immediately be obvious. In this case, you can either explicitly calculate
the horizontal component (see the spinning-top example in Section 9.7.5), or you can just do
things the formal way by finding the rate of change of L via the expression dL/dt = ω × L,
which holds for all the same reasons that v ≡ dr/dt = ω × r holds. In the present problem,
we obtain
dL/dt = (ω sin θ , ω cos θ , 0) × (1/3)m 2 ω sin θ, 0, 0
= 0, 0, −(1/3)m 2 ω2 sin θ cos θ ,
(9.36)
in agreement with Eq. (9.35). And the direction is correct, because the negative z axis points
into the page. Note that we calculated this cross product in the principal-axis basis. Although
these axes are changing in time, they present a perfectly good set of basis vectors at any
instant. ♣
• Calculate the torque: The torque (relative to the pivot) is due to gravity, which
effectively acts on the CM of the stick. So τ = r × F has magnitude
τ = rF sin θ = ( /2)(mg) sin θ,
(9.37)
and it points into the page.
• Equate τ with dL/dt: The vectors dL/dt and τ both point into the page, which is
good, because they had better point in the same direction. Equating their magnitudes
gives
m 2 ω2 sin θ cos θ
mg sin θ
=
3
2
ω
lsinθ
Fig. 9.22
ω=
3g
.
2 cos θ
(9.38)
Remarks:
ω'
θ
=⇒
1. This frequency is slightly larger than the frequency that would arise if we instead had a
mass on√the end of a massless stick of length . From Problem 9.12, the frequency in that
case is g/ cos θ. So, in some sense, a uniform stick of length behaves like a mass on
the end of a massless stick of length 2 /3, as far as these rotations are concerned.
2. As
√ θ → π/2, the frequency goes to ∞, which makes sense. And as θ → 0, it approaches
3g/2 , which isn’t so obvious.
3. As explained in Problem 9.1, the instantaneous ω is not uniquely defined in some situations. At the instant shown in Fig. 9.20, the stick is moving directly into the page. What if
someone else wants to think of the stick as (instantaneously) rotating around the ω axis
perpendicular to the stick (the x axis, in the above notation), instead of the vertical axis,
as shown in Fig. 9.22. What is the angular speed ω ?
Well, if ω is the angular speed of the stick around the vertical axis, then we may view
the tip of the stick as instantaneously moving in a circle of radius sin θ around the
9.5 Euler’s equations
vertical axis ω. So ω( sin θ) is the speed of the tip of the stick. But we may also view
the tip of the stick as instantaneously moving in a circle of radius around ω , as shown.
The speed of the tip is still ω( sin θ), so the angular speed around this axis is given by
ω = ω( sin θ). Hence ω = ω sin θ, which is simply the x component of ω that we
found above, right before Eq. (9.34). The moment of inertia around ω is m 2 /3, so the
angular momentum has magnitude (m 2 /3)(ω sin θ), in agreement with Eq. (9.34). And
the direction is along the x axis, as it should be.
Note that although ω is not uniquely defined at any instant, L ≡ (r × p) dm certainly
is.9 Choosing ω to point vertically, as we did in the above solution, is in some sense the
natural choice, because this ω doesn’t change with time. ♣
9.5
Euler’s equations
Consider a rigid body instantaneously rotating around an axis ω. This ω may
change as time goes on, but all we care about for now is what it is at a given
instant. The angular momentum is given by Eq. (9.8) as L = Iω, where I is the
inertia tensor, calculated with respect to a given origin and a given set of axes
(and ω is written in the same basis, of course).
As usual, things are much nicer if we use the principal axes (relative to the
chosen origin) as the basis vectors of our coordinate system. Since these axes are
fixed with respect to the rotating object, they will rotate with respect to the fixed
reference frame. In this basis, L takes the nice form,
L = (I1 ω1 , I2 ω2 , I3 ω3 ),
(9.39)
where ω1 , ω2 , and ω3 are the components of ω along the principal axes. In other
words, if you take the vector L in space and project it onto the instantaneous
principal axes, then you get the components in Eq. (9.39).
On one hand, writing L in terms of the rotating principal axes allows us to
write it in the nice form of Eq. (9.39). But on the other hand, writing L in this
way makes it nontrivial to determine how it changes in time, because the principal
axes themselves are changing. However, it turns out that the benefits outweigh
the detriments, so we will invariably use the principal axes as our basis vectors.
The goal of this section is to find an expression for dL/dt, and to then equate
this with the torque. The result will be Euler’s equations in Eq. (9.45).
Derivation of Euler’s equations
If we write L in terms of the body frame, which we’ll choose to be described
by the principal axes painted on the body, then L can change (relative to the lab
frame) due to two effects. It can change because its coordinates in the body frame
change, and it can also change because of the rotation of the body frame. To be
precise, let L0 be the vector L at a given instant. At this instant, imagine painting
the vector L0 onto the body frame, so that L0 then rotates with the body. The rate
9
The nonuniqueness of ω arises from the fact that Iy = 0 here. If all the moments are nonzero, then
(Lx , Ly , Lz ) = (Ix ωx , Iy ωy , Iz ωz ) uniquely determines ω, given L.
393
394
ˆ
Angular momentum, Part II (General L)
of change of L with respect to the lab frame may be written in the (identically
true) way,
dL
d(L − L0 ) dL0
=
+
.
dt
dt
dt
(9.40)
The second term here is simply the rate of change of a body-fixed vector, which
we know is ω × L0 , which equals ω × L at this instant. The first term is the rate
of change of L with respect to the body frame, which we’ll denote by δL/δt.
This is what someone standing fixed on the body measures. So we end up with
dL
δL
=
+ ω × L.
dt
δt
(9.41)
This is actually a general statement, true for any vector in any rotating frame
(we’ll derive it in another more mathematical way in Chapter 10). There was
nothing particular about L that we used in the above derivation. Also, there was
no need to restrict ourselves to principal axes. In words, what we’ve shown is
that the total change equals the change relative to the rotating frame, plus the
change of the rotating frame relative to the fixed frame. This is just the usual way
of adding velocities when one frame moves with respect to another.
Let us now make use of our choice of the principal axes as the body axes.
This will put Eq. (9.41) in a usable form. Using Eq. (9.39), we can rewrite
Eq. (9.41) as
dL
d
= (I1 ω1 , I2 ω2 , I3 ω3 ) + (ω1 , ω2 , ω3 ) × (I1 ω1 , I2 ω2 , I3 ω3 ).
dt
dt
(9.42)
The δL/δt term does indeed equal (d/dt)(I1 ω1 , I2 ω2 , I3 ω3 ), because someone
in the body frame measures the components of L with respect to the principal
axes to be (I1 ω1 , I2 ω2 , I3 ω3 ). And δL/δt is by definition the rate at which these
components change.
Equation (9.42) equates two vectors. As is true for any vector, these (equal)
vectors have an existence that is independent of the coordinate system we choose
to describe them with (Eq. (9.41) makes no reference to a coordinate system).
But since we’ve chosen an explicit frame on the right-hand side of Eq. (9.42),
we should choose the same frame for the left-hand side. We can then equate the
components on the left with the components on the right. Projecting dL/dt onto
the instantaneous principal axes, we have
dL
dt
,
1
dL
dt
,
2
dL
dt
=
3
d
(I1 ω1 , I2 ω2 , I3 ω3 )
dt
(9.43)
+ (ω1 , ω2 , ω3 ) × (I1 ω1 , I2 ω2 , I3 ω3 ).
Remark: The left-hand side looks nastier than it really is. The reason we’ve written it in
this cumbersome way is the following (this is a remark that has to be read very slowly). We
9.5 Euler’s equations
could have written the left-hand side as (d/dt)(L1 , L2 , L3 ), but this might cause confusion as to
whether the Li refer to the components with respect to the rotating axes, or the components with
respect to the fixed set of axes that coincide with the rotating principal axes at this instant. That
is, do we project L onto the principal axes to obtain components, and then take the derivative
of these components? Or do we take the derivative of L and then project onto the principal axes
to obtain components? The latter is what we mean in Eq. (9.43).10 The way we’ve written the
left-hand side of Eq. (9.43), it’s clear that we’re taking the derivative first. We are, after all,
simply projecting Eq. (9.41) onto the principal axes. ♣
The time derivatives on the right-hand side of Eq. (9.43) are d(I1 ω1 )/dt =
I1 ω˙ 1 , etc., because the I ’s are constant. Performing the cross product and equating
the corresponding components on each side yields the three equations,
dL
dt
1
dL
dt
2
dL
dt
3
= I1 ω˙ 1 + (I3 − I2 )ω3 ω2 ,
= I2 ω˙ 2 + (I1 − I3 )ω1 ω3 ,
(9.44)
= I3 ω˙ 3 + (I2 − I1 )ω2 ω1 .
We will now invoke the results of Section 8.4.3 to say that if we have chosen the
origin of our rotating frame to be either a fixed point or the CM (as we always
do), then we can equate dL/dt with the torque, τ. We therefore have
τ1 = I1 ω˙ 1 + (I3 − I2 )ω3 ω2 ,
τ2 = I2 ω˙ 2 + (I1 − I3 )ω1 ω3 ,
(9.45)
τ3 = I3 ω˙ 3 + (I2 − I1 )ω2 ω1 .
These are Euler’s equations. You need to remember only one of them, because
the other two can be obtained by cyclic permutation of the indices.
Remarks:
1. We repeat that the left- and right-hand sides of Eqs. (9.45) are components that are measured with respect to the instantaneous principal axes. Let’s say we do a problem, for
example, where τ3 has a constant nonzero value, and τ1 and τ2 are always zero (as in the
example in Section 9.4.2). This doesn’t mean that τ is a constant vector. On the contrary,
τ always points along the xˆ 3 vector in the rotating frame, but this vector is changing in
the fixed frame (unless xˆ 3 points along ω).
2. The two types of terms on the right-hand sides of Eqs. (9.44) are the two types of changes
that L can undergo. L can change because its components with respect to the rotating
frame change, and L can also change because the body is rotating around ω.
10
The former is δL/δt, by definition. The two interpretations certainly give different results. For
example, if instead of L we consider a vector fixed in the body (such as the L0 above), then
the first interpretation gives a zero result, whereas the second interpretation gives a nonzero
result. Considering what we mean by, say, the vector (ω1 , ω2 , ω3 ), I think that the more logical
interpretation of (d/dt)(L1 , L2 , L3 ) is the first one, so it should definitely be avoided.
395
ˆ
Angular momentum, Part II (General L)
396
3. Section 9.6.1 on the free symmetric top (viewed from the body frame) provides a good
example of the use of Euler’s equations. Another interesting application is the famed
“Tennis racket theorem” (Problem 9.14).
4. It should be noted that you never have to use Euler’s equations. You can simply start
from scratch and use Eq. (9.41) each time you solve a problem. The point is that
we’ve done the calculation of dL/dt once and for all, so you can just invoke the result
in Eqs. (9.45). ♣
9.6
x3
CM
Fig. 9.23
Free symmetric top
The free symmetric top is the classic example of an application of Euler’s
equations. Consider an object that has two of its principal moments equal, with
the CM as the origin. Assume that the object is in outer space, far from any
external forces. We will choose the object to have cylindrical symmetry around
some axis (see Fig. 9.23), although this is not necessary; a square cross section,
for example, would yield two equal moments. The principal axes are then the
symmetry axis and any two orthogonal axes in the cross-section plane through
the CM. Let the symmetry axis be chosen as the xˆ 3 axis. Then the moments are
I1 = I2 ≡ I , and I3 .
We’ll first look at things from the point of view of someone standing at rest
on the body, and then we’ll look at things from the point of view of someone
standing at rest in an inertial frame. The mathematics involved here isn’t so bad,
but as with most types of top problems, it’s hard to get an intuitive feel for what
all the various vectors are doing. And the intuition is even more difficult in the
following body-frame analysis because of the noninertial frame of reference. But
let’s see what we find.
9.6.1
View from the body frame
Plugging I1 = I2 ≡ I into Euler’s equations in Eq. (9.45), and using the fact that
all the τi are zero (since there are no torques, because the top is “free”), we have
0 = I ω˙ 1 + (I3 − I )ω3 ω2 ,
0 = I ω˙ 2 + (I − I3 )ω1 ω3 ,
(9.46)
0 = I3 ω˙ 3 .
The last equation says that ω3 is constant. If we then define
≡
I3 − I
I
ω3 ,
and
ω˙ 2 −
(9.47)
the first two equations become
ω˙ 1 +
ω2 = 0,
ω1 = 0.
(9.48)
Taking the derivative of the first of these, and then using the second to eliminate
ω˙ 2 , gives
ω¨ 1 +
2
ω1 = 0,
(9.49)
9.6 Free symmetric top
397
and likewise for ω2 . This is a good old simple-harmonic-oscillator equation.
We can therefore write ω1 (t) as, say, a cosine. And then Eq. (9.48) yields a sine
for ω2 (t). So we have
ω1 (t) = A cos( t + φ),
ω2 (t) = A sin( t + φ).
(9.50)
x3
We see that ω1 (t) and ω2 (t) are the components of a circle in the body frame.
Therefore, the ω vector traces out a cone around xˆ 3 (see Fig. 9.24), with frequency
, as viewed by someone standing on the body. This frequency in Eq. (9.47)
depends on the value of ω3 and on the geometry of the object (through I3 and I ).
But the radius, A, of the ω cone is determined by the initial values of ω1 and ω2 .
The angular momentum is
L = (I1 ω1 , I2 ω2 , I3 ω3 ) = IA cos( t + φ), IA sin( t + φ), I3 ω3 ,
(9.51)
so L also traces out a cone around xˆ 3 with frequency , as viewed by someone
standing on the body. This is shown in Fig. 9.24 for the case
> 0 (that is,
I3 > I ). In this case, I3 > I implies L3 /L2 > ω3 /ω2 , so the L vector lies above
the ω vector (that is, between ω and xˆ 3 ). An object with I3 > I , such as a coin,
is called an oblate top.
Figure 9.25 shows the case where < 0 (that is, I3 < I ). In this case, I3 < I
implies L3 /L2 < ω3 /ω2 , so the L vector lies below the ω vector, as shown.
And since is negative, the ω and L vectors precess around xˆ 3 in the opposite
direction (clockwise, as viewed from above). An object with I3 < I , such as a
carrot, is called a prolate top.
L
ω
x1
11
This isn’t quite correct, since the earth rotates 366 times for every 365 days, due to the motion
around the sun, but it’s close enough for the purposes here.
x2
Fig. 9.24
x3
ω
L
x1
Example (The earth): Let’s consider the earth to be our top. Then ω3 ≈
2π/(1 day).11 The bulge at the equator (caused by the spinning of the earth) makes I3
slightly larger than I , and it turns out that (I3 − I )/I ≈ 1/320. Therefore, Eq. (9.47)
gives ≈ (1/320) 2π/(1 day). So the ω vector should precess around in its cone
once every 320 days, as viewed by someone on the earth. The true value is more like
430 days. The difference has to do with various things, including the nonrigidity of
the earth, but at least we got an answer in the right ballpark. This precession of ω is
known as the “Chandler wobble.”
In practice, how can we determine the direction of ω? Simply take an extendedtime photograph exposure at night. The stars will form arcs of circles. At the center
of all these circles is a point that doesn’t move. This is the direction of ω. Fortunately,
is much smaller than ω, so the ω vector doesn’t change much during an exposure
time of, say, an hour. So the center of the circles is essentially well defined.
How big is the ω cone, for the earth? Equivalently, what is the value of A in
Eq. (9.50)? Observation has shown that the ω vector pierces the earth at a point on
view from
body frame,
Ω > 0 ( I3 > I )
Fig. 9.25
view from
body frame,
Ω < 0 ( I3 < I )
x2
ˆ
Angular momentum, Part II (General L)
398
the order of 10 m from the north pole, although this distance fluctuates over time.12
Hence, A/ω3 ≈ (10 m)/RE . The half-angle of the ω cone is therefore on the order of
only 10−4 degrees. So if you use an extended-time photograph exposure one night to
see which point in the sky stands still, and then if you do the same thing 200 nights
later, you probably won’t be able to tell that they’re really two different points.
9.6.2
View from a ﬁxed frame
Let’s now see what our symmetric top looks like from a fixed frame. Euler’s
equations won’t help much here, because they deal with the components of ω
in the body frame. But fortunately we can solve for the motion from scratch. In
terms of the (changing) principal axes, xˆ 1 , xˆ 2 , xˆ 3 , we have
ω = (ω1 xˆ 1 + ω2 xˆ 2 ) + ω3 xˆ 3 ,
(9.52)
L = I (ω1 xˆ 1 + ω2 xˆ 2 ) + I3 ω3 xˆ 3 .
L
Eliminating the (ω1 xˆ 1 + ω2 xˆ 2 ) term from these equations gives (in terms of the
defined in Eq. (9.47))
ω
x3
L = I (ω +
view from
fixed frame,
Ω > 0 ( I3 > I )
Fig. 9.26
L
ω
=⇒
ω=
Lˆ
L−
I
xˆ 3 ,
(9.53)
ˆ is the unit vector in the L direction. The linear relationship
where L = |L|, and L
among L, ω, and xˆ 3 implies that these three vectors lie in a plane. But L remains
fixed, because there are no torques on the system. Therefore, ω and xˆ 3 precess (as
we’ll see below) around L, with the three vectors always coplanar. See Fig. 9.26
for the case
> 0, that is, I3 > I (an oblate top), and Fig. 9.27 for the case
< 0, that is, I3 < I (a prolate top).
What is the frequency of this precession, as viewed from the fixed frame? The
rate of change of xˆ 3 is ω × xˆ 3 , because xˆ 3 is fixed in the body frame, so its change
comes only from rotation around ω. Therefore, Eq. (9.53) gives
d xˆ 3
=
dt
x3
view from
fixed frame,
Ω < 0 ( I3 < I )
xˆ 3 )
Lˆ
L−
I
xˆ 3 × xˆ 3 =
Lˆ
L × xˆ 3 .
I
(9.54)
But this is simply the expression for the rate of change of a vector rotating
ˆ The frequency of this rotation is |ω|
˜ = L/I .
around the fixed vector ω˜ ≡ (L/I )L.
Therefore, xˆ 3 precesses around the fixed vector L with frequency
ω˜ =
Fig. 9.27
12
L
,
I
(9.55)
This distance could theoretically be much larger or much smaller than 10 m. It happens to be of
this order due to the nature of the driving force. The present consensus for this force is pressure
changes at the bottom of the ocean and in the atmosphere; see Gross (2000). Without a driving
force, the amplitude would head to zero, due to the nonrigidity of the earth.
9.7 Heavy symmetric top
399
in the fixed frame. And therefore ω does also, because it is coplanar with
xˆ 3 and L.
Remarks:
1. We just found that ω precesses around L with frequency L/I . What, then, is wrong with
the following reasoning: “Just as the rate of change of xˆ 3 equals ω × xˆ 3 , the rate of change
of ω should equal ω × ω, which is zero. Therefore, ω should remain constant.” The error
is that the vector ω is not fixed in the body frame. A vector A must be fixed in the body
frame in order for its rate of change to be given by ω × A.
2. We found in Eqs. (9.51) and (9.47) that a person standing on the rotating body sees L (and
ω) precess with frequency ≡ ω3 (I3 − I )/I around xˆ 3 . But we found in Eq. (9.55) that
a person standing in the fixed frame sees xˆ 3 (and ω) precess with frequency L/I around
L. Are these two facts compatible? Should we have obtained the same frequency from
either point of view? (Answers: yes, no).
These two frequencies are indeed consistent, as can be seen by the following reasoning.
Consider the plane (call it S) containing the three vectors L, ω, and xˆ 3 . We know from
Eq. (9.51) that S rotates with frequency xˆ 3 with respect to the body. Therefore, the
body rotates with frequency − xˆ 3 with respect to S. And from Eq. (9.55), S rotates with
ˆ with respect to the fixed frame. Therefore, the total angular velocity of
frequency (L/I )L
the body with respect to the fixed frame is (using the frame S as an intermediate step)
ωtotal =
Lˆ
L−
I
xˆ 3 .
(9.56)
But from Eq. (9.53), this is simply ω, as it should be. So the two frequencies in Eqs. (9.47)
and (9.55) are indeed consistent.
≡ ω3 (I3 − I )/I and L/I are quite
For the earth, I3 and I are nearly the same, so
different. L/I is roughly equal to L/I3 , which is essentially equal to ω3 . On the other hand,
is roughly equal to (1/300)ω3 . Basically, an external observer sees ω precess around
its cone at roughly the rate at which the earth spins. But it’s not exactly the same rate, and
this difference is what causes the earth-based observer to see ω precess with a nonzero .
3. The fixed-frame precession of xˆ 3 around L should not be confused with the
“precession of the equinoxes” effect. See Problem 10.15 for a discussion of
the latter. ♣
9.7
Heavy symmetric top
Consider now a heavy symmetrical top, that is, one that spins on a table, under
the influence of gravity (see Fig. 9.28). Assume that the tip of the top is fixed on
the table by a pivot. We’ll solve for the motion of the top in two different ways
below in Sections 9.7.3 and 9.7.4. The first uses τ = dL/dt, and the second uses
the Lagrangian method.
9.7.1
x3
Fig. 9.28
x3
θ
φ
Euler angles
For both of these methods, it is convenient to use the Euler angles, θ , φ, ψ,
which are shown in Fig. 9.29 and defined as follows.
• θ : Let xˆ 3 be the symmetry axis of the top. Define θ to be the angle that xˆ 3 makes with
the vertical zˆ axis of the fixed frame.
fixed point
z in body
x2
ψ
y
x1
x
Fig. 9.29
400
ˆ
Angular momentum, Part II (General L)
• φ: Draw the plane orthogonal to xˆ 3 . Let xˆ 1 be the intersection of this plane with the
horizontal x-y plane. Define φ to be the angle that xˆ 1 makes with the xˆ axis in the fixed
frame. Note that xˆ 1 is not necessarily fixed in the object.
• ψ: Let xˆ 2 be orthogonal to xˆ 3 and xˆ 1 , as shown. As with xˆ 1 , xˆ 2 is not necessarily fixed
in the object. Let frame S be the frame whose axes are xˆ 1 , xˆ 2 , and xˆ 3 . Define ψ to be
the angle of rotation of the body around the xˆ 3 axis in frame S. So ψ˙ xˆ 3 is the angular
velocity of the body with respect to S. And from the figure, we also see that the angular
velocity of frame S with respect to the fixed frame is φ˙ zˆ + θ˙ xˆ 1 .
The angular velocity of the body with respect to the fixed frame is equal to the
angular velocity of the body with respect to frame S, plus the angular velocity of
frame S with respect to the fixed frame. From above, we therefore have
ω = ψ˙ xˆ 3 + (φ˙ zˆ + θ˙ xˆ 1 ).
(9.57)
It is often more convenient to rewrite ω entirely in terms of the orthogonal xˆ 1 ,
xˆ 2 , xˆ 3 basis vectors. Since zˆ = cos θ xˆ 3 + sin θ xˆ 2 , Eq. (9.57) gives
ω = (ψ˙ + φ˙ cos θ)ˆx3 + φ˙ sin θ xˆ 2 + θ˙ xˆ 1 .
(9.58)
This form of ω is generally more useful, because xˆ 1 , xˆ 2 , xˆ 3 are principal axes of
the body. (We are assuming that we are working with a symmetrical top, with
I1 = I2 ≡ I . This means that any axes in the xˆ 1 -ˆx2 plane are principal axes.)
Although xˆ 1 and xˆ 2 are not fixed in the object, they are still good principal axes
at any instant.
9.7.2
Digression on the components of ω
The above expressions for ω might look a little scary, but there is a very helpful
diagram we can draw (see Fig. 9.30) that makes it easier to see what’s going on.
Let’s talk a bit about this before tackling the original problem of the spinning top.
The diagram is rather dense (you might even say it looks scarier than the above ω),
so we’ll go through it slowly. In the following discussion, we’ll simplify things
by setting θ˙ = 0. All the interesting features of ω remain. The θ˙ xˆ 1 component of
ω in Eqs. (9.57) and (9.58) arises simply from the easily visualizable rising and
falling of the top. We will therefore concentrate on the more complicated issues,
namely the components of ω in the plane of xˆ 3 , zˆ , and xˆ 2 .
With θ˙ = 0, Fig. 9.30 shows the vector ω in the xˆ 3 -ˆz-ˆx2 plane (the way we’ve
drawn it, xˆ 1 points into the page, in contrast with Fig. 9.29). We’ll refer to this
figure many times in the problems for this chapter. There are numerous comments
to be made about it, so we’ll just list them out. The following discussion deals
with the kinematics of ω, that is, the meaning of the various components and how
they relate to each other. The discussion of the dynamics of ω, that is, why the
components take on the values they do, given a certain physical system, is the
subject of Section 9.7.3 onward.
9.7 Heavy symmetric top
z
ω
ωz
x2
Ω sin θ
x3
Ω
ω2
θ
z
x2
ω'
θ
ω3 symmetry
axis
Ω cos θ
x3
Fig. 9.30
1. If someone asks you to “decompose” ω into pieces along zˆ and xˆ 3 , what would you
do? Would you draw the lines perpendicular to these axes to obtain the lengths shown
(which we’ll label as ωz and ω3 ), or would you draw the lines parallel to these axes to
obtain the lengths shown (which we’ll label as and ω )? There is no “correct” answer
to this question. The four quantities, ωz , ω3 , , ω simply represent different things.
We will interpret each of these below, along with ω2 (the projection of ω along xˆ 2 ).
It turns out that and ω are the frequencies that your eye can see the easiest, while
ω2 and ω3 are what you want to use when doing calculations involving the angular
momentum. But as far as I can see, ωz is not of much use.
2. Note that it is true that
ω = ω xˆ 3 +
zˆ ,
(9.59)
but it is not true that ω = ω3 xˆ 3 + ωz zˆ . Another true statement is
ω = ω3 xˆ 3 + ω2 xˆ 2 .
(9.60)
3. In terms of the Euler angles, we see by comparing Eqs. (9.59) and (9.57), with θ˙ = 0,
that
˙
ω = ψ,
and
˙
= φ.
(9.61)
And we also have, by comparing Eqs. (9.60) and (9.58), with θ˙ = 0,
ω3 = ψ˙ + φ˙ cos θ = ω +
ω2 = φ˙ sin θ =
sin θ .
cos θ,
(9.62)
These are also clear from Fig. 9.30. There is therefore technically no need to introduce
the new ω2 , ω3 , , ω definitions in Fig. 9.30, because the Euler angles are quite
401
402
ˆ
Angular momentum, Part II (General L)
sufficient. But we will be referring back to this figure many times, and it is a little
easier to work with these omegas than the various combinations of the Euler angles.
4.
is the easiest of the frequencies to visualize. It is the frequency of precession of
the top around the vertical zˆ axis.13 In other words, the symmetry axis xˆ 3 traces out a
cone (assuming θ˙ = 0) around the zˆ axis with frequency . The reason for this is the
following. The vector ω is the vector that gives the speed of any point (at position r)
fixed in the top as ω × r. Therefore, since the vector xˆ 3 is fixed in the top, we can write
d xˆ 3
= ω × xˆ 3 = (ω xˆ 3 +
dt
zˆ ) × xˆ 3 = ( zˆ ) × xˆ 3 .
(9.63)
But this is precisely the expression for the rate of change of a vector rotating around the
zˆ axis with frequency . (This is exactly the same type of proof as the one leading to
Eq. (9.54).) Note that the precession frequency around the zˆ axis is not ωz . It certainly
can’t be ωz , because we can imagine grabbing the symmetry axis and holding it in
place, so that ω points along xˆ 3 . This scenario has a nonzero ωz , but no precession.
Remark: In the derivation of Eq. (9.63), we basically just stripped off the part of ω that
points along the xˆ 3 axis, because a rotation around xˆ 3 contributes nothing to the motion
of xˆ 3 . Note, however, that there is in fact an infinite number of ways to strip off a piece
along xˆ 3 . For example, we can also break ω up as, say, ω = ω3 xˆ 3 + ω2 xˆ 2 . We then obtain
d xˆ 3 /dt = (ω2 xˆ 2 ) × xˆ 3 , which means that xˆ 3 is instantaneously rotating around xˆ 2 with
frequency ω2 . Although this is true, it isn’t as useful as the result in Eq. (9.63), because the
xˆ 2 axis changes with time (it precesses around zˆ ). The point here is that the instantaneous
angular velocity vector around which the symmetry axis rotates is not well defined (Problem
9.1 discusses this issue).14 But the zˆ axis is the only one of these angular velocity vectors
that is fixed. When we look at the top (or more precisely, the symmetry axis), we therefore
see it precessing around the zˆ axis. ♣
5. ω is also easy to visualize. Imagine that you are at rest in the frame that rotates around
the zˆ axis with frequency . Then you see the symmetry axis of the top remain perfectly
still, and the only motion you see is the top spinning around this axis with frequency
ω . (This is true because ω = ω xˆ 3 + zˆ , and the rotation of your frame causes you
not to see the zˆ part.) If you paint a dot somewhere on the top, then the dot traces
out a fixed tilted circle, and the dot returns to, say, its maximum height at frequency
ω . A person in the lab frame sees this dot undergo a rather complicated motion but
must observe the same frequency at which the dot returns to its highest point. So ω is
something quite physical in the lab frame also.
6. ω3 is what you use to obtain the component of L along xˆ 3 , because L3 = I3 ω3 . ω3
is a little harder to visualize than and ω , but it is the frequency with which the top
13
14
Although we’re using the same letter, this doesn’t have anything to do with the defined in
Eq. (9.47), except for the fact that they both represent the frequency of something precessing
around an axis.
The instantaneous angular velocity of the whole body is well defined, of course. There is a definite
line of points in the body that are instantaneously at rest. But if you look at the symmetry axis by
itself, then there is an ambiguity (see Footnote 9). In short, because only one point on the axis (the
bottom end), instead of a whole line, is instantaneously at rest, the instantaneous angular velocity
vector can point in any direction.
9.7 Heavy symmetric top
instantaneously rotates, as seen by someone at rest in the frame that rotates around the
instantaneous xˆ 2 axis with frequency ω2 . (This is true because ω = ω2 xˆ 2 + ω3 xˆ 3 , and
the rotation of the frame causes the person not to see the ω2 xˆ 2 part.) This rotation is
harder to vizualize in the lab frame, because the xˆ 2 axis changes with time.
There is one physical scenario in which ω3 is the easily observed frequency. Imagine
that the top is precessing around the zˆ axis at constant θ (we’ll find in Section 9.7.5 that
this is in fact a possible motion for the top), and imagine that the top has a frictionless
rod protruding along its symmetry axis. If you grab the rod and stop the precession
motion, so that the top is now spinning around its stationary symmetry axis, then this
spinning has frequency ω3 . This is true because when you grab the rod, your torque
has no component along the xˆ 3 axis (because the rod lies along this axis, and because
it is frictionless). Therefore, L3 doesn’t change, and so neither does ω3 .
7. ω2 is similar to ω3 , of course. ω2 is what you use to obtain the component of L along xˆ 2 ,
because L2 = I2 ω2 . It is the frequency with which the top instantaneously rotates, as
seen by someone at rest in the frame that rotates around the instantaneous xˆ 3 axis with
frequency ω3 . (This is true because ω = ω2 xˆ 2 + ω3 xˆ 3 , and the rotation of the frame
causes the person not to see the ω3 xˆ 3 part.) Again, this rotation is harder to vizualize in
the lab frame, because the xˆ 3 axis changes with time. Note that by “instantaneous xˆ 3
axis,” we mean the fixed axis in space that coincides with the symmetry axis at a given
instant. The symmetry axis will therefore move away from this fixed axis, consistent
with the fact that the person in the above-mentioned rotating frame sees the top rotate
around the xˆ 2 axis.
The physical scenario that produces ω2 , analogous to the scenario that produced ω3
above, is the following. Imagine a frictionless rod glued to the top at its tip, perpendicular to the symmetry axis, so that they form a “T.” As this rod is spinning around
(ignore the fact that it has to keep passing through the table), grab it at the instant it
points along xˆ 2 . The top will then rotate with frequency ω2 around the fixed rod. This
is true for reasons analogous to the ones in the ω3 case above.
8. ωz is not very useful, as far as I can see. The most important thing to note about ωz
is that it is not the frequency of precession around the zˆ axis, even though it is the
projection of ω onto zˆ . The frequency of the precession is , as we found above in
Eq. (9.63). A true, but somewhat useless, fact about ωz is that if someone is at rest
in the frame that rotates around the zˆ axis with frequency ωz , then she sees all points
in the top instantaneously rotating around the horizontal xˆ axis with frequency ωx ,
where ωx is the projection of ω onto the xˆ axis. (This is true because ω = ωx xˆ + ωz zˆ ,
and the rotation of the frame causes her not to see the ωz zˆ part.)
9.7.3
Torque method
Let’s now finally solve for the motion of a heavy top. This first method involving
torque is straightforward, although a bit tedious. We include it here to (1) show
that the problem can be done without resorting to a Lagrangian, and to (2) get
some practice using τ = dL/dt. We’ll make use of the form of ω given in
403
404
ˆ
Angular momentum, Part II (General L)
Eq. (9.58), because there it is broken up into the principal-axis components. For
convenience, define β˙ = ψ˙ + φ˙ cos θ , so that
ω = β˙ xˆ 3 + φ˙ sin θ xˆ 2 + θ˙ xˆ 1 .
(9.64)
Note that we’ve returned to the most general motion, where θ˙ is not necessarily
zero. For our origin, we’ll choose the tip of the top, which is assumed to be fixed
on the table.15 Let the principal moments relative to this origin be I1 = I2 ≡ I ,
and I3 . The angular momentum of the top is then
L = I3 β˙ xˆ 3 + I φ˙ sin θ xˆ 2 + I θ˙ xˆ 1 .
(9.65)
We must now calculate dL/dt. What makes this nontrivial is the fact that the xˆ 1 ,
xˆ 2 , and xˆ 3 unit vectors change with time (they change with θ and φ). But let’s
forge ahead and take the derivative of Eq. (9.65). Using the product rule (which
works fine with the product of a scalar and a vector), we have
dL
d β˙
d(φ˙ sin θ)
d θ˙
= I3
xˆ 3 + I
xˆ 2 + I
xˆ 1
dt
dt
dt
dt
d xˆ 3
d xˆ 2
d xˆ 1
+ I3 β˙
+ I φ˙ sin θ
+ I θ˙
.
dt
dt
dt
(9.66)
Using a little geometry, you can show that
d xˆ 3
= −θ˙ xˆ 2 + φ˙ sin θ xˆ 1 ,
dt
d xˆ 2
= θ˙ xˆ 3 − φ˙ cos θ xˆ 1 ,
dt
d xˆ 1
= −φ˙ sin θ xˆ 3 + φ˙ cos θ xˆ 2 .
dt
(9.67)
As an exercise, you should verify these by making use of Fig. 9.29. In the
first equation, for example, show that a change in θ causes xˆ 3 to move a certain distance in the xˆ 2 direction; and show that a change in φ causes xˆ 3 to move
a certain distance in the xˆ 1 direction. Plugging the derivative expressions from
Eq. (9.67) into Eq. (9.66) gives, after some algebra,
dL
= I3 β¨ xˆ 3 + I φ¨ sin θ + 2I θ˙ φ˙ cos θ − I3 β˙ θ˙ xˆ 2
dt
+ I θ¨ − I φ˙ 2 sin θ cos θ + I3 β˙ φ˙ sin θ xˆ 1 .
(9.68)
Let’s now look at the torque on the top. This arises from gravity pulling down
on the CM. So from Fig. 9.29, τ points in the xˆ 1 direction and has magnitude
15
We could use the CM as our origin, but then we would have to include the complicated forces
acting at the pivot point, which is difficult. But see Problem 9.19 for the case where the tip is free
to slide on a frictionless table.
9.7 Heavy symmetric top
Mg sin θ , where is the distance from the pivot to the CM. Using Eq. (9.68),
the third component of τ = dL/dt quickly gives
β¨ = 0.
(9.69)
Therefore, β˙ is a constant, which we’ll call ω3 (an obvious label, in view of
Eq. (9.64)). The other two components of τ = dL/dt then give
I φ¨ sin θ + θ˙ (2I φ˙ cos θ − I3 ω3 ) = 0,
˙ sin θ = I θ¨ .
(Mg + I φ˙ 2 cos θ − I3 ω3 φ)
(9.70)
We’ll wait to fiddle with these equations until we have derived them again using
the Lagrangian method.
9.7.4
Lagrangian method
Equation (9.15) gives the kinetic energy of the top as T = ω · L/2. Using
˙ 16
Eqs. (9.64) and (9.65), we have (writing ψ˙ + φ˙ cos θ instead of the shorthand β)
T =
1
1
1
ω · L = I3 (ψ˙ + φ˙ cos θ )2 + I (φ˙ 2 sin2 θ + θ˙ 2 ).
2
2
2
(9.71)
The potential energy is
V = Mg cos θ ,
(9.72)
where is the distance from the pivot to the CM. The Lagrangian is L = T − V
(we’ll use “L” here to avoid confusion with the angular momentum “L”), and so
the equation of motion obtained from varying ψ is
d ∂L
∂L
=
dt ∂ ψ˙
∂ψ
=⇒
d
(ψ˙ + φ˙ cos θ) = 0.
dt
(9.73)
Therefore, ψ˙ + φ˙ cos θ is a constant. Call it ω3 . The equations of motion obtained
from varying φ and θ are then (making use of ψ˙ + φ˙ cos θ = ω3 )
d ∂L
∂L
=
dt ∂ φ˙
∂φ
=⇒
d
I3 ω3 cos θ + I φ˙ sin2 θ = 0,
dt
d ∂L
∂L
=
dt ∂ θ˙
∂θ
=⇒
˙ sin θ .
I θ¨ = (Mg + I φ˙ 2 cos θ − I3 ω3 φ)
(9.74)
Taking the derivative in the first equation, we see that these equations are identical
to those in Eq. (9.70).
16
It was ok to use β in the previous subsection. We introduced it only because it was quicker to write.
But we can’t use it here, because it depends on the other coordinates, and the Lagrangian method
requires the use of independent coordinates. The variational proof back in Chapter 6 assumed this
independence.
405
ˆ
Angular momentum, Part II (General L)
406
Note that there are two conserved quantities, arising from the facts that ∂L/∂ψ
and ∂L/∂φ equal zero. The conserved quantities are the angular momenta in
the xˆ 3 and zˆ directions, respectively. This is true because from Eq. (9.65), the
former is L3 = I3 ω3 and the latter is Lz = L3 cos θ + L2 sin θ = (I3 ω3 ) cos θ +
(I φ˙ sin θ ) sin θ. These angular momenta are conserved because the torque points
in the xˆ 1 direction, so there is no torque in the plane spanned by xˆ 3 and zˆ .
9.7.5
Spinning top with θ˙ = 0
A special case of Eqs. (9.70) occurs when θ˙ = 0. In this case, the first of
Eqs. (9.70) says that φ˙ is constant. The CM of the top therefore undergoes uniform circular motion in a horizontal plane. Let ≡ φ˙ be the frequency of this
motion (this is the same notation as in Eq. (9.61)). Then the second of Eqs. (9.70)
becomes
I
2
cos θ − I3 ω3
+ Mg = 0.
(9.75)
This quadratic equation can be solved to yield two possible precessional frequencies, , for the top. And yes, there are indeed two of them, provided that ω3 is
greater than a certain minimum value.
The previous pages in this “Heavy symmetric top” section have been a bit
abstract, so let’s now take a breather and rederive Eq. (9.75) from scratch. That
is, we’ll assume θ˙ = 0 from the start of the solution, and then solve things by
finding L and using τ = dL/dt, in the spirit of Section 9.4.2. In practice, this
strategy of starting from scratch is invariably the best route to take, as you’ll see
in the problems and exercises for this chapter. The procedures in Sections 9.7.3
and 9.7.4 are good to know, but the technique in the following example provides
a much more intuitive way of looking at things. This example is the classic “top”
problem. We’ll warm up by solving it in an approximate way. Then we’ll do it
for real.
Ω
θ l CM
Fig. 9.31
ω3
Example (The top): A symmetric top of mass M has its CM a distance from its
pivot. The moments of inertia relative to the pivot are I1 = I2 ≡ I , and I3 . The top
spins around its symmetry axis with frequency ω3 (in the language of Section 9.7.2),
and initial conditions have been set up so that the CM precesses in a circle around
the vertical axis. The symmetry axis makes a constant angle θ with the vertical (see
Fig. 9.31).
(a) Assuming that the angular momentum due to ω3 is much larger than any other
angular momentum in the problem, find an approximate expression for the
frequency, , of precession.
(b) Now solve the problem exactly. That is, find by considering all of th