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Fundamentals of Physics I
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THE OPEN YALE COURSES SERIES is designed to bring the depth and breadth of a Yale education to a wide variety of readers. Based on Yale’s Open Yale Courses program (http://oyc.yale.edu), these books bring outstanding lectures by Yale faculty to the curious reader, whether student or adult. Covering a wide variety of topics across disciplines in the social sciences, physical sciences, and humanities, Open Yale Courses books offer accessible introductions at affordable prices.
The production of Open Yale Courses for the Internet was made possible by a grant from the William and Flora Hewlett Foundation.
BOOKS IN THE OPEN YALE COURSES SERIES Paul H. Fry, Theory of Literature Roberto González Echevarría, Cervantes’ “Don Quixote” Christine Hayes, Introduction to the Bible Shelly Kagan, Death Dale B. Martin, New Testament History and Literature Giuseppe Mazzotta, Reading Dante R. Shankar, Fundamentals of Physics I: Mechanics, Relativity, and
Thermodynamics, Expanded Edition R. Shankar, Fundamentals of Physics II: Electromagnetism, Optics, and
Quantum Mechanics Ian Shapiro, The Moral Foundations of Politics Steven B. Smith, Political Philosophy
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Fundamentals of Physics I
Mechanics, Relativity, and Thermodynamics
Expanded Edition
R. SHANKAR

Yale
 UNIVERSITY PRESS New Haven and Lodon

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First edition 2014. Expanded edition 2019. Copyright © 2014, 2019 by Yale University. All rights reserved. This book may not be reproduced, in whole or in part, including illustrations, in any form (beyond that copying permitted by Sections 107 and 108 of the U.S. Copyright Law and except by reviewers for the public press), without written permission from the publishers.
Yale University Press books may be purchased in quantity for educational, business, or promotional use. For information, please e-mail sales.press@yale.edu (U.S. office) or sales@yaleup.co.uk (U.K. office).
Set in Minion type by Newgen North America. Printed in the United States of America.
ISBN: 978-0-300-24377-2 Library of Congress Control Number: 2019931271
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10 9 8 7 6 5 4 3 2 1
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To my students
 for their friendship and inspiration
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Deep and original, but also humble and generous, the physicist Josiah Willard Gibbs spent much of his life at Yale University. His father was a professor of sacred languages at Yale, and Gibbs received his bachelor’s and doctorate degrees from the university before teaching there until his death in 1903. The sculptor Lee Lawrie created the memorial bronze tablet pictured above, which was installed in Yale’s Sloane Physics Laboratory in 1912. It now resides in the entrance to the J. W. Gibbs Laboratories, Yale University.
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Contents
Preface to the Expanded Edition Preface to the First Edition 1. The Structure of Mechanics
1.1. Introduction and some useful tips 1.2. Kinematics and dynamics 1.3. Average and instantaneous quantities 1.4. Motion at constant acceleration 1.5. Sample problem 1.6. Deriving v2 − v20 = 2a(x − x0) using calculus 2. Motion in Higher Dimensions 2.1. Review 2.2. Vectors in d = 2 2.3. Unit vectors 2.4. Choice of axes and basis vectors 2.5. Derivatives of the position vector r 2.6. Application to circular motion 2.7. Projectile motion 3. Newton’s Laws I 3.1. Introduction to Newton’s laws of motion 3.2. Newton’s second law 3.3. Two halves of the second law 3.4. Newton’s third law 3.5. Weight and weightlessness 4. Newton’s Laws II 4.1. A solved example 4.2. Never the whole story

4.3. Motion in d = 2 4.4. Friction: static and kinetic 4.5. Inclined plane 4.6. Coupled masses 4.7. Circular motion, loop-the-loop
5. Law of Conservation of Energy 5.1. Introduction to energy 5.2. The work-energy theorem and power 5.3. Conservation of energy: K2 + U2 = K1 + U1 5.4. Friction and the work-energy theorem
6. Conservation of Energy in d = 2 6.1. Calculus review 6.2. Work done in d = 2 6.3. Work done in d = 2 and the dot product 6.4. Conservative and non-conservative forces 6.5. Conservative forces 6.6. Application to gravitational potential energy
7. The Kepler Problem 7.1. Kepler’s laws 7.2. The law of universal gravity 7.3. Details of the orbits 7.4. Law of conservation of energy far from the earth 7.5. Choosing the constant in U
8. Multi-particle Dynamics 8.1. The two-body problem 8.2. The center of mass 8.3. Law of conservation of momentum 8.4. Rocket science 8.5. Elastic and inelastic collisions 8.6. Scattering in higher dimensions
9. Rotational Dynamics I

9.1. Introduction to rigid bodies 9.2. Angle of rotation, the radian 9.3. Rotation at constant angular acceleration 9.4. Rotational inertia, momentum, and energy 9.5. Torque and the work-energy theorem 9.6. Calculating the moment of inertia
10. Rotational Dynamics II 10.1. The parallel axis theorem 10.2. Kinetic energy for a general N-body system 10.3. Simultaneous translations and rotations 10.4. Conservation of energy 10.5. Rotational dynamics using τ = dL/dt 10.6. Advanced rotations 10.7. Conservation of angular momentum 10.8. Angular momentum of the figure skater
11. Rotational Dynamics III 11.1. Static equilibrium 11.2. The seesaw 11.3. A hanging sign 11.4. The leaning ladder 11.5. Rigid-body dynamics in 3d 11.6. The gyroscope
12. Special Relativity I: The Lorentz Transformation 12.1. Galilean and Newtonian relativity 12.2. Proof of Galilean relativity 12.3. Enter Einstein 12.4. The postulates 12.5. The Lorentz transformation
13. Special Relativity II: Some Consequences 13.1. Summary of the Lorentz transformation 13.2. The velocity transformation law

13.3. Relativity of simultaneity 13.4. Time dilatation
13.4.1. Twin paradox 13.4.2. Length contraction 13.5. More paradoxes 13.5.1. Too big to fall 13.5.2. Muons in flight
14. Special Relativity III: Past, Present, and Future 14.1. Past, present, and future in relativity 14.2. Geometry of spacetime 14.3. Rapidity 14.4. Four-vectors 14.5. Proper time
15. Four-momentum 15.1. Relativistic scattering 15.1.1. Compton effect 15.1.2. Pair production 15.1.3. Photon absorption
16. Mathematical Methods 16.1. Taylor series of a function 16.2. Examples and issues with the Taylor series 16.3. Taylor series of some popular functions 16.4. Trigonometric and exponential functions 16.5. Properties of complex numbers 16.6. Polar form of complex numbers
17. Simple Harmonic Motion 17.1. More examples of oscillations 17.2. Superposition of solutions 17.3. Conditions on solutions to the harmonic oscillator 17.4. Exponential functions as generic solutions 17.5. Damped oscillations: a classification

17.5.1. Over-damped oscillations 17.5.2. Under-damped oscillations 17.5.3. Critically damped oscillations 17.6. Driven oscillator
18. Waves I 18.1. The wave equation 18.2. Solutions of the wave equation 18.3. Frequency and period
19. Waves II 19.1. Wave energy and power transmitted 19.2. Doppler effect 19.3. Superposition of waves 19.4. Interference: the double-slit experiment 19.5. Standing waves and musical instruments
20. Fluids 20.1. Introduction to fluid dynamics and statics 20.1.1. Density and pressure 20.1.2. Pressure as a function of depth 20.2. The hydraulic press 20.3. Archimedes’ principle 20.4. Bernoulli’s equation 20.4.1. Continuity equation 20.5. Applications of Bernoulli’s equation
21. Heat 21.1. Equilibrium and the zeroth law: temperature 21.2. Calibrating temperature 21.3. Absolute zero and the Kelvin scale 21.4. Heat and specific heat 21.5. Phase change 21.6. Radiation, convection, and conduction 21.7. Heat as molecular kinetic energy

22. Thermodynamics I 22.1. Recap 22.2. Boltzmann’s constant and Avogadro’s number 22.3. Microscopic definition of absolute temperature 22.4. Statistical properties of matter and radiation 22.5. Thermodynamic processes 22.6. Quasi-static processes 22.7. The first law of thermodynamics 22.8. Specific heats: cv and cp
23. Thermodynamics II 23.1. Cycles and state variables 23.2. Adiabatic processes 23.3. The second law of thermodynamics 23.4. The Carnot engine 23.4.1 Defining T using Carnot engines
24. Entropy and Irreversibility 24.1. Entropy 24.2. The second law: law of increasing entropy 24.3. Statistical mechanics and entropy 24.4. Entropy of an ideal gas: full microscopic analysis 24.5. Maximum entropy principle illustrated 24.6. The Gibbs formalism 24.7. The third law of thermodynamics
Exercises Problem Set 1, for Chapter 1 Problem Set 2, for Chapter 2 Problem Set 3, for Chapters 3 and 4 Problem Set 4, for Chapters 5, 6, and 7 Problem Set 5, for Chapter 8 Problem Set 6, for Chapters 9, 10, and 11 Problem Set 7, for Chapters 12, 13, 14, and 15

Problem Set 8, for Chapters 16 and 17 Problem Set 9, for Chapters 18 and 19 Problem Set 10, for Chapter 20 Problem Set 11, for Chapters 21, 22, 23, and 24 Answers to Exercises Problem Set 1, for Chapter 1 Problem Set 2, for Chapter 2 Problem Set 3, for Chapters 3 and 4 Problem Set 4, for Chapters 5, 6, and 7 Problem Set 5, for Chapter 8 Problem Set 6, for Chapters 9, 10, and 11 Problem Set 7, for Chapters 12, 13, 14, and 15 Problem Set 8, for Chapters 16 and 17 Problem Set 9, for Chapters 18 and 19 Problem Set 10, for Chapter 20 Problem Set 11, for Chapters 21, 22, 23, and 24 Constants and Other Data Index
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Preface to the Expanded Edition
Besides the correction of typos, the most important change is the inclusion of over 300 exercises to go with these 24 chapters. Answers are given to all exercises, but not the solutions. They were chosen to test the material covered in this book. This was done in response to calls from students and instructors. A similar inclusion will be made in the next edition of the companion volume, Fundamentals of Physics II.
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Preface to the First Edition
Given that the size of textbooks has nearly tripled during my own career, without a corresponding increase in the cranial dimensions of my students, I have always found it necessary, like my colleagues elsewhere, to cull the essentials into a manageable size. I did that in the course Fundamentals of Physics I taught at Yale, and this book preserves that feature. It covers the fundamental ideas of Newtonian mechanics, relativity, fluids, waves, oscillations, and thermodynamics without compromise. It requires only the basic notions of differentiation and integration, which I often review as part of the lectures. It is aimed at college students in physics, chemistry, and engineering as well as advanced high school students and independent selftaught learners at various stages in life, in various careers.
The chapters in the book more or less follow my 24 lectures, with a few minor modifications. The style preserves the classroom atmosphere. Often I introduce the questions asked by the students or the answers they give when I believe they will be of value to the reader. The simple figures serve to communicate the point without driving up the price. The equations have been typeset and are a lot easier to read than in the videos. The problem sets and exams, without which one cannot learn or be sure one has learned the physics, may be found along with their solutions at the Yale website, http://oyc.yale.edu/physics, free and open to all. The lectures may also be found at venues such as YouTube, iTunes (https://itunes.apple.com/us/itunes-u/physics-video/id341651848?mt=10), and Academic Earth, to name a few.
The book, along with the material available at the Yale website, may be used as a stand-alone resource for a course or self-study, though some instructors may prescribe it as a supplement to another one adapted for the class, so as to provide a wider choice of problems or more worked examples.
To my online viewers I say, “You have seen the movie; now read the book!” The advantage of having the printed version is that you can read it during take-off and landing.

In the lectures I sometimes refer to my Basic Training in Mathematics, published by Springer and intended for anyone who wants to master the undergraduate mathematics needed for the physical sciences.
This book owes its existence to many people. It all began when Peter Salovey, now President, then Dean of Yale College, asked me if I minded having cameras in my Physics 200 lectures to make them part of the first batch of Open Yale Courses, funded by the Hewlett Foundation. Since my answer was that I had yet to meet a camera I did not like, the taping began. The key person hereafter was Diana E. E. Kleiner, Dunham Professor, History of Art and Classics, who encouraged and guided me in many ways. She was also the one who persuaded me to write this book. Initially reluctant, I soon found myself thoroughly enjoying proselytizing my favorite subject in this new format. At Yale University Press, Joe Calamia was my friend, philosopher, and guide. Liz Casey did some very skilled editing. Besides correcting errors in style (such as a long sentence that began in first person past tense and ended in third person future tense) and matters of grammar and punctuation (which I sprinkle pretty much randomly), she also made sure my intent was clear in every sentence.
Barry Bradlyn and Alexey Shkarin were two graduate students and Qiwei Claire Xue and Dennis Mou were two undergraduates who proofread earlier versions.
My family, from my wife, Uma, down to little Stella, have encouraged me in various ways.
I take this opportunity to acknowledge my debt to the students at Yale who, over nearly four decades, have been the reason I jump out of bed on two or three days a week. I am grateful for their friendship and curiosity. In recent years, they were often non-majors, willing to be persuaded that physics was a fascinating subject. This I never got tired of doing, thanks to the nature of the subject and the students.
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CHAPTER 1
The Structure of Mechanics
1.1 Introduction and some useful tips
This book is based on the first half of a year-long course that introduces you to all the major ideas in physics, starting from Galileo and Newton, right up to the big revolutions of the twentieth century: relativity and quantum mechanics. The target audience for this course and book is really very broad. In fact, I have always been surprised by the breadth of interests of my students. I don’t know what you are going to do later in life, so I have picked the topics that all of us in physics find fascinating. Some may not be useful, but you just don’t know. Some of you are probably going to be doctors, and you don’t know why I’m going to cover special relativity or quantum mechanics. Well, if you’re a doctor and you have a patient who’s running away from you at the speed of light, you’ll know what to do. Or, if you’re a pediatrician, you will understand why your patient will not sit still: the laws of quantum mechanics don’t allow a very small object to have a definite position and momentum. Whether or not you become a physicist, you should certainly learn about these great strides in the human attempt to understand the physical world.
Most textbooks are about 1,200 pages long, but when I learned physics they were around 400 pages long. When I look around, I don’t see any student whose head is three times as big as mine, so I know that you cannot digest everything the books have. I take what I think are the really essential parts and cover them in these lectures. So you need the lectures to find out what’s in the syllabus and what’s not. If you don’t do that, there’s a danger you will learn something you don’t have to, and we don’t want that, right?
To learn physics well, you have to do the problems. If you watch me online doing things on the blackboard or working through derivations in the book, it all looks very reasonable. It looks like you can do it yourself and that you understand what is going on, but the only way you’re going to find out is by actually doing problems. A fair number are available, with their

solutions, at http://oyc.yale.edu/physics/phys-200, and over 300 (with answers but not solutions) in the exercises and answers sections at the back of this book. You don’t have to do them by yourself. That’s not how physics is done. I am now writing a paper with two other people. My experimental colleagues write papers with four hundred or even a thousand other people when engaged in the big collider experiments like the ones in Geneva or Fermilab. It’s perfectly okay to be part of a collaboration, but you have to make sure that you’re pulling your weight, that everybody makes contributions to finding the solution and understands it.
This calculus-based course assumes you know the rudiments of differential and integral calculus, such as functions, derivatives, derivatives of elementary functions, elementary integrals, changing variables in integrals, and so on. Later, I will deal with functions of more than one variable, which I will briefly introduce to you, because that is not a prerequisite. You have to know your trigonometry, to know what’s a sine and what’s a cosine and some simple identities. You cannot say, “I will look it up.” Your birthday and social security number are things you look up; trigonometric functions and identities are what you know all the time.
1.2 Kinematics and dynamics
We are going to be studying Newtonian mechanics. Standing on the shoulders of his predecessors, notably Galileo, Isaac Newton placed us on the road to understanding all the mechanical phenomena for centuries until the laws of electromagnetism were discovered, culminating in Maxwell’s equations. Our concern here is mechanics, which is the motion of billiard balls and trucks and marbles and whatnot. You will find out that the laws of physics for this entire semester can be written down on the back of an envelope. A central purpose of this course is to show you repeatedly that starting with those few laws, you can deduce everything. I would encourage you to think the way physicists do, even if you don’t plan to be a physicist. The easiest way to master this subject is to follow the reasoning I give you. That way, you don’t have to store too many things in your head. Early on, when there are four or five formulas, you can memorize all of them and you can try every one of them until something works, but, after a couple of weeks, you will have hundreds of formulas, and you cannot memorize all of them. You cannot resort to trial and error. You have to know the logic.

The goal of physics is to predict the future given the present. We will pick some part of the universe that we want to study and call it “the system,” and we will ask, “What information do we need to know about that system at the initial time, like right now, in order to be able to predict its future evolution?” If I throw a piece of candy at you and you catch it, that’s an example of Newtonian mechanics at work. What did I do? I threw a piece of candy from my hand, and the initial conditions are where I released it and with what velocity. That’s what you see with your eyes. You know it’s going to go up, it’s going to follow some kind of parabola, and your hands get to the right place at the right time to receive it. That is an example of Newtonian mechanics at work, and your brain performed the necessary calculations effortlessly.
You only have to know the candy’s initial location and the initial velocity. The fact that it was blue or red is not relevant. If I threw a gorilla at you, its color and mood would not matter. These are things that do not affect the physics. If a guy jumps off a tall building, we want to know when, and with what speed, he will land. We don’t ask why this guy is ending it all today; that is a question for the psych department. So we don’t answer everything. We ask very limited questions about inanimate objects, and we brag about how accurately we can predict the future.
The Newtonian procedure for predicting the future, given the present, has two parts, kinematics and dynamics. Kinematics is a complete description of the present. It’s a list of what you have to know about a system right now. For example, if you’re talking about a piece of chalk, you will want to know where it is and how fast it’s moving. Dynamics then tells you why the chalk goes up, why it goes down, and so on. It comes down due to the force of gravity. In kinematics, you don’t ask for the reason behind anything. You simply want to describe things the way they are, and then dynamics tells you how and why that description changes with time.
I’m going to illustrate the idea of kinematics by following my preferred approach: starting with the simplest possible example and slowly adding bells and whistles to make it more and more complicated. In the initial stages, some of you might say, “Well, I have seen this before, so maybe there is nothing new here.” That may well be true. I don’t know how much you have seen, but it is likely that the way you learned physics in high school is different from the way professional physicists think about it. Our

priorities, and the things that we get excited about, are often different; and the problems will be more difficult.
1.3 Average and instantaneous quantities
We are going to study an object that is a mathematical point. It has no size. If you rotate it, it will look the same, unlike a potato, which will look different upon rotation. It is not enough to just say where the potato is; you have to say which way its nose is pointing. The study of such extended bodies comes later. Right now, we want to study an entity that has no spatial extent, a dot. It can move around all over space. We’re going to simplify that too. We’re going to take an entity that moves only along the x-axis. So you can imagine a bead with a straight wire going through it, which allows it to only slide back and forth. This is about the simplest thing. I cannot reduce the number of dimensions. I cannot make the object simpler than a mathematical point.
To describe what the point is doing, we pick an origin, call it x = 0, and put some markers along the x-axis to measure distance. Then we will say this guy is sitting at x = 5. Now, of course, we have to have units and the unit for length is going to be the meter. The unit for time will be a second. Sometimes I might not write the units, but I have earned the right to do that and you haven’t. Everything has got to be in the right units. If you don’t have the units, and if you say the answer is 42, then we don’t know if you are right or wrong.
Back to the object. At a given instant, it’s got a location. We would like to describe the object’s motion by plotting a graph of space versus time. A typical graph would be something like Figure 1.1. Even though the plot is going up and down, the object is moving horizontally, back and forth along the spatial x-axis. When it is at A, it’s crossing the origin from the left and going to the right. Later, at B, it is crossing back to the left. In the language of calculus, x is a function of time, x = x(t), and the graph corresponds to some generic function that doesn’t have a name. We will also encounter functions that do have a name, like x(t) = t, x(t) = t2, x(t) = sint, cost, and so on.

Figure 1.1 Trajectory of a particle. The position, x(t), is measured vertically and the time, t, is measured horizontally.
Consider , the average velocity of an object, given by
where t2 > t1 are two times between which we have chosen to average the velocity. In the example in Figure 1.1, < 0 for the indicated choice of t1 and t2 since the final x(t2) is less than the initial x(t1).
The average velocity may not tell you the whole story. For example, if you started at x(t1) and at time t1 ended up at point C with the same coordinate, the average velocity would be zero, which is the average you would get if the particle had never moved!
The average acceleration, ā, involves a similar difference of velocities:
Now for an important concept, the velocity at a given time or instantaneous velocity, v(t). Figure 1.1 shows some particle moving a distance Δx between times t and t + Δt. The average velocity in that interval is . What you want is the velocity at time t. We all have an intuitive

notion of velocity right now. When you’re driving your car, if the needle says 60 miles per hour, that’s your velocity at that instant. Though velocity seems to involve two different times in its very definition—the initial time and the final time—we want to talk about the velocity right now. That is obtained by examining the position now and the position slightly later, and taking the ratio of the change in position to the time elapsed between the two events, while bringing the two points closer and closer in time. We see in the figure that when we do this, both Δx → 0 and Δt → 0, but their ratio becomes the tangent of the angle θ, shown in Figure 1.1. Thus the velocity at the instant t is:
Once you take one derivative, you can take any number of derivatives. The derivative of the velocity is the acceleration, and we write it as the second derivative of position:
You are supposed to know the derivatives of simple functions like x(t) = , as well as derivatives of sines, cosines, logarithms, and
exponentials. If you don’t know them, you should fix that weakness before proceeding.
1.4 Motion at constant acceleration
We are now going to focus on problems in which the acceleration a(t) is just a constant denoted by a, with no time argument. This is not the most general motion, but a very relevant one. When things fall near the surface of the earth, they all have the same acceleration, a = −9.8 ms−2 = −g. If I tell you that a particle has a constant acceleration a, can you tell me what the position x(t) is? Your job is to guess a function x(t) whose second derivative is a. This is called integration, which is the opposite of differentiation. Integration is not an algorithmic process like differentiation, though it is governed by many rules that allow us to map a given problem into others with a known solution. If I give you a function, you know how to take the

derivative: change the independent variable, find the change in the function, divide by the change in the independent variable, take the ratio as all changes approach zero. The opposite has to be done here. The way we do that is we guess, and such guessing has been going on for three hundred years, and we have become very good at it. The successful guesses are published as the Table of Integrals. I have a copy of such a table at home, at work, and even in my car in case there is a breakdown.
So, let me guess aloud. I want to find a function that reduces to the number a when I take two derivatives. I know that each time I take a derivative, I lose a power of t. In the end, I don’t want any powers of t. It’s clear I have to start with a function that looks like t2. Well, unfortunately, we know t2 is not the right answer, because the second derivative is 2, while I want to get a. So I multiply the original guess by and I know will have a second derivative a.
This certainly describes a particle with an acceleration a. But is this the most general answer? You all know that it is just one of many: for example, I can add to this answer some number, say 96, and the answer will still have the property that if you take two derivatives, you get the same acceleration. Now 96 is a typical constant, so I’m going to give the name c to that constant. We know from basic calculus that in finding a function with a given derivative, you can always add a constant to any one answer to get another answer. But if you only fix the second derivative, you can also add anything with one power of t in it, because the extra part will get wiped out when you take two derivatives. If you fixed only the third derivative of the function, you can also add something quadratic in t without changing the outcome.
So the most general expression for the position of a particle with constant acceleration a is
where b, like c, is a constant that can be anything. Remember that x(t) in the figure describes a particle going from side to
side. I can also describe a particle going up and down. If I do that, I would like to call the vertical coordinate y(t). You have to realize that in calculus,

the symbols that you call x and y are arbitrary. If you know the second derivative of y to be a, then the answer is
Let me go back now to Eqn. 1.5. It is true, mathematically, you can add bt + c as we did, but you have to ask yourself, “What am I doing as a physicist when I add these two terms?” What am I supposed to do with b and c? What value should I pick? Simply knowing that the particle has an acceleration a is not enough to tell you where the particle will be. Take the case of a particle falling under gravity with acceleration −g. Then
The formula describes every object falling under gravity, and each has its own history. What’s different between one object and another object is the initial height, y(0)≡y0, and the initial velocity v(0)≡v0. That’s what these numbers b and c are going to tell us. To find c in Eqn. 1.7 put time t = 0 on the right and the initial height of y0 on the left:
which tells us c is just the initial coordinate. Feeding this into Eqn. 1.7 we obtain
To use the information on the initial velocity, let us first find the velocity associated with this trajectory:
and compare both sides at t = 0

Thus b is the initial velocity. Trading b and c for v0 and y0, which makes their physical significance more transparent, we now write
Likewise for the trajectory x(t) when the acceleration is some constant a, the answer with specific initial position x0 and initial velocity v0 is
In every situation where the body has an acceleration a, the location has to have this form. So when I throw a candy and you catch it, you are mentally estimating the initial position and velocity and computing the trajectory and intercepting it with your hands. (The candy moves in three spatial dimensions, but the idea is the same.)
Now, there is one other celebrated formula that relates v(t), the final velocity at some time, to the initial velocity v0 and the distance traveled, with no reference to time. The trick is to eliminate time from Eqn. 1.13. Let us rewrite it as
Upon taking the time-derivative of both sides we get
which may be solved for t:
Feeding this into Eqn. 1.14 we find

which is usually written as
where v and x are assumed to be the values at some common generic time t.
1.5 Sample problem
We will work through one standard problem to convince ourselves that we know how to apply these formulas and predict the future given the present. Figure 1.2 shows a building of height y0 = 15m.
I am going to throw a rock with an initial velocity v0 = 10m/s from the top. Notice I am measuring y from the ground. The rock is going to go up to point T and come down as shown in Figure 1.2. You can ask me any question you want about this rock, and I can give the answer. You can ask me where it will be 9 seconds from now, how fast will it be moving 8 seconds from now, and so on. All I need are the two initial conditions y0 and v0 that are given. To make life simple, I will use a = −g = −10ms−2. The position y(t) is known for all future times:
Of course, you must be a little careful when you use this result. Say you put t equal to 10,000 years. What are you going to get? You’re going to find y is some huge negative number. That reasoning is flawed because you cannot use the formula once the rock hits the ground and the fundamental premise that a = −10ms−2 becomes invalid. Now, if you had dug a hole of depth d where the rock was going to land, y could go down to −d. The moral is that when applying a formula, you must bear in mind the terms under which it was derived.

Figure 1.2 From the top of a building of height y0 = 15m, I throw a rock with an initial upward velocity of v0 = 10m/s. The dotted line represents the trajectory continued back to earlier times.
If you want to know the velocity at any time t, just take the derivative of Eqn. 1.20:
Let me pick a few more trivial questions. What is the height ymax of the turning point T in the figure? Eqn. 1.20 tells you y if you know t, but we don’t know the time t∗ when it turns around. So you have to put in something else that you know, which is that the highest point occurs when it’s neither going up nor coming down. So at the highest point v(t∗) = 0. From Eqn. 1.21
So we know that it will go up for one second and then turn around and come back. Now we can find ymax:
When does it hit the ground? That is the same as asking when y = 0, which is our origin. When y = 0,

The solutions to this quadratic equation are
Why is it giving me a second solution? Can t be negative? First of all, negative times should not bother anybody; t = 0 is when I set the clock to zero, and I measured time forward, but yesterday would be t = −1 day, right? So we don’t have any trouble with negative time; it is like the year 300 BC. The point is that this equation does not know that I went to a building and launched a rock or anything. What does it know? It knows that this particle had a height of y = 15 m and velocity v = 10 m/s at time t = 0, and it is falling under gravity with an acceleration of −10 ms−2.That’s all it knows. If that’s all it knows, then in that scenario there is no building or anything else; it continues a trajectory both forward in time and backward in time, and it says that one second before I set my clock to 0, this particle would have been on the ground. What it means is that if you had released a rock at y = 0 one second before I did with a certain speed that we can calculate (v(−1) = 20m/s from Eqn. 1.21), your rock would have ended up at the top of the building when I began my experiment, with the same height y = 15m, and velocity v0 = 10 m/s. So sometimes the extra solution is very interesting, and you should always listen to the mathematics when you get extra solutions.
When Paul Dirac was looking for the energy of a particle in relativistic quantum mechanics, he found the energy E was connected to its momentum p, mass m, and velocity of light, c, by
in accord with a relation we will encounter in relativity. Now, this quadratic equation has two solutions:
You may be tempted to keep the plus sign because you know energy is not going to be negative. The particle’s moving, it’s got some energy and that’s it. This is correct in classical mechanics, but in quantum mechanics

the mathematicians told Dirac, “You cannot ignore the negative energy solution in quantum theory; the mathematics tells you it is there.” It turns out the second solution, with negative energy, was telling us that if there are particles, then there must be anti-particles, and the negative energy particles, when properly interpreted, describe anti-particles of positive energy.
So the equations are very smart. When you find some laws in mathematical form, you have to follow the mathematical consequences; you have no choice. Here was Dirac, who was not looking for anti-particles. He was trying to describe electrons, but the theory said there are two roots to the quadratic equation and the second root is mathematically as significant as the first one. In trying to accommodate and interpret it, Dirac was led to the positron, the electron’s anti-particle.
Returning to our problem, if you were only asking for the maximum height ymax, and not the time t∗ when it got there, there is a shortcut using
Using v = 0, v0 = 10m/s, and a = −10ms−2 we find
—that is, the rock reached a maximum height of 20m from the ground. You can find the speed when it hits the ground (y = 0) using

The root we should take for when it hits the ground is of course −20m/s. As mentioned earlier, the other root +20m/s is the speed with which it should have been launched upward, from y = 0 at t = −1, to follow the dotted trajectory in the figure.

1.6 Deriving v2 − v20 = 2a(x − x0) using calculus

I want to derive Eqn. 1.19, illustrates the judicious use of calculus.

, in another way that

Start with

and multiply both sides by v and write

in the right-hand side:

Now I’m going to do something that is viewed with suspicion, which is just to cancel the dt on both sides. Although I agree that you’re not supposed to cancel that d in , canceling the dt on both sides gives valid results if interpreted carefully. Doing so here gives us
This equation tells us that in an infinitesimal time interval [t, t + dt], the variables v and x change by dv and dx, and these changes are related as above in the limit dx, dv, dt → 0. Now the limit of dx → 0 or dv → 0 (as compared to their ratio) is of course trivial, and Eqn. 1.33 reduces to 0 = 0. However, the way we interpret and use Eqn. 1.33 is as follows. Suppose in the finite time interval [t1, t2], the variable v changes from v1 to v2, and x changes from x1 to x2. Let us divide the interval [t1, t2] into a very large number N of equal sub-intervals of width dt, and let dx and dv be the changes in x and v in the interval [t, t + dt]. The relation between these changes is given in Eqn. 1.33. If we sum up the N changes on both sides of Eqn. 1.33 as N → ∞, the sums converge to nontrivial limits, namely the corresponding integrals:

Thus it must be understood that the two sides of a relation like Eqn. 1.33 are to be ultimately integrated between some limits to obtain a useful

equality. Eqn. 1.19 follows upon setting
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CHAPTER 2
Motion in Higher Dimensions
2.1 Review
In the last chapter we took the simplest case, of a point particle moving along the x-axis with a constant acceleration a. What is the fate of this particle? The answer is that at any time t, the location of the particle is given by
where x0 and v0 are its initial position and velocity. If you took the derivative of this, you would get
You can easily check, by taking one more derivative, that this particle does indeed have a constant acceleration a. This equation, which gives the velocity of the object at time t, in terms of its initial velocity and acceleration can be inverted to give t in terms of v:
Feeding this into Eqn. 2.1 we obtain the result that makes no reference to time:
It is understood v and x correspond to some common time. I showed you in the end how we can use calculus to derive this result. It
is important to brush up on your calculus. When a student says, “I know

calculus,” sometimes that means the student knows it, and sometimes that means he or she once met someone who did. One solution for that is to get a copy of a textbook I wrote called Basic Training in Mathematics. This is a little awkward; I don’t want to foist my book on you. On the other hand, I don’t want to withhold relevant information. If you’re going into any science that uses mathematics—chemistry, engineering, or even economics —you should find the contents of that book useful. Don’t wait for the movie: it is not coming.
2.2 Vectors in d = 2
The next difficult thing is to consider motion in higher dimensions. Everything moves around in d = 3. However, I’m going to use only two dimensions for most of the time. Whereas the difference between one dimension and two is very great, that between two and higher dimensions is not. Later we will encounter a few concepts that make sense in d = 3 but not d < 3. String theorists will tell you that actually we need 9 spatial dimensions plus time to describe superstrings, which will be discussed in depth in Chapter 3,498 of this book.
Picture some particle that’s traveling in the x −y plane as shown in Figure 2.1. This is not an x versus t plot or a y versus t plot. It’s the actual path the particle traces out on the x − y plane. You might say “Where is time?” One way to mark time is to imagine the particle carries a clock with it, and put markers every second. Four representative markers at t = 1, 2, 33, and 34 are shown. It obviously is going much slower between 33 and 34 than between 1 and 2.
The kinematics of this particle requires a pair of numbers x and y. It’s more convenient to lump these into a single entity, called a vector. The simplest context in which one can motivate a vector and the rules for dealing with vectors is to look at movements in the plane. Let’s imagine that when I went camping I walked for 5 km from the base camp on the first day and another 5 km on the second day. How far am I from the base camp? You cannot answer that, even if I promised to move only along the x-axis. It’s not enough to say I went 5 km. I have to tell you whether I went to the right or to the left. So I could be 10 km, 0 km, or −10 km from base. If I say not just that I walked 5 km, but specified whether it was ±5 km, that takes care of all ambiguity in one dimension.

Figure 2.1 Path of a particle in d = 2. Equal intervals in time are indicated by markers on the path numbered 1, 2, . . . , 33, and 34.
But in d = 2 the options are not just left and right, but an infinity of possible directions. For example, on the first day I could leave the base camp at the origin and move along the arrow labeled A to arrive at the point labeled 1 in Figure 2.2(a). The second hike is described by the arrow B, which starts where A ended and brings me to 2. These two arrows are examples of vectors and I use them here for describing displacement, or changes in position. Vectors can be used to describe many other physical quantities, as we will see.
Figure 2.2 Adding vectors. Part (a) shows how to add vectors and that A + B = B + A. Part (b) illustrates the meaning of multiplying a vector by a number (2 in this example) and the null vector 0.

A vector is an arrow that has got a beginning and an end. This is why one says a vector has a magnitude and a direction. The magnitude is how long it is, and direction is its angle relative to some fixed direction, usually the xaxis. When you refer to a vector A in your notes, you’re supposed to put a little arrow on top like this: A. In textbooks, vectors are in boldface: A. If you don’t put an arrow on top or do not use boldface, you’re talking about just a number A. When applied to a vector A, A stands for its length.
From Figure 2.2(a), we see that there is a very natural quantity that you can call A + B. One day I moved by A and on the next by B. If I want to do it all in one shot, what is the equivalent step I should take from the start? It’s obvious that the bottom line of my two-day trip is this object C. We will call that A + B. It does represent the sum, in the same sense that if I gave you 4 bucks and then I gave you 5 bucks, you have the equivalent of a single payment of 9 bucks. Here, we are not talking about a single number, but a displacement in the plane, and C indeed represents an effective displacement due to A and B.
So here is the rule for adding two vectors that comes from a study of displacements: you draw the first one and at the end of that first one, you begin the second one, and their sum starts at the beginning of the first and ends at the end of the second.
You can verify, as illustrated in the figure, that A + B is the same as B + A where you first draw B and from where B ends you draw A. You will end up with the same point, 2, as shown by the sum of the dotted arrows.
The next thing I want to do is to define the vector that plays the role of the number 0, which has the property that when you add it to any number, it gives the same number. The vector 0 that I want to call the zero or null vector should have the property that when I add it to any vector, I should get the same vector. So you can guess who it is: a vector of no length. I cannot show you the 0. If you can see it, I’m doing something wrong.
Look at part (b) of Figure 2.2. What if I draw A, then I add to it another A to get A + A. You have to agree that if there’s any vector that deserves to be called 2A, it is this guy, A, stretched to twice its length. Now we have discovered a notion of multiplying a vector by a number. If you multiply it by 2, you get a vector twice as long and in the same direction. Then we’re able to generalize that and say, if you multiply it by 2.6, you get a vector 2.6 times as long. So multiplying a vector by a positive number means to stretch it (or shrink it) by that factor.

Let us keep going. I want to think of a vector that I can call −A. What do I expect of −A? I expect that if I add −A to A, I should get 0, which plays the role of 0 among vectors. What should I add to A so I get the null vector? It’s clear that you want to add a vector that looks like −A in part (b) of Figure 2.2, because, if you go from the start of A to the finish of −A, you end up where you started and you get this invisible 0 vector. So the minus vector is the same vector flipped over, pointing the opposite way. That’s like −1 times a vector. Once you have got that, you can do −7.3 times a vector: just take the vector, rescale it by 7.3 and flip it over. Multiplying a vector by a number is called scalar multiplication, and ordinary numbers are called scalars. You can do more complicated things. You can take one vector, multiply it by one scalar, take another vector, multiply that by another scalar, and add the two of them. We know what all those operations mean now. You don’t have to memorize the rules for all this. The only rule is: “Do what comes naturally.” Do what you normally do with ordinary numbers.
2.3 Unit vectors
Let us go back to the same x − y plane. I’m going to introduce two very special vectors. They are the unit vectors: i and j, pointing along the x and y axes and of unit length, as shown in Figure 2.3. If I had a third axis perpendicular to the page, I would draw a k, but we don’t need that yet. I claim I can write any vector you give me as a number Ax times i, plus a number Ay times j. There’s nothing you can throw at me that lies in the plane that I cannot describe as some multiple of i plus some multiple of j. It’s intuitively clear, but I will just prove it beyond any doubt. Here is some vector A. It is clear from the figure that it is the sum of the dotted horizontal vector and the dotted vertical vector, by the rules of vector addition. The horizontal part, parallel to i, has to be a multiple of i. We know that because we can stretch i by whatever factor we like. Call that factor Ax, which happens to be positive in this example. It is called the x-component of A or the projection of A along i or along the x-axis. The vertical part is likewise jAy where Ay is the y-component of A, or the projection of A along j or along the y-axis. Therefore I have managed to write A as

Figure 2.3 The unit vectors i and j and an arbitrary vector A = iAx + jAy built out of them.
We refer to the pair i and j, in terms of which any vector can be expressed, as basis vectors or as the basis.
If you gave me a particular vector A as an arrow of some length A and orientation θ relative to the x-axis, what do I use for Ax and Ay? You can see from trigonometry that
Conversely, given the components, the length and angle are
Eqns. 2.6 to 2.9 will be invoked often. So please commit them to memory. If you give me a pair of numbers, (Ax,Ay), that’s as good as giving me this
arrow, because I can find the length of the arrow by Pythagoras’s theorem and I can find the orientation from tanθ = . You have the option of either

working with the two components of A or with the arrow. In practice, most of the time we work with these two numbers, (Ax,Ay). In particular, if we are describing a particle whose location is the position vector r, then we write it in terms of its components as
The changes in r are the displacement vectors and examples are A and B in Figure 2.2 that described the two hikes.
I have not given you any other example of vectors besides the displacement vector, but at the moment, we’ll define a vector to be any object that looks like some multiple of i plus some multiple of j. If I tell you to add two vectors A and B, you have got two options. You can draw the arrow corresponding to A and attach to its end an arrow corresponding to B, and then add them, as in Figure 2.2. But you can also do the bookkeeping without drawing any pictures as follows:
so that the sum C is the vector with components (Ax + Bx,Ay + By). In the above, I have used the fact that vectors can be added in any order.
So I grouped the things involving just i and likewise j. Then I argued that since iAx and iBx are vectors along i, their sum is a vector of length Ax + Bx also along i. I did the same for j.
In summary if
then
which can be summarized as follows: To add two vectors, add their respective components.

An important result is that A = B is possible only if Ax = Bx and Ay = By. You cannot have two vectors equal without having exactly the same x component and exactly the same y component. If two arrows are equal, one cannot be longer in the x direction and correspondingly shorter in the y direction. Everything has to match completely. The vector equation A = B is actually a shorthand for two equations: Ax = Bx and Ay = By.

2.4 Choice of axes and basis vectors

I have in mind a vector whose components are 3 and 5. Can you draw the

vector for me? If you immediately said, “It is 3i + 5j,” you’re making the

assumption that I am writing the vector in terms of i and j. I agree i and j

point along two natural directions. For most of us, given that the blackboard

or notebook is oriented this way, it is very natural to line up our axes with

it. But there is no reason why somebody else couldn’t come along and say,

“I want to use a different set of axes. The x and y axes or i and j are not

nailed in absolute space. They are human constructs and we’re not wedded

to any of them.”

Quite often, it’s natural to pick the axes in a certain way to suit the

problem. If you are studying a cannon ball launched from the earth, it

makes sense to pick the horizontal as the x-axis and the vertical as the y-

axis, but, mathematically, you don’t have to. Another set of rotated but

mutually perpendicular unit vectors i′ and j′ that form another basis can also

be rescaled and added to form any given vector A in the plane. For

example, when we study objects sliding down an inclined plane, we will

choose our axes parallel and perpendicular to the incline.

If I draw an arrow A on a blank sheet of paper, it has life of its own

without reference to any axes. The same vector A can be written either in

terms of i and j, which is the old basis, or in terms of i′ and j′, the new basis.

How do the components

in the new basis relate to the components of

the old basis? It’s a simple problem, but I just want to do it so you get used

to working with vectors.

For this we need the very busy Figure 2.4. It shows the old x and y axes

and the x′ and y′ axes obtained by rotating the x −y axes counterclockwise

by an angle ϕ. The unit vectors i′ and j′ are likewise rotated versions of i

and j. The components in the two bases are shown by dotted lines and are

simply the projections of A along the various axes. We want to relate to (Ax,Ay).

Figure 2.4 The same vector A is written as iAx + jAy in one frame and as The dotted lines indicate the components in the two frames.

in the other.

First we express i′ and j′ in terms of i and j using the figure:

Here are the details. The vector i′ has got a horizontal part, which is its

length, namely, 1, times cos ϕ, and a vertical part that is 1 times sin ϕ. How

about j′? It is at an angle ϕ relative to j. So its y component is cos ϕ.

Finally, its x or horizontal component is (−sinϕ), where the minus sign

comes because it is pointing to the left, along the negative x-axis. All that

remains now is to eliminate i′ and j′ in favor of i and j in

and

equate it to A written in terms of i and j:

When we equate the coefficients of i and j on the right-hand sides of Eqns. 2.20 and 2.21, we obtain the desired expression for Ax and Ay in terms of and :
So, you can pick your basis vectors any way you like and so can I. Your basis is obtained from mine by a counterclockwise rotation by an angle ϕ. The same entity A, the same arrow which has an existence of its own, independent of axes, can be described by you and by me using different components. Your components with primes on them are related to mine by Eqns. 2.22 and 2.23. This is called the transformation law for the vector components under rotation of basis vectors.
Now, you can ask the opposite question. How do I get and in terms of Ax and Ay? The quickest way is to replace ϕ by −ϕ: if we go from the unprimed to the primed system by a rotation ϕ, then rotation by −ϕ is the way to go from the primed to the unprimed basis. The result, using cos(−ϕ) = cosϕ and sin(−ϕ) = −sinϕ, is
That turns out to be the correct answer. But I want you to think about another way to show this, which often seems to bother some students. If I told you

you certainly know how to solve for x and y, right? You have got to juggle the two equations, multiply the first by 6, the second by 5, and subtract to isolate x and so on. Why is it when some of you see Eqns. 2.22 and 2.23, you don’t realize it’s the same kind of problem, where you can multiply Eqn. 2.22 by cosϕ, Eqn. 2.23 by sinϕ and add to isolate for example? For any particular value of ϕ, sinϕ and cosϕ are just some numbers. For example, if I pick ϕ = = 60°, cosϕ = and sinϕ = The equations become (for this angle)
If you multiply the second by 3 and add it to the first you obtain
in accordance with Eqn. 2.24. So go forth and treat sinϕ and cosϕ as plain numbers and juggle Eqns. 2.22 and 2.23 to derive Eqns. 2.24 and 2.25. Along the way of course you will have to use identities like sin2ϕ + cos2ϕ = 1.
The components of the vector depend on who is looking at the vector. However, there’s one quantity that’s going to come out the same, no matter who is looking at the vector. It is the length of the vector. It is unaffected by the rotation of axes. It is an invariant under rotations. You may verify from Eqns. 2.24 and 2.25 that

The AxAy term is gone since its coefficient is 2(cosϕ sinϕ − sinϕ cosϕ). I want to conclude with one important point. We learned that a vector is a
quantity that has a magnitude and a direction. A more advanced view of vectors is that they are a pair of numbers (in d = 2) which, under rotation of axes, transform as per Eqns. 2.24 and 2.25. Anything that transforms this way is called a vector. We already know about the position vector r and the changes in it, the displacement vectors (used in describing the hike). How about more vectors? There turns out to be a very nice way to produce vectors, given one vector like the position vector. And that’s the following.
2.5 Derivatives of the position vector r
Let’s take a particle in the x −y plane that moves from r at time t to r + Δr at time t + Δt as in Figure 2.5. At time t its location is
and at t + Δt it is

Figure 2.5 The particle moving along some curved path goes from r at time t to r + Δr at time t + Δt. The velocity v is the limit of the ratio as Δt → 0, and thus parallel to Δr, which eventually becomes tangent to the curve.
When you move just along the x-axis, you wait a small time Δt and you move by an amount Δx, and their ratio gives the velocity in the appropriate limit. When you move in the plane, your position and its change are both vectors.
Can you see why the derivative of a vector is also a vector? Because Δr, the difference in the vector between two times, is itself a vector. Dividing it by Δt is like multiplying by 1/Δt, but we know that when we multiply a vector by a number, we simply rescale the vector. So the limit will be some arrow that we call the instantaneous velocity vector. It will be tangential to the curve r(t) and point toward the instantaneous direction of travel.
If I gave you the location of a particle as a function of time, you can find its velocity by taking derivatives. For example, if I say a particle’s location is
then its velocity at time t is

You can take a derivative of the velocity or the second derivative of the position to get the acceleration vector
You can then also multiply a by the mass m, which is a scalar unaffected by rotations, to get a vector ma, which Newton’s law equates to another vector, the force F.
Even though we started with one example of a vector r, we’re now finding out that its derivative has to be a vector and the derivative of the derivative is also a vector. When you learn relativity, you will find out there’s again one vector that’s staring at you, the analog of the position vector, but with four components. But more vectors can be manufactured by multiplying vectors by scalars (like mass) or taking derivatives with respect to a parameter that plays the role analogous to time.
Here is an illustration of vector addition and differentiation. Imagine an airplane in flight, as depicted in Figure 2.6. Let rpg be the location of a fixed point in the airplane, say the tail, with respect to a fixed point on the ground. Imagine that in the airplane there is a ball located at rbp as measured from this fixed point in the airplane. By vector addition the location of the ball with respect to the ground is

Figure 2.6 The position of the ball relative to (some origin on) the ground rbg is the vector sum of the position of the ball relative to the (tail of the) plane, rbp, and the position of (the tail of) the plane rpg relative to the ground.
Upon taking a time derivative and in the same notation, the law of composition of velocities follows:
which says the velocity of the ball as seen by a person on the ground is the velocity of the ball relative to the airplane plus the velocity of the airplane relative to the ground. Taking yet another derivative we may relate the accelerations:
In the special case of an airplane moving at constant velocity, apg = 0. Then we find

which means, in this case, the acceleration of the ball is the same as measured by an observer on the ground and an observer on the airplane. These results will be recalled in our study of relativity.
2.6 Application to circular motion
Now we’ll take a concrete problem where you will see how to take derivatives to obtain very useful results. I’m going to write a particular case of r(t):
where R and ω are constants. What is going on as a function of time? What’s this particle doing? Look at the length squared of this vector:
That means the particle is going around in a circle of radius R as shown in Figure 2.7. The x component is Rcosωt and the y component is Rsinωt where ω is a fixed number. As t increases, this angle ωt increases and the particle goes round and round. Let’s get a feeling for ω. As time increases, the angle increases and we can ask how long it will take the particle to come back to the starting point. Suppose the starting point was on the xaxis. As t increases, ωt increases, and the particle will come back at a time T such that

Figure 2.7 The particle moves along a circle of radius R with an angular velocity ω.
Thus ω is related to the time period T by

where

is the frequency or number of cycles per second. It is measured

in Hz, which stands for Hertz. Since in every cycle the particle rotates by

2π, and it completes f revolutions per second, ω = 2π f is called the angular

velocity and measures the radians swept out per second.

Notice that in equating a full cycle to 2π I am using radians and not degrees to measure angles. For those who have not seen a radian, it’s just another way to measure angles, wherein a full circle, which we used to think was worth 360°, now equals 2π radians. Since 2π ≃ 6.3, a radian is roughly 60°. You will see the advantages of using radians later. For now just remember that a half circle, instead of being 180°, will now be π radians, and a quarter circle will be , and so forth.
How fast is this particle moving? It’s going around a circle, the angle is increasing at a steady rate ω, and so we know it’s going at a steady speed. Let us verify that by computing the velocity
At t = 0, the velocity is v = Rωcos0 j = Rωj, so it is moving straight up at speed v = ωR. You may verify that it has the velocity as shown in the figure at later times. The magnitude of the velocity is always ωR although the direction is changing. From the figure we see it remains tangential to the circle. The constancy of the speed v at an arbitrary time may also be established by computing
Remember the tangential velocity is v = ωR. Let’s take the derivative of the derivative to find the acceleration a and its
magnitude:
That’s a very important result. It tells you that when a particle moves in a circle of radius R at constant speed v, it has an acceleration, called the

centripetal acceleration, directed toward the center and of magnitude

This acceleration at constant speed reflects the fact that velocity is a vector and you can change the velocity vector by changing its direction. For example, if a car is going on a racetrack and the speedometer says 60 miles per hour, the lay person’s view is that the car is not accelerating. But you

will say from now on that it indeed has an acceleration equal to even

though no one’s stepping on the accelerator or the brake. Suppose the particle is not moving fully around a circle but traversing
just a quarter of the circle. When it is traveling the quarter of a circle, it has the same acceleration directed toward the center of that quarter circle. In other words, you don’t have to be moving actually in a circle to have the

acceleration . At any instant, the curve you are following can be locally

approximated as part of some circle, and, in the formula

, the

acceleration is directed toward the center of that circle, R is its radius and v the instantaneous tangential velocity.

2.7 Projectile motion
I want to consider a particle for which r0 and v0 are the position and velocity at t = 0 and which has a constant vector acceleration a. What is its location at all future times? By analogy with what I did in one dimension

Once you know r0 and v0, you can find the position of the object at all future times. Let’s take one simple example. Somebody in a car has decided to drive off a cliff as shown in Figure 2.8(a). We want to know when and where the car hits the ground.
We pick our origin (0,0) at the foot of the cliff. Let the height of the cliff be h. The car is traveling with some initial speed v0x in the horizontal

direction. Equation 2.59 is really a pair of equations, one along x and one along y with a = −jg, v0 = v0xi, and r0 = hj. Separating out the components
Notice that the evolution of the two coordinates is completely independent. The time t∗ when the car hits the ground (y = 0) satisfies the equation

Figure 2.8 (a) A car flies off the cliff at (0,h) and lands at (d,0). (b) A projectile is launched with

initial velocity

The range is R = 0.35 m and the maximum height reached is

ymax = 0.15 m.

This is exactly how long it would take to hit the ground had it simply toppled over the edge from rest. The horizontal velocity does not delay the crash one bit (unless you take into account the curvature of the earth). As to
where the car lands, the location is given by (x(t∗),0) = (d,0) where

Finally the problem of projectile motion is depicted in Figure 2.8(b). You fire a projectile from (0,0) with some velocity v0 at some angle θ. It will go up and then come down, moving horizontally at the same time. Where is it going to land? What is the maximum height ymax to which it rises? With what speed will it hit the ground? At what angle should you fire your projectile so it will go the furthest?
Here are the equations that contain all the answers, namely Eqn. 2.59 written out in component form:
You can solve them but it is good to have an idea of what’s coming. Imagine you have this monster cannon to fire things. It has a fixed muzzle speed, v0, but allows you to fire at any angle. How do you aim it so the ball goes as far as possible? There are two schools of thought. One says, aim at your enemy and fire horizontally. Then the ball lands on your foot because it has zero time of flight (assuming the cannon is at zero height). The other school says, maximize the time of flight and point the cannon vertically. It goes up, stays in the air for a very long time, and lands on your head. Then it hits you: the correct answer is somewhere between 0 and 90° = . The naive guess 45° = turns out to be correct.
Now I want to show you how to use the equations to prove this. What’s the strategy for finding the range? You see how long the ball is in the air and multiply that by the constant horizontal velocity v0 cosθ. Again, let t∗ be the time when it hits the ground. The y-equation
has two solutions:

So the cannon ball is on the ground on two occasions. One is initially. We are not interested in that trivial solution. If the time you are interested in is t∗ ≠ 0, you’re allowed to divide both sides of Eqn. 2.67 by it and get

and the range

using sin 2θ = 2 sinθ cosθ. For the greatest range we must make sin2θ as

large as possible, which occurs for

. For any smaller

range you will find there are two possible angles that work since sin(2θ) =

sin(π − 2θ).

The maximum height is the y-coordinate at the halfway time

:

We could equally well find the half-time t∗ by setting the vertical velocity to zero:
How about the velocity at impact? The horizontal part is of course v0 cosθ since there is no acceleration in that direction. The vertical part starts out at v0 sinθ and decreases at a rate g:

so that at t∗ it has a value
which is just the opposite of the initial vertical speed. The magnitude of the final velocity is the same as the initial one since reversing one of the components of v does not change its length.
There are endless variations. You pick some point off the ground at (X,Y) and want the projectile to arrive there. You are given the launch angle θ and have to find the launch speed v0. How do you do that problem? You assume the projectile arrives at the destination (X,Y) at some time t∗. You go to the x equation and demand that x(t∗) = X and solve for t∗. Plug this time into y and demand y(t∗) = Y and solve for v0.
Sometimes the problems are embellished to make everyone feel involved. For example, instead of a ball dropping, nowadays there’s a monkey or horse that is falling down, so people in life sciences can say, “Hey, we should learn physics since it seems to have applications to our subject.” All those gyrating creatures are very interesting and look great in color, but, in the end, you are told, “Treat the horse as a point particle.” If you’re going to treat the horse as a point particle, why include its color picture? But I agree there are times when only a horse will do. When the Godfather wants to get the contract for Johnny Fontaine, he doesn’t tell his consigliere, “Hey, Tom, put half a point particle on Jack’s bed.” That would have been a disastrous approximation.
By the way, don’t forget to treat the falling car in Figure 2.8 as a point particle.
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CHAPTER 3
Newton’s Laws I
3.1 Introduction to Newton’s laws of motion
This is a big day in your life: you are going to learn Newton’s laws, in terms of which you can understand and explain a very large number of phenomena. It’s really amazing that so much information can be condensed into three laws.
Your reaction may be that you have already seen Newton’s laws, that you have applied them in school. I realized fairly late in life that they are more subtle than I first imagined. It’s one thing to plug in all the numbers and say, “I know Newton’s laws and I know how they work.” But as you get older and you have more spare time, you think about what you are doing. This is something I have had the luxury of doing, and I have realized the laws are more tricky. I want to share some of that understanding.
The first law, or the law of inertia, says, “If a body has no forces acting on it, it will maintain its velocity.” In other words, in the absence of external forces, a body at rest will remain at rest and a body in motion will retain its velocity. It is not surprising that a body at rest will remain so if not acted upon by a force. We see that all the time. I place an eraser on the table. It will stay put unless I do something to it.
The great discovery that Galileo and Newton made was that you don’t need a force for a body to move at constant velocity. You don’t see that in daily life—everything seems to come to rest unless you keep pushing or pulling it. But we all know that the reason things come to a halt is that there is always some friction or drag bringing them to rest. If you take a hockey puck on an air cushion and give it a push, it seems it can travel for a very long time. Galileo and Newton abstracted from this an ideal situation in which friction was totally eliminated and the bodies kept moving forever with no help. If you go to outer space, you can check for yourself that if you throw something, it will go on forever without your intervention. It’s in the

nature of things to retain a constant velocity. It is not velocity, but a change in velocity, that calls for a force.
The law of inertia is not valid for everybody. I’ll give an example from your own life. You go on an airplane and then, after the usual delays, the plane begins to accelerate down the runway. At that time, if you leave anything on the floor, you know it’s no longer yours. It’s going to slide backward and the physicist in the last row is going to collect everything. That is an example of a frame in which the law of inertia does not work: bodies accelerate with no applied force. Once the plane stops accelerating, the law of inertia becomes operative. It fails when it decelerates during landing when everything now slides to the front.
If Newton’s law of inertia works for you, you are called an inertial observer and your frame of reference is called an inertial frame. The plane that’s taking off is not an inertial frame, but the one that is cruising is. The earth seems to be a pretty good inertial frame, because if you leave something at some place, it just stays there—unless the thing is your cell phone and the place is Grand Central Station. But this is not a violation of Newton’s laws, just the laws of New York City.
Although not every frame is inertial, there are plenty of inertial frames to go around. If you find even one inertial observer, namely one person for whom this law of inertia works, I can find for you an infinite number of other people for whom this is true. Who are these people? They are people moving at constant velocity with respect to the first inertial observer. Suppose you are in a train and you’re moving past me with velocity u, and we both look at some object with no forces on it. We will not agree on its velocity or the velocity of anything: things at rest for me will be moving backward for you at velocity −u, and everything at rest in your train will be moving at velocity u according to me. In short, you and I will differ on the velocity of any object by our relative velocity. But we will agree on the acceleration of any object since it is unaffected by adding a constant to velocity. Adding a constant velocity to objects does not change the fact that those which were maintaining constant velocity still maintain a constant (but different) velocity. Thus neither is every observer inertial, nor is an inertial observer unique.
You must know the earth is not precisely inertial. The earth has an acceleration. Can you see why? Yes, it’s spinning around itself and going around the sun, both of which constitute accelerated motion. But the

acceleration due to motion around the sun at speed v and radius r is a = = .006 ms−2, which is a very small number, say compared to g. The same goes for the acceleration due to the earth’s rotation about its own axis, which is roughly .03ms−2 or roughly g/300 at the equator.
The first law might seem tautological because we never see anything that retains its velocity forever, and every time we see velocity change we say that a force is acting. But it’s not a big hoax, because you can set up experiments in free space, far from everything, where objects will, in fact, maintain their velocity forever. It’s a useful concept even on the earth, because the earth is approximately inertial.
3.2 Newton’s second law
Newton’s second law says, “If a body has an acceleration a, then you need a force
to produce that acceleration.” In this chapter we will focus on one dimension and write
where F and a are along the x-axis. A few words about units. Acceleration is measured in ms−2. Mass is
measured in kilograms or kg. So force has units kilogram meters per second squared. But we get tired of saying that long expression, so we call that a Newton, denoted by N. If you had invented mechanics, we’d be calling it by your name, but it is too late for you now.
Here is a typical problem that you may have solved in your first pass at Newton’s laws. A force of 36N is acting on a mass 4 kg. What’s the acceleration? You divide 36 by 4 and you find it is a = 9 ms−2. You say, “Okay, I know Newton’s laws.”
It’s actually more complicated than that. Take yourself back to the seventeenth century, when Newton was inventing these laws. You have an intuitive definition of force: when somebody pushes or pulls an object we say a force is acting on it. Suddenly, you are told there is a law F = ma. Are

you better off in any way? Can you do anything with this law? What does it

help you predict? Can you even tell if it’s true? Here’s a body that’s moving. Is Newton right? How are we going to check? Well, you want to

measure the left-hand side and you want to measure the right-hand side. If

they’re equal, you will say the law is working. What can you measure in this equation?

Let’s start with acceleration. What’s your plan for measuring

acceleration? What instruments will you need? If you say a watch and ruler, that would be correct provided by ruler you don’t mean Queen Elizabeth.

Here is your ruler and here is a Rolex. Tell me exactly how you plan to

measure acceleration. Everyone seems to know the answer. First, let it go a little distance, and take the distance over time. That gives you the velocity

now. Let it go a little more, and repeat the velocity measurement. Take the

difference of the two velocities and divide by the difference of the two times, and you have got the acceleration. Since the body has moved a finite

distance in a finite time, this gives the average acceleration. You want to

make these three positional measurements more and more rapidly. In the end, as all the time intervals shrink to 0, you will measure what you can say

is the acceleration now, the second derivative

defined in calculus.

Back to testing if what Newton told you is right: You see an object in motion, you measure a, and you get a certain numerical value, say 10 ms−2. But that’s not yet testing the equation, because you still have to find F and

m. What’s the mass of this object? One common idea is to take a standard

mass and balance the unknown mass on a seesaw by adjusting its position. But suppose you were in outer space. There’s no gravity. Then the seesaw

will balance even if you put a potato on one side and an elephant on the

other side. What you are doing now is appealing to the notion of mass as something that’s related to the pull of the earth on the object. You have got

to go back and wipe out everything you know. If F = ma is all you have,

there is no mention of the earth in these equations. You only know how to measure a, but not the other two. So you have a problem. You cannot say

that since F = ma, it follows that

; that is circular reasoning since you

have not told me how to measure F either.

Let me give you a hint. How do we decide how long a meter is? You

seem to know that it is arbitrary. A meter is not deduced from anything. Napoleon or somebody said, “The size of my ego is one meter.” That’s a

new unit of length. Seriously, at the National Bureau of Standards there

used to be a rod made of some special alloy in a glass case, and that defined the meter. There are fancier definitions now, but let us stick to this simple one. (See http://physics.nist.gov/cuu/Units/meter.html for more details on definitions of units.) Then I ask you, “What is two meters and what is three meters?” We have ways of handling that. You take the meter and attach it to a duplicate, and that’s two meters. You can cut it in half, using dividers and compasses; you can split the meter into any fractions you like. Likewise for mass, we will take a chunk of some material and we will call it a kilogram. That is a matter of convention, just like one second is some convention.
I’m going to give you a glass case that contains a block of some metal defined as one kilogram. Then I give you another object, an elephant. What’s the mass of the elephant? Here is a hint: I also give you a spring. We cannot do the seesaw experiment because it requires gravity. A spring, on the other hand, will exert a force even in outer space. Here’s what we do. We hook one end of the spring to a wall and we pull the other end from its equilibrium position by some amount and we attach the one-kilogram mass to it. We don’t know what force it exerts, but it will not matter. We let it go and measure the initial acceleration, a1. Then we bring the elephant (another point particle) of unknown mass mE, pull the spring by the same amount so it can exert the same force, and find the acceleration aE of the elephant. Assuming only that the same extension produces the same force in the two cases, we have
Once you have the mass of the elephant you can use it to measure any other mass mo by using
where aE and ao are produced by the same force. There are subtleties even here. For example, how do we know that when
we pull the spring the second time with the elephant, it will exert the same

force as the first time when the 1 kg was attached to it? After all, springs wear out. That’s why you change the shock absorbers in your car. So first, we have to make sure the spring exerts a fixed force every time (for a given extension). How are we going to check that? We don’t have the definition of force yet. But we can do the following. We pull the one-kilogram mass and let it go, and we note the acceleration. Then, we pull it again, by the same amount, and let it go; we do it ten times. If every time we get the same acceleration, we are convinced this is a reliable spring that is producing the same force under the same conditions. On the eleventh time we pull the spring and attach the elephant. With some degree of confidence, we can say we are applying the same force on the elephant as on the one-kilogram mass.
Why is this discussion so important? Because you need to know that everything you or I write down in the notebook or on the blackboard as a symbol is actually a measurable quantity, or, as they say in France, Les Mesurables. You should know at all times how you measure anything that enters your theory or calculation. If not, you are just doing math or playing with symbols. You are not doing physics.
This discussion also tells you that the mass of an object has nothing to do with gravitation but with how much it hates to accelerate in response to a force. Newton tells you forces cause acceleration. But the acceleration is not the same on different objects for a given force. Certain objects resist it more than others. They are said to have a bigger mass. We can be precise about how much bigger by saying, “If the acceleration of a body in response to a given force is that of a 1-kilogram mass, then the mass of
the body is 10 kilograms.”
3.3 Two halves of the second law
We have seen how all objects can be attributed a mass. Now go back to the spring. I want to know how much force F(x) it exerts when I pull it by a certain amount x, which is measured from the point when the spring is neither compressed nor expanded as shown in Figure 3.1. If x is positive, it means the spring is stretched; if it’s negative, it means the spring is compressed. Now I can measure F(x) for any given x because I can measure the acceleration it produces on a known mass m and use F = ma. So I pull it

by various amounts, measure F, and draw a graph. It will be a straight line with some slope −k for small values of x:
Figure 3.1 (Left) The mass m is attached to a spring (dotted line) of force constant k. The other end of the spring is attached to a wall. The displacement x is measured from the equilibrium position of the spring. (Right) The force F(x) as a function of x.
where k is called the force constant. The minus sign says, if you pull it to the right, so that x is positive, the spring will exert a force in the negative direction. If you compress it, then x is negative and the spring will exert a force in the direction of increasing x. All springs will have a graph like this for small enough distortions from equilibrium. Beyond that the line may bend or the spring may even snap. In any case, we now have a way to measure k for any given spring in the regime when F(x) is linear in x.
I want you to think about the two equations F = ma and F = −kx. If the first one is Newton’s law, then what’s the other one? What’s the difference between saying F = −kx and F = ma? Let me paraphrase the answer I usually hear from students: “F = ma is universally true, independent of the nature of the force acting on a body, while F = −kx is only describing the spring.” That is essentially correct, and I will now elaborate.
The cycle of Newtonian dynamics has two parts.

The first one is to find the acceleration a of a body, the force on which is

somehow known, using

The force F is the cause and the acceleration

a is the effect and

is the precise relation between them.

The second is to deduce experimentally what force F will be acting on a

given body at any given time. For example, if a mass is attached to a spring

that has been extended by x, we must do the experiment to find that the

spring force is F = −kx. Newton does not tell us that. He never tells us what

F is in a given context, with the exception of gravity where he also

furnished the left-hand side of F = ma with his law of universal gravitation.

If two like charges repel each other, we need Coulomb’s law to tell us that the force varies as 1/r2. The nuclear force, say between a proton and

neutron, falls exponentially with distance. Surely Newton did not tell us

this. But once we have experimentally determined a new force F, we can

use his second law to deduce the a it would produce, assuming classical

mechanics is applicable.

When a new force, like the electric force, is discovered, the F due to it

can be measured in one of two ways. One is to compare it to a known force

that balances it. For example, to find how the repulsive force between two

charged bodies varies with distance, we can glue them to the two ends of a

spring of known k and measure by how much it is extended in equilibrium.

Another way is to release the charges (of known mass) from rest with some

separation, measure the initial acceleration of either, and then compute ma.

So physicists are busy either finding a from F (as when computing the

orbit of a satellite given the force law for gravity) or F from a (as when

stretching springs to find k or dropping apples from trees to measure the

force of gravity).

Now, a small digression on gravity. Consider a body near the surface of

the earth, the force of gravity on it being F = −mg where g = 9.8ms−2.

That’s something you find out by dropping things from a tower. Consider a

in the field of gravity. We find it is

for all bodies. That’s a very remarkable property of the gravitational force —the cancellation of m. If you look at the electrical force, on the proton and electron for example, it’s not proportional to the mass of either object. It’s

proportional to the electric charge of the object. Therefore, when you divide by the mass to get the acceleration, the response of different bodies is inversely proportional to the mass. But gravity has a remarkable property that the pull of the earth is itself proportional to the mass of the object. So, when you divide by the same m to find a, it cancels and everything falls at the same rate on the surface of the earth. In fact, that is a property of gravitational fields anywhere, even in outer space. Everything—gold, silver, diamonds, particles, elephants—all accelerate the same way. This remarkable fact was known for a long time, but it literally took an Einstein to figure out why nature behaves in that fashion, why the two masses associated with a body are equal. One is the inertial mass, which is how much a body hates velocity change, how hard it resists acceleration, the mass in F = ma. That quality can be measured far from planets, far from everything. The other is gravitational mass, which is the measure of how much it is attracted to the earth or any other body. There’s no reason why these two attributes had to be equal. Is this just an accident or is it part of a big picture? It turns out that it’s part of a big picture called the general theory of relativity. Here is Albert Einstein’s description of gravity. Imagine a stream and some kids dropping various leaves or paper boats in it. No matter what they drop (within reason) the object’s trajectory will follow the flow lines of the water. The path of all objects is predetermined. This is what gravity does to spacetime—it defines trajectories for objects: anything you release will follow the trajectory etched in spacetime. If you oppose this flow, like a kid holding on to his paper boat, the resistance you feel is the weight of that object. What determines the flow lines at each point? The matter and energy in the universe, as dictated by Einstein’s equations.
Here’s a simple example of a complete Newtonian problem. A mass is attached to a spring. It is pulled by a certain amount x and then released. What is it going to do? Newton says F = ma, which in this case becomes
To proceed, we must know m and k, and I have already discussed how these are measured. Now we have a mathematically complete problem: find the function x(t) whose second derivative is equal to times the function
x(t). Then, we go to the math department and say, “What’s the solution to

this equation?” This is a problem in mathematics and the answer—that it’s going to be oscillating back and forth—will come from doing the math. Later we will do some of that math ourselves. For now, I am simply pointing out that once we have stated the laws in mathematical form, solving for the consequences is a mathematical problem.
Here is another example. Newton discovers a force of gravity acting on everything. Here’s the sun in Figure 3.2, orbited by a planet. At this instant, the planet may be moving at some velocity v. The acceleration of the planet is due to the force of gravity between the planet and the sun, which Newton tells you is directed toward the sun, proportional to the product of the two masses, and which decreases with distance as . This completely specifies the left-hand side of F = ma. That’s the law of universal gravitation. Then again, because the second derivative of the position is connected to the position, you go to the mathematicians and say, “What orbit is the solution to that equation?” and they will tell you it is an ellipse.
Figure 3.2 The planet is separated from the sun by r. The force F on it was determined to be always opposite to r, i.e., pointing toward the sun and falling as . (The force is slightly offset from r for clarity.)
Of course, Newton did not have mathematicians he could go to. He was the math guy. Not only did he formulate laws of gravitation, he also invented calculus and figured out how to solve the differential equation that came out of his F = ma. There has been no one like him. Here I speak of Newton the scientist; Newton the man had many flaws.
3.4 Newton’s third law

The third law says that if there are two bodies, called 1 and 2, then F12, the force on 1 due to 2, is minus the force on 2 due to 1:

Action and reaction are equal and opposite. Coulomb’s law and the law of

gravity both have this feature. You may assume it for every force in this

course.

We are now going to put the laws to work. You have to be good at

writing down the forces acting on a body. That’s what all these problems

are going to boil down to. Do not forget the existing forces and do not make

up your own forces. I have seen both happen. Every force, with the

exception of gravity, is a force due to direct contact with the body: a rope is

pulling it, a rod is pushing it, you are pushing it, you are pulling it, one

block is pushing another, and so on. Gravity is one force that acts on a body

without the source of the force actually touching it. (Later electromagnetic

forces can come in, but not in this book.) That’s it. Once more with feeling:

With the exception of gravity all forces we will discuss in this book will be

contact forces.

We are going to begin with simple problems in mechanics. They will get

progressively more difficult. Let’s start with our first triumph. There is

some object of mass 5 kilograms and I apply 10 N on it. What’s the

acceleration? Everyone knows it is

. You may have done this

before, but I hope now you understand how we know the force is 10 N and

how we know the mass is 5 kg. The algebra is, of course, very trivial here.

Next, I have a 3-kg block placed against a 2-kg block, and I’m pushing

the former with 10 N as shown in Figure 3.3. I want to know what happens.

One way is just to use your common sense and realize that these two blocks

are going to move together. You know intuitively that if they move together,

they will behave like an object of mass 5 kg and the acceleration will again

be 2ms−2. What about gravity? What about the force due to the table on

which the masses are moving? Imagine that this occurs in outer space

where there is no gravity and no need for a table.

There’s another way to do this problem, which is to draw free-body

diagrams. Here you can pick any one body that you like and apply F = ma

to it, provided you identify all the forces acting on that body. We’ll first

pick the 3-kg mass. My 10 N is certainly acting on it. What other force is

acting? The force of the 2 kg, which has a magnitude f acting to the left. Do not include the force exerted by the 3 kg on the 2 kg. Next consider the 2-kg mass. There is the same f, but acting to the right by the third law. Here is the mistake some people make: they add to that the 10 Newtons. They feel that the 2 kg will surely feel it since that is what is behind all the acceleration. That will be a mistake. That’s an example of adding a force that you should not be adding. The only force acting on this little guy is this little f.
Figure 3.3 Top: A force of 10 N acts on the two blocks, viewed as a single entity. Bottom: The freebody diagram for the two blocks showing all the forces on each block. Notice the third law is being invoked.
Now we do F = ma for these two guys:
Notice I’m using the same acceleration for both. I know that if the second mass accelerated faster than the first one, then the picture is completely wrong; it will not feel the force due to the first. If it accelerated less than the first, the first would have plowed into the second. Since that also cannot

happen, they’re moving with the same acceleration. There’s only one unknown a. Once you got a = 2, you can go back to Eqn. 3.11 and obtain f = 4 N. Now we know the full story: 4 N acting on 2 kg gives it an acceleration of 2 ms−2, while (10−4) = 6 N acting on the 3 kg gives the same a = 2ms−2.
Here’s another variation shown in Figure 3.4. I have a 3-kg mass attached by a rope to a 2-kg mass, which I pull with a force of 10 N. Again, your common sense tells you that I am pulling something whose effective mass is 5 kg; the answer is 2ms−2. Let’s confirm that systematically by using the free-body diagrams in the lower half of the figure. Now, there are really three bodies here: the two blocks and the rope connecting them. In all these examples, I assume the rope is massless. We know there is no such thing as a massless rope, so what we mean is a rope whose mass is negligible compared to the two blocks being pulled. We’ll take the idealized limit where the mass of the rope is 0. The 3 kg is being pulled by the rope to the right with a force that I’m going to call T, which stands for tension. The rope is being pulled backward by the 3 kg with a force T by the third law. What is the force on the other end of the rope? What should that be? If you said T, that is right, but not because the rope would snap otherwise. Something else will be a problem. If the two forces on its ends don’t cancel, you have a net force. What are you going to divide by to get the acceleration? Zero, right? So, a massless body cannot have a net force on it, because its acceleration would then be infinite. Massless bodies will always have equal and opposite forces on the two ends. In the case of the rope, this is called the tension on the rope. The tension is not 0 just because this T and that −T cancel. Suppose you are being pulled by my favorite animals—the elephants—from both sides by equal and opposite forces. You won’t find any consolation in the fact that these forces add up to 0 as you get subdivided.

Figure 3.4 (Top) A force of 10 N pulls the two blocks connected by a massless rope, all treated as a single entity. (Bottom) The free-body diagram for the two blocks and the rope, which experiences equal and opposite forces of magnitude T called the tension.
Now, you can invoke F = ma for the three objects, starting from the left:
We add the three equations and we get 10 = 5a or a = 2ms−2 and T = 6 N. So, the tension on the rope is 6 Newtons. This is very important: when you buy a rope, the specifications will tell you how much tension the rope can take before it will snap; if your plan is to accelerate a 3-kg mass with an acceleration of 2 ms−2, you better have a rope that can take the tension of 6 N.
If you see the rope in isolation you find its acceleration is indeterminate according to Eqn. 3.15. This is correct; it is the non-zero masses that determine a for everybody, and the rope goes along with this a for free.
3.5 Weight and weightlessness
Now let’s see what happens when you ride an elevator. In Figure 3.5 you are the stick figure standing on the weighing scale on the floor of an elevator that has a positive (upward) acceleration a. What will the scale register? We will draw free-body diagrams as in the right half of the figure for you and the massless spring. The spring is being pushed down by you

with a force W and up by the floor with the same force since it is massless. It will compress by an x such that W = kx, and x will somehow be displayed by a needle or digital readout.
(We should note a subtle thing that you may not have realized. Every [massless] spring is pushed or pulled by equal and opposite forces ±F at the two ends since otherwise it would have a = F/0 = ∞. So which of these two appears in F = −kx? Recall the mass spring system. One end of the spring is anchored to the wall. When we pull the other end by +x, we apply a force F = +kx to the right. The wall exerts a force F = −kx at the other end. In response to the +kx we apply, the spring exerts a force F = −kx on us or on the mass attached to it, and that is the F we use in writing down F = ma for the mass.)
Figure 3.5 (Left) An elevator is accelerating at a rate a and you are standing on a scale whose spring is shown by a dotted line. (Right) The free-body diagram for you and the spring. The spring is compressed at each end by a force W.
The equation describing you in the elevator is

If the elevator is standing still and a = 0 we see that W = mg is just your weight. If the elevator is accelerating upward W = m(g + a), and the needle will show a number greater than your weight. You will actually feel heavier. The spring not only has to support you from falling through the floor but also has to accelerate you counter to what gravity wants to do. That’s why we have the g + a. Say you picked up some speed and are now coasting upward at a steady speed. Then, a = 0 again and W = mg. As you come to the top of the building, the elevator has to decelerate so that it can lose its positive velocity and come to rest. So a will be negative now. If a is negative, then g + a < g. Let us write in this case a = −|a|, so that

which makes it explicit that W < mg. You will feel that your weight is reduced.
On the way down your initial acceleration is negative because you’re picking up speed toward the ground. You will feel less heavy. You can see that if |a| = g, your downward acceleration is that due to free fall under gravity and you will feel weightless. This can happen when the cable has snapped. You don’t feel any weight because your normal weight is the opposition the floor offers to keep you from falling through it; but now the floor is also falling at the same rate. It’s wrong to think that when you feel weightless you have escaped the pull of gravity. We all know that in a falling elevator you definitely do not escape the pull of gravity. It’s going to catch up with you in a few seconds. The same goes for the people floating around in space stations. They have not escaped the pull of gravity either; they have just stopped fighting it, and they are all accelerating toward the

earth at a rate

, where g∗ is the (reduced) acceleration due to gravity

at the radius r of the orbit. If they had really escaped the pull of gravity, their spaceship would be off to the far reaches of the universe.

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CHAPTER 4
Newton’s Laws II
4.1 A solved example
The goal of physics is to be able to predict something about the future, given something about the present. I’m going to provide a very simple example that illustrates how Newton’s laws are to be used for this purpose. This treatment will be brief since I will return to this problem in greater detail later.
Figure 4.1 shows a frictionless table, on which is a mass m attached to a spring of force constant k. The other end of the spring is attached to an immobile wall. The dotted outline of the mass shows it when it is displaced by an amount x from equilibrium. I want to pull the mass by some amount A and let it go. So that’s the knowledge of the present. What’s this guy going to do? That’s the typical physics problem. It can get more and more complex. You can replace the mass by a planet; you can replace the spring by the sun, which is attracting the planet; you can bring in many planets; you can make it more and more complicated. But they all boil down to a similar exercise: I have some information now and I want to be able to say what will happen next.

Figure 4.1 A mass m attached to a spring of force constant k in its equilibrium position, when the spring is neither extended nor compressed. The dotted outline of the mass shows it when it is displaced by an amount x from equilibrium.
Combining F = ma and F = −kx we obtain the equation that seals the fate of the mass:
This is a differential equation: you have to find a function x(t) whose second derivative is times itself. A differential equation tells you something about an unknown function x(t) in terms of its derivatives, and you are supposed to find it given this information. Instead of running to the mathematicians, let us solve this equation by guessing, which is a totally legitimate way to solve a differential equation.
Here is how we guess the answer. Let’s make our life simple by taking a case where k = 1, m = 1. Later on, we can put back any k and m. I’m looking for a function whose second derivative is minus that function. Now, as a word problem, it rings a bell, right? Do you know such a function? Exponential is good. Trigonometric functions are good too, provided you

mean sin and cos. I’m going to dismiss the exponential, which is actually a very good guess. If you took x(t) = et, we have

It does not help to consider x(t) = e−t since the minus sign will get squared

when you take two derivatives, and you will still end up with +x(t). Now

e±it, where

, will work, but we do not want to deal with complex

numbers yet.

We just want a function that reproduces itself when we take two

derivatives. So, I make a guess,

and you can check that it works:

Did you follow the way this very elementary problem is solved? It’s solved by making a guess. But something is not quite okay with this solution. If I put t = 0, I get x = 1. Why should it be true that I initially pulled it by exactly one meter? I could have pulled it by 2 or 3 or 9 meters. I want to be able to decide how much I pulled it by at t = 0. Suppose it was pulled to 5 m and released. I want x(0) = 5. This can be arranged by making a choice
Does the 5 screw up everything? It doesn’t, because it just comes along for the ride:

Now I have an answer that does everything I want it to do. At the initial time, it gives me 5 times cos0, which is 5; and that’s what I said was the

initial displacement. I find

sin t, which vanishes at t = 0. That too is

correct; I pulled it and let it go. So, the instant I released it, it had no

velocity. It satisfies Newton’s laws, and that’s the answer. I want to do this

simple example in totality because this is the paradigm. This is the example

after which everything else is modeled.

There are two final points about this solution.

First, there is the other option x(t) = sin(t) that occurs to some students.

There is nothing wrong with this option; it just is not needed in this

particular example. Equation 4.5 is not the final and most general solution

to the problem; it is a solution that certainly works for the one example I

had, in which a mass is pulled to 5m and released.

Second, when k and m are not both 1, we go back to the general case

The solution to this is

where A, which was 5 in our example, is arbitrary in general and is called the amplitude. We will deal with the oscillator in depth later. At this point, we just want to get a feeling for how F = ma is applied in real life, with some help from the mathematicians.
4.2 Never the whole story
Is this the whole story of the mass and spring, or is something missing? Does the mass oscillate forever as predicted by the Eqn. 4.1 we just solved? It does not: it eventually comes to rest. If you never knew about friction, and you solved for x(t) with just the spring force, you would find it didn’t work. So, you are at a fork in the road. You can say either that Newton’s laws are incorrect or that you are missing some forces. In the latter case, you will look around and eventually deduce the frictional force by insisting that F = ma works with its inclusion. We will see how this is done later in this chapter.

It turns out that even if there were no friction, Eqn. 4.1 would not be the correct law, which is given by Einstein’s relativistic dynamics. By that I mean, if you really pull a real spring by 5m on a frictionless surface and release it, its motion will not be exactly 5cost; it will be off by a very, very, very tiny amount that you probably will not discover in most laboratories. But if the mass begins to move at a velocity that is comparable to that of light, then this equation will make the wrong predictions. On hearing that even Newton can be wrong in this sense, some people say, “What kind of business are you in? Every once in a while some authority is proven wrong.” I’ll tell you right now. Everything we know is wrong at some level. Newton didn’t try to describe things moving at speeds comparable to light. He dealt with the problems he could deal with at that time. So his laws have a limited domain of validity. You can always push the frontiers of observation until you come to a situation where any given law breaks down. While the special theory of relativity does better than Newtonian mechanics for large velocities, it too fails if the mass becomes very tiny, of atomic dimensions. Then you need the laws of quantum mechanics discovered by Heisenberg, Dirac, and Schrödinger, and of course their laws also have problems in some experimental domain. Sometimes we correctly abandon the formalism; but we should not be too eager to do that. In the present case, the problem is not with formalism but with not including friction. The failure at the quantum level comes not because we did not include some forces, but because the very notions of force or trajectories x(t) seem to be invalid at that scale.
4.3 Motion in d = 2
We’re going to move on to higher dimensions, where position, velocity, acceleration, and force are all vectors. So let’s again start with simple problems in d = 2 and make them more and more complex. There is no limit to how difficult mechanics problems can be. If you go back and read some Cambridge University exams from 1700 or 1800, you will find some really difficult problems. Finally, quantum mechanics was invented and life got a lot easier.
Consider a mass m sitting on a table as shown in the left half of Figure 4.2. Because we’re in two dimensions, we need to have two axes x and y. Recall that

Figure 4.2 A mass m sitting on a table, (left) without and (right) with friction. The coefficient μ = μs if the block is at rest and μ = μk if it is moving.
is a vector equation. If two vectors are equal, then their x and y components have to match:
I’m going to apply them to this block. There are no known forces acting in the x direction, and the block is not moving in that direction. Therefore it’s a case of 0 = 0. Now I look in the y direction. In the free-body diagram there is the force of gravity mg acting down the y-axis; I will often denote this vector by mg, where g = −9.8jms−2. If that’s all you had, the block would fall through the table. Because the block is not falling, we know the table is exerting an opposing force denoted by N, where N stands for normal. And normal is a mathematical term for perpendicular. Evidently N is a positive force and mg is a negative force in the y direction. Thus
In this application of the Newtonian equation, I know the right-hand side. It is 0 because I know this block is neither sinking into the table nor flying

off the table. It’s sitting on the table; it has no velocity or acceleration in the y-direction. I come to the conclusion
4.4 Friction: static and kinetic
Now I’m ready to introduce another force, the force of friction, f. How do we infer there is a force of friction? Consider the mass on the table in the right half of Figure 4.2. What experiment will tell you there is another force called friction? You find that to keep the mass moving at constant velocity, you have to apply a force. That means the force you’re applying is canceled by something else, because there is zero acceleration. Here is another good answer: You give the mass a push and soon it stops moving; there must have been a force that produced the deceleration. But even before that, even before it starts moving, you find that, if you push it, it doesn’t move. Say I push the podium. If I push it gently, it does not move. But if I push hard enough it begins to move. What’s happening before it moves? I’m applying a force and I’m getting nothing in return for it. So I know there is another force exactly balancing mine, and that’s the force of static friction. Let us call the force I am applying Fme. I’m applying it to the right, and so there has to be another force of equal magnitude to the left. Notice it is not a fixed force: it’s whatever it takes to keep the mass from moving. It will not be less than what I apply, because then the mass will move; it cannot be more than what I apply because then it will start moving me backward. So, static friction is a force that has a range from 0 to some maximum. The maximum turns out to be
where μs is a number called the coefficient of static friction and N is the normal force. In our example you may use N = mg. So the force of friction seems to depend on how heavy the object is that’s sitting on the table. But it doesn’t depend on the area of contact. For example, it does not increase if we rest the block on another face with greater area. You might expect more friction because there’s more contact. But that is not so, for reasons not

readily explained within our elementary treatment. In reality, friction has a subtle origin at the atomic level.
If static friction provides a force equal to the force that I apply, up to a maximum of μs N, what happens when I exceed the maximum? The object I am pushing will start moving. Once it starts moving, the frictional force, which is always directed opposite to the velocity, changes to
where μk is the coefficient of kinetic friction. We find that in all situations μk < μs. Let us say μs = 0.25 and μk = 0.2 for the block on the table. If I apply a force that is 0.1 times its weight, it won’t move; at 0.2 times its weight it won’t move; at a quarter of its weight it’s a tie; at 0.26 of its weight the block will start moving. Once it starts moving, the frictional force will drop to 0.2 times its weight, and the body will accelerate if I maintain the same force as when it first begins moving.
4.5 Inclined plane
Now we are going to do the one problem that has sent more people away from physics than anything else. It is called the inclined plane. A lot of people who do not remember where they were during the Kennedy assassination say, “I remember the day I saw the inclined plane; that’s the day I decided I’m not going into physics.” This is very bad publicity for our field. You come into a subject hearing about relativity and quantum mechanics, and we hit you with this. So why am I still doing this? Because this is the entry ticket into the business. It is inconceivable to me that you could understand more advanced topics without being able to understand this one. Go ahead and mock the inclined plane, but only after you can prove you have mastered it.
Here is the notorious problem. There is a mass m sitting on a plane, inclined at an angle θ as shown in the left half of Figure 4.3. We know it’s going to slide down the hill, but we want to be more precise. The whole purpose of Newton’s laws is to quantify things for which you already have an intuition. The novel thing about the inclined plane is that for the first time we are going to pick our x and y axes not along the usual directions, but parallel and perpendicular to the incline. What are the forces on this

mass? I have told you, first deal with contact forces. But the only thing in contact with the mass is the plane. In general the inclined plane can exert a force along and perpendicular to its own surface, but I’m first going to take a case where there is no friction. By definition the plane cannot exert a force along its own length, so it can only exert a normal force N. Then there’s only one other force, the force of gravity that we agreed we have to remember. Even though the earth is not touching this block, it is able to reach out from down below and pull this block down. These are the two forces, and the mass will do what these forces tell it to do.
Figure 4.3 A mass m sitting on an inclined plane, without friction (left) and with friction (right).
First I have to take this mg pointing down and break it up into components along my x and y axes. That’s called resolving the force into various directions. Now, the key to this is to know that this angle θ of the plane is the same as the angle between vertical and the normal to the plane. You have to agree that if I draw a line perpendicular to the horizontal and I draw a line perpendicular to the incline, the angle between those two perpendiculars will be the same as the angle between the two lines with which I began. This is because “make perpendicular” means “rotate by .” If you rotate both lines by , the angle between the rotated lines will be the same. The vertical is perpendicular to the horizontal, and the normal perpendicular to the inclined plane. Once you understand that, the rest is easy.

So here are my equations:
Now we know that this block is sliding down the incline. It’s not going into the plane, and it is not flying out of the plane. That’s the reason we measure our y coordinate perpendicular to the plane: it does not change, unlike the traditional x and y, both of which do. So ay = 0 in Eqn. 4.17, which implies
Then you come to Eqn. 4.16. We cancel m and read off the answer
That’s the big result, that the mass will slide down the hill with an acceleration g sinθ. This provides a good way to measure g, because if you drop something vertically, it falls too fast for you to time it. But if you let it go down an incline, by making θ very small, you can reduce the acceleration by a factor sinθ.
Here’s another thing I should tell you right now. The professionals don’t put in numbers until the very end. So, if you are told the incline is an angle of 37 degrees, and g = 9.8 msec−2, don’t start putting numbers into the first equation. I know for some of you it’s traumatic to work with symbols. There are many reasons to work with symbols rather than with numbers till the end. First of all, if you have already put the numbers in and I suddenly tell you, “Hey, I was wrong about the slope, it’s really 39 degrees and not 37,” you are forced to do the whole calculation again. But if you use symbols to derive the formula for ax and then ask me, “What’s your θ?” you can just plug that in. Likewise, if I change the value of g because someone made a better measurement, you simply change g in the very last expression for ax. You can also see if your answer had some mistakes in it. Suppose you got ax = g2 sinθ. You will know it’s wrong because the units do not match. Maybe you have the trigonometry wrong. Maybe it’s really ax = g

cosθ? You can do a few tests. For example, as the plane becomes less and

less inclined, you have to get less and less acceleration, and when θ → 0, so

must ax, which must then be proportional to sinθ. Or, if you make the

incline almost vertical, the block is just falling under gravity with ax = g.

The result ax = g cosθ does not do that when we set

. Finally, ax = g

sinθ is independent of m. That’s an interesting property of the result you

would not notice if you kept numbers everywhere. So we will agree to work

with symbols until the end.

Now we are going to make life a little more complicated by adding

friction, as shown in the right half of Figure 4.3. Let the block be at rest and

imagine there are some hinges on the inclined plane that allow me to

increase θ from 0 to . I want to know when the block will begin to slide

down. Let me cut to the chase and write the equations:

The first equation is the same as before and tells us that since ay = 0,
In the second equation, do not write f = μsN because the frictional force is not always μsN; it is whatever it takes to keep the block still, up to a maximum of μsN. In other words, if θ is very small, the frictional force will in fact be a much smaller amount f = mg sinθ, which you get by setting ax = 0 in the right-hand side.
Let us crank up the angle to θ∗, beyond which it cannot stay still. At this angle the friction is maximum and we can assert
which gives us a way to measure μs:

Again, the mass cancels. So, it doesn’t matter how heavy a car you parked on the slope, as long as μs was the same. But g cancels too. So, whether you park your car on the earth or park it on another planet, the same restriction tan θ ≤ μs applies.
When we pass this limit, when tanθ > μs, the block will begin to slide down. It will now have a non-zero downhill acceleration because it is kinetic friction μkN that now opposes the velocity and μk < μs. The operative equations are
What’s the acceleration now? I take the y equation that says as before that N = mg cosθ since ay = 0, and plug that into the x equation to obtain the downhill acceleration of the block:
4.6 Coupled masses
Next is a problem of two masses m and M connected by a string that goes over a massless pulley mounted at the top of a frictionless inclined plane, as shown in Figure 4.4. What can we say before doing the math? If M > m am I assured it’ll go downhill? No, it depends on the angle θ because only part of its weight, Mg sinθ, is helpful in going downhill while all of mg is pulling m down. All we know for sure is that for a fixed θ, the bigger mass M will go down the slope as M → ∞ while the smaller mass m will go straight down as θ → 0. Let us figure out now when exactly the balance shifts.

Figure 4.4 (Left) Two masses linked by a massless rope and massless pulley. (Right) The free-body diagram.
Since we have two masses here, no choice of axes is going to make all their motions simple: m goes straight up and down while M glides along the slope. Let us ignore the boring N = Mg cosθ equation for the y coordinate and focus on the motion along the plane. In the free-body diagram I have shown the same tension T wherever the rope is involved. For a straight massless rope it is clear the opposing forces at the ends must have the same magnitude T since there can be no net force on a massless object. For a rope that curves around the pulley, the explanation is more subtle, as I will explain momentarily.
For M, the equation along the x-direction is
For m the equation in the up direction is
where I have deliberately used the same a as in Eqn. 4.29. I am not saying the acceleration of M and m are equal as vectors. They do not even have the same direction: m moves up and down while M moves along the plane. By a I mean the components of the two accelerations in the allowed directions of motion, defined to be positive if M is moving downhill and m is moving up. This equality reflects the inelasticity of the rope: if m moves up one

inch, M will slide down one inch along the incline. If the magnitude of the displacement is the same, so are the magnitudes of the velocity and acceleration. We can solve for a by adding Eqns. 4.29 and 4.30 to obtain
Does this solution make sense? Now you notice that for a to be positive and for M to go downhill, it’s not enough that M > m; we need M sinθ > m because only the force Mg sinθ is pulling M downhill: the force Mg cosθ is trying to ram it into the inclined plane, and that’s being countered by the normal force N.
Can we use this formula when a < 0, that is, when M moves uphill? After all, I did all my analysis assuming it’s going down. Yes, we can. All the forces I drew here—N, Mg, and so on—are not going to change when you change the body’s direction of motion. Gravity is always going to pull down whether the block moves with it or against it. So once you have a formula for positive a, you can apply it to negative a as well. However, if friction is present, you cannot do that because you have to assume a particular direction of motion before you can assign a direction to the frictional force in the equations of motion.
Now let’s consider why the forces at the two ends of the rope emerging from the pulley have equal magnitudes T. Assume that the pulley is massless and that the rope does not slip; that as the masses move and rope moves over the pulley, the pulley is forced to rotate without relative slippage. In this case the part of the rope instantaneously in contact with the pulley and the pulley itself may be viewed as one rigid body. (It is like a bicycle chain whose links mesh with the teeth on the wheel the pedal is turning.) The tensions on the rope at the two ends are clearly trying to rotate this body in opposite directions, as is clear in the free-body diagram in Figure 4.4. If these tendencies are not exactly balanced—that is, if these forces are not equal in magnitude—they would produce an infinite rotational acceleration of this massless rigid body, just like a non-zero force would produce infinite linear acceleration on a massless object. By the noslip condition, this means an infinite linear acceleration for the two masses, which is impossible because the masses are non-zero and have finite forces

acting on them. So the forces at the two ends of the rope emerging from the pulley must have equal magnitudes. (We will return to a proper study of rotations and the case of the massive pulley in Chapter 10.) Note that even though the rotational effects of the two non-parallel forces cancel, their nonzero vector sum can produce linear acceleration of the pulley. The axle about which the pulley rotates provides an equal and opposite force to prevent this. Luckily this last force does not contribute to rotations around the axle.
4.7 Circular motion, loop-the-loop
Now we’ll turn to some interesting problems in circular motion of various kinds. The first, shown in part A of Figure 4.5, describes a ride in an amusement park. You sit in these baby rockets hanging from a rope along with other petrified victims. The whole thing begins to spin, and the rope, instead of remaining vertical, starts tilting at an angle θ, which we want to determine as a function of the tangential speed v and radius R of the circular orbit. We apply F = ma and start listing the forces on the baby rocket. Gravity provides the usual mg pointing down. The rope can only exert a force T along its length. Let us trade the tension along the rope for the sum of two equivalent forces, a vertical part, Ty = T cosθ, and a horizontal part, Tx = T sinθ directed toward the center of the circular orbit. The equations are
In the first equation, we recall that by assumption the rocket’s orbit is a horizontal circle and it has no net acceleration in the vertical direction. In the second, we recall from Section 2.6 that a body moving in a circle of
radius R at speed v has a centripetal acceleration . So this is a case where we know a in F = ma. Eliminating T we find

Figure 4.5 (Left) An “amusement” ride in which you go around in a horizontal circle of radius R in a baby rocket. The rope supporting you necessarily makes an angle θ = tan−1 with the vertical. (Right) A car going around a circular racetrack of radius R at speed v. The road has a banking angle that obeys tanθ = . The normal force N has a horizontal part that gives the necessary force to bend the path into a circle. The symbol ⊗ is used in both parts to indicate that the rocket or car is going into the page, away from you. The convention is based on how an arrow with feathers would appear going away from you. Likewise ⊙ indicates an arrow coming out of the page toward you.
It is worth finding the tension on the rope since our lives may depend on it. It is
using Here is another interesting problem. You are driving on a circular
racetrack of radius R at a speed v. Suppose the plane of the road is strictly horizontal, perpendicular to g. Some agency has to exert a force on the car to bend it into this circle. It is of course the road that does this, thanks to the frictional force f ≤ μsmg directed toward the center. (I use μs, the static

coefficient, and not μk, the kinetic one, even though the car is moving, because we are discussing the force in the radial direction and the car has no velocity in that direction, unless it is skidding. Note also that the car does not really have to travel in a circle; all we need is that at this instant the trajectory is part of some circle of radius R.)
If you don’t have the requisite static friction, if μsmg , your car will not be able to make the curve; it will fly off. But there is a clever way in which you can still make the turn without any friction, and that is to bank your road by an angle θ as shown in the right half of Figure 4.5. Imagine now you’re going into the paper. The frictionless road only exerts a normal force N. Let us resolve that force into a vertical part N cosθ and a horizontal one N sinθ directed toward the center of the circle. The equations are
Eliminating N we find the banking angle to be
Let me elaborate. You want the car to go around the bend at a certain speed. If you bank your road at that angle, you don’t need any friction to make the turn. Even though the frictionless road can only exert a normal force, thanks to banking, a part of the normal force points toward the center, providing the requisite centripetal force. Of course, when you drive on a real road, you do not have to travel at exactly this speed for a given R, for any small differences will be made up by the frictional force of the tire. It is just that you do not want to rely on friction for the entire radial force.
Finally, the famous loop-the-loop problem, which defies common sense, is shown in Figure 4.6. You come down on a roller-coaster track from some height H, you go on a vertical circle, and for a while you are upside down. The eternal question is, “Why don’t you fall down?” We’ll find we can understand this phenomenon fully with Newton’s laws. The forces on the coaster are mg acting down and the force of the track, which has to be

normal to it since it is assumed to be frictionless. But it too points down! We are doomed! Why don’t we fall? The answer is that we do fall, that is, we accelerate downward, but this does not mean we get any closer to the center. The two forces mentioned combine to bend the coaster into a circular path and produce the requisite downward centripetal acceleration:
Figure 4.6 The roller coaster comes down from a height H and goes into a loop in the vertical plane. Why does it not fall down? The forces on it are mg acting down and the track force N also acting down! It does fall, as explained in the text. The three arrows forming a triangle in the inset show the initial velocity v − Δv/2 just before it reaches the top, the change Δv in a small interval near the top, and the final velocity v + Δv/2 just after it passes the top.
Solving for N we find

If N comes out positive, that is, points down in our convention, which

happens if

, we are safe. If it comes out negative, that is, if

it means the track exerts an upward force, which is impossible, unless there is some other mechanism, like a T-bracket, that goes under the track and supports the coaster even if it is just hanging upside down. I believe such things exist in real roller coasters, in case they get stuck at the top or do not go fast enough. In our idealized coaster, without any of this backup, the speed v must obey
to safely make the loop. We will figure out the minimum height from which it must be released to
satisfy this condition when we derive the law of conservation of energy. Let us be sure to understand again why accelerating down does not
always mean gaining speed toward the earth. If you drop an apple, accelerating down means really picking up speed toward the ground, toward the center of the earth. It starts with zero vertical speed and picks up speed. In our example, the coaster is also accelerating, but the tiny change in velocity in a tiny time, which points radially down in Figure 4.6 near the top of the loop, is now added to a huge horizontal velocity pointing to the left.
We will now see that this implies the velocity has a constant magnitude and changing direction. Consider the impact of adding a tiny change Δv to a velocity v on the magnitude of velocity. Assume Δv can be in any direction relative to v. The resultant velocity has a magnitude squared given by

(To find , we need to keep just the term linear in Δv.) This equation generally implies a non-zero rate of change of the magnitude of v, unless Δv and the acceleration a are perpendicular to v, as is the case at all times in circular motion.
Figure 4.7 You fire two bullets from a tower at increasing speeds, which land farther and farther away (points 1 and 2). Beyond a critical speed, the bullet would go into orbit. It has of course never ceased to accelerate toward the earth.
So a little later, the total velocity vector merely gets rotated with no change in length and becomes tangent to the circle at a slightly different point. Going around in a circle is an example of constantly accelerating toward the center but not getting any closer.

Here is another example of this phenomenon. Suppose you are on a tower and you fire a gun horizontally as shown in Figure 4.7. The bullet hits the ground at point 1, under the pull of gravity. If you fire another bullet at a greater velocity it will land a little further away, at point 2. While greater initial speed will extend the time of flight even on a flat earth, the flight is further enhanced by the earth curving under the bullet. There will be a certain speed at which the bullet will keep falling but will not get any closer to the center, because the earth is falling under it just as fast. It is in orbit, as shown by the circle. That is in fact how you launch a bullet into orbit. So, what’s the first thing you should do when you fire this gun? Move away, because it’s going to come back in about 84 minutes and get you from behind at 17,650 miles per hour. This calculation assumes that there is no atmosphere, which of course leaves you with an even more pressing problem.
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CHAPTER 5
Law of Conservation of Energy
5.1 Introduction to energy
The law of conservation of energy is a robust and powerful one. When the laws of quantum mechanics were discovered in the subatomic world, many cherished notions were abandoned. You must have heard the ugly rumors: particles do not have a definite position and definite velocity at a given time. They don’t move along continuous trajectories. You might think the particle must have had an interpolating trajectory connecting two sightings, but it does not, and to assume it does causes conflict with experiment. While many of the ideas of Newtonian mechanics were abandoned, the notion of a conserved energy survived the quantum revolution. There was a period when people were studying nuclear reactions, and the energy they began with didn’t seem to be the energy they ended up with. Niels Bohr, the father of the atom, suggested that maybe the law of conservation of energy was not valid in quantum theory. Then in 1931 Wolfgang Pauli decided to put his money on the law of conservation of energy; he postulated that some other tiny electrically neutral particle, which escaped detection, was carrying away the missing energy. That was a radical position to take in those days, when people did not lightly postulate new particles, as compared to today when if you don’t postulate a few new particles you don’t get your PhD in particle physics. Pauli’s particle, called a neutrino, was detected after many, many years in 1959 by Clyde Cowan and Frederick Reines. Nowadays, neutrinos are one of the most exciting, elusive, and mysterious things one could study, and they hold the key to many puzzles concerning the universe.
Let’s see how the notion of energy conservation arises, starting with one dimension. When a force acts on a body, it changes its velocity. In one dimension, this simply means speeding up or slowing down. We’re going to find the relation between the speed accumulated when a force acts on a body for some time and the distance the body has traveled in that time.

Consider the case when the force F is constant, not varying with time.

That produces an acceleration

. In the first chapter we learned that if a

body has constant acceleration a,

where v and x are the current velocity and position, and v0 and x0 the corresponding initial values. In those days of kinematics, we didn’t ask, “Why does it have a constant acceleration?” We were just told, “It has a constant acceleration; just find out what happens.” Now that we have learned dynamics, we know acceleration has a cause, namely some force. So I’m going to replace a by and make one more cosmetic change in
notation. All the initial (final) quantities will carry a subscript 1 (2). Then Eqn. 5.1 takes the form

where d = x − x0 ≡ x2 − x1 is the distance traveled during this interval.
5.2 The work-energy theorem and power
Equation 5.2 says that when the force acts on a body, it changes the velocity, and the change depends on how far the force has been acting, on how many meters it has been pushing the object. The change is not simply in velocity but in velocity squared. Let us move everything involving the force to one side and the particle to the other side:

The combination

is called kinetic energy and is denoted by K. The

product Fd is called the work done by the force and denoted by W. The units

of W are Newton-meters and we replace that by joules or J.

We have found the simplest version of the work-energy theorem:

What if there are 36 forces acting on the body? Which one should I use? Say I’m pulling and you’re pushing. Then, F has got to be the net force, because Newton’s law connects the net force to the acceleration. If you and I have a tug of war, and we cancel each other out to zero, then there’s no acceleration, and the body with some initial velocity will maintain its initial velocity. The work-energy theorem says, The change in kinetic energy is equal to the work done by all the forces. In this case, there is no change in energy, though there are two forces at work that cancel. It turns out that it is sensible to define the work done by me, which is equal to F times the distance traveled, and the work done by you, which will be given an opposite sign. When do we attach a plus sign and when do we attach a minus sign? If you go back and review the whole derivation, you will see it was understood that a was a positive quantity. Then everything works. If the body is moving to the right, and I’m pushing to the right, then the work done by me is positive. And if you were pushing to the left and the body still moved to the right, the work done by you is negative. In other words, if you get your way, if things move the way you’re pushing, the work done by you is positive. If the object is moving counter to your will, in the opposite direction to your force, the work done by you is negative. I’m lifting this piece of chalk at constant velocity from the ground. Its kinetic energy is not changing. That means the total work done on the chalk is zero, but not because there are no forces on it. There is gravity acting down, and I am countering gravity with exactly mg. The work done by me is positive because I want the chalk to go up, and it does. If it goes up by an amount h, the work done by me is W = mgh. The work done by gravity in the meantime is −mgh and the total work done is zero.
Let us imagine all of this occurs in a time Δt and the distance moved is d = Δx. Then if you divide both sides of Eqn. 5.4 by Δt and take all the usual limits
where P is defined as the power. So power is the rate at which work is done. For example, if I climb a twelve-story building, I have done some work, my mg times the height of the building. I can climb the building in one minute; I can climb the building in one hour. The work done is the

same, but the power is a measure of how rapidly work is done. That’s why it’s the product of force and velocity or work divided by time. The units for power are joules per second. That also has a new name, which is watts or just W. You may use a kilowatt, or kW, which is a thousand watts. So, if you have a 60 W bulb, it’s consuming energy at the rate of 60 joules per second.
Now let’s turn to the next generalization, when the force is not a constant but varies with x. Do we know any example of a force that varies with x? A spring is a good example, with F(x) = −kx. Even gravity is a good example, if we consider large distances. I think it is no secret that the force of gravity is mg only near the earth and that if you go sufficiently far, you will notice gravity itself is getting weaker. It will still look like mg′ locally but g′ won’t be a fixed 9.8ms−2; it will be decreasing as we move away from the earth.
What is the work-energy theorem when the force varies? Let’s draw ourselves a force F(x) that varies with x, as in Figure 5.1. I’m just taking any function of x that I want. Now the acceleration is not constant because the force is not a constant. We cannot apply the formula W = Fd because F varies over the distance d. So we resort to the usual trick in calculus: find an interval of width dx that is as narrow as you want, so that during that period F is essentially a constant equal to F(x), the value of F at that x. For that tiny interval I can still say that the change in K is

Figure 5.1 When a variable force F(x) acts on a body that moves by a distance dx, it does work dW = F(x)dx shown by the shaded rectangle.
Geometrically, F(x)dx is the area of the thin rectangle whose base is dx and whose height is the function F at that x. (If F(x) < 0, the area is counted as negative.) If you eventually went from x1 to x2, then the work done by the force is given by the area under that graph in the sense of calculus. In every segment you pick up the change in kinetic energy dK, you add it all up to get K2 − K1 from the left-hand side. The right-hand side is the integral of the function F(x) between x1 and x2. The general work-energy theorem now says

Even if you have never heard of an integral, if I give you a function you can still deal with this problem. You’ll come to me and say, “Give me your function. I’m going to plot it on some kind of graph paper with a grid on it, and I’m just going to count the number of tiny squares enclosed. That’s the area and that’s the change in kinetic energy.” So integration is just finding the area bounded by the function at the top, the x axis below, and two vertical lines at the starting and ending points x1 and x2.
Now here is a little digression, a three-minute introduction to a great secret for finding the area. If you give me a function and you tell me to find the area under it from x1 to x2, I can show you a trick. You don’t have to draw anything on graph paper. First, I find a function G(x) specified by its derivative
Then I claim (and will prove shortly) that

This is the opposite of taking derivatives: now we want a function whose derivative is given to be F. If I say F(x) = x3, then G(x) is that function whose derivative is x3. Now, I know I have to start with x4 because when I

take derivatives, I will lose a power. But I will also get an unwanted 4 in

front. I fight that by putting a 4 downstairs and find

. You might

point out that if you have a function whose derivative is something, adding a constant doesn’t change the derivative. Then you can say, “Well, we are in

trouble now because the world cannot agree on what G is, because if I have

one G, you can get another one with a different constant.” But the beauty is that when you take G(x2) − G(x1), this difference in the choice of c goes away. So most of the time we don’t bother with the constant. Sometimes we

make a special choice that recommends itself, as you will see when we

study gravity on a celestial scale. The choice of constant is like the choice of origin in the projectile problem: any origin will do, but it is convenient to

choose it as the point of take-off.

Why is G(x) the function whose derivative is F(x)? Let us call the area

from some arbitrary point x0 to the point x as G(x). If I add a little more

area, out to x + dx, the extra area is F(x)dx. This, by definition, is dG, the

change in G. Dividing by dx and taking the limit, we see

. How do

different G’s differing by a constant arise? By choosing different starting

points x0 from which to reckon the area. But whatever we choose for x0, the change in the area due to changing the upper limit x is always given by dG

= F(x)dx.

5.3 Conservation of energy: K2 + U2 = K1 + U1
Let us now incorporate what we have discussed so far and write

which we can rearrange to give
We now make a little cosmetic change, and introduce the function
in terms of which we obtain the standard form
This is the law of conservation of energy; E = K + U is called the total mechanical energy and U is called the potential energy. Conservation of energy in physics has a totally different meaning from “Turn the lights off when you leave the room!” Here it means that when a body is moving under the effect of this force F(x), even though it is speeding up and slowing down, a certain special combination

does not change with time. If you know the value of E at one time, you know E at all times.
Let’s consider a simple example. We take a rock, and we drop it. We know it’s picking up speed; we know it’s losing height. So, you may expect there is some quantity that is a combination of height and speed, a combination which does not change in this exchange. We can find that combination by this law. In the case of gravity where F = −mg the expression for U is
The energy conservation law takes the form
In the mass and spring system, the corresponding relations are
and the conservation law assumes the form
Let us put that to work. I’m going to pull the mass by an amount A and let it go. I want to know how fast it will be moving when it comes to some point, say, x = 0. If you go back to Newton’s laws, this is a pretty complicated problem. Think about why. You start with a mass at rest. If you pull it by an amount A, a force −kA initially acts on it. That will produce an acceleration, −kA/m, which will give it a small negative velocity by the time it moves a distance Δx to the left. But once it comes to the new location, a