university physics

University Physics I

Author:Jeffrey R. Schmidt

Affiliation:University of Wisconsin

c⃝ 2001, revision August 27, 2012

Forward

Many of us, faculty and students, are weary of being forced to migrate to a new edition of university physics textbooksevery year, differing only in price and weight. There is virtually no difference between edition X and edition X + 1of any of the standard texts, they are all very much the same, and the same from edition to edition, in all respectsexcept for price and weight (I recently got an evaluation copy of a physics 201/202 text that weighs 15 pounds, andruns over 200.00 retail price). The only way out of this charade is to provide students with a free alternative, thatI can update every year at no cost to anyone, and so here it is, University Physics I, a very bare-bones, work inprogress, but free physics text. A lot of institutions and individuals around the country are engaged in similar projects.

I make no apologies for the appearance that this text is highly mathematical. If you take any standard text andcut out all of “filler” and pointless anecdotes and superfluous photos, the results will be highly mathematical inappearance, and will be about the length of this book. I have concentrated on problem-solving and expression ofthe basic concepts, since that is what the student really needs, and that is what testing and assessment focus on. Iwill remind the student that doing the homework is not optional, but is mandatory. You can’t master something bywatching others do it.

Times are definitely changing. In 1961 the number of hours per week spent on academic study by full-timecollege students was 40 hours per week outside of class. In 2004 that number had dropped to 27 hours perweek outside of class, in both cases the full-time load was 16hours/credits. Expectations have evolved similarly:in 1961 the average grade earned in a college class was a C (earned by over 40% of the students in the course), andby 2007 the average over 40% were awarded A-grades, and only 7% were awarded C-grades, with private universitiesand colleges leading the way in grade inflation, awarding 50% A-grades on average. Polls of students conducted in2007 reveal that 40% of students expect to get B-grades for simply attending the course, without regard toperformance. This is an average over all disciplines, in the pure sciences the numbers are much less ludicrous.

College really does differ from high-school. I hear a lot of students complain that they can’t put in study timebecause of work, or because they “have lives”. Here is some shocking news for those students; everybody has a job,and everybody “has a life”, yet the ones that will succeed will be the ones that take their studies as seriously as theytake their jobs and any other aspects of their lives. The current generation of students is not unique, they don’tface hardships that previous generations did not, and nothing more is expected of them than what was expected ofprior generations. And nothing less is expected of them! In fact the current generation has a lot of time-savingtechnology at its disposal, and good use should be made of this advantage that generations before did not have. Ihave faith in you, you can do it. All it takes is discipline.

Speaking of this technological advantage, this year (2010) I integrated quite a bit of symbolic computation involvingfree software such as REDUCE, axiom, maxima, and ode into the course, partly as a way of providing math assis-tance to students, and to provide the tools for pushing the basic physics forward into other disciplines, in supportof those disciplines and the students studying them. Problems and examples labeled with ∗ are very similar toproblems appearing on the physics GRE. I have added a great deal of optional material specifically for architecture,engineering, biomedical, chemistry and geoscience/hydrology students. Most of this material will not be covered inlecture, we only have time for the basics, and much of this optional material requires some mathematical backgroundbeyond that required for the standard portion of the course. This optional material is in chapters 10, 11, 13and 14 and is clearly marked as “not covered in 201”, but is there for you if you need it. I regard therequired mathematical background for the standard 201 course to be rather minimal (one semester of calculus, whatwas called freshman math when your parents were in college), others might say one semester is inadequate, but onesemester of calculus is the nation-wide norm for this course. Math refreshers and reviews are integrated into

i

ii FORWARD

the text. The fact is that you can’t do anything of any significance in the sciences without mathematics. Youmust deal with it. Learn the mathematics, or get used to making way for those who have learned it. Harsh words,but sound advice; I am on your side, which should be evident from the fact that I wrote this book for you, and Ihave combined it with as many tools as I can to make your job easier. But easier is not easy, and you will have tospend the time outside of class, do the homework, and learn the required mathematics.

In college at the University of Wisconsin-Madison in the late 70′s-early 80′s I had a good friend and classmate, DonnHenriksen. You will always treasure the short time that you spend in college, four (or five!) very short but importantyears. For me they passed quickly, I graduated, and lost track of many good people as we began our careers andfamilies and adult lives. After two decades, I was able to re-established contact with some, Donn included, thanksto the Internet (you see, it is good for something more than copyright-violation).

In the fall of 2006 Donn was diagnosed with late-stage stomach cancer, and by May of 2007, he had passed away.Donn managed to remain a true intellectual all of his life, his thirst for learning never diminished. He read throughthese notes, solved the problems, and uncovered quite a few typos in the winter of 2006/2007 as he refreshed hisunderstanding of the physics that we learned as classmates way back in 1976, and endured his treatments. Thesenotes, which will soon become a real book (but a free one) are dedicated to the memory of my friend Donn, whosedesire to keep learning as a life-long pursuit should serve as a model to which all students should aspire.

Contents

Forward i

1 Kinematics 11.1 Motion on a line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.1.1 Motion at constant velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.1.2 Motion at constant acceleration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.2 Arbitrary motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.3 Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

1.3.1 Basic vector operations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101.3.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

1.4 Appendix. Differential calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151.4.1 Partial derivatives versus total . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161.4.2 Slopes of curves without calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161.4.3 The rules of differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181.4.4 The binomial theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181.4.5 Series expansion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191.4.6 Rational functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211.4.7 Radicals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211.4.8 Basic integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 221.4.9 The logarithm and exponential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 241.4.10 Quadratic forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 241.4.11 Complex numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 251.4.12 Calculus assistance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

1.5 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

2 Projectile motion 372.1 Basic examples of ballistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 372.2 Using the computer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

2.2.1 Using ode . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 442.3 Air resistance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 442.4 Self-propelled projectiles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 462.5 Obstacles and targets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 472.6 Problem solving suggestions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 472.7 Appendix. REDUCE code . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 482.8 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

3 Newton’s Laws. Dynamics 533.1 Free-body diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 543.2 The equations of motion (EOMS) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 553.3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

4 Frictional Forces 654.1 Problem solving strategy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 704.2 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70

iii

iv CONTENTS

5 Circular motion at constant speed 775.1 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80

6 Work and Energy 836.1 Kinetic energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 836.2 Potential energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 846.3 Total energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 866.4 The master energy equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 866.5 Computing work done by forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87

6.5.1 Gravity is conservative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 896.5.2 Friction is non-conservative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89

6.6 When does a force not have a potential? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 946.7 Advantages and techniques of the work/energy theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 95

6.7.1 Principle of infinitismal work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 976.7.2 Solving for accelerations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99

6.8 Appendix. Integral calculus techniques . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1016.8.1 Integration by parts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1016.8.2 Variable Changes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102

6.9 Appendix. Computer assistance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1046.10 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105

7 Conservation of Momentum 1137.1 Elastic and Inelastic Collisions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1197.2 Forces, momenta and impulse . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1257.3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127

8 Rotational motion 1358.0.1 The acceleration components . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1368.0.2 The case of constant tangential acceleration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136

8.1 Kinetic energy of a rotating body . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1398.1.1 Parallel axis theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1428.1.2 Rolling motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144

8.2 Rotational dynamics and torque . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1458.3 Statics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1528.4 Work and torque . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1548.5 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156

9 Angular Momentum 1659.1 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172

10 Fluids 17710.1 Velocity fields of fluids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17710.2 Newton’s laws applied to fluids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179

10.2.1 Incompressible, irrotational fluids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18010.3 Advanced topics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183

10.3.1 Incompressible flow around obstacles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18710.3.2 The divergence theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 188

10.4 Appendix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18910.5 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 190

11 Oscillations 19511.1 The simple pendulum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20411.2 Oscillations in general . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20611.3 Classical elastica. Stress and strain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 208

11.3.1 Torsional strain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20911.3.2 Complete treatment of elasticity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 210

CONTENTS v

11.3.3 Architectural examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21711.3.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 224

12 Gravitation and Planetary Motion 23312.1 Newton’s universal law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23312.2 Gravitational potential energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23512.3 Planetary orbits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 241

12.3.1 Orbital corrections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24412.3.2 High-altitude projectiles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 246

12.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 247

13 Heat and temperature 25313.1 Early experiments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253

13.1.1 Joule-Thompson experiment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25313.1.2 Calorimetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254

13.2 Phase Changes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25613.3 Heat flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25713.4 Radiative cooling and heating . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25813.5 Statistical mechanics. Kinetic theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 260

13.5.1 Kronig-Clausius kinetic theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26013.5.2 Boltzmann’s theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26113.5.3 The meaning of temperature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26213.5.4 Boltzmann entropy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263

13.6 Convection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26413.7 Appendix. Some numerical values of material properties . . . . . . . . . . . . . . . . . . . . . . . . . . 27013.8 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 271

14 Equilibrium thermodynamics 27514.1 Core definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27514.2 The laws of thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 277

14.2.1 First Law of thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27714.2.2 Kelvin’s Second Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27914.2.3 Carnot’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28014.2.4 Non-reversible Process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28114.2.5 Clausius’ Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 283

14.3 Thermodynamic Potentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28314.3.1 The meaning of F ; the F-Theorem; . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28414.3.2 The meaning of G; the G-Theorem; . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28414.3.3 The meaning of entropy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28514.3.4 The meaning of enthalpy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28714.3.5 Material properties of matter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 288

14.4 Ideal gases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28814.5 PV -diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 291

14.5.1 The Carnot cycle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29314.5.2 Arbitrary systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 297

14.6 Non-PV work. Chemical and electrical . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29914.6.1 Chemical potential for ideal gases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30114.6.2 Electrochemical work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 302

14.7 More applications of µ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30414.7.1 The centrifuge . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30514.7.2 Osmosis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30614.7.3 Boiling point elevation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 307

14.8 Non-equilibrium applications. Diffusion, Graham’s and Fick’s laws . . . . . . . . . . . . . . . . . . . . 30814.9 Freshmen chemistry crammed into the chemical potential nutshell . . . . . . . . . . . . . . . . . . . . 309

14.9.1 Ionic equilibria . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31114.9.2 Acids and bases. pH . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 312

vi CONTENTS

14.9.3 Buffering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31314.9.4 Hydrolysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31414.9.5 Chemistry glossary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 314

14.10Advanced topics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31514.10.1Maxwell relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31514.10.2The Gibbs-Duhem relation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31714.10.3The Van der Waals gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31814.10.4An application of enthalpy; refrigeration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 319

14.11Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 322

15 Answers to the problems 333

16 Appendix 33916.1 Appendix I. Mathematical formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33916.2 Appendix II. Periodic table . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33916.3 REDUCE examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 341

Chapter 1

Kinematics

Kinematics is the description of motion without any consideration as to what causes the motion. It is simply themathematical description of where an object is at time t, and how fast it is moving, and in what direction.

The information about where the moving object is can be presented in two simple ways, as a graph of its positionversus time, or as a mathematical function, position as a function of time. We will start with the very simplest typeof motion, that which takes place on a straight line, which we will call one-dimensional motion.

1.1 Motion on a line

The mathematical description of motion in one dimension is incredibly simple, we specify the position x(t) of anobject on a coordinate line as a function of time. This function can be arbitrary, but it must at least be unicursal(what we think of intuitively as continuous), meaning that it can be drawn without lifting the pen from the paper

limdt→0

(x(t+ dt)− x(t)

)= 0, ∀ t (1.1)

so that if you are at x = 1.0m (for example) at t = 0 s, you can’t spontaneously jump to some other point in aninfinitismal time (no teleportation, no transporter beams, and so forth). We do not necessarily require x(t) to besmooth, which would to say

limdt→0

(x(t+ dt)− x(t)

dt

)∀ t (1.2)

exists and is bounded. For example the motion

x(t) =

1.0 + 2.0 t 0 ≤ t ≤ 13 + 4(t− 1) 1 ≤ t ≤ 27 + 3(t− 2) 2 ≤ t

(1.3)

is just fine, it describes motion at constant velocity,then at t = 1 s and again at t = 2 you change velocity.Motion graphs of x(t) versus t show this.

This brings us to a set of important ideas. The dis-placement ∆x of an object between times t1 and t2 isthe change in its position

∆x = x(t2)− x(t1) (1.4)

and the average velocity of the object over this inter-val is the rate with which its displacement changed 0.0 0.5 1.0 1.5 2.0 2.5 3.0

0

2

4

6

8

10

t, sec

x(t)

, m

1

2 CHAPTER 1. KINEMATICS

v(t1, t2) =x(t2)− x(t1)

t2 − t1(1.5)

This is positive if the object moves to the right (x(t2) > x(t1)) and is negative if it moves left, and so has a magnitudecalled the average speed

s(t1, t2) =∣∣∣x(t2)− x(t1)

t2 − t1

∣∣∣ (1.6)

and a direction indicated by its sign. The best way to think of average speed is as arc-length traveled per unit time,in one and two dimensions respectively this will be

s(t1, t2) =

∫ t2

t1

√v2 dt, s(t1, t2) =

∫ t2

t1

√v2x + v2y dt (1.7)

and you might need to be a bit careful in your calculation of√v2.

Instantaneous velocity is the limit of average velocity as t2 = t1 + dt→ t1, or as dt→ 0

v(t1) = limdt→0

x(t1 + dt)− x(t1)

dt=dx

dt(t1) (1.8)

in other words, the derivative of x(t1) evaluated at t1. The end of this chapter contains a comprehensivedifferential calculus review.

Acceleration measures how fast the velocity changes. The average acceleration is

a(t1, t2) =v(t2)− v(t1)

t2 − t1(1.9)

and instantaneous is

a(t1) = limdt→0

v(t1 + dt)− v(t1)

dt=dv

dt(t1) (1.10)

A position function must be continuous and smooth if you want to find its instantaneous velocity at any time.

Example 1. For the motion described by Eq. 1.3 find the displacement and average speed between t = 0 and t = 3.Your first step is to get v(t) = dx

dt (t) for each segment

v(t) =

2.0 0 ≤ t ≤ 14 1 ≤ t ≤ 23 2 ≤ t

(1.11)

and integrate from ti = 0 to tf = 3 , breaking up the integral (see the Chapter 1 appendix, Eq. 1.131) into partsover which v(t) is constant

∆ =

∫ 3

0

v(t) dt =

∫ 1

0

2 dt+

∫ 2

1

4 dt+

∫ 3

2

3 dt = 9m, v =∆

T=

9

3= 3

m

s(1.12)

(can you see why s = v for this example?).

Example 2. Lets extend this example, suppose that

v(t) =

2.0 0 ≤ t ≤ 1−4 1 ≤ t ≤ 23 2 ≤ t

(1.13)

Then

∆ =

∫ 3

0

v(t) dt =

∫ 1

0

2 dt−∫ 2

1

4 dt+

∫ 3

2

3 dt = 1m, v =∆

T=

1

3

m

s

s =1

3 s

(∫ 3

0

v(t) dt =

∫ 1

0

2 dt+

∫ 2

1

4 dt+

∫ 3

2

3 dt)= 3

m

s(1.14)

1.1. MOTION ON A LINE 3

Example 3. Consider the one-dimensional motion

x(t) =

3.0 t2 0 ≤ t ≤ 13 + 6(t− 1) 1 ≤ t ≤ 415 3 ≤ t

(1.15)

To get v(t), we differentiate this once;

v(t) =

6.0 t 0 ≤ t ≤ 16 1 ≤ t ≤ 40 3 ≤ t

(1.16)

Lets use gnuplot as described in problem 1.35 to graph both functions

f1(x)=(x<=1) ? 3*x**2 : 0

f2(x)=(x>1) ? 3+6*(x-1) : f1(x)

f3(x)=(x>3) ? 15 : f2(x)

plot[0:4][-1:7] f3(x)

0

2

4

6

8

10

12

14

16

0 0.5 1 1.5 2 2.5 3 3.5 4

f3(x)

f1(x)=(x<=1) ? 6*x : 0

f2(x)=(x>1) ? 6 : f1(x)

f3(x)=(x>3) ? 0 : f2(x)

plot[0:4] f3(x)

-1

0

1

2

3

4

5

6

7

0 0.5 1 1.5 2 2.5 3 3.5 4

f3(x)


1.1.1 Motion at constant velocity

Motion at constant velocity is very simple, the position function has the generic form

x(t) = x0 + v0 t (1.17)

Compute the average velocity

v(t1, t2) =(x0 + v0t2)− (x0 + v0t1)

t2 − t1= v0 (1.18)

The object moves at a constant rate (speed) in the same direction, forever. The instantaneous velocity is exactly thesame

v(t1) = limdt→0

(x0 + v0(t1 + dt))− (x0 + v0t1)

(t1 + dt)− t1= v0 (1.19)

and the accelerations (average and instantaneous) are of course zero. There is nothing more to this type of motion.

What is the meaning of x0? It is the position when the clock reads zero, x(0) = x0 + v0 · 0 = x0.

1.1.2 Motion at constant acceleration

Motion at constant acceleration is far more interesting (it is the motion occurring when constant forces are appliedto a body), the position function has the generic form

x(t) = x0 + v0 t+1

2at2 (1.20)

Compute the average velocity

v(t1, t2) =(x0 + v0t2 +

12at

22)− (x0 + v0t1 +

12at

21)

t2 − t1= v0 + a(

t2 + t12

) (1.21)

and instantaneous

v(t1) = limdt→0

(x0 + v0(t1 + dt) + 12a(t1 + dt)2)− (x0 + v0t1 +

12at

21)

(t1 + dt)− t1= v0 + a t1 (1.22)

The average and instantaneous accelerations are the same

a(t1, t2) =(v0 + at2)− (v0 + at1)

t2 − t1= a =

d

dt(v0 + at)

∣∣∣t=t1

(1.23)

What is the meaning of x0? It is the position when the clock reads zero, x(0) = x0 + v0 · 0 + 12a · 0

2 = x0. What isthe meaning of v0? It is the velocity when the clock reads zero v(0) = v0 + a · 0 = v0.

One of the most important generic classes of problems in one-dimensional kinematics is the problem of when, whereand if two moving objects with prescribed trajectories will meet or collide.

Example 4. A ball is thrown from ground level straight up into the air. A time T later another is thrown up at thesame speed. At what height do they collide.The first object, launched at t = 0, will have height versus time of

y1(t) = v0t−1

2gt2 (1.24)

The second ball travels the same path, but reaches each height at a time T after the first ball reaches that sameheight, and so

y2(t) = v0(t− T )− 1

2g(t− T )2 (1.25)

1.1. MOTION ON A LINE 5

0 2 4 6 80

10

20

30

40

50

60

t, (s)

y 1(t

), y

2(t)

, (m

)

Plots of these for v0 = 30ms and T = 2s areillustrated.From the graph we can immediately obtain auseful bit of data; the collision takes place whenthe first projectile is on its way back, having al-ready peaked in height. The algebraic solutionis gotten by finding the time at which bothprojectiles reach the same height;

y1(t∗) = y2(t∗)

or v0t∗ −1

2gt2∗ = v0(t∗ − T )

− 1

2g(t∗ − T )2 (1.26)

solving, we discover that the collision occurs at

t∗ =v0g

+T

2, at a height y∗ = v0t∗ −

1

2gt2∗ =

v20T

2g− gT 2

8(1.27)

0 2 4 6 80

50

100

150

200

t, (s)

y 1(t

), y

2(t)

, (m

)

Example 5. A car stopped at a traffic lightbegins at rest. At time t = 0 the light turnsgreen and the car accelerates with accelerationa. At just that instant a car traveling at speedv0 speeds through the light.How much time does it take the acceleratingcar to catch the speeder, and how far from thelight does this occur?A graph of both position versus time curves forthe vehicles

y1(t) =1

2at2 (1.28)

andy2(t) = v0t (1.29)

are below for v0 = 30ms and a = 10ms2 .

We can see from the slope of the graph that the accelerating car is traveling much faster than the speeder when theyfinally meet, which takes place when they arrive at the same point at the same time;

y1(t∗) = y2(t∗), or v0t∗ =1

2at2∗ giving t∗ =

2v0a, and location y∗ = v0t∗ =

2v20a

(1.30)

at which time the accelerating car has velocity

v1(t∗) = at∗ = 2v0 (1.31)


which we could see from the graph.

Example 6. Train A traveling at speed vA is a distance D from a switch in the track. At that instant train Benters the track traveling in the same direction but at a greater speed vB . What is the minimal acceleration thatthe engineer of train B must apply in order to avoid a collision?

0 2 4 6 80

50

100

150

200

t, (s)

x A(t

), x

B(t

), (

m)

The position versus time of the trains is

xA(t) = D + vAt, and xB(t) = vbt−1

2at2

(1.32)

graphs of these for vB = 2vA = 30ms and a =2ms2 ,D = 50 are illustrated.Our new condition is that when

xA(t∗) = xB(t∗)

that vA(t∗) = vB(t∗) (1.33)

so that there is no relative velocity between thetrains, hence no collision. This means

vA = vB − at∗)

or t∗ =vB − vA

a(1.34)

0 2 4 6 80

50

100

150

200

t, (s)

x A(t

), x

B(t

), (

m)

inserting this into

D + vAt∗ = vbt∗ −1

2at2∗ (1.35)

results in

a =(vB − vA)

2

2D(1.36)

The slopes of the two curves must also matchup when the trains meet, they will be tangentto one another as in the figure.

1.2. ARBITRARY MOTION 7

1.2 Arbitrary motion1

We will study motion at constant acceleration (due to constant forces acting on a body) for the first half of thecourse, but you have to realize that acceleration may not be constant, it can be arbitrary. The problem of describingthe position of a body given its acceleration (given the forces acting on it) is one of the fundamental problems of thiscourse. Mathematically, it amounts to solving differential equations, in fact this very problem is what calculuswas invented for. Let me illustrate two rather simple and universal approaches based on the idea that the motionwill be smooth, meaning that all of the derivatives of x(t) will exist.

Suppose that the acceleration of a body depends on its position; a simple example is motion caused by a spring orrestoring force

a(t) = −ω2 x(t) (1.37)

in which ω is a constant with units of 1/s, subject to x(0) = 1 and v(0) = 0. We will use one of the most powerfultools of calculus, the series expansion (see the appendix to this chapter), assume that all of the derivatives of xexist, and write

x(t) = x0 + x1 t+1

2x2 t

2 +1

3!x3t

3 +1

4!x4 t

4 +1

5!x5 t

5 + · · · (1.38)

in which xi are all constants. Note first that

x(0) = x0 = 1

Then

v(t) =dx

dt(t) = x1 + x2t+

1

2!x3 t

2 +1

3!x4t

3 + · · · (1.39)

and again examine things at t = 0

v(0) = x1 = 0

and compute a(t)

a(t) =d2x

dt2(t) = x2 + x3t+

1

2!x4 t

2 +1

3!x5t

3 + · · · (1.40)

and simply put these expansions into the equation of motion and solve in succession for the constantsby matching coefficients of like powers of t

a(t) = −ω2 x(t)

x2 + x3t+1

2!x4 t

2 +1

3!x5t

3 + · · · = −ω2(1 + 0 t+

1

2x2 t

2 +1

3!x3t

3 +1

4!x4 t

4 +1

5!x5 t

5 + · · ·)

(1.41)

We obtain

coefficient of t0; x2 = −ω2

coefficient of t1; x3 = 0

coefficient of t2;1

2!x4 = −ω2 1

2x2 = ω4

coefficient of t3;1

3!x5 = −ω2 1

3!x3 = 0

· · · (1.42)

and if we put things together, and actually paid attention in our calculus courses, we might even recognize that theseries adds up to some well-known function

x(t) = 1− ω2

2!t2 +

ω4

4!t4 + · · · = cos(ωt) (1.43)

This technique almost never fails, and is very simple and mechanical to carry out, its only drawback is that theanswer is a series, which you might not recognize how to sum up.

1This section is optional, but I strongly suggest reading it.


The second approach (very powerful and universal, see Chapter 6) makes use of the product rule of calculus. Beginwith the same equation,

a(t) = −ω2 x(t)

multiply by v(t) a(t) v(t) = −ω2 x(t) v(t)

both sides are t derivativesd

dt

(12v2(t)

)= −ω2 d

dt

(12x2(t)

)integrate

∫ t

0

d

dt

(12v2(t)

)dt = −ω2

∫ t

0

d

dt

(12x2(t)

)dt

1

2

(v2(t)− v2(0)

)= −ω

2

2

(x2(t)− x2(0)

)(1.44)

and solve for v(t)

v(t) =dx

dt=

√v2(0)− ω2

(x2(t)− x2(0)

)∫dt =

∫dx√

v2(0)− ω2(x2(t)− x2(0)

) (1.45)

and you have reduced the problem to doing an integral to get x(t). This will be our preferred technique (theenergy method) which we will expand upon when the concept of energy is introduced. Suppose that v(0) = 0 andx(0) = 1; ∫ t

0

dt =

∫ x

1

dx√ω2(1− x2

) , t =1

ωcos−1 x, x = cos(ωt)

We can reduce both of these approaches to a simple formula provided the acceleration is an explicit function of t(as opposed to implicit such as a = a(x) = a(x(t))). All we do is integrate;

dv

dt= a(t),

∫ v(t)

v(0)

dv =

∫ t

0

a(t′) dt′

v(t) =dx

dt= v(0) +

∫ t

0

a(t′) dt′∫ x(t)

x(0)

dx =

∫ t

0

v0 dt+

∫ t

0

(∫ t′

0

a(t′′) dt′′)dt′

x(t) = x(0) + v(0) t+

∫ t

0

(∫ t′

0

a(t′′) dt′′)dt′ (1.46)

and all you need to do is the integrals, for example let a(t) = α t in which α is a constant

x(t) = x(0) + v(0) t+

∫ t

0

(∫ t′

0

α t′′ dt′′)dt′ = x(0) + v(0) t+

∫ t

0

α

2t′2 dt′ = x(0) + v(0) t+

α

6t3

or suppose a(t) = a0, a constant

x(t) = x(0) + v(0) t+

∫ t

0

(∫ t′

0

a0 dt′′)dt′ = x(0) + v(0) t+

∫ t

0

a0 t′ dt′ = x(0) + v(0) t+

a02t2

Formulas! Hurrah, your favorite things! You can’t make an omelet without breaking some eggs, and you can’t doscience without a little bit of math.

1.3. VECTORS 9

1.3 Vectors

Most of the quantities that you will work with in this course have direction as well as a magnitude associated withthem, since we will want to describe motion in two or three dimensions. Such quantities are called vectors. Aquantity with no direction associated with it, that is unaltered by any changes in coordinate referential, is called ascalar. The most important scalars in classical physics are mass and temperature.

Typical vectors are forces, positions,velocities and momenta. The commonrepresentation of a vector is as a pointin space such as in the figure.

The point is specified by giving its x andy coordinates, either as an ordered pair(triple) or as multiples of unit vectors.We could specify the vector in the figureby giving Cartesian coordinate of its tip

v = (vx, vy) = vxi+ vyj

v = (vx, vy, vz) = vxi+ vyj+ vzk

i = (1, 0, 0)

j = (0, 1, 0)

k = (0, 0, 1)

(1.47)

from the Pythagorean theorem we obtain its length (magnitude)

|v| =√v2x + v2y, or |v| =

√v2x + v2y + v2z (1.48)

and its direction relative to the x-axis

θ = tan−1 vyvx

(1.49)

(thats inverse tangent, not one over tangent). In terms of the magnitude and direction we can express the compo-nents as

vx = |v| cos θ and vy = |v| sin θ (1.50)

Vectors as algebraic objects add component by component

a = (ax, ay), b = (bx, by), then a+ b = (ax + bx, ay + by) (1.51)

and similarly in three dimensions a = (ax, ay, az), b = (bx, by, bz)


a+b = (ax+ bx, ay+ by, az+ bz) (1.52)

It will prove to be very useful to be ableto compute the angle between two vec-tors, from the figure below we see that

θab = θa − θb = tan−1 ayax

− tan−1 bybx

(1.53)take the cosine of both sides of this equa-tion and use

cos(x− y) = cos(x) cos(y) + sin(x) sin(y)

cos(θab) = cos(θa − θb)

= cos(θa) cos(θb) + sin(θa) sin(θb)

(1.54)

cos(θab) =AxBx|A||B|

+AyBy|A||B|

(1.55)

Define the dot product of two vectors,which produces a scalar

a · b = axbx + ayby

a · b = axbx + ayby + azbz

(1.56)

and we see that the cosine of the anglebetween two vectors is

cos(θab) =a · b|a||b|

(1.57)

which is very useful since it involves onlythe components of the two vectors.

1.3.1 Basic vector operations

Remember at all times that there are only two allowed vector arithmetic operations that are legal (at this point);

1.3. VECTORS 11

Scalar multiplication. When you multiply a vector by a scalar, each component is amplified, changing the lengthof the vector but not its direction.

s r = s (rx, ry, rz) = (srx, sry, srz) (1.58)

forr a vector, s a scalar (1.59)

Vector addition. Vectors add component by corresponding component

a+ b = (ax, ay, az) + (bx, by, bz) = (ax + bx, ay + by, az + bz) (1.60)

The length of a vector is computed via the Pythagorean theorem;

|a| =√a2x + a2y + a2z (1.61)

Find a vector ⊥ to another vector Given a vector in two dimensions such as

a = (ax, ay) = axi+ ayj (1.62)

any vector b perpendicular to it is a vector such that

cos θab = 0 =a · b|a||b|

= 0, a · b = 0 (1.63)

Write this equation out in componentsaxbx + ayby = 0 (1.64)

This is one equation in two unknowns bx, by, and so cannot be uniquely solved. As long as both components ax anday are non-zero, we simply pick a value for bx and proceed;

Let bx = 1, ax + ayby = 0, by = −axay

(1.65)

and thereforeb = (1,−ax

ay) (1.66)

However if b ⊥ a, then so is sb for s any scalar, and we could even pick s = ay, making

sb = (ay,−ax) ⊥ a = (ax, ay) (1.67)

In three dimensions there is a whole plane of vectors ⊥ to

a = (ax, ay, az), ax, ay, az = 0 (1.68)

and so we proceed in steps. Let

b = (bx, by, bz), a · b = axbx + ayby + azbz = 0, if b ⊥ a (1.69)

There will be many such vectors, so for example let bx = 0, by = 1;

0 + ay + azbz = 0, bz = −ayaz

(1.70)

and then any vector

sb = (0, s,−sayaz

) ⊥ a (1.71)

This vector lies in the yz−plane. We could also set by = 0, bx = 1; to get

sb = (s, 0,−saxaz

) ⊥ a (1.72)


another vector perpendicular to a but lying in the xz−plane.

Linear independence. Suppose that you wish to solve the equation y− ax = 0 for a; it is easy enough, you simplydivide by x, since numbers form a field, and division by anything but zero is legitimate in a field. But what aboutsolving v1 + av2 = 0? You can’t divide vectors because they do not form a field. It might be possible thatthe only solution to

av1 + bv2 = 0 (1.73)

is a = b = 0. If this is true, then we say that v1,v2 are linearly independent. For example

a (1, 2) + b (3, 4) = 0 (1.74)

implies that a+3b = 0 and 2a+4b = 0, solving we obtain −2b+3b = b = 0, which makes a = 0 too, the vectors arelinearly independent. The most important sets of linearly independent vectors are the bases i, j in two dimensions,and i, j,k in three.

Write a vector as a sum of vectors. This is tantamount to using a set of vectors as a basis, and so requires thatthe set of vectors is a linearly independent set. Let

a = (ax, ay), b = (bx, by) (1.75)

be two non-collinear vectors. They are non-collinear if

a = sb (1.76)

Pick another vectorc = (cx, cy) (1.77)

and determine the two scalars α, β such thatc = α a+ β b (1.78)

We call this resolving vector c into a and b. Write it out in components

(cx, cy) = α(ax, ay) + β(bx, by) (1.79)

use the scalar multiplication rule(cx, cy) = (αax, αay) + (βbx, βby) (1.80)

use the vector addition rule(cx, cy) = (αax + βbx, αay + βby) (1.81)

and look at the two resulting equations

cx = αax + βbx, cy = αay + βby (1.82)

which is two equations in two unknowns, which we solve; in the first solve for β

β =cx − αax

bx(1.83)

and put it into the second

cy = αay +(cx − αax

bx

)by (1.84)

and get α by itself;

cybx − cxby = α(aybx − axby), α =cybx − cxbyaybx − axby

(1.85)

and go back and get β;

β =cyax − cxayaybx − axby

(1.86)

You will make extensive use of your abilities to solve two equations in two unknowns in this course.

1.3. VECTORS 13

Dot products. The dot product (or scalar product) can be neatly summarized by the rules

i · i = j · j = k · k = 1, i · j = j · i = 0

i · k = k · i = 0, k · j = j · j = 0 (1.87)

and the fact that it is distributive;

a ·(b+ c

)= a · b+ a · c (1.88)

Demonstration of these facts from the definition

a · b = axbx + ayby + azbz (1.89)

I will leave to you.

Cross products. The cross product is another vector product invented as a tool for computing areas. We knowthat the area of a rectangle is base × height, and the same is true for a parallelogram. What if we want to computethe area of a parallelogram whose sides are given by two vectors A = axi + ayj and B = bxi + byj? Of course wecan do it the long way and resolve the vectors into components, find the angle between them and construct the baseand height, but the cross product is designed to do all of this from just the components themselves.We want a vector product that represents the area of the parallelogram made from A and B, we call it A×B. Sincethere will be no area at all if A = B, we can ensure that the product gives us zero when we apply it to two copiesof the same vector if we define × such that

A×B = −B×A (1.90)

since then A×A = −A×A = 0. Then use the distributive property

A×B = (axi+ ayj)× (bxi+ byj)

= axbxi× i+ axbyi× j + aybxj × i+ aybyj × j = (axby − aybx)i× j (1.91)

Is this the area in question? It is, simply resolve the expression into magnitudes and phases;

(axby − aybx) (i× j) = |A||B|(cos θa sin θb (i× j)

− cos θb sin θa (i× j))

= |A||B| sin(θa − θb) (i× j))

(1.92)

which from the figure we see is precisely base × height.We now finish the job by assigning the cross product thestatus of a vector, so that i× j is a vector, by defining

i× j = k (1.93)

In order to have a consistent definition and make Eq. 1.91 true, we can show that this requires

i× i = j × j = k × k = 1, i× j = −j × i = k

k × i = −i× k = j, j × k = −k × j = i (1.94)

The cross product of two vectors lets us not only compute the area of the parallelogram made from the vectorsentirely in terms of its components, but also the sine of the angle between the vectors

sin(θa − θb) =|A×B||A||B|

(1.95)


and provides us with a test of whether or not two vectors are parallel to one another;

A×B = 0 if A ||B (1.96)

Perpendicularity

Two vectors a and b are ⊥ to each other if there is a θab = 90 degree angle between them, making

cos θab = 0, a · b = 0 (1.97)

Collinearity

Two vectors a and b are collinear if there is a scalar s such that

a = sb, (ax, ay, az) = (sbx, sby, sbz) (1.98)

Solve these three equations

s =axbx

=ayby

=azbz

(1.99)

To test for collinearity, we compute all three ratios, if they are not all the same, the vectors are not collinear.Remember that you can divide numbers, such as individual components of vectors, but you cannot divide by thevectors themselves.

1.3.2 Examples

Example 7. Let

v = (1, 1, 1) = i+ j+ k, u = (1,−2, 1) = i− 2j+ k, w = (1, 0,−1) = i− k

Find the lengths of these vectors.

|v| =√v · v =

√1 + 1 + 1 =

√3, |u| =

√1 + 4 + 1 =

√6, |w| =

√1 + 1 =

√2

Example 8. Show that these vectors have 90o angles between any pair.

u · v = 0 = (1)(1) + (1)(−2) + (1)(1), w · v = 0 = (1)(1) + (−1)(1), w · u = 0 = (1)(1) + (−1)(1)

Example 9. Show that these vectors are linearly independent.

0 = av + bu+ cw = (a+ b+ c, a− 2b, a+ b− c) = (0, 0, 0)

The middle one says a = 2b, the las one says c = 3b, and the first says (1+2+3)b = 0, so b = 0 making a = b = c = 0the only solution.

Example 10. Let

z = (1, 2, 3), Show that z = αu+ βv + γw, and find α, β, γ

We know this can be done uniquely since u,v, c are linearly independent. Because we can use the dot-productsabove, note

z · u = (1)(1) + (−2)(2) + (3)(1) = 0 =(αu+ βv + γw

)· u = α|u|2 = 6α, we find α = 0

z · v = (1)(1) + (1)(2) + (1)(3) = 6 =(αu+ βv + γw

)· v = β|v|2 = 3β, we find β = 2

z ·w = (1)(1) + (−1)(3) = −2 =(αu+ βv + γw

)·w = γ|w|2 = 2γ, so that γ = −1

1.4. APPENDIX. DIFFERENTIAL CALCULUS 15

1.4 Appendix. Differential calculus

It is very important that you be able to perform the basic calculus operations;1. computations of limits and derivatives, including partial derivatives2. expansions of functions, parameterization of curves3. simple integrations.

In this review we will make sure that you are up to speed on all of these basic skills, and we will review trigonometrywhile we are at it.

There is no substitute for a full year of calculus instruction, and so math 221 is an absolute prerequisite forphysics 201, and math 222 is an absolute prerequisite for physics 202. There is no back door.

A smooth function y = f(x) of a variable x is differentiable at x if its derivative

f ′(x) =df(x)

dx= limdx→0

f(x+ dx)− f(x)

dx

exists (is a number) at x.

The limit-taking process is very simple; weexpand the numerator in ascending powers ofdx, perform the division, and take the limit bysetting dx = 0 afterwards (in a nutshell).The geometrical interpretation of the deriva-tive of f(x) at x0 is that f ′(x0) is the slope ofthe line tangent to f(x) at x0;

Therefore a useful formula for translating cal-culus to geometry is

tan θ =(df(x)

dx

)x=x0

(1.100)

for the tangent to the curve at x0.θ

x0

y=f(x)

The limit taking process is most easily handled for polynomial functions such as

f(x) =N∑n=0

an xn

by use of the following simple rule;

d

dx

(f(x) + g(x)

)=df(x)

dx+dg(x)

dx(1.101)

which is easy to prove from the definition above. This says that the derivative of a (finite) sum is the sum of thederivatives of the summands.

Example 11. Let f(x) = x3, the steps taken in computing the derivative are;Step 1. Write out the fraction


f(x+ dx)− f(x)

dx=

(x+ dx)3 − x3

dx

=(x3 + 3x2 dx+ 3x (dx)2 + (dx)3)− x3

dx

Perform the division(x3 + 3x2 dx+ 3x (dx)2 + (dx)3)− x3

dx=

3x2 dx+ 3x (dx)2 + (dx)3

dx

= 3x2 + 3x dx+ (dx)2

Perform the limit (set dx = 0);dx3

dx= lim

dx→0

(3x2 + 3x dx+ (dx)2

)= 3x2 (1.102)

Example 12. Let f(x) = a+ b x2, in which a and b are constants.Step 1. Write out the fraction

f(x+ dx)− f(x)

dx=

(a+ b(x+ dx)2)− (a+ bx2)

dx

=(a+ b(x2 + 2x dx+ (dx)2))− (a+ bx2)

dx

Perform the division;(a+ b(x2 + 2x dx+ (dx)2))− (a+ bx2)

dx=

2bx dx+ b(dx)2

dx= 2bx+ b dx

Perform the limit (set dx = 0);d

dx

(a+ bx2

)= lim

dx→0

(2bx+ b dx

)= 2bx (1.103)

which we can see is the sum of the derivatives of the two terms a and bx2, the derivative of a constant being zero.

1.4.1 Partial derivatives versus total

Technically nearly all derivatives computed in this course are partial derivatives, derivatives of a function of morethan one variable with respect to one variable. If the function depends on one variable, these derivativesare the same. The definition is very simple(∂f(x, y)

∂x

)y= limdx→0

f(x+ dx, y)− f(x, y)

dx,

(∂f(x, y)∂y

)x= limdy→0

f(x, y + dy)− f(x, y)

dx(1.104)

so that partial derivation with respect to x regards everything but x as a constant. For example

f(x, y) = x+ 2y,(∂f(x, y)

∂x

)y= 1,

(∂f(x, y)∂y

)x= 2

f(x, y) = 3x y(∂f(x, y)

∂x

)y= 3y,

(∂f(x, y)∂y

)x= 3x

f(x, y) =x

y,

(∂f(x, y)∂x

)y=

1

y,

(∂f(x, y)∂y

)x= − x

y2(1.105)

in which the basic rules for differentiating powers, products and quotients (which we are about to list) apply topartial as well as total derivatives.

1.4.2 Slopes of curves without calculus

Using nothing more that basic algebra, one can easily compute the slopes of lines tangent to arbitrary planar curves,you simply need to make a few simple observations about the curve in question, or even better yet, be able to graph it.


-60

-40

-20

0

20

40

60

-4 -3 -2 -1 0 1 2 3 4

oo

Consider for example the curve y = ax2 + bx+ c. Suppose that you want to find the slope of the line tangent to thecurve at some point x0. You can draw three types of straight lines relative to this curve; ones that don’t intersectthe curve, ones that intersect it at one point, and lines that intersect the curve at two points. “Generic” lines willintersect the curve at two points, but only vertical lines (infinite slope) and tangent lines intersect at one point. Thisis the key observation, since we know that the slope of the curve is not infinite for any finite value of x.

To find the slope of the curve (slope of tangent line) at point (x, y) = (u, v) on the curve, notice that v = au2+bu+c,and rewrite the curve equation as

y − (ax2 + by + c) = 0

The parametric equation of a line of slope m through (u, v) is

x = u+ t, y = mt+ v, y = m(x− u) + v

(notice that when the parameter t = 0, you are at the point (u, v) on the curve). Substitute this into our equation

(mt+ v)− (a(t+ u)2 + b(tu) + c) = 0, (mt+ v)− (at2 + 2atu+ au2 + bt+ bu+ c) = 0

Use v − (au2 + bu+ c) = 0;

mt− (at2 + 2atu+ bt) = 0, at2 + (2au+ b−m)t = 0

If this line is tangent to the curve at (u, v) then t = 0 must be a double-root of this equation! In otherwords

m = 2au+ b

which is exactly the derivative of our curve at the point (u, v)

d

dx

(ax2 + bx+ c

∣∣∣x=u

= 2au+ b

Example. Find the slope of the line tangent to a circle x2 + y2 = R2 at the point (x0, y0) on the circle.


Once again you see three types of lines, those that intersect the circle at 0, 1, 2 points. The lines through a singlepoint of the circle are the tangent (or the vertical, which we don’t care about) lines. Play the same game;

y2 + x2 −R2 = 0, (mt+ v)2 + (u+ t)2 −R2 = 0, (m2t2 + 2mvt+ t2 + 2ut) + (v2 + u2 −R2) = 0

and look for the condition making the root a double root;

(m2 + 1)t2 + (2mv + 2u)t = 0, t = 0 is a double root iff m = −uv

Verify this with calculus;

y2 = R2 − x2,dy2

dx= 2y

dy

dx= 0− 2x,

dy

dx

∣∣∣x=u,y=v

= −xy

∣∣∣x=u,y=v

= −uv

These methods work (the ideas are from the field of algebraic geometry) because the derivative has a geometricalmeaning; it is the slope of the line tangent to the curve, the slope of the line through two points that are taken tobe the same. The methods work just as well for rational functions as for polynomial.Example. Find the slope of the line tangent to curve y = 1/x at the point (u, 1/u).Once construct the line x = u + t, y = 1

u + mt of slope m that passes through both points (u, 1/u) and (u, 1/u)(that’s not a typo), the first point corresponding to t = 0, the second t = 0 (a double root);

xy − 1 = 0, (u+ t)(1

u+mt)− 1 = 0, mt2 + (mu+

1

u)t = 0

a quadratic equation with a double-rrot t t = 0 if m = − 1u2 , the correct derivative dy

dx = − 1x2 at x = u.

1.4.3 The rules of differentiation

Derivatives of a product f(x) g(x) can be easily computed from the definition,

d

dx

(f(x)g(x)

)= limdx→0

f(x+ dx)g(x+ dx)− f(x)g(x)

dx

by replacing f(x+ dx) with f(x+ dx)− f(x) + f(x) and rearranging

d

dx

(f(x)g(x)

)= lim

dx→0

(f(x+ dx)− f(x) + f(x)

)g(x+ dx)− f(x)g(x)

dx

= limdx→0

(f(x+ dx)− f(x)

)g(x+ dx) + f(x)

(g(x+ dx)− g(x)

)dx

= limdx→0

(f(x+ dx)− f(x)

dxg(x+ dx)

)+ f(x) lim

dx→0

(g(x+ dx)− g(x)

dx

)=

df(x)

dxg(x) + f(x)

dg(x)

dx(1.106)

If f(x), f ′(x), g(x) and g′(x) all exist.We state this as being the product rule for derivatives.

d

dx

(f(x)g(x)

)=df(x)

dxg(x) + f(x)

dg(x)

dx(1.107)

1.4.4 The binomial theorem

This theorem is of great antiquity, and is extremely useful for both algebraic and calculus applications. It says that

(a+ b)N =

N∑m=0

(N

m

)ambN−m (1.108)


where the number (N

m

)=

N !

m!(N −m)!

is a binomial coefficient, and the factorial of an integer N is

N ! = N · (N − 1) · (N − 2)...2 · 1, 1! = 1, 0! = 1

(the last relation is a definition). For example

(a+ b)2 = a2 + 2ab+ b2

(a+ b)3 = a3 + 3a2b+ 3ab2 + b3

(a+ b)4 = a4 + 4a3b+ 6a2b2 + 4ab3 + b3 (1.109)

1.4.5 Series expansion

This last calculation was a little on the tricky side, but there exists a powerful tool for performing most of theoperations of calculus in a simple way, the series expansion.

We suppose that the function f(x) exists at the point x0, and for that matter that it exists near x0. Let x− x0 besmall, so that x is close to x0. The idea of the series expansion is that in the neighborhood of x0, we could replacef(x) with a polynomial

Pf (x) =

N∑n=0

ann!

(x− x0)n (1.110)

in which n! = n · (n− 1) · (n− 2)...2 · 1 is our factorial of the integer n.

The number of terms N that we need to calculate to get Pf depends on what we want to do with it, and is based onthe following concept: the function f(x) and polynomial Pf (x) agree at x0, and have the same derivativeat x0, and the same second derivative, and so on up to the N th derivative. We would call Pf (x) an N

th orderseries expansion of f(x) about the point x0.Step 1. Both f(x) and Pf (x) agree at x0;

f(x0) = Pf (x0) = a0 + a1(x0 − x0) +a22!(x0 − x0)

2 + ...+aNN !

(x0 − x0)N = a0

requires that a0 = f(x0).Step 2. Both f ′(x) and d

dxPf (x) agree at x0;

f ′(x0) = 0 + a1 + 2a22!(x0 − x0) + 3

a33!(x0 − x0)

2 + ...+NaNN !

(x0 − x0)N−1 = a1

requires that a1 = f ′(x0) =dfdxx=x0

.

Step 3. Both f ′′(x) and ddxPf (x) agree at x0;

f ′′(x0) = 0 + 0 + 2a22!

+ 3 · 2a33!(x0 − x0) + ...+N · (N − 1)

aNN !

(x0 − x0)N−2 = a2

requires that a2 = f ′′(x0) =d2fdx2 x=x0

.

For ninety percent of all of the calculus applications in our physics text, this is enough; the polynomial Pf (x) thatagrees with f(x) up to two derivatives in the neighborhood of x0 is

Pf (x) = f(x0) + f ′(x0) (x− x0) +1

2f ′′(x0) (x− x0)

2 + ... (1.111)

This is called the Euler-Maclaurin or Taylor series for f(x) near x0, and it may be substituted in place of f(x)in the neighborhood of x0.


What do we use it for? For starters it can be used to get formulas for derivatives of products and quotients. Inmost applications you only need to keep one or two terms in a Taylor series. For example, let x = t + dtand x0 = t, then

Pf (t+ dt) = f(t) + f ′(t) dt+1

2f ′′(t) (dt)2 + ... (1.112)

and you can replace any occurrence of f(t+dt) in a formula that involves taking the limit dt→ 0 with this expression.

Example 13.

d

dt

(f(t)g(t)

)= lim

dt→0

(f(t+ dt)g(t+ dt)− f(t)g(t)

)dt

= limdt→0

(f(t) + f ′(t) dt+ ...

)(g(t) + g′(t) dt+ ...

)− f(t)g(t)

dt

= limdt→0

g(t) f ′(t) dt+ g′(t) f(t) dt+ ...

dt= g(t) f ′(t) + g′(t) f(t) (1.113)

and we are done quickly and cleanly, all of the terms in (...) contain at least two factors of dt, and so in the limitbecome zero.

Example 14. l’Hospitals rule is a formula for computing the limit of the ratio of two functions that both vanishat x0,

limx→x0

f(x) = 0 = limx→x0

g(x)

The limit of the ratio is then

limx→x0

f(x)

g(x)= limx→x0

f(x0) + f ′(x0) (x− x0) +12f

′′(x0)(x− x0)2 + ...

g(x0) + g′(x0) (x− x0) +12g

′′(x0)(x− x0)2 + ...

but both f(x0) = 0 and g(x0) = 0, so

limx→x0

f(x)

g(x)= limx→x0

f ′(x0) (x− x0) +12f

′′(x0)(x− x0)2 + ...

g′(x0) (x− x0) +12g

′′(x0)(x− x0)2 + ...

divide out the factor (x− x0) from numerator and denominator:

= limx→x0

f ′(x0) +12f

′′(x0)(x− x0) + ...

g′(x0) +12g

′′(x0)(x− x0) + ...

and in the limit x→ x0, (x− x0) → 0,

limx→x0

f ′(x0) +12f

′′(x0)(x− x0) + ...

g′(x0) +12g

′′(x0)(x− x0) + ...=f ′(x0)

g′(x0)

We restate this as l’Hospitals rule;

limx→x0

f(x)

g(x)=f ′(x0)

g′(x0)(1.114)

provided g′(x0) is non-zero. If g′(x0) and f ′(x0) are in fact both zero, we simply repeat the process notingthat in

limx→x0

f ′(x0) +12f

′′(x0)(x− x0) + ...

g′(x0) +12g

′′(x0)(x− x0) + ...

the first term in both numerator and denominator are zero and we can divide out another factor of (x− x0);

limx→x0

f(x)

g(x)= limx→x0

0 + 12f

′′(x0) + ...

0 + 12g

′′(x0) + ...=f ′′(x0)

g′′(x0), if f(x0) = g(x0) = f ′(x0) = g′(x0) (1.115)


1.4.6 Rational functions

Consider a function that is the ratio of two functions, both of which you can differentiate;

h(x) =f(x)

g(x)

To compute the derivative of h(x) we take these algebraic steps, first

h(x) g(x) = f(x)

now apply the product rule

d

dx

(h(x) g(x)

)=

d

dxf(x) = f ′(x), h′(x) g(x) + h(x) g′(x) = f ′(x)

and rearrange

h′(x) =f ′(x)− h(x) g′(x)

g(x)=f ′(x)− f(x)

g(x) g′(x)

g(x)=f ′(x) g(x)− f(x) g′(x)

g2(x)(1.116)

which we will call the quotient rule.

1.4.7 Radicals

Consider the radicalf(x) = n

√x

To compute its derivative, first raise both sides to the nth power

fn(x) = f(x)fn−1(x) = x

Now differentiate and apply the product rule repeatedly to the left side

d

dxfn(x) =

d

dxx = 1, n fn−1(x)f ′(x) = 1

solve for f ′(x);

f ′(x) =1

nfn−1(x)=

1

nxn−1n

=1

nx

1n−1

We have shown thatd

dxn√x =

d

dxx

1n =

1

nx

1n−1 (1.117)

and therefore for any power a, integral, rational or otherwise

d

dxxa = a xa−1

which we call the power rule for differentiation.

Example 15. Find a series expansion for f(x) =√x valid near x0 = 4.

The first step is to compute a few derivatives, using the power rule with a = 12 ,

d

dx

√x =

1

2

1√x,

d2

dx2√x =

d

dx

1

2

1√x= − 1

221√x3

and so inserting this all into Eq. 5 we find that

√4 + dx =

√4 +

1

2√4dx+

1

2

(− 1

221√43

)(dx)2 + ... = 2 +

dx

4− (dx)2

64+ ...


This should be written using x = 4 + dx, as

√x = 2 +

(x− 4)

4− (x− 4)2

64+ ...

and in any formula involving√x that will be used for x near 4, this is a valid replacement.

In particular, this can be used to calculate square roots of numbers close to 4, such as 5 for which

√5 ≈ 2 +

1

4− 1

64+ ... = 2.234375...

which gives quite good accuracy (we are off in the third decimal place) with only these three terms in the series.

1.4.8 Basic integration

Integration is the anti-derivative; this is how we will define it. The process of integration must undo the process ofdifferentiation, and so we define ∫ b

a

df(x)

dxdx = f(b)− f(a) (1.118)

A Riemann sum will do the trick; consider∫ b

a

f(x) dx = limN→∞

N∑n=0

f(a+(b− a)

Nn)

(b− a)

N(1.119)

which is the area under the curve f(x) from a to

b, being made up of little strips of width (b−a)N of

height f(a+ n (b−a)N ).

How does this undo the derivative? Let

∆ =(b− a)

N(1.120)

and write the Riemann sum as∫ b

a


N∑n=0

f(a+ n∆) ∆ (1.121)

and insert into this a differentiated function

f(x) =dg(x)

dx= lim

∆→0

g(x+∆)− g(x)

∆

a b∫ b

a


∆(f(a) + f(a+∆) + ...+ f(a+ (N − 1)∆) + f(a+N∆)

)(1.122)

= limN→∞

∆(g(a+∆)− g(a)

∆+g(a+ 2∆)− g(a+∆)

∆+ ...

+g(a+ (N − 1)∆)− g(a+ (N − 2)∆)

∆

+g(a+N∆)− g(a+ (N − 1)∆)

∆

)(1.123)


You can see that consecutive terms partially cancel, and so does the ∆ in the denominator, leaving

= limN→∞

(g(a+N∆)− g(a)

)= limN→∞

(g(a+N

(b− a)

N)− g(a)

)= g(b)− g(a) (1.124)

Integration is a harder problem than differentiation, since the only procedure for performing integrals is to either dothe Riemann sum directly, or find a function whose derivative is the integrand, or to perform variable changes thatput the integrand into a more readily recognized form.

Example 16. It is not hard to show thatN∑n=1

n =N(N + 1)

2(1.125)

To do it lay out all numbers 1 through N in a row

1 + 2 + 3 + 4 + 5 + 6 + ...+ (N − 2) + (N − 1) +N (1.126)

and below it all numbers in reverse order

N + (N − 1) + (N − 2) + ...+ 6 + 5 + 4 + 3 + 2 + 1 (1.127)

and add the two rows

(N + 1) + (N + 1) + (N + 1) + ...+ (N + 1) + (N + 1) + (N + 1) = N(N + 1) (1.128)

but this is each number counted twice.We can use this to integrate∫ b

a

x dx = limN→∞

N∑n=0

(a+ n∆)∆ = limN→∞

((N + 1)a∆+∆2

N∑n=0

n)

(1.129)

with ∆ = b−aN . Put this in;∫ b

a

x dx = limN→∞

((N + 1)

a(b− a)

N+

1

2N(N + 1)

(b− a

N

)2)= a(b− a) +

1

2(b− a)2 =

1

2b2 − 1

2a2 (1.130)

By considering more difficult Riemann sums we can establish∫ b

a

xn dx =1

n+ 1

(bn+1 − an+1

), n = −1 (1.131)

This formula is valid for any n = −1, even irrational values.

Like differentiation, integration is a linear operation∫ b

a

(f(x) + g(x)) dx =

∫ b

a

f(x) dx+

∫ b

a

g(x) dx (1.132)

and another very useful property inherited from the Riemann sum definition is∫ c

a

f(x) dx =

∫ b

a

f(x) dx+

∫ c

b

f(x) dx (1.133)

What is integration used for in 201? Suppose that you know the value of a function, such as the position of a body,at time t0, and the velocity at all times. Integration is used to recover the position at any time t from

v(t) =dx

dt, x(t) = x(t0) +

∫ t

t0

v(t) dt (1.134)


1.4.9 The logarithm and exponential

In Eq. 1.131 the case n = −1 is special. Consider the integral

g(x) =

∫ x

1

dx′

x′(1.135)

since for such a function ∫ ab

1

dx

x=

∫ a

1

dx

x+

∫ ab

a

dx

x(1.136)

and the variable changex = ay (1.137)

turns this into ∫ ab

1

dx

x=

∫ a

1

dx

x+

∫ b

1

dy

y, or g(ab) = g(a) + g(b) (1.138)

This is the law of exponents, recall that xa · xb = xa+b, from which we conclude that g(a) is an anti or inverseexponentiation, we need only to establish the base, which we will call “e”. In other words f(x) = ex and g(x) areinverses of one another,

g(f(x)) = x = f(g(x)) (1.139)

From our integral representation

lnx =

∫ x

1

dy

y(1.140)

we see thatd

dxlnx =

1

x(1.141)

Apply the chain rule to Eq. 1.139

d

dx

(g(f(x)

)=dx

dx= 1,

1

f(x)

df

dx= 1,

df

dx= f(x) (1.142)

which is the law of differentiation of the antilog or exponential function. We can use our power series methods todetermine e;

f(x) =∞∑n=0

1

n!

(dnfdxn

)x=0

xn

=

∞∑n=0

1

n!

(f(x)

)x=0

xn =

∞∑n=0

1

n!xn

e = f(1) =∞∑n=0

1

n!1n = 1 +

1

2!+

1

3!+ · · · = 2.71828 · · · (1.143)

To summarize we call g(x) = ln(x), f(x) = ex and we have

ln(x) =

∫ x

1

dx′

x′, eln(x) = ln(ex) = x,

d

dxex = ex,

d

dxeax = a eax (1.144)

1.4.10 Quadratic forms

Solving an equation such asax2 + bx+ c = 0

for x is relatively simple. We will follow a procedure that actually generalizes to cubic and even quartic equations.First divide by a and write the equation in the following way

x2 +b

ax+

c

a= 0 = (x− r1)(x− r2)


where r1 and r2 are the two roots that we are looking for. Then let

r1 = u+ v, r2 = u− v

so that

(x− r1)(x− r2) = x2 − (r1 + r2)x+ r1r2

= x2 − 2ux+ (u2 − v2) = x2 +b

ax+

c

a

and we find that

−2u =b

a, u = − b

2a

and

u2 − v2 =b2

4a2− v2 =

c

a, v2 =

b2 − 4ac

4a2

and the roots are

r1 =−b+

√b2 − 4ac

2a, r2 =

−b−√b2 − 4ac

2a(1.145)

which is our familiar Quadratic Formula.

1.4.11 Complex numbers

It may happen that the discriminant b2 − 4ac of the quadratic form is negative. Denote the imaginary numberi =

√−1. Then 4ac− b2 is real, and

r1 =−b+ i

√4ac− b2

2a, r2 =

−b− i√4ac− b2

2a

Let R and I be real numbers (R, I ∈ R). We call C = R + i I a complex number, with real part R = ReC andimaginary part I = ImC. The real and imaginary parts of a complex number are always real.Complex numbers do not make math more complicated, but just the opposite. All things become much simpler whenone is willing to use complex numbers, especially calculus. Complex numbers enable one to perform practically anydefinite integral (one with specified limits of integration) in a simple, almost mechanical way.Let x, y be real variables (x, y ∈ R), and f(x, y) be a function that is complex-valued. This means that f(x, y) iscomplex (f(x, y) ∈ C), and so

f(x, y) = U(x, y) + i V (x, y), U(x, y), V (x, y) ∈ R

and given some known-to-be-complex function f(x, y), finding the two functions that are its real and imaginary partsis an important first step in working with f .Example

f(x, y) = (x+ iy)2 = (x2 − y2) + i(2xy), U = x2 − y2, V = 2xy

Example

f(x, y) = (x+ iy)3 = (x3 − 3xy2) + i(3x2y − y3), U = x3 − 3xy2 V = 3x2y − y3

Example

f(x, y) =1

x+ iy=

1

x+ iy

x− iy

x− iy=

x

x2 + y2− i

y

x2 + y2

If f = U + iV , then f = U − iV is called its complex conjugate. Note that f f = U2 + V 2 ∈ R.


1.4.12 Calculus assistance

You should be able to deduce that we will be using calculus extensively, physics 201 is a calculus-based course. Ifyou are rusty, one option is to use the computer to re-build your math skills on an as-needed basis. The physicsdepartment research support servers are running REDUCE, axiom and maxima as CGI programs, and of coursethese symbolic math programs are free software and can be downloaded from your Internets. Both programs areexcellent, and can help you out by performing the basic calculus operations for you while you re-learn the ropes.Don’t become over-reliant on the computer to solve your problems. Someday soon you will be in direct competitionwith people who have the math skills for the job, and they will whip you like the family mule if your skills fall short.

I suggest downloading gnuplot to draw graphs, and either REDUCE (http://sourceforge.net/projects/reduce-algebra/) or maxima (http://maxima.sourceforge.net/), or open-axiom http://www.open-axiom.org/, and installthem on your own computer. If you don’t want to do this, you can run all four fromhttp://azazelo.uwp.edu/cgi-bin/gnuplot.cgihttp://azazelo.uwp.edu/cgi-bin/reduceweb.cgihttp://azazelo.uwp.edu/cgi-bin/maxima.cgihttp://azazelo.uwp.edu/cgi-bin/axiom.cgiHere is how to perform basic calculus operations in both CAS systems;

Define a function

REDUCE

depend f,x;

f:=(q1*k/(x-a)-q2*k/x);

maxima

f: 1/(a*a+x*x);

axiom

f:= 1/(a*a+x*x)

print f

Differentiate a function

REDUCE

depend f, x;

f:=1/(x^2+a^2);

df(f,x);

df(f,x,2);

maxima

f: 1/(a*a+x*x);

diff(f,x);

diff(f,x,2);

axiom

f:=1/(x^2+a^2)

print f

print D(f,x)

Partial derivatives

REDUCE

depend f, x, y;

f:=1/(x^2+y^2);

df(f,x);

df(f,y);

maxima

f: 1/(y*y+x*x);

diff(f,x);

diff(f,y);

axiom

f:= 1/(y*y+x*x)

print D(f,x)

print D(f,y)

Expand a function is a series in x (about point x = 2)

REDUCE

load_package taylor;

f:=1/(x^2+a^2);

taylor(f,x,2,4);

maxima

taylor(1/(x-1),x,2,4);

axiom

q:=series(1/(x-1),x=0)

print q


Solve a system of linear equations

REDUCE

solve(R1*I1+R3*I3-V0,

R1*I1-R2*I2,

I1+I2-I3,I1,I2,I3);

maxima

f: 3*x+3*y-6*z=1;

g: x-z=1;

h: 2*x+4*y=1;

solve([f,g,h],[x,y,z]);

axiom

S==[a*x+b*y-f,c*x+d*y-g]

print solve(S,[x,y])

Perform indefinite integral

REDUCE

int(1/(a^2+x^2),x);

maxima

integrate(1/(a^2+x^2),x);

axiom

f:=integrate(1/(x+a),x)

print f

Perform definite integral∗

REDUCE

load_package defint;

int(x*e^(-x),x,0,infinity);

maxima

integrate(1/(a*a+x*x),x,0,inf);

nonzero;

positive;

axiom

f:=integrate(1/(x^2+1),x= -1 .. 1)

print f

f:=integrate(1/(x^2+1),

x= %minusInfinity .. %plusInfinity)

print f

Extremize a function

REDUCE

depend f,x;

f:=(q1*k/(x-a)-q2*k/x);

solve(df(f,x),x);

maxima

f: q1/(x-1)-q2/x;

solve(diff(f,x),x);

axiom

f:=q/(x-1)-qp/x

print f

print radicalSolve(D(f,x),x)

∗ Maxima will ask you questions to resolve cases, if your session is interactive you can answer them. In batch modeyou can anticipate them. You can see similarities in the syntax; both CAS systems are programmed in common LISP.

I strongly recommend that you re-build your calculus skills if they are not where you need them to be. You willnot be able to use any computer aids on exams, and everything in the sciences will be easier for you if your math isstrong, a fact well-understood by those people that you will soon be competing with for jobs.


1.5 Problems

1. Suppose that you are planning a trip (starting at x = 0 at t = 0) in which you will accelerate for time t1 atacceleration a, cruise at the constant speed that you will have achieved at the end of your acceleration, for a timet2, and finally you will brake to rest with acceleration −a for time t1.Draw a picture of x(t) versus t, using a = 1.0ms2 , t1 = 25 s, and t2 = 100 s.

2. Compute the maximal speed achieved during the trip of the previous problem, and compute the total distancetraveled. Compute the average velocity v for the entire trip. Make a graph of the instantaneous velocity v(t) versust for the trip. Find the area under the graph of v(t) versus t. How is this area related to v?

3. Another driver plans to race you, in an identical car (each car can accelerate/brake at the same rate ±a only).This driver will accelerate/brake for t3, and will coast at the constant maximal speed attained during the accelerationfor t4. That makes her trip-time 2t3 + t4. Find the acceleration time t3 that makes the total trip time minimal. Forhow long does this driver move at constant speed?

4. Suppose that a car consumes gasoline at a rate dGdt = 1.4× 10−5 gal·s

m |v|, where |v| is the speed of your car. Thisis pretty much how things would work if only mechanical friction was responsible for consumption. Compute theamount of fuel used in traveling for a time 1hr at a speed of 20ms . Compute the amount of fuel used in acceleratingfrom rest with a = 2.0ms2 until you reach 20ms .

5. An object moves at constant acceleration

x(t) = x0 + v0t+1

2at2

Find the average velocity v between the times t1 and t2, with t2 > t1. Find the time t∗ between t1 and t2 such thatthe instantaneous velocity at t∗ equals the average velocity between t1 and t2. This is one reason why such motionis special.

6. An object moves at constant “jerk”

x(t) = x0 + v0t+1

2at2 +

1

6Jt3

Find the average velocity v between the times t1 and t2, with t2 > t1. Find the time t∗ between t1 and t2 such thatthe instantaneous velocity at t∗ equals the average velocity between t1 and t2.

7. In the previous problems you needed to compute v(t) from x(t) by differentiation. What if you have v(x), thevelocity of an object at position x? You can recover x(t) from this by using v(t) = dx

dt , re-arranging and integrating

v dt = dx, dt =dx

v,

∫ t

t0

dt = t− t0 =

∫ x

x0

dx

v

For example let v(x) =√2a(x− x0) with v(x0) = 0. Then

t− t0 =

∫ x

x0

dx√2a(x− x0)

= 2

√x− x02a

∣∣∣xx0

=

√2(x− x0)

a

Now invert by squaring and solving for x;

(t− t0)2 =

2(x− x0)

a, x− x0 =

1

2a(t− t0)

2

and note that x = x0 when t = t0.

Suppose that at t = 0 we have x = x0, and v(x) =√x20 − x2. Find x(t). This is an extremely important type of

motion in which the acceleration is not constant.

1.5. PROBLEMS 29

8. Let

a = 4i+ 5j+ 3k, b = 3i+ 4j+ 5k, c = 5j− k

Find a+ b. Find a− 3c. Find a · b.Find b · c. Find the angle between a and b.Find the angle between b and c.Find b× c.Find the area of the parallelogram whose sides are formed by a and b.Find a vector lying in the xy-plane that is ⊥ to b.

Compute ddt

(a+ b t

)at time t = 3. Compute d

dt

(a+ b t3

)at time t = 3.

9. You will see partial derivatives such as ∂f∂x scattered throughout your book. These are computed just like

ordinary derivatives, but are simpler. If f = f(x, y, z), the x-partial derivative treats y and z like constants, andso forth. In other words ∂x

∂y = ∂x∂z = ∂y

∂z = 0. For example

∂

∂x

(x2z + y +

x

y+z

x

)= 2xz + 0 +

1

y− z

x2

Compute ∂∂x

(y + 1

x2+y2

). Compute ∂

∂y

(1√x2+y2

)Compute ∂

∂z

(e−xy + x e−zy

)

z

yx

4

3

5

a

b

c

d

10. A rectangular box of dimensions 3.0, 4.0, 5.0moriented as illustrated has four corners labeled as vectorsa, b, c, d.

A. Find the x, y and z components of all four vectors,list them using unit vector notation.

B. Find the magnitudes of all four vectors.C. Compute d · b.D. Find the angle between c and b.E. Find a× c.

y

x

h

h’

1

1

11. A mass (at the origin) is connected by strings to points (1, 1)and (1,−1) so that right now h = h′ =

√2. The string of length h is

decreasing in length at rate dhdt = −3 m

s and the string of length h′

is decreasing in length at rate dh′

dt = −2 ms . At this instant find the

velocity vector of the mass.


h

d

12. Imagine a boat on the water being pulledtowards the shore by a winch mounted a heighth above the water. The winch reels in cable(shortening the hypotenuse) at constant rateu in m

s . When the boat is a distance d fromthe shore, find its speed and acceleration.This is an example of motion at non-uniformacceleration.

13. Prove that the vectors a = (3, 4, 5), b = (4, 5, 3) and c = (5, 3, 4) are linearly-independent.

14. An object has position vector r = R cos(ωt) i + R sin(ωt) j, in which R,ω are constants. Find its velocity andacceleration vectors. Show that

dr

dt= ωk× r

by explicitly calculating both sides.Show that

d2r

dt2= ωk×

(ωk× r

)by explicitly calculating both sides.

15. An object has position vector r = R cos(ωt+ 12αt

2) i+R sin(ωt+ 12αt

2) j, in which R,ω are constants. Find itsvelocity and acceleration vectors. Show that

dr

dt=(ω + αt

)k× r

by explicitly calculating both sides.

d2r

dt2= ωk×

(ωk× r

)+ αk× r

by explicitly calculating both sides. This is a decomposition of acceleration into radial and tangential components(respectively).

16. Find a vector perpendicular to both a = i− 3j and b = j+ 3i+ 2k.

17. Find a unit vector perpendicular to both a = i− 3j and b = 2k.

18. Compute the angle between the vectors a = i− 3j and b = j+ 3i+ 2k.

19. Prove that if a(t) is a unit vector whose components are functions of t, then the vector b(t) = da(t)dt is perpen-

dicular to a(t).

20. Suppose that you travel along a straight line for 2.25hr at 30mihr , 3.5hr at 45mihr , and finally 1.25hr at 60mihr .Find your displacement and average velocity.

21. Suppose that you travel along a straight line for 2.25hr at 30mihr , 3.5hr at −45mihr , and finally 1.25hr at 60mihr .Find your displacement and average velocity.

22. Suppose that you travel along a straight line (West) for 2.0hr a (directed) distance of 60mi, 3.5hr a (directed)distance 210mi, and finally 2.5hr a distance 30mi. Find your displacement and average velocity.

1.5. PROBLEMS 31

23. Find a unit vector the lies in the xy-plane and makes a 30o angle above the x-axis.

24. Find a unit vector the lies in the xy-plane and makes a 120o angle above the x-axis.

25. You travel with velocity v = (25 − 35 t)ms i from t = 0 to t = 10 seconds. Find your average velocity and youraverage speed.

26. Show from Eq. 1.1 that the motion

x(t) =

1.0 + 2.0 t 0 ≤ t ≤ 13 + 4(t− 1) 1 ≤ t ≤ 27 + 3(t− 2) 2 ≤ t

(1.146)

is smooth at t = 1 and at t = 2. Is it smooth at any 1 < t < 2?

27. Determine from Eq. 1.1whether or not the motion

x(t) =

1.0 + 2.0 t 0 ≤ t ≤ 14 + 4(t− 1) 1 ≤ t ≤ 28 + 3(t− 2) 2 ≤ t

(1.147)

is smooth at t = 1 and at t = 2. Is is smooth at any 1 < t < 2? Draw motion graphs to illustrate the motion. Seeproblem 1.35 for assistance.

28. Computed

dx

√x2 + 1,

d

dx3√x2 + x,

d

dx

( 1

x3 + x

),

d

dx

( 1− x2

x3 + x

)and check your results with REDUCE, or axiom or maxima.

29. Find a series expansion for

f(x) =1

(1− x)2

valid near x = 0 that agrees with f(x) up through the third derivative at x = 0.and check your results with REDUCE, or axiom or maxima.

30. Use l’Hospital’s rule to compute

limx→1

x30 − x−30

x− x−1

and check your results with REDUCE, or axiom or maxima

%Here’s how to do it in REDUCE

load_package limits;

limit(sin(x)/x,x,0);

%Here’s how to do it in maxima

limit(sin(x)/x,x,0);"

%Here’s how to do it in axiom

print limit(sin(x)/x,x=0)

31. Use l’Hospital’s rule to show that ( ddx

20∑n=0

xn)∣∣∣x=1

=20∑n=0

n = 210

32. Suppose that you can find two functions f(x) and g(x) such that

d

dxf(x) = g(x),

d

dxg(x) = −f(x)


for any point x. Prove that

f2(x) + g2(x) is a constant for any x

Can you think of two functions for which this is true?

33. Find a series expansion for

f(x) = x13

valid near x = 1 that agrees with f(x) up through the third derivative at x = 1, and check your results with RE-DUCE, or axiom or maxima.

34. Solve the equation v(t) = dxdt =

√a2 − x2 using REDUCE, and determine the value of the constant of integration

that will make x(0) = a and v(0) = 0.

load_package odesolve;

odesolve(df(x,t)-sqrt(a^2-x^2),x,t);

35. Making motion graphs with gnuplot. Log into gnuplot at http://azazelo.uwp.edu/cgi-bin/gnuplot.cgi, ordownload your own copy. To make a graph of continuous functions is pretty easy in gnuplot, but gnuplot wantsto use x as the independent variable, whereas we want t to be the independent variable, so you must pretendthat x is t.Make a graph of a continuous motion such as motion at constant acceleration, plotting position, velocity and accel-eration all together

y0=1.0

v0=2.0

a=2.0

plot[0:3] y0+v0*x-0.5*a*x*x,

v0+a*x,a

-2

0

2

4

6

8

0 0.5 1 1.5 2 2.5 3

y0+v0*x-0.5*a*x*xv0+a*x

a

That was fun. In order to graph a piece-wise continuous function such as that on page 1, we use gnuplot’s abilityto graph a function whose form is determined by a condition, the syntax is

f(x) = (condition C) ? shape if C true : shape if C false

for example;

1.5. PROBLEMS 33

f1(x)=(x<=1) ? 1+2*x : 0

plot[0:3] f1(x)

-1

0

1

2

3

4

0 0.5 1 1.5 2 2.5 3

f1(x)

These can be strung together to create graphs of piece-wise continuous functions (which will have discontinuousderivatives). For example try this one

f1(x)=(x<=1) ? 1+2*x : 0

f2(x)=(x>1) ? 3+4*(x-1) : f1(x)

f3(x)=(x>2) ? 7+(x-2) : f2(x)

plot[0:3] f3(x)

1

2

3

4

5

6

7

8

0 0.5 1 1.5 2 2.5 3

f3(x)

Here is another example; that of problem 1.1

x(t) =

1.0 t2 0 ≤ t ≤ 25625 + 50 ∗ (t− 25) 25 ≤ t ≤ 1004375 + 50 ∗ (t− 100)− (t− 100)2 100 ≤ t ≤ 125

(1.148)


f1(x)=(x<=25) ? x**2 : 0

f2(x)=(x>25) ? 625+50*(x-25) : f1(x)

f3(x)=(x>100) ?

4375+50*(x-100)-(x-100)**2 : f2(x)

plot[0:125] f3(x)

0

1000

2000

3000

4000

5000

0 20 40 60 80 100 120

f3(x)

Create the velocity graph for the previous two examples.

36. Train #1 starts at the origin and travels east at constant velocity v. Train #2 starts at x = D with D > 0 andtravels west at constant velocity v. A bird flies from train to train at constant speed 2v, starting when train #1 is atthe origin. How much distance does the bird fly before the trains crash together? Draw a motion graph using gnuplot.

37. Traveler #1 starts at city A on a trip to city B at a constant velocity. Traveler #2 starts at city B at the sametime on a trip to city A at a constant velocity as well. They pass each other at 12:00, and traveler #1 arrives at Bat 4:00 PM, traveler #2 arrives at A at 9:00 PM. At what time did the trips begin?

38. Three vectors are linearly independent if all three do not lie in the same plane. Explain why a,b, c are linearly

independent if a ·(b× c

)= 0.

39. Show by direct calculation that

a ·(b× c

)= b ·

(c× a

)= c ·

(a× b

)40. Parenthesis are important. What does a× b× c mean? What about a · b× c? Both are ambiguous if notincalculable if you don’t put in parenthesis. Show that(

a× b)× c = a×

(b× c

)(the cross product is not associative). Show that

a×(b× c

)+ b×

(c× a

)+ c×

(a× b

)= 0

for any three vectors a = (ax, ay, az), b = (bx, by, bz), and c = (cx, cy, cz). Explain why(a · b

)× c is meaningless

(incalculable) and why a ·(b× c

)is calculable, and is a scalar.

41. Make REDUCE do it. Type the following (or cut and paste it from the REDUCE examples page) intohttp://azazelo.uwp.edu/cgi-bin/reduceweb.cgi, and use it to verify or solve all of the previous vector problems.

% cut and paste all of this into REDUCE

1.5. PROBLEMS 35

infix dot;

procedure v1 dot v2;

begin

return(for i:=1:3 sum part(v1,i)*part(v2,i));

end;

infix cross;

procedure v1 cross v2;

begin

v3x:=part(v1,2)*part(v2,3)-part(v1,3)*part(v2,2);

v3y:=part(v1,3)*part(v2,1)-part(v1,1)*part(v2,3);

v3z:=part(v1,1)*part(v2,2)-part(v1,2)*part(v2,1);

retval:=v3x,v3y,v3z;

return(retval);

end;

% try it on the 201 hw

1,-3,0 dot 1,1,1;

1,-3,0 cross 3,1,2;

42. Lets do some financial math. Suppose that every time period (say a year) you add amount of money A toyour account, your yearly contribution. Suppose that each time period the account gains a certain fraction C of itscontents through interest. Suppose the amount in your account after n years is f(n). Show that in one more year

f(n+ 1) = f(n) +A+ C f(n)

This is called a recursion relation. Show by direct substitution that the solution is exponential growth

f(n) = f0 + f1ξn

for some unknown constant ξ, and by substituting this into the equation show that if f(0) = 0

f0 = −A/C, ξ = (1 + C), f1 = A/C

The importance of starting to save as early as possiible: suppose that you put $100.00 into your account per month(A = $1200.00 per year) and your fund pays an interest rate of 4% = 0.04 = C. After 25 years, how muchmoney do you have? What if you double your yearly contribution, how does this affect your 25-year savings? Whatif you find a different fund that pays 5%, how does that affect your savings? Use REDUCE to perform the arithmetic.

43. Elliptic curves of the form y2 = x3 + ax + b are used very extensively in cryptography, because it is possibleto associate a group with the curve, in the sense that a generic line will intersect a cubic in three points, call themp, q, r, and for elliptic curves of this type a group operation ⊞ can be defined for which p⊞ q⊞ r = 0, or p⊞ q = ⊟r.Give the curve and two points on it, a third can be found. How this is used in cryptography is discussed in MATH367, Number Theory.Use the aglebraic-geometric method of section 1.4.2 to find the slope of the line tangent to this curve at a point(u, v0 that lies upon it.

Chapter 2

Projectile motion

2.1 Basic examples of ballistics

Projectile motion is a simple and familiar type of motion; a body is thrown from some point with some initial velocity,and gravity alone determines what happens to it. This chapter is important because we will do case-studies of solvingequations of motion.

x

y

ymax

xmax=R

θ

(x0, y0)

(vx0, vy0)

We now study a series of examples thatillustrate the variety of projectile problemsand the methods for extracting data fromthe equations that describe the motion ofprojectiles in two dimensions.

Every problem begins the same way; we writea generic vector version of the position versustime for a body undergoing motion at constantacceleration;

r(t) = (x(t), y(t)

= r0 + v0 t+1

2a t2

= (x0, y0) + (v0x, v0y) t+1

2(ax, ay) t

2

= (x0 + v0xt+1

2axt

2, y0 + v0yt+1

2ayt

2)

(2.1)

and we specialize it to motion of a projectile above the earth’s surface by using the constant acceleration of gravity;

a = (ax, ay) = (0,−g), g = 9.8m

s2(2.2)

The next set-up step is to specify the four constants x0, y0, v0x, v0y for each particular problem. We finish eachproblem by stating the word-problem in precise mathematical terms and by performing a few algebraic steps.

Example 1. A football is kicked from ground level at initial speed v0 at angle θ. Compute the flight time, rangeand maximum height attained.

We first get the vectorv0 = (v0x, v0y) = (|v0| cos θ, |v0| sin θ) (2.3)

37

38 CHAPTER 2. PROJECTILE MOTION

obtaining the two components in terms of the launch angle and speed. This results in

x = v0 cos θt, y = v0 sin θt−1

2gt2

vx = v0 cos θ, vy = v0 sin θ − gt (2.4)

The trajectory or equation in space of the path of the ball is gotten by eliminating t;

t =x

v0 cos θ(2.5)

insert this into the y-equation to get

y(x) = x tan θ − gx2

2v20 cos2 θ

(2.6)

which is a parabola. We can find the range by looking for the values of x at which y = 0, the ball is on the groundat these times

x = 0 and x =2 sin θ cos θv20

g(2.7)

the second of which can be rewritten as

R =v20 sin 2θ

g(2.8)

Notice that the range is maximal when θ = π4 rad, and that the two angles π

4 + θ and π4 − θ both result in the same

range. By launching the projectile at 45o (π4 rad) a maximum range of

Rmax =v20g

(2.9)

is attained. Actually with air resistance taken into account, the range of a projectile in actuality is much less thanthis.The maximum height reached can be gotten by finding the time at which the velocity in the y-direction goes to zero,and inserting this into the y-equation;

vy = v0 sin θ − gt∗ = 0, t∗ =v0 sin θ

g, ymax =

v20 sin2 θ

2g(2.10)

this reaches it’s largest value for θ = π2 or ninety degrees. You can see this in the figure above. The total time of

flight is twice the time needed to get to the top of the flight

tflight =2v0 sin θ

g(2.11)

Example 2. Find the speed of the projectile, and the inclination of the velocity to the horizontal.The speed is very simple;

s = |v| =√v2x + v2y =

√v20 + g2t2 − 2gtv0 sin θ (2.12)

and calling the velocity at any time v = (vx, vy) = (s cosϕ, s sinϕ) we obtain

tanϕ =vyvx

= tan θ − g

v0 cos θt (2.13)

which tells us which way the projectile is headed; down if this is negative, up if positive, so we can tell from thisquantity whether or not the projectile is ascending or descending. Note that from Eqs. 2.6, 2.13 this is the trajectoryslope

dy

dx= tanϕ (2.14)

(just eliminate t using Eq. 2.4).

2.2. USING THE COMPUTER 39

x

y

xmax=R

(x0, y0)

(vx0, vy0)

Example 3. Suppose that a projectileis launched horizontally from a height Hat speed v0 . What must be the value ofv0 for it to clear an obstacle of height hplaced a distance d away?

x = v0t, y = H − 1

2gt2 (2.15)

are the equations describing the flight.The trajectory is

y(x) = H − gx2

2v20(2.16)

We want the point (x, y) = (d, h) to beon the trajectory and so

h = H − gd2

2v20(2.17)

giving us a speed

v0 =

√gd2

2(H − h)(2.18)

2.2 Using the computer1

We can use the computer to graph the motion of a projectile by either writing a program to integrate the accelerationequations, using

df(x)

dx=f(x+ dx)− f(x)

dx= g(x) (2.19)

or

f(x+ dx) = f(x) + dx ∗ g(x) = f(x) + dx ∗ dfdx

(2.20)

For example if

vy(t) =dy

dt= v0y − gt (2.21)

theny(t+ dt) = y(t) + dt ∗ (v0y − gt) (2.22)

or alternatively to use the cgi interface to ode, a sophisticated program that solves differential equations that youprovide together with initial data. I will describe to you can use ode lastly in this section. You can either use myweb interface at http://azazelo.uwp.edu/cgi-bin/cgi.ode, or you can download your own copy from

Lets write a program (for example in C) to write data that we can use to plot the path.

/* prog1.c */

/* This line is a comment. Insert directives to */

/* look up library functions declared in two ‘‘header’’ files */

/* for input/output and math functions */

#include <stdio.h>

#include <math.h>

/* Declare any double precision variables */

1This section is optional


double v0x, v0y, x, y, t, g, dt;

/* Declare integer variables used in looping */

int n;

main()

/* Set values for variables that do not change */

v0x=20.0, v0y=20.0, g=9.8, dt=0.01;

/* Set intial values of x, y */

x=0.0, y=0.0, t=0.0;

printf("%f\t%f\n", x,y);

/* begin the loop */

do

t=t+dt;

x=x+v0x*dt;

y=y+(v0y-g*t)*dt;


while(y >= 0.0);

/* end of program */

Output of prog. 1

0 20 40 60 80 100−5

0

5

10

15

20

25

x, (m)

y, (

m)

In order to use the program to draw thegraph, we need to first compile the codeinto an executable file. This requires thatyou have a C-compiler installed in yourcomputer. If you are a Linux user, youalready have one. If you are a Windowsuser, you do not. Windows users canget a free C/C++/FORTRAN compiler fromhttp://www.mingw.org/download.shtml.You will need two files, download bothMSYS-1.0.10.exe, and MinGW-3.1.0-1.exe fromthe web-site, or copy them from the CD-ROMput on reserve in the library for your use.Install MinGW first by double-clicking on it,and then install MSYS. Afterwards you willget an icon that starts up a shell (like theDOS window).

You can open the file in Excel and plot thedata, or download the free function plottingsoftware gnuplot and open the file in gnuplot,which will graph it for you. You can printthe output, which will look like what you seeabove.

The data can be output into a file which can be plotted by some computer package such as Mathematica or gnuplot.We have formatted our output for gnuplot, and the result looks like this.

Suppose that we want to be able to change the launch angle θ and produce a succession of graphs with different θbut the same v0. We alter the program so that it accepts parameters;


/* prog2.c */




#include <stdio.h>

#include <stdlib.h>

#include <math.h>

#define PI 3.1415926


double v0x, v0y, x, y, t, g, dt, v0, theta;


int n;

main(int argc, char*argv[])

/* get the angle theta */

if(argc != 2)

printf("./prog2.exe theta(deg)\n");

exit(1);

theta=(PI/180.0)*atof(argv[1]);


v0=20.0, g=9.8, dt=0.01;

/* set initial velocity */

v0x=v0*cos(theta), v0y=v0*sin(theta);


x=0.0, y=0.0, t=0.0;



do

t=t+dt;

x=x+v0x*dt;

y=y+(v0y-g*t)*dt;


while(y >= 0.0);



θ=20, 30, 45, 60, 70o

0 10 20 30 40 50−5

0

5

10

15

20

x (m)

y (m

)

and we compile the program to ex-ecutable prog2.exe and run it bytyping things like ./prog2.exe 20> D20, ./prog2.exe 30 > D30,./prog2.exe 45 > D45 to createthe data files D20, D30, D45, ...which we then plot together.

Suppose that we want to take into account air resistance, this provides a resistance that opposes the velocity vector,and gets bigger as the speed gets bigger. The equations can be complicated

dvydt

= −g − b√v2x + v2y vy,

dvxdt

= −b√v2x + v2y vx (2.23)

and solving these could be very hard, but the computer can do it numerically with no big changes over our oldprogram.

/* prog3.c */




#include <stdio.h>

#include <stdlib.h>

#include <math.h>

#define PI 3.1415926


double vx, vy, x, y, t, g, dt, v0, theta, b, rad;


int n;

main(int argc, char*argv[])

/* get the angle theta */

if(argc != 3)

printf("./prog2.exe theta(deg) b\n");

exit(1);


theta=(PI/180.0)*atof(argv[1]);

b=atof(argv[2]);


v0=28.4, g=9.8, dt=0.01;

/* set initial velocity */

vx=v0*cos(theta), vy=v0*sin(theta);


x=0.0, y=0.0, t=0.0;



do

t=t+dt;

rad=sqrt(vx*vx+vy*vy);

vx=vx-b*dt*vx*rad;

vy=vy-g*dt-b*dt*vy*rad;

x=x+vx*dt;

y=y+vy*dt;


while(y >= 0.0);


b=0.0, 0.01, 0.02, 0.03, 0.04

0 20 40 60 80 100−5

0

5

10

15

20

25

x (m)

y (m

)

The results are plotted below forvarious values of the wind resis-tance constant b. You can see thatthe trajectory falls off very fastafter reaching the top of the flight,and even mild wind resistance hasa dramatic effect on the range.These are for initial launch angleof 45 degrees and speed 28.4ms .

Use of the computer to solveproblems is strictly optional,but understand that many seem-ingly innocuous problems inphysics may be difficult to solveanalytically. Practically anyproblem can be solved or simu-lated numerically, and although anumerical solution to a problemis technically no solution at all,it may help you to visualize thebehavior of the system understudy.You are encouraged to use thistool if you are interested.


2.2.1 Using ode

Ode is extremely easy to use, you simply log into http://azazelo.uwp.edu/cgi-bin/ode.cgi, read through a few of theexamples, and try it out. You do not need to know any programming, all you do is have list (define) your constants,

your equations of motion (in the required syntax y′ = dydt = vy, v

′y =

dvydt ), and your initial t = 0 values for all of your

variables, in that order. Note that ode only solves first order equations, so you must trick it thusly

a =d2x

dt2= x′′ = F (t), −→ x′ = v, v′ = F (t)

(if you eliminate v, you are back to x′′ = F (t)).

Ode does not do what you want if your equations of motion are trivial, it does a great job if your EOMS are toocomplicated to solve by hand. Don’t be surprised if you get unexpectedly straight-line graphs coming out of it ifyour equations are so simple that you could solve them by hand!Here are some examples

# define constants

g=9.8

b=0.1

# give equations, prime denotes t derivatives

x’=vx

vx’=-b*sqrt(vx*vx+vy*vy)*vx

y’=vy

vy’=-g-b*sqrt(vx*vx+vy*vy)*vy

# give initial data

x=0.0

vx=40.0

y=0.0

vy=40.0

# we plot trajectory

print x,y

# we could plot y vs t with ‘‘print t,y’’

step 0,3 # solve for 0 <= t <= 3

y(x) for v0=40√2, b=0.1, θ=45

0 5 10 15 20 25−2

0

2

4

6

8

10

12

x(m)

y (m

)

2.3 Air resistance

The meaning of the constant b in the examples of the previous section can be simply elucidated; drop an object fromheight H. Its velocity will be

v = (0,−vy), and√v2x + v2y = |vy| (2.24)

giving

ay =dvydt

= −g + b v2y (2.25)

the object reaches terminal velocity when it stops accelerating and so

ay = 0 = −g + bv2term, or b =g

v2term(2.26)

If you jump from a plane with no parachute, you can minimize your terminal velocity by increasing the surface areathat you present to the wind, getting it down to about 22.0ms . This makes b in this case equal to

b = 0.02021

m(2.27)

2.3. AIR RESISTANCE 45

for an arrow or bullet with a higher degree of streamlining, b could be ten times smaller.

It is not difficult to solve this equation of motion taking air resistance into account for a falling body, but it is alittle harder for a projectile. An exact approach is this; start with

ay =dvydt

= −g + bv2y,

dt = − dvyg − bv2y

t− 0 = −∫ vy

0

dvyg − bv2y

= − 1

2√bg

ln(√ g

b + vy√gb − vy

)(2.28)

solve;

vy = −√g

b

(e√bgt − e−√bgt

e√bgt + e−

√bgt

)= −

√g

btanh(

√gb t) (2.29)

which reaches vterm =√

gb as t→ ∞.

Here is a successive approximations method that is comparatively low-tech mathematically, and is very useful in abroad variety of circumstances. We solve the equations order by order by neglecting the smallest terms in theequation of motion, typically any term that is not a known function of t.

Start with the equationdvydt

= −g + bv2y,

Neglect vy;dv

(0)y

dt= −g, v(0)y (t) = −gt+ v0 = −gt

Subst. into original;dv

(1)y

dt= −g + b(−gt)2,

Solve; v(1)y (t) = −gt+ bg2

3t3,

Subst. into original;dv

(2)y

dt= −g + b(−gt+ bg2

3t3)2

Solve; v(2)y (t) = −gt+ bg2

3t3 − 2b2g3

15t5 (2.30)

and just continue until the solution is reached that has the desired accuracy. Note that

−√g

btanh(

√gb t) = −gt+ bg2

3t3 − 2b2g3

15t5

of course the two methods agree perfectly, but the second method is very easy to apply.

Lets try our hand at this casedvydt

= −g − b vy,dvxdt

= −b vx (2.31)

with initial conditions x(0) = y(0) = 0, vx(0) = v0x, vy(0) = v0y.

dv(0)y

dt= −g, dv

(0)x

dt= 0

v(0)y = v0y − gt, v(0)x = v0x

dv(1)y

dt= −g − b(v0y − gt),

dv(1)x

dt= −b v0x

v(1)y = v0y − (g + bv0y)t+bg

2t2, v(1)x = v0x − bv0xt (2.32)


We have enough to get the lowest-order correction to the motion due to this form of air-resistance, so lets proceedto the trajectory. Integrate again

y(2) = y0+v0yt−(g + bv0y)

2t2+

bg

6t3 = v0yt−

(g + bv0y)

2t2+

bg

6t3, x(2) = x0+v0xt−

bv0x2t2 = v0xt−

bv0x2t2 (2.33)

What is the time of flight?

y(2)(t∗) = 0, 0 = v0y −(g + bv0y)

2t∗ +

bg

6t2∗ (2.34)

We know that this is going to be close to the time of flight without air-resistance, so lets get an approximate valueby the same process by which we got this far

t∗ =2v0yg

− bv0ygt∗ +

b

3t2∗

neglect the b term t(0)∗ =

2v0yg

Substitute t(1)∗ =

2v0yg

− bv0yg

2v0yg

+b

3

(2v0yg

)2=

2v0yg

− b2v20y3g2

(2.35)

so that is why the range is so badly affected; the time of flight is reduced by a lot for a fast projectile! Wemight be able to compensate by adjusting the launch angle to maximize the time of flight.

Lets get the range

R = x(t∗) ≈ v0x

(2v0yg

− b2v20y3g2

)− bv0x

2

(2v0yg

− b2v20y3g2

)2≈ 2v0xv0y

g− b

8v0xv20y

3g2(2.36)

this is a big short-fall for a fast projectile. You can see that these methods of approximation are simple, but you canattack real-world problems with them. They also apply to any other discipline in which one must solve complicateequations.

2.4 Self-propelled projectiles

A self-propelled projectile is a rocket, it provides constant thrust, which gives it acceleration in addition to gravity.it’s mass also decreases as it uses up fuel, but we can’t handle this case until we have discussed forces. lets study asimple case in which a projectile is able to provide a small constant thrust directed at a 45o during its flight, so that

ax = w, ay = −g + w (2.37)

and we can try to figure out how this affects its range. Integrate

vx = v0x + wt, vy = v0y − (g − w)t

x = x0 + v0xt+1

2wt2, y = y0 + v0yt−

1

2(g − w)t2 (2.38)

How long is it flight? Solve the equation y(t∗) = 0, with x(0) = y(0) = 0

t∗ =2v0yg − w

≈ 2v0yg

+2v0yw

g2+ · · · (2.39)

so it stays in the air a bit longer than a ballistic projectile. Now get the range

R = x(t∗) = v0x2v0yg − w

+1

2w( 2v0yg − w

)2(2.40)

which is quite a bit more than R =2v0xv0y

g , the range of a ballistic projectile. If the “rocket” can keep its nose up,so the thrust stays at a 45o angle throughout the flight, the projectile can get a terrific boost in range with a small

2.5. OBSTACLES AND TARGETS 47

thrust w.

Example 4. A rocket such as I have described has a thrust w = 0.05 g, by how much is its range improved if thelaunch angle is 450?Under these circumstances v0x = v0 cos θ = v0y = v0 sin θ = v0/

√2, so

Rrocket =2g v0xv0y(g − w)2

,RrocketRballistic

=( g

g − w

)2= 1.11 (2.41)

a rather substantial 11# increase.

2.5 Obstacles and targets

Problems such as hitting a target placed at some point (xT , yT ) can be easily solved using the trajectory equation,since the target is assumed to be on the trajectory if it is going to be hit, so y(xT ) = yT . Obstacles such as hills aretreated the same way.

x

y

ymax

xmax=R(x0, y0)

(vx0, vy0)

φ

Example 5. You stand at thebase of a hill of slope ϕ and throwa ball at speed v0 at an angle θwith respect to horizontal. Howfar up the hill will it go beforehitting the ground?

The equation for the trajectory is

y(x) = x tan θ − gx2

2v20 cos2 θ

(2.42)

and the equation of the hill is

yhill(x) = tanϕx (2.43)

where these intersect, that is thex-value of the place where the balllands on the hill

yhill(xmax) = y(xmax), xmax tanϕ = xmax tan θ −gx2max

2v20 cos2 θ

(2.44)

or

xmax =2v20 cos

2 θ(tan θ − tanϕ)

g(2.45)

2.6 Problem solving suggestions

You will find that the trajectory equation y = y(x) gotten by eliminating t from x(t) and y(t) provides you with thesimplest way of solving most elementary problems that do not involve velocity questions. My advice is startwith the EOMS and obtain the range formula for a given scenario. If you need to answer questions about velocity,directions or speeds, stick to the once-integrated EOMS. For more complicated problems, such as the addition ofair-resistance, your best bet is always the approximation method applied to the EOMS.


2.7 Appendix. REDUCE code

In REDUCE, t and T are reserved keywords for a boolean “true”, so we switch this off by putting ! in front of thesymbol. Here we solve the projectile with air resistance. We could create a loop, instead we repeat the iteration cycle.

% initialize

on revpri;

on div;

ax := 0;

ay := -g;

% zeroth order

vx:=vx0+int(ax,!T);

vy:=vy0+int(ay,!T);

% first order

ay:=-g-b*vy;

ax:=0-b*vx;

vx:=vx0+int(ax,!T);

vy:=vy0+int(ay,!T);

% second order

ay:=-g-b*vy;

ax:=0-b*vx;

vx:=vx0+int(ax,!T);

vy:=vy0+int(ay,!T);

y:=y0+int(vy,!T);

y:=0;

% we get for y

1 1 1 2 1 2 2 1 3 2

T*(vy0 - ---*T*g - ---*T*b*vy0 + ---*T *b*g + ---*T *b *vy0 - ----*T *b *g)

2 2 6 6 24

Now lets try a very simple but labor intensivemethod for getting the time of flight Tf whichis the time for which y(Tf ) = 0. We let

T = A+B b+ C b2 + · · ·

substitute and solve for each coefficient of agiven power of b;

!T:=!A+!B*b+!C*b^2;

C0:=coeffn(y,b,0);

solve(C0,!A);

% gives

1

c0 := A*(vy0 - ---*A*g)

2

let first(solve(C0,!A));

% selects first of

-1

A=2*g *vy0,A=0

C1:=coeffn(y,b,1);

let solve(C1,!B);

!B;which is giving you exactly Eq. 2.35. If you enjoy working with the computer, perhaps using REDUCE to solvephysics problems will let you enjoy physics as well.

2.8. PROBLEMS 49

2.8 Problems

y

x

R

v

1. A golfer drives a ball horizontally withinitial velocity v = (50ms , 0) from a tee at theorigin, down a 20o below-horizontal slope asillustrated above.A. How far from the tee measured alongthe slope does the ball land on the slope?B. With what speed does it land?

v0

45o

D

h

2. A golfer drives a ball with initial angleθ = 45o from a tee at the origin. The ballhits a window at height h = 10m a distanceD = 80m away.

a. Find the initial launch speed v0 of the ball.

b. Suppose a second golfer tees off from thesame spot at an unknown angle and speed,and miraculously her shot also hits the windowat the same height 1.8 s after launch. Withwhat speed did it hit the window?

c. Was her ball ascending or descending or in level flight when it hit?

3. Suppose that a projectile is launched horizontally from a height of h = 10m, at what speed must it be shot inorder to barely clear (pass over) an obstacle of height 5.0m placed on the ground 20m away?

4. If a projectile is launched horizontally with initial speed v0 = 30 ms , from a height h = 20m, with what speed

does it hit the ground? What is the angle of impact?

x

y

v0

α

R

θ

5. A projectile is launched with initialspeed v0 at angle θ with respect to thehorizontal up a hill of angle α (slopeis tanα). Compute the range of theprojectile R (find how far from thelaunch point it lands on the hill.

6. In the previous problem, compute the launch angle that maximizes the range.Ans. θ = π

4 + α2 .


7. If a projectile is launched horizontally (in direction of increasing x) with initial speed v0, from a height h above

the origin, find its speed s =√v2x + v2y as a function of x.

8. If a projectile is launched with initial speed v0, at angle θ from the origin, find its speed s =√v2x + v2y as a

function of x.

9. If a projectile is launched with initial speed v0, at angle θ from the origin, find the angle its velocity vector makeswith the horizontal as a function of x.

10. You can see that taking into account air resistance is pretty difficult if the motion is truly two or three-dimensional. If it is not, things may not be too complicated. Suppose that the effects of air resistance is to modifyyour acceleration by amount α|v|, always in a direction opposing your velocity. This is a good approximation, andα is usually small. Assuming α is small, a falling body will reach terminal velocity, it will reach a point where itfalls at constant speed. Find this speed.

11. Suppose that the effects of air resistance is to modify your acceleration by amount α|v|, always in a directionopposing your velocity. Assuming α is small, find the velocity of a body as a function of time (let it be released att = 0). If you get stuck, try using REDUCE.

12. Find the maximum height and range R attained by a projectile launched from the ground at v0 = 50ms , angleθ = 60o.

13. Find the speed and angle that v makes with the horizontal at x = R/4, R/2, 3R/4 by a projectile launchedfrom the ground at v0 = 50ms , angle θ = 60o.

14. A projectile is launched horizontally at v0, initial height h. Obtain a formula for its velocity vector as a functionof time, and as a function of x, how far it has traveled from where it was launched.

15. A projectile is launched from ground level at v0, initial angle θ, from the origin. Obtain a formula for its velocityvector as a function of time, and as a function of x, how far it has traveled from where it was launched.

16. A golfer launches a ball from ground level at angle 45o to the horizontal, towards a wall 40m away. After hittingthe wall, the ball lands right on the tee from which it was launched! Find the speed with which it was launched.

17. A golfer launches a ball from ground level at angle 45o to the horizontal, towards a wall 40m away. After hittingthe wall, the ball lands 10m from the wall. Find the speed with which it was launched. Note that when a projectilericochets from a surface, the component of its velocity normal to (⊥ to) the surface gets reversed.

18. Find the maximum range and speed attained by a projectile launched horizontally at v0 = 50ms , initial heighth = 20m before hitting the ground.

19. Find the maximum height, speed and range attained by a projectile launched from height h = 20m at v0 = 50ms ,angle θ = 60o before hitting the ground.

20. A golf ball launched from the origin at angle θ = 30o must clear a fence of height 20m a distance 40m from thetee. Find the minimal launch speed required.

21. A golf ball launched from the origin at initial speed v0 = 80ms must clear a fence of height 20m a distance 40mfrom the tee. Find the minimal launch angle θ required.

22. Find the velocity as a function of time for a body falling from rest under the influence of gravity and air resistancesuch that ay = −g + b|vy|.

23. Use ode to find the maximum range of a projectile launched at 45o, v0 = 50ms with air resistance ay =−g − 0.05 vy, ax = −g − 0.05 vx.

2.8. PROBLEMS 51

24. In the example with air resistance solved in Eq. 2.35, find the angle θ of launch that maximizes the range. Theresults will depend on b, g, v0. Hint; v0y = v0 sin θ.

25. An ecological/biological model. Suppose that you have an isolated island with two species, bunny rabbitsand wolves. Each year the bunny population grows in proportion to the number of bunnies, and shrinks because ofwolf predation. The wolf population grows in proportion to the number of wolves, but shrinks if the ratio of wolvesto rabbits gets to be too big. Use ode to study the populations as a function of initial populations, watch for crashesin populations. For example

# a is bunnies

a’=0.06*a-0.05*b

# b is wolves

b’=0.05*b-0.02*b/a

# those wolves are breeding nearly

# as fast as the bunnies, not good!

a=20

# try these wolf pops.

b=6

# b=10

# b=15

# b=20

# print wolf population

print t,b

# print bunny pop.

# print t,a

step 0,25

Note that if either population goes to zero, that species becomes extinct on the island, and you would need to modifythe EOMS accordingly. Can you figure out how to make the bunny population explode if the wolf population goesdown? We have assumed that only predation limits population, how could you model disease or overcrowding impacton the silly wabbits? Hint, try this, it produces a stable population

# a is bunnies

a’=0.06*a-0.0025*a*a

a=20

print t,a

step 0,50

The new term describes detrimental effects of over-crowding. What is the asymptotic bunny population? Do yousee that this is fundamentally the same equation of motion that we studied for air resistance?

26. Plotting trajectories with gnuplot is very easy; we simply tell gnuplot to graph x(t) and y(t) parametri-cally, such as in this example


set parametric

x(t)=2*t

y(t)=4.0-t**2

plot[0:1] x(t),y(t)

3

3.2

3.4

3.6

3.8

4

0 0.5 1 1.5 2

x(t), y(t)

Lets try it on the example with air resistance illustrated in Eq. 2.31

set parametric

v0y=25

v0x=25

g=9.8

b=0.1

x(t)=v0x*t-0.5*b*v0x*t**2

y(t)=v0y*t-0.5*(g+b*v0y)*t**2

+0.1666*b*g*t**3

plot[0:4.6] x(t),y(t)

0

5

10

15

20

25

30

0 10 20 30 40 50 60 70 80 90

x(t), y(t)

Graph the case with v0 = 40ms but with different launch angles (let b = 0.1) and by making graphs try to findthe launch angle that gives the greatest range for this v0, g, b. Then graph the trajectory of the rocket Eq. 2.38with v0x = v0y = 50, w = g/10, and find the range of the rocket by making graphs.

27. By graphing (with gnuplot) both trajectories Eq. 2.42, 2.43 with θ = 45o, ϕ = 10o, v0 = 50 determine xmax, theplace where the ball hits the hill.

Chapter 3

Newton’s Laws. Dynamics

Up until now we have studied motion without any regard to the cause of the motion. This is called kinematics. Thecause and effect relationship between forces and motion upgrades our studies to that of dynamics.These hypothesiswere first laid down by Isaac Newton in the 17th century. Newton invented calculus as a means to solve the resultingequations of motion.Newton’s three dynamical laws, established empirically (via observation) are as follows;1. A body in uniform motion will remain so unless acted upon by forces.2. An agent applying a force on a body experiences an equal and opposite reaction force exerted upon it by thebody.3. The vector sum of the forces on a body equal it’s mass times acceleration∑

i

Fi = ma (3.1)

The third law has the first as a consequence, if no forces act on a body then

0 = ma = mdv

dt(3.2)

which can be integrated

dv = 0 dt,

∫ v(t)

v0

dv = 0, v(t) = v0 (3.3)

which is precisely what is meant by uniform motion; the velocity has no change over time, the direction and magni-tude remain constant.

There are two main categories of application-problems; given a collection of forces acting on a mass m, determineits acceleration vector, and given a partial set of forces and a bodies acceleration, find all forces acting on it. Thestrategy for solving both types is the same;Step 1 set up a coordinate system of your choice into which we resolve the components of all vectors;Step 2 Write Newton’s third law for every mass involved in the problem in component form using the laws ofvector mathematics;

ma = m (ax, ay, az) = (max,may,maz)

=∑i

Fi =∑i

(Fi,x, Fi,y, Fi,z) = (∑i

Fi,x,∑i

Fi,y,∑i

Fi,z) (3.4)

finally obtaining three equations for each mass

max =∑i

Fi,x, may =∑i

Fi,y, maz =∑i

Fi,z (3.5)

Step 3 we simply solve these equations for any unknowns present in them.

53

54 CHAPTER 3. NEWTON’S LAWS. DYNAMICS

Most of the cases that we will study early in the course will involve forces that are constant in magnitude anddirection. These will always result in motion with constant acceleration. Next we study circular motion, in whichthe applied forces will be constant in magnitude and direction in polar coordinates, and again the motion will befamiliar. We finally study restoring forces that vary in magnitude with position, these will lead to periodic or simpleharmonic motion.

3.1 Free-body diagrams

In every case the solution is expedited by drawing a free-body diagram for every mass involved in the problem.This figure will illustrate the body embedded in the coordinate system of your choice and will show only thoseforces applied directly to the body exerted through physical contact. The only way to apply a force to a bodyat this stage in the game is through physical contact.We have three rules that are applied to the determination of forces applied to a body;

Rule 1. A rope under tension pulls on each object that it is attached to with equal force, the magnitude of whichis the string tension, the direction is tangent to the rope.

M1 M2

T (exerted on M2)

M1 M2

T (exerted on M1)

Rule 2. Gravity acts on a body of mass M with force

Fg = mg, g = |g| = 9.8m

s2, g = (0,−g) (3.6)

Rule 3. The force between any two solid bodies in physical contact (the normal force) is a push on each directednormal to (perpendicular to) the surface of contact.

N ’−N ’

(Force on M2

exerted by M1)

N, exerted by table

M M1 M2

3.2. THE EQUATIONS OF MOTION (EOMS) 55

3.2 The equations of motion (EOMS)

M1

M2

M3

x

y

Example 1. Three masses hang on ropes as shown in the figurebelow. Find the tensions in the ropes.

The first step is to draw all of the force vectors acting on each bodyin the figure. Remember that a rope pulls both ends towardsthe center by application of the second law, and all vectors haveonly y-components. Therefore for each mass

M1a1,y =

onM1∑i

Fi,y, M2a2,y =

onM2∑i

Fi,y (3.7)

M3a3,y =

onM3∑i

Fi,y (3.8)

We then write Newton’s law for each body in turn by inserting onlythose forces actually acting on the body in each sum. Since no objectaccelerates, all accelerations are zero and we find for example

M1; M1 (0, a1,y) = (0, T1) + (0,−T2) + (0,−M1g) (3.9)

orT1 − T2 −M1g =M1a1,y = 0 (3.10)

andM2; M2 (0, a2,y) = (0, T2) + (0,−T3) + (0,−M2g) (3.11)

orT2 − T3 −M2g =M2a2,y = 0 (3.12)

andM3(0, a3,y) = (0, T3) + (0,−M3g), T3 −M3g =M3a3,y = 0 (3.13)

M1 M2 M3

T1

T2 M1 g M2 g T3

T2

M3 g

T3

We have taken up to be positive, all vectors such as forces and accelerations, are positive if they point up, negativeif they point down. Adding these equations we find that

T1 = (M1 +M2 +M3)g, T2 = (M2 +M3)g (3.14)

andT3 =M3g (3.15)


Example 2. A force F is applied horizontally to a rope attached to two masses as shown. Find the resultingaccelerations and tension in the connecting rope. Ignore friction with the table.

M1 M2

F

Again apply Newton’s law to each body in turn, only considering forces applied directly to that body, and usingright as the positive x-direction.

M2(a2,x, 0) = (F, 0) + (−T1, 0) + (0, N ′) + (0,−M2g), F − T1 =M2a2,x, N ′ =M2g (3.16)

M1(a1,x, 0) = (T1, 0) + (0, N) + (0,−M1g), T1 =M1a1,x, N =M1g (3.17)

where we use a1,x = a2,x = a since the masses coaccelerate if the ropes do not stretch. The free-body diagramssay it all;

T1

N

M1 g

T1

N’

M2 g

M1 M2

F

Eliminate T1

F = (M1 +M2)a, a =F

M1 +M2(3.18)

and substituting this back into the system of equations gives us the string tension

T1 =M1F

M1 +M2(3.19)

For many body systems Newton’s laws always results in systems of simultaneous linear equations.

FN

M g

θ Fx

FyExample 3. A box slides freely withno friction on a tabletop. A force F,insufficient to lift the box, is applied asshown. Find the acceleration and theforce exerted on the table.Using up at positive y-direction andright as positive x, draw all forces actingon the box. The table exerts a forceN upward, which by the second lawequals the force exerted on the tabledownwards.

3.2. THE EQUATIONS OF MOTION (EOMS) 57

Decompose F into components

F = (|F| cos θ, |F| sin θ) = (Fx, Fy) (3.20)

then by noting that if the block accelerates in contact with the table a = (ax, ay) = (a, 0), then

Ma = N+ Fg + F, M(a, 0) = (N, 0) + (0,−Mg) + (F| cos θ, |F| sin θ) (3.21)

and so

|F| cos θ =Ma, we find that a =|F| cos θM

and N =Mg − |F| sin θ (3.22)

Example 4. A box slides down an in-clined plane of angle θ under the in-fluence of gravity, but with no friction.Compute its acceleration and the normalforce exerted by the incline.It is this normal force or contact force,applied perpendicular to the plane ofcontact between the two bodies, thatkeeps the box sliding along the inclinerather than moving into it.

A good choice of coordinate axes makes the vector arithmetic simple; choose x to point up the incline, yto point perpendicular to it. Only one vector in the whole problem will have both components non-zero, and that isthe gravitational force on m;

a = (ax, 0), N = (0, N) (3.23)

Fg = (−mg sin θ,−mg cos θ) (3.24)

and so

m(ax, 0) = (0, N) + (−mg sin θ,−mg cos θ) (3.25)

and we find that

ax = −g sin θ, N = mg cos θ (3.26)

the negative acceleration indicates that it is directed to the left along the x-axis.


Example 5. A massless, frictionless pulley has two masses suspended from it.Compute their accelerations and the rope tension.A pulley simply redirects a strings tension. If it is massless, the tension in therope must be everywhere the same, and remember that a rope pulls both ends intowards the center. Draw all of the forces acting on the individual bodies and useup as the positive direction. The two masses have equal and opposite accelerationsif the rope does not stretch or break.First examine M1; the forces acting on it are

T = (0, T ), Fg = (0,−M1g) (3.27)

and we assume it accelerates downwards

a1 = (0, a1,y) = (0,−a) (3.28)

and so

M1a1 = T+ Fg, M1(0,−a) = (0, T ) + (0,−M1g), −M1a = T −M1g (3.29)

Now examine M2; the forces acting on it are

T = (0, T ), Fg = (0,−M2g) (3.30)

and we assume it accelerates upwards

a2 = (0, a2,y) = (0, a) (3.31)

and so M2a2 = T+ Fg

M2(0, a) = (0, T ) + (0,−M2g), M2a = T −M2g, solving; a =M1 −M2

M1 +M2g, T =

2M1M2

M1 +M2g (3.32)

Example 6. Here is a very interesting way to look at the problem; let x1and x2 be the coordinates of the two masses relative to the fixed-point ofthe center of the pulley, then we have three equations, since the length ofthe string is constant (ℓ) if it does not stretch

M1a1 = M1g − T

M2a2 = M2g − T

ℓ = x1 + x2, 0 = v1 + v2 =d

dt(x1 + x2), 0 = a1 + a2

This means that a2 = −a1, and therefore

M1a1 = M1g − T

−M2a1 = M2g − T

a1 =M1 −M2

M1 +M2g (3.33)

This method allows you to easily solve problems with multiple pulleys andblock/tackle devices.

3.3. PROBLEMS 59

3.3 Problems

1. Three forces

F1 = 4N j, F2 = 5N i+ 5N j, F3 = −2N i− 5N j

are applied to a body of mass m = 10 kg. Find its acceleration.

2. Three forces

F1 = 4N j, F2 = 5N i+ 5N j, F3 = −2N i− 5N j

are applied to a body of mass m = 10 kg whose acceleration is a = 2ms2 i+ 3ms2 j. Find the fourth force that must beacting.

3. A body of mass m = 10 kg initially at rest has force F = 20N i+ 30N j applied to it. Find its velocity at t = 3 s.

FN

M g

θ Fx

Fy

4. A block of mass m = 10kg rests on a smooth surface. Youpull on a rope attached to it with a force of |F| = 10N at a 30o

angle to the horizontal. Find its acceleration a and the normalforce N.

5. A body of mass m = 10 kg has force F = 20N i + 30N j applied to it. It moves at constant velocity v = 3.0ms i.There must be at least one other force on the body, what is it?

M1F

6. A block (m = 10 kg) resting on a surface has gravity anda normal force only acting on it, unless you push with F asshown, then friction acts in a direction that would opposethe sliding of the box on the surface. Suppose that you pushwith F = 3.0N i and the box does not move. Find the force offriction. Suppose that you push with F = 24.0N i and the boxaccelerates at a = 2.0ms i, Find the force of friction.

M1

M2

7. Two blocks suspended as illustrated have massesM1 = 5.0 kg, M2 = 5.0 kg. The table is frictionless.

A. Draw a free-body diagram for each mass.B. Find the tension in the string and the accelerations.


M3 gM1 g

M2 g

8. Consider three masses interconnected by a rope of length ℓ loopedover three pulleys as illustrated. Find the accelerations of all threemasses. Note that the accelerations must be inter-related ifthe rope does not stretch or break. If you use the method of example6;

M1a1 = M1g − T

M2a2 = M2g − 2T

M3a3 = M3g − T

ℓ = x1 + 2x2 + x3

0 = a1 + 2a2 + a3

Can you see why all of these are true?

M2 g

M1 g

9. This device is a block and tackle a simple machine that uses mul-tiple pulleys to gain a mechanical advantage. Find the accelerationsof both masses, and the tensions in all of the ropes. Hint, the ropes donot stretch or break, let the long rope have length ℓ, use the previousproblem.

M2

M1

10. This device is a block and tackle a simple machine that usesmultiple pulleys to gain a mechanical advantage. Find the accelera-tions of both masses, and the tensions in all of the ropes. Hint, theropes do not stretch or break, let the long rope have length ℓ.

3.3. PROBLEMS 61

M1

M2

θ

L

11 M2 is not enough to lift M1 off of the ground. Find theacceleration of M1 and M2 when the string between M1

and the pulley has length L and makes angle θ with respectto the horizontal, and both masses are released from rest.

θ

M

m

12. A box of mass M = 20 kg is placed on atable as illustrated. A second box m = 20 kgis suspended as illustrated with θ = 60o. Fric-tion holdsM in place, and nothing accelerates.

Find the tensions in all three strings.

Find the force of friction.

M1

M2

13. M1 = 120 kg, M2 = 20 kg.Suppose that accelerations do occur, find a1and a2 in terms of M1 and M2. If you use themethod of example 6;

M1a1 = T

M2a2 = M2g − 2T

ℓ = x1 + 2x2

Note that as x2 increases, x1 must shrink.


M2

M1

Θ

14. A mass M1 = 10 kg rests on an incline ofangle θ = 60o. Find the largest M2 such that noaccelerations occur.

M1

M2

15. Find the tensions in all of the strings. If the top string is cut, find the(instantaneous) acceleration of M1. If the bottom string is cut, find the(instantaneous) acceleration of M2 ( a GRE problem).

16. Return to problem 3.8, and solve it using REDUCE, maxima or axiom,for example

% in REDUCE

EOMS:=M1*a1+T-M1*g,

M1*a2+2*T-M2*g,

M3*a3+T-M3*g,

a1+2*a2+a3;

VARS:=a1,a2,a3,T;

solve(EOMS,VARS);

This gives you a way to check your calculations if you have trouble withbasic algebra. Of course you still have to do the physics (get the EOMS).

M1 g

M2 g

17. Find the accelerations of both masses and the tension in the rope.

3.3. PROBLEMS 63

M3 g

M1 g

M2 g

x1

x2

y1y2

18. Show that the equations of motion for this system are

M1x1 = M1g − T1

M2(x2 + y2) = M2g − T2

M3(x2 + y1) = M3g − T2

T1 = 2T2

ℓ1 = x1 + x2

ℓ2 = y1 + y2

Hint; regard the lower pulley as a body of mass M4 = 0. Solve this systemof equations.

19. A body of mass 10 kg starts at rest at time t = 0 and is acted upon by a force

F = (Fx, Fy) = (3N, 3 tN

s)

Find its acceleration, velocity and speed after 4 seconds.

20. A body of mass 10 kg is found to have velocity v =(4.0 m

s +3.0t3ms4

)i+2.0 tms2 j at time t. Find its acceleration

and the total force acting on it at t = 3 s.

21. Carefully examine the figure for problem 1.12, in which a boat is pulled along the negative x-axis by a rope.Suppose that the water exerts a force Fw = −1000N i on the boat. At the instant shown, where h = 5m andd = 10m, suppose that the tension in the rope is T = Txi+ Ty j, and the boat moves right at constant speed. Theonly other force on the boat is the combined force of gravity and bouyancy which is purely vertical Fg,b = Fy j. FindTx, Ty, Fy.

Chapter 4

Frictional Forces

Friction will be handled in basic Newton’s laws problems by a phenomenological method. Friction between twosurfaces is caused by two things. There will be mild electrostatic forces between the molecules that comprise the twosurfaces. There also will be partial interlocking of the surfaces, since in the molecular level even the smoothest ofsurfaces is jagged and contains high and low points like teeth on a saw.The greater the normal force pushing the surfaces together, the greater will be the degree of interlocking and thestrength of these weak chemical bonds. For this reason we propose that the magnitude of the frictional forcebetween two surfaces is proportional to the normal force between them and the proportionality constant is called thecoefficient of friction

|F|

|Ff |

µs |N|

µk |N|

µs |N|

|a|

|F|µs |N|−µk |N|/m

N

F Ff

Friction is complex. Consider a block of mass m at rest on a smooth surface. we apply a force F to it, slowlyincreasing its magnitude from zero. The block does not move. The force of friction is variable, being justenough to counteract applied forces and prevent slipping until a critical value is reached. The critical value is

65

66 CHAPTER 4. FRICTIONAL FORCES

the maximum value that the force of friction can exert before it is overcome, and slipping occurs. this value dependson a constant µs and the normal force between the mass and the surface applying the friction.

|Ff,static| ≤ µs|N|, no slipping (4.1)

Once F exceeds this value, the force of friction relaxes to a constant magnitude force

Ff,kin = −µk|N| v

|v|, v = 0 (4.2)

and the body begins to accelerate.The direction of the force of kinetic friction opposes any relative velocity vector between the two surfaces. Onceslipping begins to occur, friction acting on a given object opposes the velocity of that object measured with respectto the body exerting the frictional force.

The constants µs, µk are dimensionless, and typically

0 ≤ µk < µs < 1 (4.3)

for most smooth surfaces.

MFapp

M g

N

Ff

a, v, x Example 1. A block of mass M rests on atable, the coefficient of friction between thetwo surfaces is nonzero. A horizontal forceFapp is applied, find the resulting acceleration.

Solve the “critical” problem first;Break up all forces into components paralleland perpendicular to the x-axis (points right)and y-axis (up). If the box slips, v pointsright

Fapp = (F, 0)

N = (0, N)

Fg = (0,−Mg)

Ff = (−Ff , 0) (4.4)

and if the acceleration and velocity are to the right (and clearly the box is slipping if v = 0)

a = (a, 0), v = (v, 0) (4.5)

We apply Newton’s laws

Ma = N+ Fg + Fapp + Ff

M(a, 0) = (0, N) + (0,−Mg) + (F, 0) + (−Ff , 0) (4.6)

and solve both the x and y-equations to get

N =Mg, a =F − FfM

(4.7)

Now decide where on the friction graph you are; let Ff = Ff,max = µsN and a = 0 (the critical case).This will give you the largest F that you could apply without having the box slip. For example let M = 10 kg,µs = 0.1 and F = 8N . We find F = 8N < Ff,max = 9.8N , and so the box won’t slip, a = 0, and Eq. 4.7 givesFf = F = 8N .Suppose that F = 11N , then F > Ff,max and the box does slip, making a = 0 and

Ff = (−µk|N|, 0) = (−µkN, 0)

N =Mg, a =F − µkMg

M(4.8)

67

and so the acceleration is slowed down by the presence of friction between the surfaces.

M

Fapp

θ

M g

N

Ff

a, v, xExample 2. Consider the same problem, butwith the force Fapp supplied by a rope pullingup at an angle θ to the horizontal.Decompose all forces, and again we study thecase of motion and acceleration to the right;

Fapp = (F cos θ, F sin θ)

N = (0, N), Fg = (0,−Mg)

a = (a, 0), v = (v, 0) (4.9)

then the force of friction is

Ff = (−Ff , 0) (4.10)

because if slipping occurs, the box moves to the right. Solve the critical problem first; Newton’s laws equationsbecome

Ma = N+ Fg + Fapp + Ff

M(a, 0) = (0, N) + (0,−Mg) + (F cos θ, F sin θ) + (−Ff , 0) (4.11)

solving we find

N =Mg − F sin θ, and a =F cos θ − Ff

M(4.12)

Let Ff = µsN and a = 0; this gives the conditions for a critical state;

Fc cos θc = µs(Mg − Fc sin θc), Fc =µsMg

cos θc + µs sin θc(4.13)

If this condition is exceeded, thenFf = (−µk|N|, 0) = (−µkN, 0) (4.14)

and

N =Mg − F sin θ, and a =F cos θ − µk(Mg − F sin θ)

M(4.15)

if instead this condition is not met, then a = 0 and

Ff = F cos θ (4.16)

Example 3. If we place a block on a flathorizontal surface, and slowly tip the surfaceat ever increasing angle to the horizontal,the block does not slip until a critical angleis reached, where gravity overcomes friction.Find the angle.

If we assume no slipping, the force of frictionis static and points up the incline, parallel tothe surface. Its value is just enough toexactly cancel the sum of all other forcesparallel to the incline.


When the sum of these other forces has a magnitude that reaches or exceeds the largest magnitude force that staticfriction can apply, µs|N|, the grip of static friction is broken and the block begins to slip.In the coordinate axes-choice illustrated;

N = (0, N), Fg = (−mg sin θ,−mg cos θ) (4.17)

and so force of friction exactly opposes the sum of all forces in the x-direction. When the critical angle θc is reachedwhere

Ff = (µs|N|, 0) (4.18)

we have the point where the block will begin to slip, a = (0, 0), v = (0, 0), and

ma = N+ Fg + Ff

m(0, 0) = (0, N) + (−mg sin θc,−mg cos θc) + (µsN, 0) (4.19)

Solving we get

tan θc = µs (4.20)

and we are at the threshold of breaking friction’s grip.If the angle is increased past this value, the block accelerates down the incline at a non-zero rate a, with Ff = µkN =µkmg cos θ, and a = mg sin θ − µkmg cos θ since

m(−a, 0) = (0, N) + (−mg sin θ,−mg cos θ) + (µkN, 0) (4.21)

If θ < θc, then a = 0 and Ff = mg sin θ since

m(0, 0) = (0, N) + (−mg sin θ,−mg cos θ) + (Ff , 0) (4.22)

Example 4. For the problem below, now includ-ing frictional forces, compute the acceleration of themasses.Consider mass M1, and let right and up be positivedirections.

N = (0, N), T = (T, 0), Fg = (0,−M1g)(4.23)

and we see that acceleration and velocity is to theright;

a1 = (a, 0), v1 = (v, 0) (4.24)

making the force of friction that acts on M1 be

Ff = (−µkN, 0) (4.25)

For mass M2, we have

T = (0, T ), Fg = (0,−M2g) (4.26)

anda2 = (0,−a) (4.27)

If the rope does not break or stretch we must have coacceleration

|a1| = |a2| = a (4.28)

Newton’s laws become

M1(a, 0) = (0, N) + (T, 0) + (−µkN, 0) + (0,−M1g), M2(0,−a) = (0, T ) + (0,−M2g) (4.29)

69

Solving we find that

N =M1g, a =(M2 − µkM1)g

M1 +M2(4.30)

M1

M2FExample 5∗. In the figure below, there is only friction betweenthe two blocks, none with the floor. Find the maximal force Fthat we can apply to the top block and still not have the topblock slip against the bottom.Pay careful attention to the direction of friction acting on M2.The top block would slip to the right over the top ofM1, so friction acts to the left on the top block.

This frictional force is a contact force exerted on M2 by M1, and so M2 exerts an equal magnitude, oppositelydirect force of friction on M1. The same can be said of the normal force N2 exerted upwards on M2 by M1. Theupper block exerts an equal magnitude, oppositely directed normal force on M1.The free-body diagrams say it all;

M2 M1

N2

M2 g

FFf

Ff

M1 g

N2

N1

For M2;

a2 = (a, 0) Fg = (0,−M2g) N2 = (0, N2)

F = (F, 0), Ff = (−µsN2, 0) (4.31)

and soM2(a, 0) = (0,−M2g) + (0, N2) + (F, 0) + (−µsN2, 0) (4.32)

For M1;

a1 = (a, 0) N1 = (0, N1) N2 = (0,−N2)

Fg = (0,−M1g), Ff = (µsN2, 0) (4.33)

andM1(a, 0) = (0, N1) + (0,−N2) + (0,−M1g) + (µsN2, 0) (4.34)

Solving these two equations we arrive at

N2 =M2g, M2a = F − µsM2g

N1 = N2 +M1g = (M1 +M2)g, M1a = µsN2 = µsM2g (4.35)

and the final result is

a = µsgM2

M1, F = µsg(M1 +M2)

M2

M1(4.36)


4.1 Problem solving strategy

Always solve the “critical value” problem first (example 3), in which Ff | = µs|N| and a| = 0. For an incline problem,this will give you the conditions (such as the angle) at which the system is ready to break free of static friction andbegin to move. If your actual angles or masses in the given problem exceed the critical values, the system acceleratesand Ff | = µk|N|. If they are less than the critical values, the force of friction can simply be solved for in terms ofthe other forces. See problem 2 and 5.

4.2 Problems

FM1

M2

Θ

1. In this problem the coefficients of frictionbetween wedge M1 and the horizontal tableit rests on are zero, but the friction betweenmoving wedge M1 and the block M2 hascoefficients µs = 0.3 and µk = 0.2. LetM1 = 10.0 kg and M2 = 5.0 kg. Θ = 30o.

A. Draw free-body diagrams for each mass below this line.B. What maximal co-acceleration can M1 and M2 undergo without any slipping of M2 on M1? Show all work.C. Compute the required force F (applied to the wedge, pulling left) such that the blocks have this acceleration.

M

Θ

2∗. A box of mass M = 20.0 kg is placedon a variable-angle ramp as illustrated. Thecoefficients of friction between the ramp andbox are µk = 0.3 and µs = 0.4.

A. If θ = 18o determine the magnitude of the force of friction acting on the box.B. The ramp is raised until the box begins to slip. At what angle does slipping first occur?C. The ramp is raised until the box begins to slip. Find the acceleration of the box when θ is increased to 40o.

FM1

M2

3∗. In this problem the coefficients of friction betweenblocks M1 and M2 are µs = 0.5 and µk = 0.4. Let M1 =10.0 kg and M2 = 5.0 kg. There is no friction between M1

and the horizontal surface on which it slides.A. Draw a free-body diagram for each mass separatelybelow this line.B. The magnitude of F is 60N . Find the acceleration(vector) of M2.C. The magnitude of F is 60N . Find the acceleration(vector) of M1.

4.2. PROBLEMS 71

4. Another block and tackle device; M1 sits on a surfacewith µs, µk = 0, and M2 hangs freely. Find the acceler-ations of both masses, assuming that they do accelerate(see problem 14) and the tensions in all of the ropes. Hint,the ropes do not stretch or break, let the long rope havelength ℓ.

M1

M3

M2

5. Three blocks suspended as illustrated have massesM1 = 5.0 kg, M3 = 5.0 kg and the coefficient of frictionbetween M1 and M3 is µk = 0.1, µs = 0.15. The table isfrictionless.A. Find the maximum value of M2 such that the massescoaccelerate.B. Find the tension in the string if all masses coaccelerate.C. Now let M2 = 1.0 kg, and let all three masses startout at rest. After M2 has descended 1.0m from rest,determine the speeds of all three masses.

θ

M

m

µk, µs

6. A box of mass M is placed on a table as illustrated.The coefficients of friction between the table and boxare µk = 0.5 and µs = 0.6. A second box m = 20 kg issuspended as illustrated with θ = 60o. There are noaccelerations.A. Draw all force vectors properly labeled acting on bothboxes on the figure.B. Find the smallest mass M such that no accelerationsoccur.C. Instead let M = 80 kg. Compute the force of friction(magnitude) acting on it.


M1

M2F

7. In this problem the coefficients of friction betweenblocks M1 and M2 are µs = 0.5 and µk = 0.4. LetM1 = 10.0 kg and M2 = 5.0 kg. There is no frictionbetween M1 and the horizontal surface on which it slides.A. The magnitude of F is 10N . Find the acceleration(vectors) of M2, of M1 and the force of friction acting onM2.B. The magnitude of F is 30N . Find the acceleration(vector) of M1, of M2 and the force of friction acting onM2.

FM1

M2

Θ

x

y

8. In this problem the coefficients of friction betweenstationary wedge M1 and the block M2 are µs = 0.3 andµk = 0.2. Let M1 = 10.0 kg and M2 = 5.0 kg. Θ = 30o.

A. If the block moves at constant speed |v| = 2.0ms upthe incline, find |F|.B. Compute the minimal force |F| will prevent anyacceleration of the block.

ML v0

9. You are driving at a speed of v0 in a flat-bed truckwith a bed of length L. At the tail end of the bed thereis a crate of mass M , with a coefficient of friction of µsor µk between crate and bed. You hit the brake, findthe minimal stopping distance for the truck such that thecrate does not slip at all.

ML

10. You are driving at a speed of v0 in a flat-bed truckwith a bed of length L. At the tail end of the bed thereis a crate of mass M , with a coefficient of friction of µsor µk between crate and bed. You hit the brake, findthe minimal stopping distance for the truck such thatthe crate slips the full length of the bed L in the sametime that it takes the vehicle to stop. The figure for theprevious problem shows the crate prior to braking, thefigure above shows the truck once it stops.

M

θ

F 11. Wingus and Dingus get jobs mopping up the poopdeck on a large ship. The poop deck is not what youthink it is. Wingus has taken physics, and so he adjuststhe angle θ of Dingus’s mop at such that no matter howhard Dingus pushes, the mop-head won’t move. The mop-head has mass M = 1.0 kg, and the coefficients of frictionbetween the wet deck and mop-head are µs = 0.4 andµk = 0.3. Find θ.

4.2. PROBLEMS 73

M

θ

F

12. You need to hold a book shelf in place against a wallby lifting vertically with a force equal and opposite to itsweight, but you have taken physics, and realize that fric-tion can assist you if you push on it with force F at angle θ.The case has massM = 60.0 kg, and the coefficients of fric-tion between the wall and case are µs = 0.6 and µk = 0.4.Find minimum force magnitude |F| and θ needed to holdthe case in place against the wall.

M2

M1

Θ

13. A very standard problem; a mass M1 = 5 kg rests onan incline of angle θ = 30o. The coefficients of frictionwith the incline are µs = 0.3, µk = 0.1.A. Find the extremal (largest/smallest) mass M2 suchthat the box M1 does not slip down the incline.B. Find the extremal mass M2 such that the box M1 doesnot slip up the incline.C. Suppose that M2 is so great that M1 is pulled up theincline with constant acceleration. Find that accelerationin terms of the symbols given in the problem (a formulasolution, not numerical).

M1

M2

14. There is friction with coefficients µk, µs betweenM1 and the table. Find the maximal M2 such thatno accelerations occur. If M2 less than this value,find the force of friction acting on M1. If M2 isgreater than this amount find the accelerations ofboth masses.


M1

FM2

15. I take my penguin friend out for a walk, its abeautiful day (−20C). However this is an obstinate(but trim at M2 = 20 kg) penguin who immediatelyhops onto a block of ice of mass M1 = 100 kg. Ofcourse penguins have very rough feet, and so there isfriction of coefficients µk = 0.1, µs = 0.2 betweenpenguin and block, but no friction between blockand ground.

If I pull with |F| = 30N , find the accelerations of the block and penguin, and the magnitude of the force of friction.With what largest force can I pull without the penguin slipping? If I pull with 60N , find the acceleration of blockand penguin.

16. The coefficient of friction between loose grains of sand is µs. Show that the largest conical pile of sand that youcan create with a fixed circular base of radius R has volume V = µsπ

3 R3.

M1 M1

M2

17∗. The coefficient between table-top andtriangles is µs. Find the largest M2 that canbe placed as illustrated without the trianglesslipping.Ans. M2 = 2M1g

1−µs

θ

M1

M2

µk, µs

18. A box of mass M1 = 20 kg is placed on a tableas illustrated. The coefficients of friction between thetable and box are µk = 0.2 and µs = 0.3. A second boxM2 = 10 kg is suspended as illustrated. There are noaccelerations.A. Draw all force vectors properly labeled acting on bothboxes on the figure.B. Find the largest angle θ such that no accelerationsoccur.C. If θ has this value and you tap M1 gently to break thebond of static friction, find the instantaneous accelerationof M1.

4.2. PROBLEMS 75

FM

Θ

19. The ramp cannot move; there is frictionwith coefficients µs = 0.3, µk = 0.2 betweenM = 50 kg and the ramp of angle θ = 45o.What is the smallest horizontal force F thatyou must apply to keep the block from slippingdown the ramp?

What is the largest horizontal force F that youmust apply to keep the block from slipping upthe ramp?

Chapter 5

Circular motion at constant speed

θ

R cos(θ)

R sin(θ)

sR

Consider a mass m tied to a string,forced to travel in a circle horizon-tally at constant speed v, radius R.If in time t the mass travels arclength s along the circle, it sweepsout an angle θ.Then since

s = Rθ, θ in radians (5.1)

differentiate

v =ds

dt= R

dθ

dt= Rω (5.2)

where ω is the angular speed orfrequency and has units of radi-ans per second. Remember that

360o = 2π rad (5.3)

Impose a coordinate system on the figure and write down the position vector after integrating the above equation to

θ = ωt (5.4)

thenr(t) = (R cosωt,R sinωt) (5.5)

and the velocity is

v(t) =d

dtr(t) = (−Rω sinωt,Rω cosωt) (5.6)

Notice thatr · v = 0 (5.7)

This is a consequence of the fact thatr(t) = R (cosωt, sinωt) = Rer (5.8)

and er =r|r| is a unit vector;

er · er = cos2 ωt+ sin2 ωt = 1 (5.9)

77

78 CHAPTER 5. CIRCULAR MOTION AT CONSTANT SPEED

Calculate derivatives;d

dt

(er · er

)= 2er ·

( ddt

er

)=

d

dt1 = 0 (5.10)

x

y

r(t)a(t)

v(t)

θ=ω t

The moral of the story; the derivative of aunit vector is perpendicular to the vectoritself.Therefore the position vector is radial and thevelocity vector is tangential to the circle. Nowcompute the acceleration

a(t) =d

dtv(t)

= (−Rω2 cosωt,−Rω2 sinωt)

= −ω2r(t) (5.11)

This acceleration, called centripetal acceleration, points towards the center of the circle. In any problem involvingcircular motion at constant speed, the acceleration is given by this expression, which has magnitude

ac = ω2R =v2

R(5.12)

Example 1. In the problem below, M2 slideson a table-top, presumed frictionless; find thespeed thatM2 must move in a circle at in orderto supply enough string tension to support M1

against gravity.Write Newton’s laws, we want M1 to have noacceleration so

Fg = (0,−M1g) = −M1gj

T = (0, T ) = T j (5.13)

and soM10 = Fg +T (5.14)

(0, 0) = (0,−M1g) + (0, T ), T =M1g (5.15)

and for M2 all relevant vectors point in the ±er direction;

T = −Ter, a = −acer = −v2

Rer (5.16)

andFg = −M2gj, N = N j, and therefore M2a = Fg +N+T (5.17)

−M2v2

Rer = N j−M2gj− Ter (5.18)

so that

M2v2

R= T, and elimination of T results in v =

√M1gR

M2(5.19)

79

y

x

N

Mg

Ff

ac

Example 2∗. A disc rotates at rate ωcounter-clockwise about the y-axis. A blockof mass M sits on it at a radius R from thecenter. Static friction holds it in place. Givena value for µs find the maximum spin-rate ωsuch that the block does not slide radially.If the spin-rate is too great, the maximalforce that friction can apply, µs|N|, will notbe enough to provide the required centripetalacceleration to hold it at radius R.

Therefore the block slips radially away from the center of the disk. This means that the force of friction pointsradially (in the x-direction at this instant) towards the center. Resolve all forces

a = (−ac, 0) = (−ω2R, 0) = (−v2

R, 0) (5.20)

N = (0, N), Fg = (0,−Mg), Ff = (−µsN, 0) (5.21)

and thereforeMa = N+ Fg + Ff , M(−ω2R, 0) = (0, N) + (0,−Mg) + (−µsN, 0) (5.22)

solving these equations

N =Mg, Mω2R = µsN = µsMg, makes ω =

√µsg

R(5.23)

y

x

Ff

Mg

N

ac

Example 3. Pictured is one of those awfulamusement park rides; a circular, open-topdrum that spins about the y-axis at ω. Passen-gers are held in place against the circular wallby friction. When the ride is up to speed, thefloor falls away. Amusing, hey? If the drumhas radius R, and the coefficient of frictionis µs, we compute the minimal spin-rate ωneeded to keep the passengers in place againstthe wall.

If they slip, they slip downwards under the influence of gravity. The wall exerts a normal force on them, and theacceleration is prescribed; at this instant

a = (−ac, 0) = (−ω2R, 0) (5.24)

The other forces areN = (−N, 0), Fg = (0,−Mg), Ff = (0, µsN) (5.25)

Put all the parts into place;

Ma = N+ Fg + Ff

M(−ω2R, 0) = (−N, 0) + (0,−Mg) + (0, µsN) (5.26)

and we obtain the equationsMω2R = N, Mg = µsN (5.27)

solving we find that

ω =

√g

µsR(5.28)


5.1 Problems

θR

m

1. A tiny metal ball is thrown intoa smooth (frictionless) sphericalbowl of radius R = 20 cm. Theball travels in a horizontal circle asillustrated, the path it makes is atan angle of 30o as measured fromthe center of the bowl and the ver-tical axis.

Find the required speed v of the ball such that it travels in such a path. Of course gravity is acting on it.

θ

ω

R

m

2. Consider a penguin egg of radius R that is placed on asmooth surface covered with a layer of water, and spun atrate ω about an axis through its center and the contact point.Droplets of water will creep up the sides of the spinning egg to amaximum angle θ determined by the adhesive force f (a normalforce of attraction) between egg-shell and droplet, gravity andthe centripetal force. Find θ.

θ

h

m

3. Find the speed with which a marble of radiusr = 0.01m, mass m = 0.01 kg must slide on a frictionlesscone of angle θ = 30o in order to travel in a horizontalcircular path (under the influence of gravity) a heighth = 0.5m above the apex of the cone.

5.1. PROBLEMS 81

4. In the problem illustrated,M2 is swung freely, the stringsweeping out a cone of angle θ. Find the angular speed ωthatM2 must move in a circle at in order to supply enoughstring tension to supportM1 against gravity if θ = 30o andr = 2m. Alternatively find θ given the speed v0, M1,M2.

y

x

R M

v0

v0

5. You and and I and a third person (I am the blocklabeled “M” on the right, that’s you screaming interror in the upper-center) get into an amusementpark ride consisting of a rotating drum of radius R =5m. There is a coefficient of friction between ourbacks and the drum walls of µk = 0.3, µs = 0.4.Once the drum reaches a minimal safe rotation rateω0, the floor drops away and we are held in placeagainst the walls by various forces. Find ω0 (notethat we travel at speed v0 = ω0R).

6∗. A “conical pendulum” of mass M “orbits” in acircle of radius R with the string making an angle ofθ with the vertical. Show that the magnitude of thestring tension is

T =√(Mg)2 + (Mω2R)2

v

N

mg

a

7. If a car traveling at constant speed v0 rounds thecrest of a hill of radius of curvature R too fast, thecar can become airborne, losing contact with theroad. In terms of R and g, find the largest speedv0 such that this does not happen. Then watch themovie “Bullitt”.


v

Nmg

a

8. There is a circus routine in which a motor-cyclist rides inside ofa steel cage, traveling fast enough to negotiate the top of the loopwithout losing contact between wheels and cage. In terms of thecage radius R and g, find the smallest speed v0 such that the cyclistcompletes the loop without falling off at the apex.

9∗. Suppose that you travel a plane curve along a parameterized pathr = x(t) i + y(t) j. Your velocity and acceleration are (introducing adot notation for t-derivatives)

v(t) =dx

dti+

dy

dti ≡ x i+ y j, a(t) =

d2x

dt2i+

d2y

dt2i ≡ x i+ y j

Your velocity is tangent to the trajectory, so lets create two unit vectors t and n that are respectively tangent toand normal to your trajectory

t =v

|v|=

x√x2 + y2

i+y√

x2 + y2j, n = k× t =

y√x2 + y2

i− x√x2 + y2

j

Now decompose a into a part tangential to and a part normal to the trajectory, by simply changingvector basis (see Chapter 1)

a = x i+ y j = at t+ an n

Prove that

at =xx+ yy√x2 + y2

=d

dt

√x2 + y2, an =

yx− xy√x2 + y2

(5.29)

The first result tells you that if you move along an arbitrary path at constant speed, your only acceleration isnormal.

Define the radius of curvature of the trajectory curve R(t) at point r = x(t) i+ y(t) j from

|an(t)| =|v(t)|2

R(t),

1

R(t)=

|an(t)||v(t)|2

(5.30)

Does this all look familiar from your math classes? You can see the obvious applications.

Suppose that round the top of a parabolic hill y(x) = −k2x

2, what maximum speed can you negotiate the top of thehill at without losing contact with it? Hint, parameterize the hill at that instant (t = 0) as

x(t) = v0 t, y(t) = −kv20

2t2

10∗. A car travels at constant tangential speed v0 over a succession of hills y(x) = h cos(

2πxλ

). Find the maximal

speed v0 such that the car never loses contact with the hill-tops.

11. A scale reads out the normal force needed to support you in a stationary position on the earth’s surface. Show

that at the North Pole a scale reads Mg but at the equator M(g −Reω

2)where M is your mass, RE the radius of

the earth, and ω its rotation rate.

Chapter 6

Work and Energy

The dynamical equations relating cause (forces) to effect (acceleration) in particle dynamics are technically secondorder differential equations. Let F be a force applied to a body of mass m

ma = F, or md2r(t)

dt2= F(r) (6.1)

and in general second order differential equations are much harder to solve than first order equations. The processof transforming this second order equation into a first order one involves a single integration, which gives rise to aconstant of integration called the energy. This provides the definitive answer to the question “what is energy”.It is a constant of integration produced when Newton’s laws are integrated. Being a constant (over time), it is aconserved quantity.

6.1 Kinetic energy

Consider

ma = F (6.2)

Now dot the velocity vector into both sides

ma · v = F · v (6.3)

and use the identity

a · v = (d

dtv) · v =

d

dt(1

2v · v) (6.4)

We have manipulated Newton’s law into

d

dt

(12mv · v

)= mv · a = F · v = F · dr

dt(6.5)

This can be integrated ∫ tf

t0

d

dt

(12mv · v

)dt =

∫ tf

t0

F · drdtdt (6.6)

let

r0 = r(t0), rf = r(tf ), v0 =(drdt

)(t0), vf =

(drdt

)(tf ) (6.7)

be the initial and final positions (and velocities) of the body as it is displaced along some path, while being actedupon by the force F, then changing variables∫ tf

t0

F · drdtdt =

∫ rf

r0

F · dr (6.8)

83

84 CHAPTER 6. WORK AND ENERGY

in the integration leads us to the Work-Energy theorem;∫ tf

t0

d

dt

(12mv · v

)dt =

1

2mv(tf ) · v(tf )−

1

2mv(t0) · v(t0) =

∫ rf

r0

F · dr (6.9)

The left hand side is called the change in kinetic energy, the right is called the work done by the force F.

KE(t) =1

2mv(t) · v(t), W =

∫ r

r0

F · dr (6.10)

Many types of dynamical problems can be solved by either the use of Newton’s laws or by the use of this formula.The advantage of the work-energy theorem is that it does not require much vector arithmetic, and certain types offorces are incapable of doing work. For example consider a body sliding along a surface. The normal force is alwaysperpendicular to the displacement dr, and so

N · dr = 0 (6.11)

6.2 Potential energy

Recalling the definition of the integral as the operation that reverses the process of differentiation∫ b

a

df

dxdx = f(a)− f(b) (6.12)

imagine that you are at position x = a at time ta, and at x = b at time tb, and the force applied to a body m is thederivative of some function V (x), in other words let

Fx = −dVdx

(6.13)

(this is not always true, forces could depend on velocity as well). Then substituting into Eq. 6.12 the left side is thework done by Fx

W =

∫ b

a

Fx dx = −∫ b

a

dV

dxdx = −

(V (b)− V (a)

)(6.14)

We call a force that possesses such a function V conservative, and V is its potential function. The work done bya conservative force depends only on the value of its potential function at the beginning and end of the displacement.

Lets generalize this to two and three dimensions with the chain rule of calculus applied to the integral on the right-sideof the work-energy theorem for a conservative force.

− d

dtV (x, y, z) = −∂V

∂x

dx

dt− ∂V

∂y

dy

dt− ∂V

∂z

dz

dt

= Fxdx

dt+ Fy

dy

dt+ Fz

dz

dt= F · v (6.15)

This will mean that all of the components of F can be gotten from the same V ;

F =(− ∂V

∂x,−∂V

∂y,−∂V

∂z

)= −∂V

∂xi− ∂V

∂yj− ∂V

∂zk = −∇V (6.16)

This may not be possible to do for a given force. Friction is a good example of a non-conservative force; there is nosuch function V (r) because technically the force of friction depends on both position r and velocity v of an object.There is a simple test for whether or not a given force has a potential function.

Before you jump on me and say that “we haven’t gotten to partial derivatives yet in calculus”, letme remind you that derivatives are derivatives, and there is nothing here that can’t be handled by going back to

6.2. POTENTIAL ENERGY 85

the basics of calculus 221. Lets just work out a collection of simple examples to illustrate my point, using the basicdefinitions of derivatives, but of two-variable functions;

∂f(x, y)

∂x= lim

dx→0

f(x+ dx, y)− f(x, y)

dx∂f(x, y)

∂y= lim

dy→0

f(x, y + dy)− f(x, y)

dy(6.17)

For example let f(x, y) = α · x · y where α is constant;

∂f(x, y)

∂x= lim

dx→0

(α(x+ dx) y

)−(αx y

)dx

= limdx→0

α y dx

dx= limdx→0

α y = αy

∂f(x, y)

∂y= lim

dy→0

(αx (y + dy)

)−(αx y

)dy

= limdy→0

αxdy

dy= limdy→0

αx = αx (6.18)

let f(x, y) = α · (x2 − y2) where α is constant;

∂f(x, y)

∂x= lim

dx→0

(α((x+ dx)2 − y2

))−(α(x2 − y2)

)dx

= limdx→0

2αxdx

dx= limdx→0

2αx = 2αx

∂f(x, y)

∂y= lim

dy→0

(α(x2 − (y + dy)2

)−(α (x2 − y2)

)dy

= limdy→0

−2α y dy

dy= limdy→0

−2α y = −2αy (6.19)

If you need some help checking your work, log into REDUCE

f:=a*x*y;

df(f,x); % reduce just needs you to tell it what to differentiate

df(f,y); % and what to diff. with respect to (x or y)

clear f;

f:=a*(x^2-y^2);

df(f,x);

df(f,y);


6.3 Total energy

For a conservative force,d

dt

(12mv · v

)= F · v = − d

dtV (x, y, z) (6.20)

which we can rearrange tod

dt

(12mv · v + V (r)

)= 0 (6.21)

which can be integrated. Let the position of the moving body at time t1 be

r1 = r(t1), and r2 = r(t2) (6.22)

respectively. Integrate both sides of the above equation accordingly∫ t2

t1

d

dt

(12mv · v

)dt = −

∫ t2

t1

d

dtV (r(t)) dt

1

2m(v(t2) · v(t2)− v(t1) · v(t1)

)= −

∫ r2

r1

dV (r)

1

2m(v(t2) · v(t2)− v(t1) · v(t1)

)= −

(V (r(t2))− V (r(t1))

)= −

(V (r2)− V (r1)

)(6.23)

Get all of the stuff involving t2 on one side, the stuff involving t1 on the other

1

2mv(t2) · v(t2) + V (r2) =

1

2mv(t1) · v(t1) + V (r1) (6.24)

Since the quantity

E =1

2mv2 + V (x, y, z) (6.25)

has the same value at t2 as it does at time t1, and these times were arbitrary, it must in fact be a constant. Thisis an integration constant, since we integrated F = ma once to get it. The constant is called the energy E, andconsists of two parts; kinetic

KE =1

2mv2, v2 = |v|2 = v · v (6.26)

energy associated with motion. The potential energy V (r) is energy associated with spatial position, and can bewritten several different ways. The derivatives of the potential function represent the components of conservativeforce associated with it

Fx = −∂V∂x

, Fy = −∂V∂y

, Fz = −∂V∂z

(6.27)

Suppose that the only forces present are conservative, then the total energy is conserved, it merely changesform, from kinetic to potential or kinetic to work, as the particle undergoes its prescribed motion.

6.4 The master energy equation

Lets divide forces into conservative ones Fc that possess potentials V and non-conservative ones (like friction) Fncthat do not. Starting with Newton’s laws we run through the steps above

ma = m v = Fc + Fnc Newton’s laws

mv · v = v · Fc + v · Fnc, dot velocity into Newton’s laws

d

dt

(12mv · v

)= −dV

dt+ v · Fnc Use Eq. 6.20

d

dt

(12mv · v + V

)= v · Fnc Isolate the derivative of E

d

dt

(E)

= v · Fnc∫ tf

t0

d

dt

(E)dt =

∫ tf

t0

v · Fnc dt, integrate over time

E(tf )− E(t0) = Wnc, the master equation (6.28)

6.5. COMPUTING WORK DONE BY FORCES 87

The Master energy equation says that the energy is conserved unless there are non-conservative forces around, theywill change the energy (1

2mv2 + V

)tf

−(12mv2 + V

)t0

=Wnc (6.29)

6.5 Computing work done by forces

Lets calculate the right side of Eq. 6.9 for some simple forces.

x

y

Example 1. Compute the work done bythe force

F = kx i (6.30)

moving a particle from (0, 0) to (1, c)along the path

y = cx2 (6.31)

against this force.This is our first example of a force thatis not constant, it varies with position.The figure shows the curve, the velocity(tangent) vectors in blue, and the forcevector at each point in red.

The first step in performing the integral is to parameterize the path. This step turns what appears to be amany-variable integral into a single variable integral. We will use the parameterization

x = t, y = ct2 (6.32)

If we eliminate t we get y = cx2. Now we write the r vector for the points on the curve, and compute the velocityvector

r = (x, y) = (t, ct2), v =d

dtr = (1, 2ct) (6.33)

Next we calculate the force vector at every point of the path

F = kx i = (kx, 0) = (kt, 0) (6.34)

and finally the dot-productF · v = kt · 1 + 2ct · 0 = kt (6.35)

Now we integrate. ∫ (1,c)

(0,0)

F · dr =

∫ 1

0

F · drdtdt =

∫ 1

0

kt dt =1

2k(12 − 02

)(6.36)

The parameterization method is virtually fool-proof, and you can easily make REDUCE, axiom or maxima do theresulting integrals if you get stuck (see the problems). For conservative forces, you will get the same result no matterwhat parameterization you use, so you cannot make the wrong choice. Try that out, let

x = t2, y = ct4, r = (t2, ct4), v = (2t, 4ct3) (6.37)

andF = (kx, 0) = (kt2, 0) (6.38)

put this into the integral and churn out an answer; it will again be 12k.


x

y

(D,H)

Example 2. Compute the work done inmoving a mass of weight Fg = (0,−Mg)by the force of gravity along the pathshown below, a ramp of length D andheight H.

The path is a straight line, which we pa-rameterize with t ∈ [0, 1]

x = Dt, y = H t, r = (Dt,Ht)(6.39)

along which the velocity vector is

v =d

dt(Dt,Ht) = (D,H) (6.40)

Now calculate

F · v = (0,−Mg) · (D,H) = −MgH(6.41)

and the work done against gravity is

W =

∫ (D,H)

(0,0)

F · dr =

∫ 1

0

F · v dt =∫ 1

0

−MgH dt = −MgH (6.42)

Notice that no work was done moving the weight horizontally, only vertically by gravity. This lesson is driven homeby computing the work done in moving the weight first horizontally from (0, 0) to (D, 0), then vertically to (D,H)

Step 1; from (0, 0) to (D, 0)The force is at all times

F = (0,−Mg) (6.43)

Parameterize the path with

x = Dt, y = 0, t ∈ [0, 1] (6.44)

then points along the path are

r = (Dt, 0), v = (D, 0) (6.45)

and

F · v = (0,−Mg) · (D, 0) = 0 (6.46)

and so

Wstep1 =

∫ 1

0

0 dt = 0 (6.47)

No work is done since the displacement is ⊥ to the force all along the path.

Step 2; from (D, 0) to (D,H)Again the force is at all times

F = (0,−Mg) (6.48)

Parameterize the path with

x = D, y = Ht, t ∈ [0, 1] (6.49)

then points along the path are

r = (D,Ht), v = (0,H) (6.50)

and

F · v = (0,−Mg) · (0,H) = −MgH (6.51)


and so

W =

∫ 1

0

−MgH dt = −MgH (6.52)

All of the work is done on the lift.The sum of the work done on these two steps is precisely the net result gotten along the path from (0, 0) directly ina straight line to (D,H). This is because gravity is a conservative force. The force itself is path-independent,and therefore so is the work done. Compute the work done in moving the weight from (0, 0) to (D,H) along thepath

y =H

D3x3 (6.53)

parameterized byx = Dt, y = Ht3, t ∈ [0, 1] (6.54)

thenr = (Dt, Ht3), v = (D, 3Ht2), F = (0,−Mg) (6.55)

the relevant dot-product isF · v = (0,−Mg) · (D, 3Ht2) = −3MgHt2 (6.56)

and integrate

W =

∫ 1

0

−3MgHt2 dt = −31

3MgHt3

∣∣∣10= −MgH (6.57)

again the same result.

6.5.1 Gravity is conservative

Return to the work done by gravity in Eq. 6.42 and note that

Fg = −mg j = −∇(mg y

)= −

(i∂

∂x+ j

∂

∂y+ k

∂

∂z

)(mg y

)(6.58)

and so we could perform the Eq.6.42 calculation using the potential V = mg y

W = −(Vf − Vi

)= −

(mgH −mg 0

)= −mgH (6.59)

6.5.2 Friction is non-conservative

Frictional forces are forces that always oppose the velocity, hence are path dependent, in fact always pointing tangentto the differential arc length of the path. To see this, construct the force of friction for a moving body

F = −µk|N(r)| v|v|

= µk|N(r)| t (6.60)

(using the tangent unit vector to the path) but

v =dr

dt,

v

|v|=

drdt

|drdt |=

dr

|dr|(6.61)

The work done by friction will then be

W =

∫path

(− µk|N(r)| dr

|dr|

)· dr = −µk

∫path

|N(r)| |dr| (6.62)

becausedr · dr = |dr|2 (6.63)

The fact that this is negative means that when friction acts on a system, it reduces its mechanical energy, so thatEf < Ei.


Example 3. A block of mass M on a horizontal surface is given an initial speed v = vi to the right. How far doesit slide before friction brings it to rest?

All of its kinetic energy will be used up doing work against friction, so by energy conservation

1

2M v2f −

1

2M v2i =W =

∫ 1

0

Ff · v dt (6.64)

Parameterize the path by (note I am using dimensionless time t, its just a parameter)

r = (Dt, 0), t ∈ [0, 1] (6.65)

find v;

v =d

dtr = (D, 0) (6.66)

The force of friction isFf = (−µkMg, 0) = (−µk|N|, 0) (6.67)

calculate the work

Wf =

∫ 1

0

(−µkMg, 0) · (D, 0) dt = −µkMgD (6.68)

apply the theorem;

0− 1

2M v2 = −

(µkMgD

)(6.69)

and we find

D =v2

2µkg(6.70)

is the distance it slides before friction has expended all of its kinetic energy.

x

y Example 4. A box of mass M slides down ahill whose shape is

y = H −H( xD

)2(6.71)

under the influence of gravity (red vectors) andfriction with coefficient µk. The normal forceis shown in green. How much work is done byall forces involved?The velocity (tangent to hill) is illustrated inblue. Gravity is conservative but friction is not.Parameterize the curve of the hill;

x = Dt, y = H(1− t2), t ∈ [0, 1] (6.72)

Compute the tangent vector (velocity vector)for the hill at every point

r = (Dt,H(1− t2)), v = (D,−2Ht) (6.73)

We are ready to rock; lets explicitly calculate all of the vectors involved.Work done by gravity

Fg = (0,−Mg), Fg · v = (0,−Mg) · (D,−2Ht) = 2MgH t (6.74)


WFg =

∫ 1

0

2MgHt dt =1

22MgH =MgH (6.75)

This is positive, so by Eq. 6.9 it will increase the kinetic energy (speed up the mass).

Work done by the normal forceFirst compute a vector normal to the hill, this will be the vector ⊥ to v since v is tangent to the hill;

v = (D,−2Ht), v⊥ = (2Ht,D) (6.76)

now get the unit vector n normal to the hill

n =v⊥

|v⊥|= (

2Ht√4H2t2 +D2

,D√

4H2t2 +D2) (6.77)

The normal force is opposite to the component of the force of gravity that points in this direction.We will not calculate∗ the normal force; it is of magnitude N in the n-direction

N = Nn = N( 2Ht√

4H2t2 +D2,

D√4H2t2 +D2

)(6.78)

Looks icky, right? Now dot this into v;

N · v = N( 2Ht√

4H2t2 +D2,

D√4H2t2 +D2

)· (D,−2Ht) = 0 (6.79)

and so

WN =

∫ 1

0

N · v dt = 0 (6.80)

because N is ⊥ to v.

Work done by FrictionThe force of friction has magnitude µk|N| and points in the direction opposing the velocity vector;

Ff = −µk|N| v

|v|

= −µk∣∣∣( 2HtMgD

4H2t2 +D2,

MgD2

4H2t2 +D2

)∣∣∣ (D,−2Ht)√D2 + 4H2t2

= −µk(Mg

D√D2 + 4H2t2

) (D,−2Ht)√D2 + 4H2t2

(6.81)

Dot this into v;

Ff · v = µk

(Mg

D√D2 + 4H2t2

) (D,−2Ht)√D2 + 4H2t2

· (D,−2Ht) = −µkMgD (6.82)

and so despite the intermediate complexity of the forces themselves, once they are dotted into the velocity theintegrands of the work function are very simple;

WFf=

∫ 1

0

Ff · v dt =∫ 1

0

−µkMgDdt = −µkMgD (6.83)

which is negative, its presence will cause a slow-down.

∗A note; If the box slides on an arbitrary curve, one with a non-zero curvature, the normal force would not havemagnitude mg cos θ, since there would be a component of the acceleration normal to the curve. This is a fine pointthat we do not need to address in problems concerning work done by the normal force, since the result will bezero simply because N ⊥ dr. Of course one could find it necessary to compute this force, for example consider thefollowing problem;


θ v

dx

dy

mg

FF N

Example 5. Consider a wire bent into the shape of anarbitrary curve y = y(x). A bead with a hole through thecenter slides on the hill without friction, so only gravityand the normal force affect its motion. Find the workdone by friction as it slides to the bottom.

Solution. Examine the box at some point along the hill,where is moves from (x, y) to (x + dx, y(x + dx)) = (x +dx, y + y′ dx). As it moves right by dx, it descends by dy,so this part of its path is a little right triangle with angleθ and hypotenuse

ds =√(dx)2 + (dy)2 =

√(dx

dt)2 + (

dy

dt)2 dt = |v| dt

(6.84)

As it moves along the curve of the wire, its velocity v stays tangent to the wire, and the normal force N stays normalto the wire, so in passing from (x, y) to (x+ dx, y(x+ dx)) the normal force does work

dW = N · v dt = 0 (6.85)

and gravity does work

dW = −mg j · v dt = 0

= −mg |j||v| cos(90− θ) dt

= −mg ds sin(θ) since |v| dt = ds

= −mg dy (6.86)

Note that this calculation did not use the explicit equation of y(x) for the hill, you get this result for anyhill. You can integrate this to get the total work done descending the hill.

Friction does work

dW = −(µk|N| v

|v|

)· v dt = 0

= −µk |N| |v| dt= −µk |N| ds (6.87)

You can make the next step hard or easy; remember that if the hill has curvature, there will be centripetalforce on the sliding mass, and this will contribute to the normal force, but you can neglect this centripetalforce if the curvature is gradual, in that case the result is simple

dW ≈ −µk |N| ds≈ −µkmg cos θ ds

≈ −µkmg dx (6.88)

The punch-line is that if the curvature is gradual or the speeds are low then the work done by frictiononly depends on the horizontal displacement of the mass, and not on the shape of the hill. This gives the falseimpression that the result is path independent (conservative), but it is only approximate.

If the centripetal force associated with the hill’s curvature can’t be neglected, use Eq. 5.29, and you may need to dosome heavy calculus juju to compute the work (clearly not conservative since an = an(v))

dW = −µk |N| ds = −µk(mg cos θ +man

)ds (6.89)


F(θ)

Mg=Fg

T(θ)

Mg=Fg

T(θ)

F(θ)

θ

dr

Consider a crate hanging from a ropefrom the ceiling of mass M . A horizon-tal force F = F(θ) is applied to it, suffi-cient to move it at constant (very slow)speed to a new height h above its previ-ous height.Compute the force that needs to be ap-plied as a function of θ and compute thework done against gravity, against thetension in the rope, against the verticalcomponent of the rope tension, the hori-zontal rope tension, and the force F.Break T = T(θ) up into horizontal andvertical components

T cos θ −Mg = 0,

T sin θ − F = 0 (6.90)

since there are for all practical pur-poses no accelerations (The centripetal

acceleration v2

ℓ ≈ 0 if the motion is veryslow; a subtle point clarified by DonnHenriksen). Then

Ty = T cos θ =Mg, Tx = Fh = T sin θ =Mg tan θ (6.91)

The work done by the rope tension is

WT =

∫T · dr (6.92)

but at every step of the way you can see that T is perpendicular to the differential displacements dr and so

WT =WTy +WTx = 0 (6.93)

so that

WTy = −WTx (6.94)

We can further decompose the displacements

dr = i dx+ j dy = (dx, dy) (6.95)

and since

Fg = −Mgj (6.96)

the work done by gravity is

Wg =

∫ r(tf )

r(ti)

−Mg j · (i dx+ j dy) = −∫ h

0

Mg dy = −Mg h (6.97)

Compute the work done by the vertical part of the tension;

WTy =

∫ r(tf )

r(ti)

Ty · dr =

∫ h

0

+Mg j · (i dx+ j dy) =Mg h = −WG (6.98)


Suppose we know that in swinging up a height h the crate moves horizontally a distance D, the work done by thehorizontal component of the tension is then

WTx =

∫ r(tf )

r(ti)

Th · dr (6.99)

With our choice of origin above the apparatus, at the pivot point, and defining r to point away from the origin andθ to be counterclockwise, we find (with the origin at the point of attachment for the string)

r = r sin θ i− r cos θj, dr =(r cos θ i+ r sin θj

)dθ (6.100)

and

Tx = −Mg tan θ i (6.101)

putting all of this into the integral we obtain

WTx =

∫ θf

0

−Mg tan θ i ·(r cos θ i+ r sin θj

)dθ

= −∫ θf

0

Mgr sin θ dθ =Mgr(cos θf − 1) = −Mgh (6.102)

and this clearly is what we expect from

WTy= −WTx

(6.103)

6.6 When does a force not have a potential?

If one cannot find a function V such that

Fx = −∂V∂x

= −(∂V∂x

)y, Fy = −∂V

∂y= −

(∂V∂y

)x

(6.104)

then the force is not conservative and has no potential. Because the order in which partial derivatives are taken isimmaterial

∂2V

∂x∂y=

∂2V

∂y∂x, more precisely

( ∂∂x

(∂V∂y

)x

)y=( ∂∂y

(∂V∂x

)y

)x

(6.105)

we can get both Fx and Fy from the same V if and only if

∂Fx∂y

=∂Fy∂x

=∂2V

∂x∂y(6.106)

If Fx and Fy are such that ∂Fx

∂y = ∂Fy

∂x , then there is no potential and the force is non-conservative.

A good example is

F = −yi+ xj = (−y, x) (6.107)

We attempt to solve

Fx = −y = −∂V∂x

, −V =

∫−y dx = −yx+ C(y) (6.108)

(since we integrate with respect to x, our “constant” of integration could depend on anything but x, including y).This potential does not give the correct Fy;

Fy = x = −∂V∂y

= − ∂

∂y

(yx− C(y)

)= −x+ C ′ (6.109)

On the other hand the force

F = yi+ xj = (y, x) (6.110)

6.7. ADVANTAGES AND TECHNIQUES OF THE WORK/ENERGY THEOREM 95

is conservative and does have a potential function;

Fx = y = −∂V∂x

, −V =

∫y dx = yx+ C(y) (6.111)

checking that this gives the correct Fy;

Fy = x = −∂V∂y

= − ∂

∂y

(− yx− C(y)

)= x+ C ′ (6.112)

which works fine if C(y) = C so C ′ = 0. The potential that correctly gives both force components is

V (x, y) = −xy + C (6.113)

Example 6. Find the work done by

F = yi+ xj = (y, x) (6.114)

when we move a mass from (0, 0) to (a, b) along the curve

y = b(xa

)2(6.115)

Solution

W = −(V (a, b)− V (0, 0)

)= −

((−ab+ C)− (−0 + C)

)= ab (6.116)

(The work done against the force by the mover of the object is the opposite of this).

Example 7. Find the work done by

F = −yi+ xj = (−y, x) (6.117)

when we move a mass from (0, 0) to (a, b) along the curve

y = b(xa

)2(6.118)

Solution We have no choice but to integrate. Parameterize the curve

x(t) = a t, y(t) = b t2, 0 ≤ t ≤ 1 (6.119)

Find the velocity

vx(t) =dx

dt= a, vy(t) = 2bt (6.120)

find the force at any point on the curve

Fx = −y = −bt2, Fy = x = at (6.121)

and put it together; t0 = 0, tf = 1;

W =

∫ 1

0

(− bt2, at

)·(a, 2bt

)dt =

∫ 1

0

abt2 dt =ab

3(6.122)

6.7 Advantages and techniques of the work/energy theorem

The gradient of a function

∇V =(i∂V

∂x+ j

∂V

∂y+ k

∂V

∂z

)(6.123)


is a vector that points in the direction of the most rapid increase in V . This is easy to see. Examine V (x, y) at somepoint, and find the place (xθ, yθ) = (x+ r cos θ, y + r sin θ) for r very small where the difference V (xθ, yθ)− V (x, y)is the greatest

∂

∂θ

(V (xθ, yθ)− V (x, y)

)= 0

=∂V

∂x

dxθdθ

+∂V

∂y

dyθdθ

= −∂V∂x

r sin θ +∂V

∂yr cos θ

= r(cos θ i+ sin θ j

)·(∂V∂y

i− ∂V

∂xj)

(6.124)

The solution is

cos θ i+ sin θ j = α(∂V∂x

i+∂V

∂yj)

(6.125)

where α = 1/|∂V∂x i+∂V∂y j| (the left side is a unit vector in the direction that we must go to see the biggest change in

V , so the right side must be a unit vector). This means F = −∇V says that a conservative force points in thedirection of the most rapid decrease of its potential function.

The units of work, potential and kinetic energies are all N ·m = J (Joule). One Joule or one Newton-meter is not alot of work.

The smartest way to use the work-energy theorem is to think of it as saying that works change the total energy ofa system, and in many applications you may have formulas for work done by various agents that can work on (addenergy to/take energy from) a system. This will change its energy content.

Example 8. How much work must you do to move a mass m vertically from height H to H + dy? Let it be at restat the beginning and end.

We have a formula for the potential energy, and the work that you do will change it(0 +mg(H + dy)

)−(0 +mgH

)=W (6.126)

Suppose that you did this by applying a constant force Fyou. Find this force.

W = mg dy = Fyou · dr = Fyou ·((H + dy)j−Hj

)= Fyou,y dy, Fyou,y = mg (6.127)

We recover only the vertical part of the force, it points opposite to the gravitational force acting on the mass.

Example 9. Given that the potential energy of one plate of a parallel plate electrical capacitor carrying charge q inthe electric field of the other plate is

PE =q2d

2Aϵ0(6.128)

in which A is the plate area, d is the plate separation, and ϵ0 is just a physical constant, and the charge on such aplate while connected to a battery of voltage V0 is

q =V0ϵ0A

d(6.129)

and the work done by a battery when charge ∆q flows out of it is

Wbatt = ∆q V0 (6.130)

compute the work needed to pull the plates apart from separation d to 2d while they are connected to the battery.


You need to know nothing about electromagnetism to solve this problem. It is a straightforward applicationof basic principles. At separation d the plates have charge qi =

V0ϵ0Ad , at separation 2d they have charge qf = V0ϵ0A

2d .

That means gf − qi = −V0ϵ0A2d flowed out of the battery, and therefore

Wyou +Wbatt = PEf − PEi

Wyou + V0

(− V0ϵ0A

2d

)=

q2f (2d)

2ϵ0A− q2i d

2ϵ0A

Wyou = V0

(V0ϵ0A2d

)+

(V0ϵ0A2d )2(2d)

2ϵ0A−

(V0ϵ0Ad )2d

2ϵ0A

=1

2V0

(V0ϵ0A2d

)(6.131)

6.7.1 Principle of infinitismal work

If you compute the potential of a function at a point nearby x

V (x+ dx) ≈ V (x) + dxdV

dx(x) + · · · = V (x)− dxFx + · · · (6.132)

and the second term is the work done by the conservative force in question as the object is displaced from x to x+dx,according to −(Vf − Vi) = W (Eq. 6.14). This result is very handy, from it one can find the force exerted by someagent that does work by acting on the system, by comparing the work done to displace the system to position x+dxwith the work done in displacing it to x

W (x+ dx)−W (x) =

∫ x+dx

x

Fx dx ≈ Fx dx (6.133)

Example 10. Find the force that you must exert in order to pull the capacitor plates apart by amount dx whentheir separation is D.From the previous example

W (D + dx)−W (D) = −(12V0

( V0ϵ0AD + dx

)− 1

2V0

(V0ϵ0AD

))=

1

2

V 20 ϵ0A

D(D + dx)dx = Fx dx (6.134)

making the required force

Fx =1

2

V 20 ϵ0A

D2(6.135)

Example 11. The potential energy of a spring stretched by amount x is V = 12kx

2 in which k is a constant.Compute the force needed to stretch the spring from x to x+ dx.(1

2k(x+ dx)2 − 1

2kx2)=W (x+ dx)−W (x) = Fx dx, Fx = kx (6.136)

The spring pulls back in the opposite direction with restoring force F = −kx.

The real power of the principle of infinitismal work is in its simplifying treatment of forces of constraint, such astensions and normal forces. The idea here is that forces of constraint, forces that prevent motion in certain directions,can’t do work since there are no displacements in the directions of these forces.

Example 12. Consider a box resting on a straight incline y = − tan θ x, with no friction. Solve its equations ofmotion.What we do is to pretend that the incline is not there at all, and instead add in whatever force of constraintis needed to make it move along the incline resulting in

ma = mg + Fc

or md2x

dt2= Fcx

md2y

dt2= −mg + Fcy (6.137)


and the virtual work statement

dW = Fcxdx+ Fcydy = 0 (6.138)

and the constraint

y = − tan θ x (6.139)

Differentiate the constraint and put it into both the EOMS and the VW statement

dy = − tan θ dxd2y

dt2= − tan θ

d2x

dt2

md2x

dt2= Fcx

−m tan θd2x

dt2= −mg + Fcy

Fcxdx− Fcy tan θ dx = 0 (6.140)

The last equation says that

cos θFcx − sin θFcy = 0, Fc = (Fc cosϕ, Fc sinϕ) (6.141)

in other words

Fc(cos θ cosϕ− sin θ sinϕ) = Fc cos(θ − ϕ) = 0, ϕ = 90− θ (6.142)

the force of constraint is ⊥ to the incline; it must be the normal force! We now solve for this force

d2x

dt2= Fcx/m

d2x

dt2= (g − Fcy/m) cot θ = (g − Fcx cot θ/m) cot θ

equating LHS’s; Fcx =mg cot θ

1 + cot2 θ= mg sin θ cos θ

Fcy = Fcx cot θ = mg sin2 θ, |Fc| = mg cos θ (6.143)

You might look at this as swatting a fly with a sledgehammer, but the method has the advantage of replacing vector-based geometric problem solving with purely algebraic work, of a highly mechanical nature, so that if geometryand vectors are not your favorite game, you have other options.

Example 13. Consider a box resting on a straight incline y = −12 x

2, with no friction. Find the normal force andacceleration when the box reaches (x,− 1

2 x2) traveling with velocity v = (v0, vy) with v0 given, and you should find vy!

This could be quite hard by vector geometry, but in the new approach is no more complicated than the previousexample.

max = Fcx

may = −mg + Fcy

y = −1

2x2

vy = −x vxay = −v2x − xax

0 = Fcx dx+ Fcy dy =(Fcx − xFcy

)dx (6.144)


Note that vx = dxdt and ax = d2x

dt2 . Lets solve it;

Fcy =Fcxx

ax =Fcxm

ay = −v2x − xax = −v20 − xax = −g + Fcym

= −g + Fcxmx

so we get Fcx =m(g − v20)x

1 + x2

and Fcy =m(g − v20)

1 + x2

and ax =(g − v20)x

1 + x2, ay = −v20 − x ax = − gx2

1 + x2

(6.145)

(note that at this point vy = −x v0). This was not hard to solve, but try to do it geometrically using vectors. Thetechnique described here is the next step in the evolution of dynamics beyond Newtonian methods, the divorce ofvectors from dynamics, and the “algebrization” of dynamics. These last two examples are common Physics GREquestions.

6.7.2 Solving for accelerations

Suppose that we apply the principle of infinitismal work, allowing our various forces to cause a displacement dx;start at z at ti and end at x+ dx at time tf

1

2mv2(x+ dx) + V (x+ dx)−

(12mv2(x) + V (x)

)=

∫ x+dx

x

Fncdx ≈ Fnc dx

1

2m(v2(x+ dx)− v2(x)

)= Fnc dx+ V (x)− V (x+ dx)

1

2m(v2(x+ dx)− v2(x)

dx

)= Fnc −

dV

dx=∑

Fx = max

d

dx

(12v2(x)

)= ax (6.146)

This is extremely useful, since it means that when you solve a problem by the work-energy theorem, you actuallysolve two problems simultaneously, obtaining v as a function of displacement and the acceleration.

M1 M2

Example 14. Consider thisAtwood machine (two masses anda simple pulley). The masses beginat the same height, at rest. Theyare allowed to move distances x,at which time their speeds are v.

1

2M1(v

2 − 02) = (T j) · (xj)

+ (−M1gj) · (xj)1

2M2(v

2 − 02) = (T j) · (−xj)

+ (−M2gj) · (−xj)(6.147)

Note that internal forces coupled by Newton’s second law do no work on the entire system; add theseequations

1

2(M1 +M2)v

2 = (M2 −M1)gx,1

2v2(x) =

(M2 −M1

M1 +M2

)gx (6.148)


This gives us v, but also a;

a =d

dx

(12v2(x)

)=(M2 −M1

M1 +M2

)g (6.149)

M1

M2F

Example 15. Find the maximal|F| that causes coacceleration.Let both blocks codisplace by xi tothe right. Clearly it is friction caus-ing M1 to move at all;

1

2M2 v

2 = F x− Ff x

1

2M1 v

2 = Ff x (6.150)

with friction “maxed out” atFf = µsM2 g

Differentiate each equation with respect to x;

a =d

dx

(12v2(x)

)=F − µsM2g

M2, a =

d

dx

(12v2(x)

)= µs

M2

M1g (6.151)

giving you both

a = µsM2

M1g, F =M2 a+ µsM2 g (6.152)

Take note of the fact that friction does no work on the combined system.

6.8. APPENDIX. INTEGRAL CALCULUS TECHNIQUES 101

6.8 Appendix. Integral calculus techniques

You will need to do very few types of integrals for 201/202, most can be handled by two methods.

6.8.1 Integration by parts

This technique is used to rewrite integrals of products oftwo functions. Consider functions u(x) and v(x) and inte-grals of the form∫ uf

u0

v du =

∫ u(xf )

u(x0)

v(x)du(x)

dxdx = C +D (6.153)

which is numerically equal to the area of the regions C andD in the figure. Notice that the shaded region has area

ufvf − u0v0 = A+B + C +D (6.154)

Using this our original integral can be rewritten as∫ uf

u0

v du = u(x) v(x)∣∣∣f0−∫ vf

v0

u dv

= (A+B + C +D)− (A+B)

(6.155)

u0 uf

v0

vf

A

D

B

C

which is the entire integration by parts scheme.

Example 16. Consider integration by parts for

I =

∫ π

0

x2 sinx dx (6.156)

We let v = x2 and u = − cos(x) to get after one application

I = −x2 cosx∣∣∣π0+

∫ π

0

2x cosx dx (6.157)

and we apply again to the last integral with v = 2x and u = sinx to get

I = −x2 cosx∣∣∣π0+ 2x sinx

∣∣∣π0−∫ π

0

2 cosx dx (6.158)

The last integral is elementary;

I = −x2 cosx∣∣∣π0+ 2x sinx

∣∣∣π0− 2 sinx

∣∣π0

(6.159)

A common situation is to have integration by parts apparently take you around in circles;

Example 17. apply partial integration to

I =

∫ ∞

0

e−ax sinx dx (6.160)

with v = e−ax and u = − cosx;

I = −e−ax cosx∣∣∣∞0

− a

∫ ∞

0

e−ax cosx dx (6.161)

and now integrate the last integral by parts again with v = e−ax and u = sinx;

I = 1− ae−ax sinx∣∣∣∞0

− a2∫ ∞

0

e−ax sinx dx = 1− a2I (6.162)


and so

I =

∫ ∞

0

e−ax sinx dx =1

1 + a2(6.163)

This is a useful integral, it is the Laplace transform of a sine function, and so will appear in classical mechanics andelectronics in the Newtonian equations of motion for a particle with a sinusoidal (in time) force applied to it, or forthe current in a circuit with a sinusoidal applied voltage.

6.8.2 Variable Changes

Some of the simplest and most useful are;1. Simple polynomial, for example consider ∫

x dx

x2 + 2(6.164)

Let y = x2 + 2, then dy = 2x dx so that x dx = dy2 and our integral can be written as∫

x dx

x2 + 2=

1

2

∫dy

y=

1

2ln |y|+ C =

1

2ln |x2 + 2|+ C (6.165)

2. Trigonometric. Consider ∫dx√1− x2

(6.166)

Let x = sin θ, then dx = cos θ dθ and√1− x2 = cos θ and our integral becomes∫

dx√1− x2

=

∫dθ = θ + C = sin−1 x+ C (6.167)

which gives us a rather nice definition of the arcsine or inverse-sine function.

3. Trigonometric. Consider ∫dx

x2 + 1(6.168)

Let x = tan θ, then dx = dθcos2 θ (prove this, it is a nice exercise), and 1 + x2 = 1 + sin2 θ

cos2 θ = 1cos2 θ and our integral

becomes ∫dx

x2 + 1=

∫dθ = θ + C = tan−1 x+ C (6.169)

which, again, provides us with a nifty representation of an inverse trig function.

Example 18. Determine a power series expansion for tan−1 x valid near x = 0.We can use the previous problem ∫ x

0

dy

y2 + 1= tan−1 x (6.170)

and if we recognize that in this expression 0 ≤ y ≤ x and x is very small, then y is very small and we can replace1

1+y2 with its power series for y ≈ 0 in the integrand;

1

1 + y2= 1− y2 + y4 − y6 + · · · (6.171)

which you should prove. We now integrate term by term;∫ x

0

(1− y2 + y4 − y6 + · · · ) dy = x− x3

3+x5

5− x7

7+ · · · (6.172)

and so very near x = 0 we find that

tan−1 x = x− x3

3+x5

5− x7

7+ · · · (6.173)

6.8. APPENDIX. INTEGRAL CALCULUS TECHNIQUES 103

4. Trigonometric. Consider ∫dx√1 + x2

(6.174)

There is a standard substitution to use whenever your integrand has a radical containing 1±x2; try x = cot θ. Thendx = − dθ

sin2 θand 1 + x2 = 1

sin2 θand the integral becomes∫

dx√1 + x2

= −∫

dθ

sin θ= −

∫dθ

2 sin θ2 cos θ2

(6.175)

At this stage we notice (after we have become sufficiently experienced) that

d

dθln tan

θ

2=

1

2

1

cos2 θ2

1

tan θ2

=1

2 sin θ2 cos θ2

(6.176)

and so the integrand is the derivative of something;∫dx√1 + x2

= −∫

dθ

sin θ= − ln tan

θ

2+ C (6.177)

We now need to recover the answer in terms of x, but

x = cot θ,1

x= tan θ =

2 tan θ2

1− tan2 θ2(6.178)

which we solve for

tan2θ

2+ 2x tan

θ

2− 1 = 0 (6.179)

or

tanθ

2= −x±

√x2 + 1 (6.180)

and so ∫dx√1 + x2

= − ln | − x±√x2 + 1| = ln |x±

√x2 + 1| (6.181)

since1

−x±√x2 + 1

=x±

√x2 + 1

(x±√x2 + 1)(−x±

√x2 + 1)

=x±

√x2 + 1

−x2 + (x2 + 1)= x±

√x2 + 1 (6.182)

This was a rather complicated example, however once an integral has been done, you may archive it in your ownpersonal integral table and re-use the result without having to go through all of the trouble of re-deriving it.

5. ∫dx√

(x2 + 1)3(6.183)

is an integral that will come up once or twice in Physics 202. We use the x = tan θ variable change on it, to get∫dx√

(x2 + 1)3=

∫cos θ dθ = sin θ + C

=x√

x2 + 1+ C (6.184)

using the triangle figure to resolve the trig functions.

6. ∫ √1− x2 dx (6.185)


This integral comes up if we want to compute the area of a unit circle, or at least the part of the circle that lies inthe first quadrant. We use x = sin θ to get ∫ √

1− x2 dx =

∫cos2 θ dθ (6.186)

which we simplify by applying a trigonometric identity∫cos2 θ dθ =

1

2

∫(1 + cos 2θ) dθ =

1

2(θ +

1

2sin 2θ) (6.187)

and in terms of x, cos θ =√1− x2, so∫ √

1− x2 dx =1

2(sin−1 x+

1

22x√1− x2) + C (6.188)

In all likelihood these few examples and methods will get you through 201 and 202. If you get stuck, I reiterate thataxiom, REDUCE and maxima do very well with definite integrals, even the tough ones, but be prepared to provideadditional information about parameters. See problem 6.30 for an example.

6.9 Appendix. Computer assistance

When you get stuck, you can make the computer give you a cold metallic hand.

%REDUCE code

% check two force components to see if F

% is conservative

procedure IsItConservative(Fx,Fy);

begin

if df(Fx,y)-df(Fy,x)=0 then write "yes" else write "no";

end;

% Try to create the potential of conservative force

procedure FindPotential(Fx,Fy);

begin

off msg;

operator C; % create a generic data type, a function

vee:=-int(Fx,x)+C; % this gives F_x correctly

soln:=odesolve(-df(vee,y)-Fy,C,y); % Construct C(y) to get correct F_y

let soln; % Set C(y) equal to the solution of this DE

return(vee);

end;

% Lets try it on an HW problem

Fx:=a*x;

Fy:=-a*y;

IsItConservative(Fx,Fy);

Pot:=FindPotential(Fx,Fy);

% Check it

eff_x:=-df(pot,x);

eff_y:=-df(pot,y); % It works!

6.10. PROBLEMS 105

6.10 Problems

1. A. For the force

F1 =a x

(x2 + y2)2i+

a y

(x2 + y2)2j

determine the potential function (if there is one) and the work done in moving from (0, 1) to (1, 0) along the curvex2 + y2 = 1 (let a = 5.0).B. For the force

F2 =a y

(x2 + y2)2i− a x

(x2 + y2)2j

determine the potential function (if there is one) and the work done in moving from (0, 1) to (1, 0) along the curvex2 + y2 = 1 (let a = 5.0).

2. Consider a force

F = Fxi+ Fyj = 8.0N

my i− 8.0

N

mx j

If this force is conservative, find its potential and use it to compute the work done by the force in moving a bodyalong the curve y = 2− 2x2 from (0, 2) to (1, 0). If the potential does not exist, find the work by integration.

3. Consider a force

F = Fxi+ Fyj = 8.0N

mx i− 8.0

N

my j

If this force is conservative, find its potential and use it to compute the work done by the force in moving a bodyalong the curve y = 2− 2x2 from (0, 2) to (1, 0). If the potential does not exist, find the work by integration.

4. A test for the existence of V ; for a force in two dimensions, if

∂Fx∂y

=∂Fy∂x

then there is a V since if

Fx = −∂V∂x

, Fy = −∂V∂y

then∂Fx∂y

=∂Fy∂x

= − ∂2V

∂x∂y

Apply this to all of the forces above and test for a potential.

F

θM

5. A. A block of mass M dragged adistance d by force F along a horizontalsurface with coefficient of friction µk.The force is just sufficient to pull it atconstant speed. Find |F|, and the workdone by all of the forces acting on themass.B. Suppose that the block begins atrest, and a much larger magnitude force(but not enough to lift it off of the table)is applied so that the block accelerates.Find its speed after it is dragged adistance d.

6. Imagine a chain of N links, each of mass m (total chain mass is Nm) resting stretched-out on a smooth table-top(no friction), with one link hanging over the edge, but with you holding the chain in place by its left end (so v = 0).If you release the left end, the chain begins to slide off of the table. Each link has length ℓ (total chain length Nℓ).


v2

Use the work-energy theorem to find the speed with which the whole chain is moving after n < N links have beenpulled over the side. I have illustrated the situation with two links pulled over. For further fun, solve the problem ifthere is friction µk with the table-top.

θ=60omF

vf

d

7. A block of mass m = 5.0kg is pushed up an inclineof width d = 4.0m with coefficient of friction µk = 0.2up to the top of the incline by a constant applied force ofmagnitude 60N parallel to the incline. The block beginsat rest.A. Determine the work done by the force of friction.B. Determine the work done by the normal force.C. Determine the work done by gravity.D. Determine the work done by F ,E. Determine the final speed of the block at the top.

Hint. Put the origin at the bottom left corner of theramp, then find all vectors in component form.

8. Consider this Atwood machine (two masses and a simple pulley). The masses begin at the same height, at rest.They are allowed to move distances d, at which time their speeds are v.

M1 M2

A. Find the work done by stringtension on M1.B. Find the work done by stringtension on M2.C. Find the work done by stringtension on the entire system.D. Find the work done by gravityon the entire system.E. Find v in terms of M1,M2, d, g.

F∗. Suppose that the pulley has a tiny amount of friction in it, so that each time it rotates it does work −W on thesystem. Its radius is R. Compute v in terms of M1,M2, d, g,W,R after the masses move distance d.

6.10. PROBLEMS 107

F

M

9. You are already familiar with the block and tackle, a simple machinethat uses multiple pulleys to gain a mechanical advantage. Is the mechanicaladvantage the amount of work you need to do to lift a mass? To answerthis, take these steps;A. Compute the force |F| with which you must pull down on the rope (atthe point where the knot is) in order to lift the box M at constant speed.B. If you pull the knot downwards a distance d with this force, compute thework that you do (supplying |F|) on the system. This is energy that youexpend, and the system (the mass M) gains.C. Compute the work done by string tension on the mass M .D. Compute the work done by gravity on the the mass M .E. The total energy possessed by the box is its kinetic energy plus its poten-tial energy (if there are conservative forces present). Use the work energytheorem to show that the increase in the energy of the box equals the workdone on it by the tension. Show that this gain is exactly the amount ofenergy that you lost.A block and tackle is not a work-saver, it is a force-amplifier.

N

v

mgR

θ

10. A small object of mass m is placed at the top ofa frictionless hill of radius R. As it begins to slidedown, it reaches a point where it loses contact withthe hill. Find the angle at which that happens, andthe speed of the object at that point.

L0

v

d

11. Favorite toys from yesteryears,before we became the nanny-state.This happy little fellow is calleda Wrist-Rocket, beloved toy of alllittle boys. It is a sling-shot witha perforated handle through whicha ball-bearing of mass m is shotafter being propelled by powerfullatex tubing or bands.

Imagine that the bands have an unstretched length of L0 = 0.25m. When rubber is stretched by amount x, itpulls back with tension |T| = k x in the opposite direction like a spring. The greater the stretch, the greaterthe tension. Suppose that it takes 222N (which is fifty pounds) to pull the rubber bands to a total lengthL0 + xmax = L0 + d = 1.0m, compute the “force constant” k.Compute the amount of work done by the precious little darling who accomplishes this dubious feat. Now computethe velocity that this will give a m = 0.05 kg ball-bearing used as a projectile. Do so in feet per second if you wanta real surprise.

12. A. Consider a force

F = Fxi+ Fyj = 2N

m2xy i+ 1

N

m2(x2 − y2) j

compute the work done by the force in moving a body along the curve y = 2− 2x from (0, 2) to (1, 0).


B. Consider a force

F = Fxi+ Fyj = 2N

m2xy i+ 1

N

m2(y2 − x2) j

Determine whether or not this force is conservative.

13. Consider a force F that has a potential function

V =4

x2 + y2Jm2

Find F and the work done in moving a 1.0 kg mass from (1.0m, 0) back to (1.0m, 0) along a circle of radius 1.0mcentered on the origin. Hint, you could parameterize the path as (x(t), y(t)) = (1.0m cos( 2πs t), 1.0m sin( 2πs t)).

14. A. Friction is technically velocity-dependent since it always opposes v. Such a force cannot be conservative. Aunit vector in the direction of the velocity is u = v

|v| . Consider a force

F = −9.8Nv

|v|

Compute the work done in moving a 1.0 kg mass from (1.0m, 0) back to (1.0m, 0) along a circle of radius 1.0mcentered on the origin. It is best to parameterize the curve as in the previous problem.

B. Magnetic forces are exerted on charged particles of electrical charge q by a magnetic field B only if the chargesare moving, with velocity v

F = qv ×B

Prove that magnetic forces do no work on charged particles, no matter what path they take. This means thatmagnetic fields cannot slow-down or speed-up particles, they can only deflect them.

M1

M2F

15. The work-energy theorem can often pro-vide solutions to complex dynamic problemswith startling ease. Re-consider problem 19,in which we seek to find the maximal |F| suchthat M1 and M2 coaccelerate. The coefficientsof friction between the two blocks are µs andµk.

Let both blocks start at rest. Compute the work by F if the blocks are pushed (coaccelerating) by a distance d, anduse the work-energy theorem to find their final speeds v. Assume static friction is maxed out. Compute the workdone by the force of friction on M1 for this displacement, along with the work done on M1 by gravity and the normalforce, and apply the work-energy theorem to M1 alone to obtain its final speed v. Use these results to compute themaximal |F|. That was easy, no?

6.10. PROBLEMS 109

θmF

vf

d=L cosθ

L

h=L sinθ

16. Do for problem 24 what you did inthe previous problem. You will compute theforce needed to push a block up a stationaryincline (with friction µk) at constant speed vf .Compute the work done by horizontal forceF in pushing the block from bottom to top.Compute the work done by gravity and thenormal force as the block moves from bottomto top. You must find the normal force fromthe fact that the block moves along the incline.Compute the work done by friction as theblock moves from bottom to top. Put it allinto the work-energy theorem and compute|F| in terms of M, g, θ, µk.

h

R

a

b

c

17. An amusement park roller-coaster ofheight h has a loop-the-loop of radius R. Africtionless car starts at the top. Find its speedat each of the points a, b, c. Find the normalforce (vector) exerted on it at points a and b.Find the minimal h-to-R ratio that will enablethe car to negotiate the loop without losingcontact at point b.

h

x

Mg

v(x)

y18. Consider a large ski-hill in the shape of acircle of radius h. There is kinetic friction of co-efficient µk between ski and snow. A skier be-gins at the top at rest. Compute her speed v(x)once she reaches the position illustrated, rightof the starting point by x, neglecting hercentripetal acceleration. Find her speed atthe bottom of the hill neglecting the cen-tripetal acceleration of the skier.

19. Show that in the previous problem you are not moving fastest at the bottom of the hill, and find the x-valuewhere you do reach your highest speed neglecting the centripetal acceleration of the skier.

20. Velocity dependent forces not only do work that depends on the path taken, but also the speed with which thepath is traversed. Lets examine the price to be paid for exceeding the speed limit.Air resistance is a velocity dependent force

Fa = −0.2Ns

mv

Imagine traveling in a circle of radius R at constant speed v0. Compute the work WFa done on you by this forceif you travel for time t. This is negative, which means that you lose mechanical energy at this rate, and you must


replenish this energy at the same rate (by burning more gas) in order to maintain your speed v0. Your car’s engineconverts chemical energy in the gas to mechanical energy, and there is a conversion factor that depends on your carengine’s efficiency. The power ℘ supplied by your engine is the rate with which it must supply the energy neededto compensate for the loss incurred against air-resistance, ℘ = −dWFa

dt . This is proportional to the rate with whichyou burn gas. Compute the power required to travel at v0 with air resistance. Show that if you raise v0 by fivepercent, you burn gas ten percent faster (the power requirement increases by ten percent)! This result is completelyindependent of engine efficiency or type, it affects Suburbans, Humvees and Prius’s equally.Most of the fuel consumption of your car is due to working against air resistance to maintain constant speed, someto work against gravity to go up hills, and some to work against energy loss due to friction between moving parts incontact in the engine and drive-train. Timely oil changes and lubes can really impact fuel economy by minimizingthis last energy loss.

21. Consider a path in space specified by the curve y = 8.0− 12x

2, between points (0, 8) and (4, 0) (all distances inmeters).A. Compute the work done by force F1 = 2N i+ 3N j as an object is moved along the path from (0, 8) to (4, 0).

B. Compute the work done by force F2 = 2Nm x i+ 3Nm y j as an object is moved along the path from (0, 8) to (4, 0).

C. Compute the work done by force F3 = 2Nm y i+ 3Nm x j as an object is moved along the path from (0, 8) to (4, 0).

D. Compute the work done by force F4 = −0.1N v|v| as an object is moved along the path from (0, 8) to (4, 0).

M

L

F(θ)

θ 22. A box of mass M hangs from the ceil-ing by a rope of length ℓ. You push with suf-ficient force F(θ) horizontally to deflect thebox/rope to angle θ. Find the sufficient forceF(θ), the tension in the rope, and the work thatyou have done to move the box. How muchwork has the rope-tension performed? Notethat the tension and F(θ) are not constant, butchange with θ.

23. As you will eventually learn later in the course, the force of gravity is constant only near the surface of theearth. In fact the force of gravity exerted by the earth (mass Me) on a body of mass m has magnitude FG = mMeG

r2

where r is the distance from the body to the center of the earth, and G = 6.6× 10−11Nm2

kg2 (Newton’s gravitational

constant). This force is conservative, and points towards the center of the earth. Its potential is V (r) = −mMeGr .

With what speed would an object hit the surface of the earth if it fell from rest from a distance of 100 km above theearth’s surface? (You will need to look up the earth’s radius and mass).The escape velocity is the speed ve with which a ballistic projectile must be shot from the earth’s surface in orderto reach r → ∞, finally arriving at zero speed. Compute ve.

24. You have a ramp of length ℓ and angle θ = 45o, and you wish to measure the coefficient of friction µk between theramp and a box of penguins. You put the box at the top of the ramp, and let it slide to the bottom. The box arrivesat the bottom of the hill with speed v. You repeat the experiment, but you install tiny wheels on the box so that it canroll without friction down the ramp. This time the box arrives at the bottom of the hill with speed 1.5 v. Compute µk.

6.10. PROBLEMS 111

FM1

M2

25. Use the work energy theorem tocompute the smallest force with whichyou must push on M1 in order to supplysufficient friction to keep M2 fromslipping vertically. Hint; compute thework done by the normal force on M2.There is no friction between the groundand M1.

26. When a perfectly elastic rubber ball hits the ground at speed v, it rebounds at exactly the same speed v butin the opposite direction. A real rubber ball does not do this; if it hits the ground at v it rebounds at ϵ v in whichthe coefficient of restitution 0 ≤ ϵ ≤ 1. The lost mechanical energy is dissipated as sound and heat as the balldeforms.Imagine taking such a rubber ball and dropping it from height h. With what speed will it hit the ground? Withwhat speed will it rebound? How high will it ascend after this bounce? How high will it ascend after n bounces?A rubber ball dropped from a height of one meter will rebound after one bounce to a height of 90 cm, compute itscoefficient of restitution.

27. Suppose that air resistance worked something like ordinary friction, doing work −ϖ for each meter that youtravel against it (so after traveling x air resistance has done work W = −ϖx on you).Under such circumstances, drop a ball from height h, with what speed would it hit the ground? If it rebounds atthat speed, how high will it travel after the first bounce?

La Lc

h

vc

vb

Mc

mp

28. This happy little fellow is called atrebuchet, a machine brought back fromthe Middle East by French crusaders dur-ing the early Middle Ages. It reachedmammoth size and mechanical perfectionin the ensuing centuries, and ultimatelywas capable of throwing stone balls ofmass 250 kg well over 300m.Early trebuchets were anchored firmly tothe ground, so only the arm, here oflength La + Lc could rotate. A mas-sive counterweight (let Mc = 10, 000 kg)would fall, swinging the arm and launch-ing the projectile (mp = 100 kg) with asling from a height 2h+La. Let h = 3m,La = 15m, Lc = 3m. The only operat-ing constraint is that ω = vb

ℓa+h= vc

ℓb.

Compute the range of the projectile.

29. Consider a box resting on a straight incline y(x) =√R2 − x2, with no friction. Find the normal force and

acceleration when the box reaches (x, y(x)) traveling with velocity v = (v0, vy) with v0 given, and you should findvy!

30. So you are having some trouble doing work integrals. Here is how you can use REDUCE to check your work

%define our force function

Fx:=4*x;


Fy:=3*x*y;

% define our straight path from (x0,y0) to (x1,y1)

let x=x1+(1-t)*x0;

let y=y1+(1-t)*y0;

% velocity for path

vx:=df(x,t);

vy:=df(y,t);

% compute the work

dW:=-Fx*vx-Fy*vy;

W=int(dW,t,0,1);

Re-do problem 6.21, make REDUCE do all of the work, for example in 6.21.D

% define our straight path from (x0,y0) to (x1,y1)

let x=2*t;

let y=8-2*t^2;

% velocity for path

vx:=df(x,t);

vy:=df(y,t);

%define our force function

Fx:=-0.1*vx/sqrt(vx^2+vy^2);

Fy:=-0.1*vy/sqrt(vx^2+vy^2);

% compute the work

dW:=-Fx*vx-Fy*vy;

W=int(dW,t,0,2);

31∗. Suppose that a mass m has speed v(x) = v0 + β x−n where v0, β, n are some suitable constants. Find itsacceleration, and the potential function for the force exerted on it.

32. A skier of mass m starts at the top of a hill at (x, y) = (0, h) of shape y = hd2 (x− d)2 and slides to the bottom

(x, y) = (d, 0) under the combined influences of gravity and the forces of friction with coefficient µk. Find his speedand acceleration when he has traveled a distance x (measured horizontally) from the starting point.

33∗. Suppose that a mass m has speed v(x) =

√β(

1x − 1

x0

)where x0, β are some suitable constants. Find its

acceleration, and the potential function for the force exerted on it.

Chapter 7

Conservation of Momentum

Consider a system of particles, three for example, in which some forces (F1,ext,F2,ext,F3,ext) are externally applied,and all other forces are internal (inter-particle). The force exerted in M1 by M2 is F12. By Newton’s second law thisis equal and opposite to the force F21 exerted on M2 by M1.

m1 at r1

m2 at r2

m3 at r3

F2,1

F1,2

F2,3 F2, ext

F3,1

F3,2

F3,ext

F1,ext

Write the equation of motion for each mass

M1a1 = F1,ext + F12 + F13

M2a2 = F2,ext + F21 + F23

M3a3 = F3,ext + F32 + F31 (7.1)

Add all three of these, and note that all of the inter-particle forces cancel pairwise since

Fij + Fji = 0 (7.2)

113

114 CHAPTER 7. CONSERVATION OF MOMENTUM

we find that

M1a1 +M2a2 +M3a3 = F1,ext + F2,ext + F3,ext =∑i

Fi,ext (7.3)

and since a1 = d2r1dt2 we can rewrite this as

d2

dt2(M1r1 +M2r2 +M3r3) = F1,ext + F2,ext + F3,ext =

∑i

Fi,ext

=(M1 +M2 +M3

) d2

dt2

(M1r1 +M2r2 +M3r3M1 +M2 +M3

)= Mtotal

d2

dt2rCOM =

∑i

Fi,ext (7.4)

In other words, the acceleration of the center of mass of a collection of particles is due only to the sum of externalforces acting on the collection. The center of mass is the ”balancing point” of the collection, it can be found byapplying a gravitational force to the collection, regarded as a rigid body, and locating the single point of supportwhere a force can be applied to hold the body stationary against gravity, with no rotation or acceleration of thecollection. The formula for the center of mass is

rCOM =

∑iMiri∑iMi

=

∫rdm∫dm

(7.5)

What if there are no external forces acting on the collection? All forces are inter-particle and cancel in the sum, andwe find

d2

dt2(M1r1 +M2r2 +M3r3) = 0 (7.6)

which can be integrated once ∫ t2

t1

d2

dt2

(M1r1 +M2r2 +M3r3

)dt = 0 (7.7)

to give

M1v1(t1) +M2v2(t1) +M3v3(t1) =M1v1(t2) +M2v2(t2) +M3v3(t2) (7.8)

where vi =ddtri.

This is in the form of a conservation law, a quantity evaluated at one time equals its value at all times. This isprobably the most important conservation law in physics, and is called conservation of momentum. Define themomentum of a mass m with velocity v to be

p = mv (7.9)

then in the absence of externally applied forces, a collection of particles evolves subject to the constraint∑i

pi(t1) =∑i

pi(t2) (7.10)

and Newton’s law for the center of mass motion if external forces are present can be written as

d

dt(∑i

pi) =∑i

Fi,ext (7.11)

Example 1. A railroad car with a cannon attached rigidly to it has mass M . It fires a cannonball of mass m withvelocity w to the left with respect to the ground. Find the recoil velocity of the cannon.

115

w

m

v

M

All forces between the cannon and cannonball are internal, there are no external forces involved. The gun pushesthe ball, the ball pushes the gun. Taking right to be the positive x direction, conservation of momentum says

Pbefore = Pafter, 0 + 0 =Mv −mw (7.12)

and so

v =mw

M(7.13)

The muzzle velocity or velocity of the ball with respect to the gun is

vm = w + v =(m+M)w

M(7.14)

Note that all velocities should be measured with respect to the same fixed point. In typical applicationsit will be the muzzle velocity that is a known factor, and so the velocity of the ball relative to the ground is computedfrom the velocity of the ball relative to the gun vmuzzle−vgun, and the velocity of the gun relative to the ground by

(w, 0) = vball = (vm, 0) + (−v, 0) = vmuzzle + vgun = (vm − v, 0) (7.15)

The momentum conservation law is written in terms of the velocities relative to a single fixed point, so

0 =Mvgun +mvball, v =mvmm+M

(7.16)

gives the recoil speed of the gun in terms of the muzzle velocity of the cannon ball.

M

m

Example 2. A penguin of mass m stands at theright end of an ice floe of mass M , length L. Don’tlook at me like that, this is really how you spell floe,F-L-O-E. Look it up. She walks to the far left end.How far does the floe move through the water withrespect to the shore?


M

m

x

Again momentum is conserved. Let the floe moveright at speed V , the penguin walks with speed v0with respect to the floe. She is moving at speedv0−V with respect to the shore; all speeds should bemeasured with respect to the same reference point.Conservation of momentum says

0 + 0 =MV −m(v0 − V ) (7.17)

and soV =

mv0m+M

(7.18)

Now multiply by T , the time it takes the penguin to walk a distance L at speed v0 on the floe, also the time neededfor the floe to travel a distance x at speed V

x = TV =mv0

m+MT =

mL

M +m(7.19)

Now a few simple examples of center of mass determinations.

Example 3. Compute the center of mass (COM) of the system consisting of M1 at (0, 0), M2 at (0, b), M3 at (a, b)and M4 at (a, 0).Simple application of the formula results in

rCOM =M1(0, 0) +M2(0, b) +M3(a, b) +M4(a, 0)

M1 +M2 +M3 +M4(7.20)

so that

xCOM =aM3 + aM4

M1 +M2 +M3 +M4, and yCOM =

bM2 + bM3

M1 +M2 +M3 +M4(7.21)

dx

dy

r=(x,y)

a

b

Example 4. Compute the center of mass of a sheet ofplywood of dimensions a by b with one corner at the origin.Let the plywood have a mass per unit area ρ. Then adifferential area of it has mass dm = ρdA. Break theboard up into small elements, located at (x, y) of width dxand height dy. Add up all of the contributions

xCOM =

∫ a0dx∫ b0dy x ρ∫ a

0dx∫ b0dy ρ

=12ρ a

2 b

ρab=a

2

yCOM =

∫ a0dx∫ b0dy y ρ∫ a

0dx∫ b0dy ρ

=12ρ b

2 a

ρ ab=b

2(7.22)

which places the COM where we expect it to be, in thecenter of the board.

Example 5. Find the center of mass of a wedge of plywood of radius R with apex at the origin, subtending an angleof π4 rad.

Again break up the object into elements and sum their contributions. An area element subtending angle dθ at radiusr with a length dr has mass

dm = ρ r dr dθ (7.23)

117

if it is located at x = r cos θ and y = r sin θ we find

xCOM =

∫ π4

0dθ∫ R0dr r cos θρr∫ π

4

0dθ∫ R0drρ r

, yCOM =

∫ π4

0dθ∫ R0dr r sin θρ r∫ π

4

0dθ∫ R0drρr

(7.24)

and these integrals are trivial, the denominator is ∫dm = ρ

1

8πR2 (7.25)

and the numerator of the x integral is

ρ

∫ π4

0

cos θdθ

∫ R

0

r2dr = ρ sin(π

4)R3

3(7.26)

and so

xCOM =sin(π4 )8R

3π(7.27)

and the y component can be gotten similarly.

dx

r=(x,0)

Example 6. Compute the COM for a rod of length Lwith one end at the origin whose linear mass density isρ = a x in kg

m .again subdivide the body into bits suitable for summation.A fragment of length dx a distance x from the origin hasmass dm = a x dx and so

xCOM =

∫ L0x ax dx∫ L

0ax dx

=2L

3(7.28)

The COM is closer to the densest end of the stick.

Example 7. Illustrated here is a fundamental cell of a NaCl crystal, with the origin at the top foreground corner,z-direction down, and x-direction left, y-direction right (and into the paper a bit). The picture is stereo-graphic;look straight at it and focus your eyes on infinity, and you can merge the two figures into a single three-dimensionalpicture (or break your brain, one or the other).

Sodium atoms are at

(0, 0, 0), (1

2,1

2, 0), (

1

2, 0,

1

2), (0,

1

2,1

2) (7.29)


and Chlorines at

(1

2,1

2,1

2), (0, 0,

1

2), (0,

1

2, 0), (

1

2, 0, 0) (7.30)

in conventional Cartesian coordinates, in multiples of the fundamental unit cell dimension (the side of the cube,5.63 A). Can you see that this places three sodium atoms at the centers of the faces of three sides of the cube?Chlorines occupy midpoints of three edges of the cube.The center of mass of the four Sodium atoms is therefore at

rsodium =(14,1

4,1

4

)(7.31)

What is it for the four Chlorines? For all eight atoms? In all three cases it is at the same point.

Here is an ammonia molecule stereo-gram. The atomic coordinates are

H; (x, y, z) = (0.866,−0.5,−1.0), (−0.866,−0.5,−1.0), (0.0, 1.0,−1.0) (7.32)

and

N ; (0.0, 0.0, 0.1) (7.33)

and so with masses in AMU, we obtain

rcom =1.0(0.866,−0.5,−1.0) + 1.0(−0.866,−0.5,−1.0) + 1.0(0.0, 1.0,−1.0) + 14.0(0, 0, 0.1)

14 + 3= (0, 0,−0.0842) (7.34)

For you amusement here is a stereo-gram of hexane

Back in the day, all of the organic chemistry books had figures like these. You can download a kit for making themfrom my 201 web-site, along with some molecular XYZ files. Linux only, sorry.

7.1. ELASTIC AND INELASTIC COLLISIONS 119

7.1 Elastic and Inelastic Collisions

Consider now a completely elastic collision in one dimension between two simple objects of mass m1 and m2. elasticis taken to mean perfect kinetic energy conserving. Such collisions are rare in nature, usually some energy is lostin a collision, used to make noise and deformation in the colliding objects. The initial velocities are all taken to bepositive (right) and so are the final velocities. We now solve the momentum and kinetic energy conservation laws

M1

M1

M2

M2

v1

v1’

v2

v2’

m1 v1 +m2 v2 = m1 v′1 +m2 v

′2

1

2m1 v

21 +

1

2m2 v

22 =

1

2m1 (v

′1)

2 +1

2m2 (v

′2)

2 (7.35)

eliminate one of the final velocities, for example using

v1,2 = (v1,2, 0), v′1,2 = (v′1,2, 0) (7.36)

v′1 =m1v1 +m2v2 −m2v

′2

m1(7.37)

eliminate the speed v′1 from the kinetic energy equation to obtain

m1v22 −m2v

22 = (m1 +m2)(v

′2)

2 + 2m1v1v2 − 2(m1v1 +m2v2)v′2 (7.38)

after canceling out several terms and dividing by m1. This is a quadratic equation with solutions

v′2 =2(m1v1 +m2v2)±

√4(m1v1 +m2v2)2 − 4(m1 +m2)(m2v22 + 2m1v1v2 −m1v22)

2(m1 +m2)

=m1v1 +m2v2 ±

√m2

1(v1 − v2)2

m1 +m2(7.39)

or

v′2 =2m1

m1 +m2v1 +

(m2 −m1)

m1 +m2v2, v′1 =

(m1 −m2)

m1 +m2v1 +

2m2

m1 +m2v2 (7.40)

This result applies to both masses initially moving. We will usually only be concerned with the case

v1 = v0 and v2 = 0 (7.41)

so that

v′1 =(m1 −m2)

m1 +m2v0 and v′2 =

2m1

m1 +m2v0 (7.42)

Collisions in two dimensions between point particles are only slightly more complicated due to the additional com-ponent of conserved momentum. However we will find that the precise outcome of a collision between point particlesis indeterminate unless some additional geometrical factors about the collision are known.Begin with the target particle at rest and the projectile, of equal mass, moving on a collision course at speed v0


m

v0

θ

φ

v1

v2

x

y

The momentum conservation law reads

m (v0, 0) = m (v1 cos θ, v1 sin θ) +m(v2 cosϕ,−v2 sinϕ) (7.43)

and kinetic energy conservation is1

2mv20 =

1

2mv21 +

1

2mv22 (7.44)

Solving the momentum conservation y-component we find that

v1 sin θ = v2 sinϕ (7.45)

insert this into the x-component equation to get

v0 = v1(cos θ +cosϕ sin θ

sinϕ) = v1

sin(θ + ϕ)

sinϕ(7.46)

and

v0 = v2sin(θ + ϕ)

sin θ(7.47)

Inserting it into the energy equation results in

v20 = v21 + v21sin2 θ

sin2 ϕ(7.48)

Comparing the two we find

sin(θ + ϕ) =

√sin2 θ + sin2 ϕ (7.49)

square this after using the law for the sine of a sum

sin2 θ + sin2 ϕ = cos2 θ sin2 ϕ+ 2 sin θ cos θ sinϕ cosϕ+ sin2 θ cos2 ϕ (7.50)

which can be reduced to

sin2 ϕ(1− cos2 θ) + sin2 θ(1− cos2 ϕ) = 2 sin θ cos θ sinϕ cosϕ (7.51)

orsin θ sinϕ = cos θ cosϕ (7.52)

the solution of which iscos(θ + ϕ) = 0 or θ + ϕ =

π

2(7.53)


in other words the total angle between the outgoing particle trajectories after the collision is 90 degrees for equalmass particles. We can then list the results

θ + ϕ =π

2, v1 = v0 sinϕ, v2 = v0 sin θ (7.54)

but the individual angles cannot be determined further without some separate geometrical data, data not determinedby the conservation laws.

For point particles the precise outcome of a two dimensional collision is indeterminate; conservation laws give threeequations in four unknowns, the two final speeds and angles. If a geometrical constraint can be added in, as in thecase of colliding spheres, then an additional equation allows us to find all of the final speeds and angles.

Example 8. For colliding spheres with no rotation, geometry alone determines the exit angle of the target.

m

v0

θ = arcsin(s/2R)

x

y

In this case the projectile moves along a line that is a perpendicular distance s from the center of thetarget sphere. This is called the impact parameter and is related to the angular momentum of the projectile.The target will receive an impulse directed along the line connecting the centers of the spheres, and so will exit thecollision in direction

sin θ =s

2R, sinϕ = cos θ =

√4R2 − s2

2R(7.55)

since θ + ϕ = π2 is required by energy and momentum conservation. Now the outcome is completely determined.

Example 9. Consider a triple collision between a billiard ball and two identical balls in contact.


m

v0

θθ

w

v1

v2

x

y

The collision is perfectly symmetric, the projectile strikes the two targets simultaneously. Because of this symmetry,the exit angles of the two targets must be the same. Since it is a multiple collision, there will be no constraint onthe sum of exit angles. All exit angles are determined by symmetry alone. The top target receives impulses fromboth of the other balls, sending it into a direction determined by symmetry to be 30 degrees north of the x−axis,the path of the projectile.Assuming that the projectile continues along this path with speed w after the collision, the conservation laws give us

mv0 = 2mv1 cosπ

6+mw, 0 = mv1 sin

π

6−mv1 sin

π

6(7.56)

1

2mv20 =

1

2mw2 + 2

1

2mv21 (7.57)

from which we can extract

v1 = v2 =2√3

5v0 and w =

1

5v0 (7.58)

The most basic applications of the collision equations involve several steps. Imagine a system that is evolving nor-mally, is suddenly interrupted by a collision, and then continues to evolve normally after the collision.

M v0

Θ

mh

Example 10. A pendulum of mass m is at rest. It is struckelastically by a mass M traveling at speed v0 perpendicular tothe string. To what final height does it swing?Since the collision is elastic, the target (pendulum bob) will betraveling at speed

v′ =2M

m+Mv0 (7.59)

instantaneously after the collision. From this point on energyis conserved and so all of the kinetic energy of the bob will beexpended working against gravity to raise the bob to a height hgiven by

1

2m(

2Mv0m+M

)2 = mgh, and so it will swing to a final height h =1

2g(2Mv0m+M

)2 (7.60)


M v0

Θ

m+Mh

Example 11. Suppose that in the previous problem the collisionwas totally inelastic. How high will the contraption swing afterthe collision?The collision itself conserves only momentum, and so

Mv0 = (M +m)w (7.61)

where w is the speed of the combined pendulum bob and pro-jectile after the collision. After the collision, the system evolvesnormally and mechanical energy is again conserved, so that allof the kinetic energy

KE =1

2(M +m)w2 (7.62)

will be converted into gravitational potential energy at the topof the swing, so that

1

2(M +m)w2 = (m+M)gh =

1

2(m+M)(

Mv0m+M

)2, h =1

2g(Mv0m+M

)2 (7.63)

A very generic conservation of momentum problem would be...

Example 12. Consider an Antarctic penguin crossing. A penguin of massM speeds towards the crossing at velocity(0, w), while a smaller, less butch penguin of mass m < M races there at velocity (v, 0). They arrive at the originsimultaneously and stick together (did I mention they are both wearing Velcro?).

M

m

v

w

The combined momentum vector before the collision was

pbefore = m (v, 0) +M (0, w) (7.64)

This equals the momentum vector of the resulting super-penguin

pafter = (m+M) (Vx, Vy) = (m+M)V (cos θ, sin θ) (7.65)


and so

mv = (m+M)V cos θ, Mw = (m+M)V sin θ (7.66)

dividing these two to eliminate V , the super-penguin speed, we find that

tan θ =Mw

mv(7.67)

Example 13. A pair of blocks slide on a frictionless surface. The left block collides inelastically with the right one,initially at rest. After the collision they slide to the right, encountering a surface with coefficient of friction µk .Howfar do they slide before coming to rest on this surface?The collision, being inelastic conserves only momentum so that

mv0 = (m+M)w (7.68)

where w is the speed of the combined wreck after the collision. Afterwards, total mechanical energy is conserved.All of the kinetic energy of the mass will be expended doing work against friction

(0− 1

2(m+M)w2

)=

∫ d

0

−µk(m+M)gi · i dx = −µk(m+M)gd (7.69)

giving a distance of

d =1

2gµk(mv0

m+M)2 (7.70)

M2

M1

v0

d

w

Example 14. Two blocks rest upon a smoothfrictionless surface. Block M1 rests upon blockM2, and the coefficient of friction between thetwo is µk. The top block is given a sudden jolt,resulting in initial speed v0. It slips on thebottom block, causing it to slide to the right.Eventually the two end up sliding to the righttogether at the same speed w. How far willthe top block slide on the bottom one beforethey begin to move together at the same speed?

The total momentum is always the same, sothat

M1v0 = (M1 +M2)w (7.71)

Apply the work-energy theorem to each, letM2 be displaced by ℓ

1

2M1w

2 − 1

2M1v

20 = WG +WN +WF =

∫ ℓ+d

0

(− µkM1g

)i · i dx = −µkM1g(ℓ+ d)

1

2M2w

2 − 0 = WG +WN +WF

=

∫ ℓ

0

(− µkM1gi

)· i dx = −µkM1gℓ (7.72)

Add to eliminate ℓ and solve for d.

7.2. FORCES, MOMENTA AND IMPULSE 125

θθ

m 2m

Example 15. Two pendula of masses m and2m are initially displaced by equal angles θ andreleased. They collide elastically. Find the fi-nal angles that they swing to.Let

h0 = ℓ(1− cos θ) (7.73)

where ℓ is the string length be the originalheights of the two pendula. Once released theright pendulum will pass the low point of theswing with velocity

(−v0, 0) = (−√2gh, 0) (7.74)

and the left pendulum arrives there with velocity

(v0, 0) = (√2gh, 0) (7.75)

Since the collision is elastic, after the collision the mass m has velocity

m− 2m

m+ 2m(v0, 0) +

2 · 2mm+ 2m

(−v0, 0) =−5

3(v0, 0) (7.76)

and so is moving left. The mass 2m will have velocity

2m

m+ 2m(v0, 0) +

2m−m

m+ 2m(−v0, 0) =

1

3(v0, 0) (7.77)

after the collision and so is moving right. After the collision takes place energy is again conserved and so

φψ

m

2m

mgℓ(1− cosϕ) =1

2m(

−5

3v0)

2

2mgℓ(1− cosψ) =1

22m(

1

3v0)

2

(7.78)

Notice that if the bottom of the swing is theorigin, the initial center of mass of the systemhas a height of

ycom =mℓ(1− cos θ) + 2mℓ(1− cos θ)

m+ 2m=

1

2gv20 (7.79)

and if the two pendula reach their respective maximum heights simultaneously, then the final center of mass will beat height

y′com =ml(1− cosϕ) + 2ml(1− cosψ)

m+ 2m=

1

3mg

(12m(−5

3v0)

2 +1

22m(

v03)2)=

1

2gv20 (7.80)

and so the center of mass rises to the same height after the collision that it began at!We will discover that the two pendula do reach their maximum heights at the same time, no matter what anglesthey begin at.

7.2 Forces, momenta and impulse

The most correct way to view Newton’s third law is not as F = ma but rather as

F =dp

dt=

d

dt

(mp)

(7.81)


This allows you to correctly take into account the fact that mass might change, and gives you a more direct inter-pretation of how forces alter the momentum

p(tf )− p(ti) = ∆p =

∫ tf

ti

F dt (7.82)

The right side of this formula is called the impulse of the force over the time interval [ti, tf ]

Example 16. Find the change in momentum delivered to a mass m by the force F = 10N i from t = 0 to t = 5s.

p(tf )− p(ti) = ∆p =

∫ 5s

0

10N i dt = 50kgm

si (7.83)

A fairly common MCAT or GRE problem is to get a graph of the function versus time, and to have to compute theintegral as the area under the curve of the graph.

The classic problem in which the mass of the system changes is a rocket, which propels itself by ejecting mass at afixed rate and velocity relative to the rocket ejection nozzle. The acceleration is gotten by conservation of momentum.

Let the rocket have initial mass m0 and expel exhaust at rate dmdt = α (in kg/s), so m(t) = m0 − αt. The exhaust

has speed u relative to the rocket. Conservation of momentum dictates that when the rocket has speed v

m(t)v =(m(t)− dm

)(v + dv

)− dm

(u− v

)(7.84)

or

m(t)dv

dt= u

dm

dt= αu (7.85)

(note that in any conservation of momentum problem, momenta and speeds are calculated with respect to a stationaryorigin). The rocket has speed at time t

v(t) = −u ln(m0 − αt

m0

)(7.86)

The quantity uα is called the thrust of the rocket, and has units of Newtons. For rockets used to launch spacevehicles this can be over a million Newtons.

Example 17. Consider a chain coiled up by the edge of a table. If a small bit is allowed to slip over the edge, itsweight will pull more links over, and the portion of the chain in motion will have a variable mass; if x is over theedge, moving at speed v, the mass of this amount of chain is m = m(x) = ρx = m

ℓ x.

A uniform chain of mass m and length ℓ is coiled up at the edge of a table, with a miniscule length hanging over theedge, but held in place. If it is released, when length x is over the edge, the velocity v of moving chain obeys

d

dt

(m(x) v

)= v

d

dx

(m(x) v

)= m(x) g

v2 + xdv

dt= xg

v2 + xvdv

dx= xg (7.87)

and this can be integrated to (using (v(0) = 0)

v =

√2xg

3

7.3. PROBLEMS 127

7.3 Problems

M1 v0

M2

v1v2

1. Consider a box M2 = 1.0kg resting motion-less on a friction-free surface. It is struck by afast-moving bullet M1 = 0.05kg, v0 = 100msthat passes right through it. Find its speedv2 after the collision if v1 = 60ms . Was itan elastic or inelastic collision? If inelastic,compute the kinetic energy loss.

M1 M2

x=0

2. Two blocks M1 and M2 rest motionless ona surface of friction coefficient µk. A powerfulspring of force constant k, compressed byamount x, is in a cavity inside of them. Ahook holds them together. When the hook isreleased, they are flung apart. How far apartwill they be after they stop sliding? Is xcom(initially at x = 0) conserved?

M

v0

x=0

3. A block slides on a surface that is friction-free for x > 0, but has coefficient of friction µkand µs for x ≤ 0. , at speed v0. At the originit encounters a spring of force constant k (andfriction). How far will the block compress thespring before it is stopped by the spring andfriction combination?

4. In the previous problem there is a threshold velocity v0 such that if the block is traveling at or below this speed,after compressing the spring, the system will “lock up”, the spring will not be able to exert sufficient force to over-come the static friction that engages when the block stops. The block will come to a permanent rest. Compute this v0.


M1 M2

v1 v2

5. A block of mass M1 = 0.5kg and a block ofmass M2 = 0.8kg compress a spring of forceconstant k = 100 N

m by amount x = 0.1m.When released and disengaged from the springthe blocks slide without friction at v1 and v2.Find both v1 and v2.

6. A. A force F = 30N i + 10N j acts on a mass 1.0 kg that is at the origin, at rest at t = 0. Find its momentumand kinetic energy at time t = 4 s.B. A force F = 30Ns ti + 10Ns2 t

2 j acts on a mass 1.0 kg that is at the origin, at rest at t = 0. Find its momentumand kinetic energy at time t = 4 s.C. A conservative force whose potential is V = −3x J

m acts on a mass 1.0 kg that is at (1, 0) (both in meters), atrest at t = 0. Find its position and velocity at time t = 2 s

θ θ

θ φ

v0

v0

v0

v1

7. Lets explore how physics influences thebehavior of the various types of balls used inathletics. Ignore gravity in this problem.Suppose that the ball is impossible to deform(like a steel ball) and it hits a similarly hardsurface. The surface simply reverses its normalcomponent of momentum. The incoming ballhas momentum (mv0 sin θ,−mv0 cos θ) andoutgoing has momentum (mv0 sin θ,mv0 cos θ),since there is no mechanism for taking awaykinetic energy and turning it into some otherform. The ball and surface are in contactfor an instant. Is momentum conserved?Explain.Suppose the ball can deform, so that ball andsurface are in contact for time T . Ignoreany spinning of the ball. The ball slipson the surface against friction (coefficient µk)for time T . Which way does this forceof friction on the ball point? Show thatthe y-component of the balls momentum isreversed by the bounce.

Show that friction reduces the balls x-component of momentum. Let v0 = 20ms , θ = 45o, µk = 0.1, and T = 0.001 s.Find the speed of the rebounding ball v1 and the angle ϕ at which it rebounds. This is not a conservation ofenergy problem but rather ∆p =

∫F dt, the relationship between force and momentum. The right-side of this

equation is called the impulse delivered by F.

8. In the previous problem assume that the normal force N was constant (equals its average) during the entirecollision time T . If the ball has mass m = 1.0 kg, compute the magnitude of the normal force. Note that

F = ma =dp

dt, ∆p =

∫F dt implies that

pf − pitf − ti

= F

7.3. PROBLEMS 129

9. Re-consider the previous two problems, but incorporate the affects of top-spin. We will neglect the rotationalenergy of the ball for now. Suppose that the ball is spinning clock-wise (top-spin) at rate ω = v0 sin θ

R (its radius isR) throughout the entire collision process. Explain why you would get v1 = v0 = 20ms , and ϕ = θ = 45o.

Suppose that the ball is spinning clock-wise (top-spin) at rate ω = 2v0 sin θR (its radius is R) throughout the entire

collision process. Recalculate both v1 and ϕ, and show that now v1 > v0 and tanϕ = tan θ + 2µk.

It appears that friction is giving the ball kinetic energy, contrary to what we know friction should do (it robs you ofmechanical energy, and turns it into heat). We will see in a few weeks that friction is robbing you of energy in thisproblem, reducing the rotational kinetic energy and slightly increasing the ordinary kinetic energy.

m

v0

θθ

w

v1

v2

x

y10. Consider a triple collision be-tween a billiard ball and two iden-tical balls in contact. This cue-ballinitial speed is v0 = 5.0 m

s . Findthe speed w of its recoil, the angleθ that the two target balls move offat, and their speeds v1 after col-lision. Neglect rotation of theballs, they don’t rotate.

m3m1 m2 x

11. The Giant Martian Black-Widow penguin is famous for the habit of the female devouring the male after mating.A female penguin of mass m1 = 100 kg at the left end of an dry-iceberg of mass m3 = 400 kg and length ℓ = 5mgives a “come hither” glance towards the amorous (and tasty) m2 = 50 kg male at the other end. They both waddletowards the center of the iceberg, and its all over in 60 seconds. How far x does the dry-iceberg move?


θ

M2

M1

v1

θ

M2

M1

12. A box of mass M1 begins at the top of a frictionless ramp of height h which rests on a frictionless surface. Thebox slides to the bottom of the ramp, after which both objects slide horizontally at respective speeds v1 and v2.Explain why only the x-component of momentum is conserved. Compute both v1 and v2.Try the whole problem again if there is friction between M1 and M2 of coefficient µk. Is the x-component of mo-mentum still conserved?

v1

θ

M2

M1θ

v2

M2

M1

13. A box of mass M1 speeds towards a frictionless ramp which rests on a frictionless surface. The box slides to thetop of the ramp, momentarily coming to a stop relative to M2, at which time both objects slide horizontally at v2. Explain why only the x-component of momentum is conserved. Compute both v2 and the height h of the ramp.Try the whole problem again if there is friction between M1 and M2 of coefficient µk. Is the x-component of mo-mentum still conserved?

M1 v1 M2v2

M3

14. The theory of Relativity shows that energy can be trans-formed into matter (mass), and vice-versa. This is the mecha-nism by which new forms of matter (elementary particles) arecreated in nature. The transformation law is E = mc2, amountof energy E can be transformed into mass m.Consider this center of momentum collision in which the newparticle of mass M3 is created at rest by the collision betweenM1 and M2. Compute M3 by conserving total energy and mo-mentum.

15. Find the center of mass (COM) position of each of these mass arrangements relative to the given xy-axes.

7.3. PROBLEMS 131

1.00 kg2.00 kg

3.00 kg

4.00 kg 5.00 kg

6.00 kg

1.0 m

1.0 kg, 1.0 m

1.5 kg, 1.5 m1.5 kg, 1.5 m

2.0 m

1.0 m 5.0 kg

1.0 kg, radius 0.05 m

m, v 2m

φφ

16. Find the center of mass of a solid hemisphere ofbase-radius R.

17∗. Find the center of mass of a right-circular cone ofbase-radius R and height h.

18. A particle of mass m and velocity v = v i collideswith a stationary mass 2m as in the upper figure. Af-ter the collision the mass m is found to be stationary,and the target mass has split into two equal portions mmoving off at identical speeds w at identical angles ϕ.Prove that

w >v

2

Is the collision elastic? If not, find the energy loss (asa function of m, v and ϕ).


m, v 2m

φφ

w

w

u

19. A particle of mass m and velocity v = v i collideswith a stationary mass 2m as in the upper figure.After the collision the mass m is found to be movingat u = u i, and the target mass has split into twoequal portions m moving off at identical speeds w atidentical angles ϕ. The collision is elastic. Prove thatif ϕ > 45o the projectile mass ends up movingleftwards after the collision.

20. In the previous problem prove that

w = 2v( cosϕ

1 + 2 cos2 ϕ

)

-1

-0.5

0

0.5

1

1.5

2

2.5

3

0 0.5 1 1.5 2 2.5 3

F(t

), N

t (s)

f3(x)

21∗. A particle of massm = 1.0 kg at rest at t = 0 issubjected to this force. Findits momentum at t = 1, 2, 3 s.

Find the acceleration of theparticle for any time t in theinterval 0 ≤ t ≤ 1 s.

The power ℘ delivered by aforce is the rate with whichit does work ℘ = dE

dt = F · v.Compute the power deliveredto the particle for any time tin the interval 0 ≤ t ≤ 1 s.

22. A big tank of water is pushed from rest by a constant force F = F0 along a frictionless track. Water escapesfrom the tank at rate α, so that m(t) = m0 − αt. Explain why under these circumstances that

F0 = m(t)dv(t)

dt, and not F0 =

d

dt(m(t) v(t))

Show that at time t, the velocity of the tank is

v(t) = −F0

αln(1− α

m0t)

23∗. What force does a fire-hose spewing water at rate dmdt = 10kgs at nozzle-speed 20ms exert on the fireman directing

the water?

24. Find the final speed of a rocket that has initial mass m0 = 700, 000 kg (the first stage of a Russian UR-500proton rocket) whose mass is 80% fuel mass, with exhaust thrust of 8.5 × 106N (thats 1.9 million pounds) and atotal burn time of 126 s.

7.3. PROBLEMS 133

25. A uniform chain of mass m and length ℓ is stretched out perpendicular to the edge of a table, with a length ahanging over the edge. You hold it in place, and let go at t = 0. Show that the velocity of the chain when length xis over the edge is

v =

√g

ℓ(x2 − a2)

26. Compute the time it takes the chain in example 17 to completely slip over the edge of the table.

27. How far below the apex of an isosceles triangle of base w and height h is the center of mass?

28. A railroad car has mass M including the mass of a large cannon. There is a heap of N cannon balls on the car,each of mass m, so the total initial mass of the system is M + Nm. How fast is the railroad car moving after allcannon balls have been fired off horizontally by the cannon with muzzle speed u?

29. A frog of mass mf sits on a block of ice of mass m1. The thing about frogs is that they don’t like to have coldbutts, so the frog leaps at speed u relative to the block onto another nearby block of ice of mass m2. Find the finalspeeds of both blocks.

30. Find the initial acceleration (at launch) of a rocket of total thrust 2× 106N , burn rate 2000 kgs , and initial mass

5× 105 kg. Find its acceleration at the instant that all of its fuel has been consumed at the end of 200 s.

Chapter 8

Rotational motion

x

y

r(t)

ac(t)

v(t)

θ=θ(t)

a T(t)Up until now we have dealt exclusively withrectilinear motion, and all bodies have beenstructure-less. Of course nearly any but the simplestobjects in nature have spatial dimension and extent.We now consider motion of an extended body aboutit’s center of mass, regarded as a pivot point. Con-sider a mass m tethered to the origin with a rigid,massless tie-rod. The axis of rotation is the positivez-axis k and the sense is counter clock-wise.

The object has coordinates

r(t) = (x(t), y(t)) = (R cos θ(t), R sin θ(t))

= R cos θ(t) i+R sin θ(t) j (8.1)

if the tie-rod has length R. Compute the velocity and acceleration of the object;

v(t) =d

dtr(t) =

(−R

dθ

dtsin θ i+R

dθ

dtcos θ j

)=dθ

dtk×

(R cos θ(t) i+R sin θ(t) j

)(8.2)

(by the use of k× i = j, k× j = −i) or

v(t) =dθ(t)

dtk× r(t)

a(t) =d

dtv(t) =

d2θ(t)

dt2k× r(t) +

dθ

dtk× d

dtr(t)

=d2θ(t)

dt2k× r(t) +

(dθdt

)2k×

(k× r(t)

)(8.3)

the acceleration can be divided into two terms; if r ⊥ k as we intend in the figure, then (and you should check this)

k×(k× r(t)

)= −r(t) (8.4)

so that

a = ac + at = −(dθdt

)2r+

d2θ

dt2k× r(t) (8.5)

This is the decomposition of the acceleration vector into it’s polar components. The first term is the centripetalacceleration, towards the center of the circle (normal to the circle), that we have studied previously

an = ac = −(dθdt

)2r(t) (8.6)

135

136 CHAPTER 8. ROTATIONAL MOTION

The second term at (a tangential acceleration) is entirely new, and contains the angular acceleration of the object,or the rate of change of it’s angular velocity. This is perpendicular to both position vector r and the axis of rotationk.

8.0.1 The acceleration components

In Chapter 5 we studied circular motion at constant speed, but of course you can have motion on a circle withtangential as well as radial (normal or centripetal) acceleration. The simplest such example is the pendulum.

θ

ar

at

The position of the pendulum “bob” or massm is typicallylabeled by the angle θ that the string (of length ℓ) makeswith the vertical.

If the pivot-point is the origin, the coordinates of the bobare

x = ℓ sin θ, y = −ℓ+ ℓ cos θ (8.7)

Note that the bob has both radial and tangential acceler-ation

x = ℓθ cos θ

y = −ℓθ sin θ

x = ℓθ cos θ − ℓθ2 sin θ

y = −ℓθ sin θ − ℓθ2 cos θ

an = acent =yx− xy√x2 + y2

= ℓ θ2 (8.8)

at =xx+ yy√x2 + y2

=d

dt

(√x2 + y2

)= ℓ θ

I have drawn the total acceleration vectors a = ann + att in the lower figure to illustrate the net accelerations.Recall from the second to last problem of Chapter 5; only when the speed is constant will the accelerationbe purely radial (normal) to your trajectory, and only when the pendulum stops will it be tangential. Afalling body such as the pendulum bob is picking up speed as potential energy is converted into kinetic, and so onlyat one point, the bottom of the swing, is the acceleration normal.

8.0.2 The case of constant tangential acceleration

Consider a case in which the angle θ has a constant second derivative

d2θ

dt2= α (8.9)

integrate once; suppose at t = 0 the angular speed is dθdt |t=0 = ω0, then∫ t

0

d

dt

(dθdt

)dt =

∫ t

0

dω

dtdt

=

∫ ω

ω0

dω = ω − ω0

=

∫ t

0

αdt = αt

so that ω =dθ

dt= ω0 + αt (8.10)

137

now integrate again, using initial condition

θ(t = 0) = θ0∫ θ(t)

θ0

dθ =

∫ t

0

(ω0 + αt)dt

or θ(t) = θ0 + ω0t+1

2αt2 (8.11)

this is the most common case to encounter for a rotating body. The purely angular part of the acceleration of themass m tethered to the origin in the first example becomes under these circumstances

aT =(αk)× r(t) (8.12)

or in magnitudeaT = Rα (8.13)

analogous tovtangential = Rω, vtangential = ωk× r(t) (8.14)

Eliminate t from

ω = ω0 + αt and θ = θ0 + ω0t+1

2αt2 (8.15)

to get

θ = θ0 + ω0(ω − ω0

α) +

1

2α(ω − ω0

α)2 (8.16)

and rearrange to2α(θ − θ0) = ω2 − ω2

0 (8.17)

which is the rotational analog of2a(x− x0) = v2 − v20 (8.18)

We will not dwell much on angular kinematics, but move directly to dynamics and work both types of problemstogether.

Lets refresh ourselves on the cross-product, which is a central ingredient in the description of rotations. Return tothe case of a mass m traveling counterclockwise around a circle of radius R in the xy- plane

r(t) = (R cosωt,R sinωt)

then v(t) =d

dtr(t) = (−Rω sinωt,Rω cosωt) (8.19)

Define an axial angular velocity vector ωk with magnitude ω but pointing in the z-direction, which is along theaxis of rotation

ω = (0, 0, ω) = kω (8.20)

then the velocity vector of the mass can be written as

v = ω k× r (8.21)

using the cross product. Recall the basic rules pertaining to the cross product;

a× b = −b× a

a× (b+ c) = a× b+ a× c

i× j = k, j× k = i, k× i = j

and finally |a× b| = |a| |b| sin θab (8.22)

where θab is the angle between the two vectors when placed tail to tail, equal to the area of the parallelogram whosesides are copies of the two vectors.


In the figure

b = |b|(cos θbi+ sin θbj

)a = |a|

(cos θai+ sin θaj

)b× a = |a||b|

(cos θb sin θai× j+ cos θa sin θbj× i

)= |a||b|

(cos θb sin θa − cos θa sin θb

)k

= |a||b| sin(θa − θb)k (8.23)

This defines the time derivative of the r vector in a rotating coordinate system. The acceleration of our particlewhich moves in the plane ⊥ to the rotation axis can be gotten from the chain rule

a =d

dt(ωn× r) =

dω

dtn× r+ ωn× dr

dt

=dω

dtn× r+ ωn× (ωn× r)

= α× r− ω2r (8.24)

where we have used the easily provable identity

n× (n× r) = −r, if n ⊥ r (8.25)

and the definition of the angular acceleration. This is our previous result for the acceleration.

We will now concisely define the angular velocity and acceleration vectors as being vectors of respective magnitudesω and α pointing in the direction of the axis of rotation n.

8.1. KINETIC ENERGY OF A ROTATING BODY 139

8.1 Kinetic energy of a rotating body

ω

n

r||

r⊥

dm

We can compute the kinetic energy of abody undergoing rotation at a fixed rateω about a fixed point by placing the ori-gin at that point. Let the axis of rotationpoint in the n direction (a unit vector).Then a small mass dm in the body willhave position vector r

r = r⊥ + r|| (8.26)

in whichn · r⊥ = 0 (8.27)

this is the part of the position vec-tor perpendicular to the angularvelocity vector.

It will have velocity (for example the little blue arrow)

v = ωn× r = ωn× r⊥, ω ≡ ω n (8.28)

and kinetic energy

dKE =1

2dm |ω × r|2 =

1

2dmω2 |n× r⊥|2 =

1

2dm |r⊥|2 ω2 (8.29)

and integration gives

KE =1

2

∫dm(r) r2⊥ ω

2 =1

2I ω2 (8.30)

where I is the moment of inertia.What is the moment of inertia? Mass (the amount of matter possessed by an object) is a measure of its dynamicalinertia; the extent to which an object will “resist” changes in its velocity. You know that out in space where thereis no gravity, everything is in free-fall, and is “weight-less”. On the moon the surface gravity is 1

16 times that on thesurface of the earth. However if you want to accelerate an object horizontally from rest on the moon, you need topush with the same force to get it going as you would on the earth. The body has the same inertia, or resistance toacceleration, no matter where it is.The moment of inertia is the measure of a body’s resistance to changes in its rotational velocity.Because a body can rotate about a variety of axes, each with different degrees of inertia associated with them, themoment of inertia is more complex than the mass. It is in fact an example of a tensor, which is a vector-like objectvery commonly encountered in physics and engineering applications.

Example 1. Compute the kinetic energy of a rotating hoop of radius R, mass m with constant angular speed ωabout an axis perpendicular to the loop through it’s center of mass.A segment of the ring located at

r = R cos θi+R sin θ j (8.31)

subtending angle dθ has mass

dm = (mass of ring) · (length of segment)(length of ring)

= mRdθ

2πR= m

dθ

2π(8.32)


ω n

θ

R dθ

r

v=ω n×r First we compute

ωn× r = ωk×(R cos θi+R sin θ j

)(8.33)

= Rω (− sin θi+ cos θ j)

(8.34)

Our segment has energy

dKE =1

2dm |ωk× r|2 =

1

4πmR2ω2 dθ

(8.35)and we add up all of the fragments composingthe ring;

KE =

∫ 2π

0

1

4πmR2ω2 dθ =

1

2(mR2)ω2

(8.36)establishing that

Iring = mR2 (8.37)

for an axis through the center perpendicular to the plane of the ring.

Example 2. Compute the kinetic energy of a disk of mass m radius R about the same axisWe already know that a ring of mass dm and radius r will have energy

dKE =1

2dmr2 ω2 (8.38)

Cut the disk up into concentric rings, each of mass dm(r) and radius r. Each ring has differential thickness dr andcircumference 2πr, and so represents a fraction

dm (r) =2πr dr

πR2m (8.39)

of the total disk-mass. Add up the kinetic energies of all of the concentric rings;

KE =

∫ R

0

1

2dm(r) r2 ω2 =

∫ R

0

1

2

2πr dr

πR2m r2 ω2 =

1

4mR2 ω2 (8.40)

and so we have established that

Idisk =1

2mR2 (8.41)

for an axis through the center perpendicular to the plane of the disk.


ω

z

dz√r2−z2

Example 3. Compute the rotational kineticenergy of a solid ball of mass m, radius rrotating about an axis through it’s center ofmass at angular rate ω.

Since we already know what the moment of in-ertia of a disk of mass dm and radius r′ is, wecarve up the sphere into a stack of pancakesof thickness dz and appropriate radii; the ra-dius of the pancake that is distance z from theequatorial plane of the sphere is

r′(z) =√r2 − z2 (8.42)

and it has mass

dm(z) = m(volume of pancake)

(volume of sphere)= m

πr′2(z) dz4π3 r

3(8.43)

The moment of inertia about the axis skewering the whole stack is

dI(z) =1

2dm(z) r′2(z) =

3m

8r3(r2 − z2)2 dz (8.44)

from which we can recover the moment of inertia of a sphere rotating about it’s center of mass axis;

Isphere =

∫ r

−r

3m

8r3(r2 − z2)2 dz =

2

5mr2 (8.45)

and its kinetic energy;

KE =1

2

2

5mr2 ω2 (8.46)

x

y

x

ym

v0

Example 4. Compute the rota-tional kinetic energy of a rod ofmass m, length ℓ rotating aboutan axis through it’s end (andperpendicular to it) at angularrate ω.Locate a sub-mass dm(x) atr = r⊥ = (x, 0, 0). Let this masshave length dx and mass

dm(x) = m(length of segment)

(length of stick)= m

dx

ℓ(8.47)

It has velocity (into the paper)v = (0, 0, ω)× (x, 0, 0) = (0, ωx, 0) = (ωx) j (8.48)

and kinetic energy

dKE =1

2dm(x) v2 =

1

2mdx

ℓ(ωx)2 (8.49)


and the kinetic energy of the stick is

KEstick =

∫ ℓ

0

1

2mdx

ℓ(ωx)2 =

1

6mℓ2 ω2 (8.50)

and so for rotation about a perpendicular axis through its end

Iend =1

3mℓ2 (8.51)

x

y

x

ym

v0 v1

Example 5. As an illustrationthat a single body will have manymoments of inertia, depending onthe axis about which it rotates,compute the kinetic energy of adisk of mass m and radius r spunat ω about an axis along a diame-ter of the disk.Carve the disk up into rods of vary-ing lengths. The mass of a rod ⊥to the rotation axis of thickness dzand length

x(z) =√r2 − z2 (8.52)

is

dm(z) = m(√r2 − z2 dz)

(πr2)(8.53)

and we add up the kinetic energy of all of the rods comprising the disk;

KE = 2

∫ r

−r

1

2

(13m

(√r2 − z2 dz)

(πr2)

)(√r2 − z2

)2ω2 =

1

2

1

4mr2 ω2, and so Idisk,dia =

1

4mr2 (8.54)

8.1.1 Parallel axis theorem

In many applications of rotation, for example rolling motion, we will need to be able to compute the moment ofinertia of a body with respect to an arbitrary axis, not just an axis through the center of mass. For this purposewe will develop the Parallel Axis Theorem. Consider a planar body with a coordinate origin placed at thecenter of mass. Then under these circumstances

xcom =

∫dm x = 0, ycom =

∫dm y = 0 (8.55)


ω, C.O.M.

ω

dm

r

d

r’

Run a rotation axis parallel to the rotationaxis through the center of mass both axes ⊥to the plane of the body (to simplify themath), and rotate the body about this new axisat angular speed ωLet a small mass element located at r with re-spect to the origin have position r′ with respectto the point where the rotation axis hits theplanar body, with

r = d+ r′ (8.56)

where d points from the center of mass axis tothe new axis. The kinetic energy of this smallmass element when the body is rotated aboutthe new axis is

KE =1

2

∫dm |ω n× r′|2 =

1

2

∫dm |ω n× (r− d)|2

=1

2

∫dm |ω n× r|2 −

∫dm (ω n× r) · (ω n× d) +

1

2

∫dm |ω n× d|2 (8.57)

but(ω n× r) · (ω n× d) = ω2 r · d = ω2|d||r| cos θ = ω2|d|x (8.58)

(a vector arithmetic identity valid for n ⊥ d, r) and so∫dm (ω n× r) · (ω n× d) = ω2|d|

∫dm x = 0 (8.59)

and we find

KE =1

2

∫dm |ω n× r|2 + 1

2

∫dm |ω n× d|2

=1

2

∫dm |r|2ω2 +

1

2

∫dm |d|2ω2

=1

2(Icom +m|d|2)ω2 (8.60)

and so we discover that the moment of inertia about an axis parallel to one through the center of mass, but a distance|d| from it is

Iparallel = Icom +m |d|2 (8.61)

Example 6. Compute the moment of inertia of a stick of mass m, length ℓ being rotated about one of its end points.What is the moment of inertia for rotation about an axis through its center (the C.O.M.)?We know that

Iend =1

3mℓ2 (8.62)

The new axis is a distance ℓ2 from the center of mass and so

ICOM =1

3mℓ2 −m(

ℓ

2)2 =

1

12mℓ2 (8.63)

Example 7. Compute the total kinetic energy of a disk of radius R mass m that rolls without slipping on a smoothsurface, its center of mass moving at speed v.


8.1.2 Rolling motion

Notice that if the disk rolls without slipping, the contact point between surface and disk cannot be slipping andtherefore has zero velocity with respect to the surface. Divide rolling up into two parts, pure translation of all pointsat speed v, plus rotation at angular speed ω. We take right to be the x-direction, up to be y, and out of the paperz. The center of mass translational velocity of all points on the disk is

vtrans = (v, 0, 0) = vi (8.64)

The axis of rotation is

n = (0, 0,−1) = −k (8.65)

Place the center of the disk at (0, 0, 0).For rotation, the bottom point of the disk is moving left at speed Rω;

vtotal = vtrans + (−ωk)× (−Rj) = (v − ωR)i (8.66)

×

v v

Rω

Contact point at rest

The net speed of the bottom point of the diskwith respect to the surface when the two sep-arate motions are combined is

v −Rω (8.67)

yet this must be zero. We discover that for thedisk to be rolling

vcom = Rω, acom = Rα (8.68)

These are known as the rolling conditions.

The total kinetic energy of the disk has two contributions, one from the center of mass motion and one for rotationabout the center of mass

KE =1

2mv2 +

1

2(1

2mR2)ω2 (8.69)

but when the rolling condition is inserted to eliminate v we find

KE =1

2m(Rω)2 +

1

2(1

2mR2)ω2 =

1

2(3

2mR2)ω2, and Icontact =

1

2mR2 +mR2 =

3

2mR2 (8.70)

which is equivalent to saying that rolling is the same as pure rotation about the contact point, since themoment of inertia about the point of contact is

KE =1

2Icontactω

2 =1

2

(Icom +md2

)ω2 =

1

2Icomω

2 +1

2mv2 (8.71)

8.2. ROTATIONAL DYNAMICS AND TORQUE 145

8.2 Rotational dynamics and torque

m1 at r1

m2 at r2

F2,1

F1,2

F2, ext

F1,extC.O.M.

r1’

r2’

Consider a two or more body system,with internal forces F12 = −F21 and ex-ternal forces F1,ext and F2,ext

The equations of motion are

m1a1 = F12 + F1,ext

m2a2 = F21 + F2,ext (8.72)

We already know that the sum of theseequations results in the expression forthe motion of the center of mass of thesystem

(m1 +m2)d2

dt2(m1r1 +m2r2m1 +m2

)

= F1,ext + F2,ext (8.73)

Now decompose the position vectors intocenter of mass position, and positionsrelative to the center of mass

r1 = rcom + r1′, r2 = rcom + r2

′ (8.74)

and cross the relative position into each equation of motion in turn

r1′ × (m1acom +m1

d2

dt2r1

′) = r1′ × F12 + r1

′ × F1,ext (8.75)

but since r1′ and F12 are parallel, their cross product is zero. In the frame of reference comoving with the center of

mass acom = 0 and so we find

r1′ ×m1

d2

dt2r1

′ = r1′ × F1,ext, r2

′ ×m2d2

dt2r2

′ = r2′ × F2,ext (8.76)

add these two together, and use the fact that

v1′ = ω n× r1

′ (8.77)

d

dt

(m1r1

′ × (ω n× r1′) +m2r2

′ × (ω n× r2′))

= r1′ × F1,ext + r2

′ × F2,ext

=d

dt

((m1(r

′1)

2 +m2(r′2)

2)(ω n))

=d

dt

(Icomω n

)= Icom αn (8.78)

This is called Newton’s second law for rotational motion. We define angular momentum to be

L = mr× v = Iω n (8.79)


and torque of force F about point p to be

τ p = τpn = Ip αn = r× F (8.80)

where r, called the lever arm, is the vector from point p to where on the body the force is applied. Then

d

dt(L) = Iα =

∑i

τ i (8.81)

analogous to

ma =∑i

Fi (8.82)

a

F

X

Example 8. A spool of mass m radius R of moment ofinertia I has thread wrapped around its circumference. Itfloats freely in space.If the thread is pulled with a force F , describe the resultingmotion of the spool.The center of mass of the spool will move in a straight linewith acceleration a given by

F = ma (8.83)

Let x point right, y point up, and z point out of the paper. Place the origin at the disk center. The location wherethe force is applied to the disk has

r = R j, F = F i (8.84)

and so the applied torque about the center of mass is

τn = r× F = RF j× i = RF (−k) = Icom αn (8.85)

and the spool will spin about its center of mass axis with angular acceleration determined by

α =RF

Icom, about axis n = −k (8.86)

Example 9. A spool of mass m radius R of moment of inertia I has thread wrapped around it as illustrated. Itfloats freely in space.

a

Fα r

If the thread is pulled with a force F , describe the resultingmotion of the spool.The center of mass of the spool will move in a straight linewith acceleration a given by

F = ma (8.87)

Let x point right, y point up, and z point out of the paper.Place the origin at the disk center. The location where theforce is applied to the disk has

r = −r j, F = F i (8.88)and so the applied torque about the center of mass is

τn = r× F = −rF j× i = rF (k) = Icom αn (8.89)

and the spool will spin about its center of mass axis with angular acceleration determined by

α =rF

Icom, about axis n = k (8.90)

The rotation is in the opposite sense to that in the previous example.


a

Fα r

R

Ff

Example 10. A spool of mass m radius R of moment ofinertia I has thread wrapped around its circumference. Itrests on a surface of unknown coefficient of friction.If the thread is pulled with a maximal force F (assumeits value is given) that results in rolling without slipping,find µs, and describe the resulting motion of the spool.The center of mass of the spool will move in a straightline with acceleration a given by

F+ Ff = (F − Ff )i = ma (8.91)

We saw in the previous example that the disk will move right, and rotate counterclockwisewithout the interventionof the ground and the friction it applies. When the ground is introduced, friction will cause the contact pointto be stationary, forcing Ff to point left. The disk will therefore rotate clockwise (about n = −k) as it rolls withoutslipping (unless we pull too hard). Using the same coordinates as in the previous examples

(−rj)× (F i) + (−Rj)× (−Ff i) = Icomα(−k), or RFf − rF = Icomα (8.92)

but the disk rolls, and so the rolling condition is satisfied

RFf − rF = Icomα = Icoma

R(8.93)

Solving for acceleration we find that

a =R(R− r)F

Icom +mR2(8.94)

and we can compute the static coefficient of friction (it rolls, it doesn’t slide)

Ff = µsmg = F −ma =(Icom +mrR)

(Icom +mR2)F (8.95)

Example 11. A massive spool of moment ofinertia I has thread wrapped around it. Sus-pended at the end of the thread is a mass m.Gravity acts on the mass. Compute all accel-erations.The mass m will accelerate downwards

m(−aj) = T j+ (−mg)j (8.96)

the string is tangent to an axle of radius r2at the point r = r2j as illustrated, and with xto the right, y up and z out; compute torquesabout the axis through the center of mass ofthe disk

(r2j)× (−T i

)= τn = r2Tk (8.97)

is the torque applied by the string tension, rotational acceleration has magnitude α and direction n = k.

τn = r2Tk = Icomαn = Icomαk (8.98)

Since the center of the spool is fixed, the thread rolls off of it and a point on the thread accelerates with acceleration

a = r2 α (8.99)


This rolling condition is inserted by hand. We solve the three simultaneous equations above by eliminating T andalpha to get

a =mg

m+ Ir22

, α =a

r2(8.100)

Ff

a

F

X

Example 12. A spool of moment of inertia I = 12mR

2

is wound with thread and set on a smooth table. Thecoefficient of friction is unknown and therefore cannot beused in the problem. The thread is pulled with enoughforce to cause the spool to accelerate with no slipping.Find the acceleration.If the spool were not resting on a table, it would acceleratewith

F i = mai, F = ma (8.101)

and rotate clockwise with acceleration

(Rj)× (F i) = −FRk =1

2mR2 αn (8.102)

and so it spins with angular acceleration

α =2FR

mR2, about axis n = −k (8.103)

Then if it is brought near the horizontal surface, the bottom point of the spool would have acceleration with respectto a fixed point on the surface of

ai+ (−αk)× (−Rj) = F

mi+ (− 2F

mRk)× (−Rj) = −F

mi (8.104)

and so the contact point would slip to the left on the surface when the two are brought into contact. The forceof friction will oppose this slipping and therefore point right! The equations of motion including this force and itstorque are

(F + Ff )i = ma i, (Rj)× (F i) + (−Rj)× (Ff i) =1

2mR2 αn (8.105)

since F and Ff exert torques in the opposite sense. We find that n = −k is the rotation axis, and we have the rollingcondition

a = αR (8.106)

Solving these by eliminating α and Ff we find

a =4F

3m(8.107)

This problem illustrates some of the subtleties of rotational dynamics problems.

Example 13. Find the acceleration of a disk yo-yo.The equations of motion are

ma(−j) = mg(−j) + T j, (Ri)× (T j) = Icomαn =1

2mR2 αn

(8.108)(we compute torques about the disk center of mass)and therolling condition

a = αR (8.109)

resulting in n = k and

a =2

3g (8.110)

We finish our discussion of rotational dynamics with some nontrivial examples, and an illustration of how carefullyapplied conservation of energy methods can solve even dynamical problems without the use of Newton’s laws. This


is possible since the energy function is the constant of integration obtained when Newton’s laws are integrated once.

×

T1

T2

m1 g

m2 g

Example 14. Consider the Atwood machine below with amassive frictionless pulley. Compute the acceleration.

Since only an imbalance of torques can cause an object with anonzero moment of inertia to accelerate angularly, the tensionin the two sides of the rope cannot be the same. Labelingthem as in the figure and assuming the the disk rolls on therope without slipping and has radius R, we find that the usualforce equations

T1 j+M1g (−j) = M1a j,

M2g (−j) + T2 j = M2a (−j) (8.111)

(−Ri)× (−T1j) + (Ri)× (−T2j) = Icomαn = Icomα (−k)

and the condition that the string does not slip

a = Rα (8.112)

which can be solved for a

a =(M2 −M1)g

M1 +M2 +IcomR2

(8.113)

×

T

T

mg

Example 15. Compute the acceleration of a double yo-yo asshown in the figure. Both disks are identical.

This proceeds pretty much the same way, for the hanging disk

mg (−j) + T j = ma (−j)

(Ri)× (T j) = TRk = Iα1 k

(Ri)× (−T j) = TR (−k) = Iα2 (−k) (8.114)

from which we conclude that the angular accelerations have thesame magnitudes

α1 = α2 (8.115)

but opposite directions.

This would not be true if the disks had different radii. The two torque equations were gotten by computingtorques about the centers of mass of each disk, which is why mg has no contribution. It has no lever arm aboutthe center of mass of the lower disk, since the force of gravity always acts on the center of mass, as do allforces. Each spool unrolls string at rate Rω and so the total rate at which the string length increases is

v = 2Rω, differentiate; a = 2Rα (8.116)


which results in accelerationa =

mg

m+ I2R2

(8.117)

Newton’s laws are simple to apply if one pays attention to the fact that they are vector equations, and so directionsof forces must be carefully taken into account. The only inconvenience is that they result in several equations inseveral unknowns, which are accelerations and constraint forces such as rope tensions, normal forces and frictionalforces. Many times conservation of energy can be used to solve dynamical problems for the acceleration without everhaving one of these constraint forces appear in the analysis. Consider an object that accelerates from rest for a timet. It will cover distance

d =1

2at2 (8.118)

and end up with speedv = at, eliminate t, v2 = 2ad (8.119)

This together with conservation of energy can solve dynamics problems.

×

T

T

m g

Example 16. Find the acceleration of the massm in the picturebelow. The rope “unrolls” without slipping from the massivedisk.The mass begins at rest. Let it fall a distance h under gravity,to ground level. By time it does so it has speed v and the spoolhas angular speed ω. Since the string unwinds without slippingv = Rω. Apply conservation of energy

mgh =1

2mv2 +

1

2Iω2, or mgh =

1

2mv2 +

1

2

I

R2v2 (8.120)

put this into the form with v2 alone on one side of the equation

v2 = 2( mg

m+ IR2

)h (8.121)

and you can read off the acceleration

a =mg

m+ IR2

(8.122)

This is a quick method for checking your answers and canreplace Newton’s laws methods in most cases if applied carefully.

N

v

mgR

θ

Example 17. A small object of mass m isplaced at the top of a frictionless hill of radiusR. As it begins to slide down, it reaches a pointwhere it loses contact with the hill. Find theangle at which that happens, and the speed ofthe object at that point.Apply conservation of energy. When at angleθ the object has speed v;

mgR = mgR cos θ +1

2mv2 (8.123)

Examine the radial force equation. At any time the mass is in instantaneous circular motion and so its radialacceleration is centripetal;

mg cos θ −N = mac = mv2

R(8.124)


It loses contact with the hill when N → 0 which happens at θc

mg cos θc = mv2

R(8.125)

substitute this into the energy balance equation to get

cos θc =2

3, and v2(θc) =

2Rg

3(8.126)

is the speed of the mass when this happens.

N

v

mgR

θ

ω×

Example 18. Consider the same basicproblem, but now with a rolling object.

The energy equation is

mg(r +R) = mg(r +R) cos θ +1

2mv2 +

1

2Iω2

(8.127)rolling condition is

v = rω (8.128)

and the radial force balance is

mg cos θ −N = mac = mv2

r +R(8.129)

again this is solved by noting that the mass loses contact with the hill when the normal force goes to zero.

Ff

v0 vω×

d

Example 19. A bowling ball is thrown ontothe alley with initial speed v0.If the coefficientof friction between the ball and alley is µk,compute the distance traveled by the ballbefore it begins to roll without slipping.

The ball will slip to the right and so theforce of friction at the contact point with thealley points left. This will cause the ball toaccelerate left, slowing it down

Ff = ma = mdv

dt, integrate

∫ v

v0

dv = −∫ t

0

µkgdt, or v = v0 − µkgt (8.130)

This also provides a torque that causes the ball to begin to roll

FfR = Iα = Idω

dt(8.131)

integrate ∫ ω

0

dω =

∫ t

0

µkmgR

Idt, ω =

µkmgRt

I(8.132)

rolling occurs whenv = Rω (8.133)

which we can solve for the time it takes for rolling to begin

t =v0

µkg(1 +mR2

I )=

2v07µkg

(8.134)


for a sphere. Insert this into the kinematic relation

d = v0t−1

2at2, with a = µkg (8.135)

and we see that the ball slips for

d =12v2049µkg

, before it begins to roll. (8.136)

M

θh

Example 20. A disk rolls down an incline ofheight h, starting at rest. Find its speed androtation rate at the bottom of the hill.

If it rolls, friction will do no work since there is no relative motion (hence no displacement) of the contact point.Therefore energy is conserved;

(12mv2 +

1

2

(12mR2

)( vR

)2+mg 0

)−(0 + 0 +mg h

)= 0, v =

√2mgh32m

(8.137)

Example 21. Suppose that the hill is so steep that the disk cannot initially roll, but begins slipping. How far willit slip, and once it begins rolling, how far will it have descended?You will discover in the homework that if θ > 36.870o (tan θc = 3µs), the disk will slip, if θ < 36.870o, it will roll. Ifit slips, we discover that αR < a, ωR < v at all times, and it slips all the way down;

mg sin θ − µkmg cos θ = ma

a = g(sin θ − µk cos θ)

v = v0 + at = g t (sin θ − µk cos θ)

µkmg cos θR =1

2mR2 α

α = 2µkg

Rcos θ

ω = ω0 + αt = 2µk tg

Rcos θ (8.138)

Rewrite v, if it slips it is because θ > θc so tan θ > tan θc

v = gt cos θ(tan θ − µk) > gt cos θ(3µs − µk) > 2gtµk cos θ = ωR (8.139)

since µs > µk. It never attains rolling motion, are you surprised by this?

8.3 Statics

Statics is a field of simple applications of Newton’s laws. The generic statics problem consists of finding all forcesacting on a body subject to the constraint that it is not accelerating; its center of mass undergoes no rectilinearacceleration, and it exhibits no angular acceleration about any axis or pivot point, and so∑

i

Fi,ext = 0,∑i

ri × Fi,ext = 0 (8.140)

8.3. STATICS 153

N2

mp g

mL g

Ff

N1

θ

Example 22. A penguin of mass mp standsa distance d (measured along the ladder) fromthe center of a ladder of length 2ℓ mass mℓ

inclined at angle θ. The coefficient of frictionwith the ground is µs, but with the verticalwall is zero. Find the smallest value of θ forwhich the ladder does not slip and collapse.

The sum of external forces equals zero, this leads to

Ff = µsN1 = N2, N1 +N2 −mpg = 0 (8.141)

We obtain the final equation by computing torques on the ladder with respect to the contact point with the ground;

(ℓ cos θi+ ℓ sin θj)× (−mℓgj) + (d cos θi+ d sin θj)× (−mpgj) + (2ℓ cos θi+ 2ℓ sin θj)× (−N2i) = 0 (8.142)

orℓ cos θmℓg + d cos θmpg = 2ℓ sin θN2 (8.143)

which immediately gives N2. We can now recover the other forces.

T

mp g mL g

Ff

N1

θ

Example 23. A penguin of mass mp standsa distance d (measured along the beam) fromthe left end of a beam of length 2ℓ mass mℓ

supported by a rope with tension T at angle θ.Compute T in terms of mℓ,mp, d, ℓ, g, θ suchthat the beam does not collapse. The hingewith the wall exerts whatever forces Ff , N1

are necessary for stability.

This is a good example of a mechanical engineering statics problem; we would compute the forces supplied by thehinge, and obtain the rope tension. Horizontal and vertical equilibria imply

N1 − T cos θ = 0, Ff + T sin θ −mpg −mℓg = 0 (8.144)

Torques about the left end of the beam but add up to zero or the beam rotates

(di)× (−mpgj) + (ℓi)× (−mℓgj) + (2ℓi)× (−T cos θi+ T sin θj) = 0 (8.145)

or

T =dmpg + ℓmℓg

2ℓ sin θ, N1 =

cos θ(dmpg + ℓmℓg)

2ℓ sin θ(8.146)


N2

mp gmL g

N1

Example 24. Two small penguins support their emperorpenguin on a beam of length 2ℓ and mass mℓ. Theemperor, whose mass is mp, stands off-center by amountd closer to the right penguin. Find the forces N1 and N2

exerted by the two supporting birds.

Vertical force balancing requiresN1 +N2 = (mℓ +mp)g (8.147)

Torques about the left end

(ℓi)× (−mℓgj) + ((ℓ+ d)i)× (−mpgj) + (2ℓi)× (N2j) = 0 (8.148)

gives us N2;

N2 =ℓmℓg + (ℓ+ d)mpg

2ℓ(8.149)

N2θmp g

T

N1

Example 25. A penguin of mass mp = 20.0 kg heroicallyclimbs an A-frame ladder (hinged at the top) to a distanced = 7.5m (measured along the ladder from the bottom leftcorner) , each side of which has length 2ℓ = 10.0m. Theangle θ = 60o. The ladder rests on ice and is stabilized bya wire connecting the centers of the two halves. Computethe values of the two normal forces N1, N2 and the tensionT in the wire.

The sum of vertical forces is zero;N1 +N2 −mpg = 0 (8.150)

Take torques of external forces on the whole structure about left leg contact point with ground;

0 = (7.5m cos 60i+ 7.5m sin 60j)× (−mpgj) + (10.0mi)× (N2j), N2 = mpg3

4cos 60 =

3mpg

8(8.151)

This gives

N1 =5mpg

8(8.152)

Take torques on the right ladder-leg about the top;

(5m cos 60i− 5m sin 60j)× (−T i) + (10m cos 60i− 10m sin 60j)× (N2j) = 0 (8.153)

gives

T = 2 cot 60N2 =

√3

4mpg (8.154)

and thus we conclude our most penguiniferous section.

8.4 Work and torque

Energy methods gave us a useful alternative to pure vector-based force determination from Newton’s third lawF = ma, and the same is true for Newton’s third law for rotational acceleration τ = Iα. Examine Eq. 8.8. For

8.4. WORK AND TORQUE 155

a rotating body (about some pivot point or axis p) we can always recover the angular speed v = ωR = θR fromconservation of energy

1

2Ip ω

2 =∑i

Wi (8.155)

θ

ar

at

For example, if we begin with the pendulumillustrated in the horizontal position, and let itswing down to angle θ shown, then

1

2

(mℓ2

)ω2 =

1

2

(mℓ2

)θ2 (8.156)

= −(0−mgℓ sin θ

)Among other things, this gives you the centripetal part of the acceleration

ac = Rω2 = ℓθ2 = 2g sin θ (8.157)

which vanishes at the top of the swing, and maxes out at 2g at the bottom.

From Eq. 8.8 we know that the tangential acceleration is at = ℓθ which we can recover from our energy calculationin two ways; first by simple derivation using the chain rule

d

dt

(12ℓ2θ2

)=

(ℓθ)(ℓθ)

= v at

=d

dt

(gℓ sin θ

)= gℓ cos θ θ

=(g cos θ

)(v), at = g cos θ (8.158)

or alternatively from the torque

Ipα =(mℓ2

)(θ k)

=(− ℓ cos θ i− ℓ sin θ j

)×(−mg j

)= mgℓ cos θ k, ℓθ = at = g cos θ (8.159)

and now we have all acceleration components.

Consider a force that exerts a torque with lever arm r on a body constrained to rotate about a fixed point, so thatthe velocity of the point where the force is applied is v = ω × r;

℘ =dW

dt= F · v

= F ·(ω × r

)= ω ·

(r× F

)= ω · τ

W =

∫τ · ω dt =

∫τ · n dθ (8.160)

is the work don by a torque.


8.5 Problems

M

θh

1. A. This ball rolls down the ramp withoutslipping. Find the largest angle θ = θc forwhich this is possible in terms of µs. Once theball reaches the bottom of the hill, determineits speed in terms of the hill height h.

B. If the angle θ < θc, the ball still rolls allthe way. Under these circumstances compute|Ffric|.

Mg

h

w

T

D.Wagenheim

Dentist

2. A sign of dimensions illustrated and massm = 10.0kg hangs from two cords. If the rightcord breaks, find the tension in the left cordinstantly afterwards. Let h = w = 0.9m.

Double or nothing if you figure out the literaryreference.

b

Ff

Nmg

θ

L

3. A disk of massm = 0.005kg and radius a = 0.01mrolls without slipping on a smooth table. It rollsin a circle of radius b = 0.2m. There are threeforces acting on it; gravity, friction, and the normalforce. It spins about its center of mass at rateω = 20.0 rads . As it rolls around on its “orbit”, itsspin angular momentum vector L makes and angleθ with the vertical If we watch the disk, we see thatits contact point with the ground travels around thecircle of radius b at speed Ωb, and its spin angularmomentum vector tip moves around on a circle atangular rate Ω. The angular momentum vector is

L = Lspin + Lorbital = Icomω(cos θk+ sin θ sin(Ωt)i+ sin θ cos(Ωt)j

)+ |Lorbital|k

Ω n

L

Ω tL⊥

The position vector of the center of mass of the diskis

r = (b− a cos θ)(sin(Ωt)i+ cos(Ωt)j

)+ a sin θk

From all of this information compute Ω, and themagnitude of the normal force N. From thisformula for the angular momentum vector andfrom its time derivative (and the torque equationL =

∑i ri × Fi,ext), determine the rate ω as a func-

tion of θ such that the disk rolls without slipping.From this information compute the magnitude ofthe force of friction that keeps the disk traveling ina circle.

8.5. PROBLEMS 157

mg

N

θ4. A rod of mass m and length 2ℓ rests on-end on asmooth frictionless table. It begins to fall, its contactpoint with the table sliding out from beneath it. At whatangle of inclination θ does it lose contact with the table?Note that with gravity and the normal force being theonly forces acting on the rod, its center of mass can onlymove vertically.

F

α ri

ro

5. A spool with an inner hub of radius ri = 0.2m andan outer hub of radius ro = 0.3m and mass m = 10 kgand moment of inertia Icom = 3

5 mr2o rests on a smoothhorizontal surface of unknown coefficients of friction.A string wound around its inner hub is pulled straightupwards with force F = 25N j. The disk rolls withoutslipping. Compute the magnitude and directionof the acceleration of the disk and the force offriction exerted by the horizontal surface.

aFα ri

ro

θ

6. A spool with an inner hub of radius ri = 0.2m andan outer hub of radius ro = 0.3m and mass m = 10 kgand moment of inertia Icom = 3

5 mr2o rests on a smoothhorizontal surface of unknown coefficients of friction.A string wound around its inner hub is pulled gentlyat angle θ yet the disk does not move! Find the angle θ.

M1

M2

F

R

7. A block of mass m1 = 4.0 kg slides on a frictionless surface. A ball (sphere) of mass m2 = 1.0 kg and radiusR = 0.08m rests on it. A force F = F i = 4.0N i is applied to the block as shown. The coefficients of frictionbetween ball and block are unknown. Find the acceleration a1 of the block and a2, α2n of the sphere if the ballrolls without slipping on the block.


M

r2

r1

8. A spool with an inner hub of radius r1 = 0.2m and an outer hub ofradius r2 = 0.3m and mass m = 10 kg and moment of inertia Icom =35 mr22 rests on a smooth horizontal surface of unknown coefficients offriction. A string wound around its inner hub is attached to a hangingmass M = 1.0 kg. The spool rolls without slipping. Find a for spooland mass, and the force of friction.

M1

M2

F

r2r1

9. A block of mass M1 slides on africtionless surface. A spool of mass M2

and radii r1, r2 rests on it. A stringwound around the spool’s inner radius isgently pulled with force F as shown.The coefficients of friction betweenspool and block are unknown. Find theacceleration a1 of the block and a2, α2nof the spool if the spool rolls withoutslipping on the block.

M1

M2

Fr2

r1

10. A block of mass M1 slides on africtionless surface. A spool of mass M2

and radii r1, r2 rests on it. A stringwound around the spools inner radius isgently pulled with force F as shown.The coefficients of friction between balland block are unknown. Find the accel-eration a1 of the block and a2, α2n ofthe sphere if the sphere rolls withoutslipping on the block.

FM1

M2

Θ

11. In this problem the coefficientsof friction between wedge M1 and thehorizontal table it rests on are zero,but the friction between moving wedgeM1 and the spool M2 has coefficientsµs = 0.3 and µk = 0.2. LetM1 = 10.0 kgand M2 = 5.0 kg, spool (disk) radiusR = 0.1m, Θ = 30o.

A. What co-acceleration can M1 and M2 undergo without any slipping or rolling of M2 on M1?B. Compute the magnitude of the force of friction FF acting on the spool.We obtained Ff = 0 in part A.C. Compute the required force F such that the blocks have this acceleration.

8.5. PROBLEMS 159

M

r2

r1 12. A disk with an inner hub of radiusri = 0.2m and an outer hub of radiusro = 0.3m and mass m = 10 kg and momentof inertia Icom = 3

5 mr2o rests on a smoothhorizontal surface of unknown coefficients offriction. A string wound around its inner hubis pulled straight downwards by a hangingmass M = 1.0 kg. The disk rolls withoutslipping. Compute the magnitude anddirection of the acceleration of the disk.Compute the force of friction exerted by thehorizontal surface.

T

M, R

µs, µk

13. This spool of mass M = 3.0 kg and radiusR = 0.1m and moment of inertia 3

5MR2 isbeing pulled by a rope of tension T connectedto its axle. The coefficients of friction withthe ground are µk = 0.2, µs = 0.3. Find thelargest T for which the spool rolls withoutslipping.

14. Suppose that the same spool is pulled by a rope of tension T = 30N connected to its axle. Find Ff , a and α forthe spool.

15. What if T = 12N for the spool of problem 13, find a, α and Ff .

16∗. Find the moment of inertia of these seven disksof mass m (each) and radius R about the axis throughthe center of the middle disk.

17∗. Find the moment of inertia of this frame madeof four rods of mass m length ℓ about an axis throughthe geometric center of the square interior.


18∗. A rod of mass m and length ℓ connected to theground with a frictionless hinge is released from rest andallowed to fall. Find the speed with which the upper endhits the ground.

M

θh

19∗. This spool of mass M and radius R rollswithout slipping to the bottom of the hill,upon reaching it its center of mass speed is

measured to be v =√

8Hg7 . Find its moment

of inertia.

N2θ

T

N1

20. A penguin of mass mp = 30.0 kg heroically climbs anA-frame ladder (hinged at the top) each side of which haslength ℓ = 5.0m. The angle θ = 45o. The ladder rests onice and is stabilized by a wire connecting the centers ofthe two halves. Find the tension T in the wire, andthe forces N1, N2 .

v

21. This disk of mass m = 2.0 kgand radius R = 0.1m compresses aspring of force constant k = 20 N

m byamount x = 0.1m. When released anddisengaged from the spring the disk rollswithout slipping at speed v. Find v.Find its acceleration at the time it isreleased and begins to move.

v1 v2ω

22. A sphere of radius R = 0.1m, mass m2 = 2.0 kgand a block of mass m1 = 3.0 kg compress a springof force constant k = 200 N

m by amount x = 0.1m.When released and disengaged from the spring thesphere rolls without slipping at v2 and the blockslides without friction at v1. Find both v1 and v2.

8.5. PROBLEMS 161

M M M M

m m m

23. Compute the moment of inertia about the originof this gadget (four masses M at intervals of ℓ ona rod of length 3ℓ mass 3m), and find its kineticenergy when rotated at rate ω. Find the speed ofeach mass M .

M M M M

m m m

24. Compute the moment of inertia aboutthe origin of this gadget (four masses M atintervals of ℓ on a rod of length 3ℓ mass 3m),and find its kinetic energy when rotated atrate ω. Find the speed of each mass M .

F

25. A stick of length ℓ and mass m ordinarilyhangs vertically from its hinge point. Whatforce F must be applied vertically at its centerin order to hold it horizontally? What force(magnitude and direction) must the hingeexert?

F

26. A stick of length ℓ and mass m ordinarilyhangs vertically from its hinge point. Whatforce F must be applied vertically at its endin order to hold it horizontally? What force(magnitude and direction) must the hingeexert?


27. A stick of length ℓ and mass m ordinarilyhangs vertically from its hinge point. Whatforce F must be applied horizontally at itscenter in order to hold it at an angle θ withrespect to vertical? What force (magnitudeand direction) must the hinge exert?

28. A stick of length ℓ and mass m ordinarilyhangs vertically from its hinge point. Whatforce F must be applied normally at its centerin order to hold it at an angle θ with respectto vertical? What force (magnitude anddirection) must the hinge exert?

29. For the stick in the previous problem, held in place at rest ant angle θ with respect to vertical; suppose that itis released, find its instantaneous angular acceleration at the moment of release.

30. For the stick in the previous problems, held in place at rest at angle θ with respect to vertical; suppose that itis released, find its speed as it passes through the vertical position.

31. What fraction of a rolling sphere’s kinetic energy is translational? What fraction of a rolling disk’s kinetic energyis translational?

32. A disk of mass m and radius R is given an initial spin rate ω0 about its central axis, and is set down on its edgeon a table with which it has frictional coefficient µk. Once it begins to roll, find its spin rate and COM speed. Howfar did it slip before rolling began?

33∗. In Eq. 8.8 show that unit vectors normal to and tangent to the path of the pendulum bob are, respectively

n = cos θ j+ sin θi, t = − sin θ j+ cos θi

and thatan = a · n = ℓθ2, at = a · t = ℓθ

and (this is very useful)vn = v · n = 0, vt = v · t = ℓθ

Using the figure for problem 6.32, show that after falling from θ0 = 0 to θ, a pendulum bob of mass m has speedgiven by

1

2m(ℓθ)2

= mgℓ sin θ

and tangential acceleration given by

m(ℓθ)= mg cos θ

Hint; draw a free-body diagram to get at.

8.5. PROBLEMS 163

θ

ar

at

34∗. A pendulum of mass m length ℓbeing at rest in the horizontal position,and is allowed to fall to the point θ illus-trated. Show that the magnitude of itsacceleration at that point is√

a2n + a2t = g√1 + 3 sin2 θ

Hint, use

at = ℓθ = ℓτ/I, an = ℓθ2

and energy conservation.

35∗. A body of mass m moves on a circle R = 10m. At a certain time its instantaneous speed is 10m/s, and theinstantaneous rate of change of its speed is 10m/s2. At that instant find the angle between its velocity vector andacceleration vector.

36. A body of mass m moves on a circle R. At a certain time its instantaneous speed is ωR, and the instantaneousrate of change of its speed is αR. At that instant show that the angle between its velocity vector and acceleration

vector is tan−1(ω2

α

).

37. When θ = 45o in problem 8.34, find the angle between the total acceleration vector and the instantaneousvelocity vector of the falling pendulum.

38. Consider a heavy rotating flywheel of mass m = 100 kg and radius R = 0.25m spinning at ω0 = 100 rad/s. Abrake applies a force of friction of 100N to its rim tangentially. Through what total angle θ does it rotate beforestopping?

x

y

m, L

39. Wind a string of length 5m around the rim of a wheelof mass m = 1.0 kg and radius R = 0.25m with its axleheld fixed. Pull the string off tangentially with a constantforce of 10N . By time the string has rolled off completely,how fast is the wheel rotating?

40∗. A equilateral triangular frame is made with threesticks each of mass m = 1.0 kg and length L = 1.0m.One corner is at the origin. Find the cent er of massposition (xcom, ycom) and compute the moment of inertialfor rotation about the origin.

(Fh1, Fv1)

(Fh2, Fv2)ω 41. A disk of mass m and radius R rotates at speedω while in contact with both walls in a corner, withcoefficients of friction µk. Find the torque on the disk.


42. A massless level table rests on three equally spacedlegs placed on its perimeter, forming an equilateraltriangle of side a = 0.5m. A ball rests on the table. Theforces exerted upwards by the legs are 10, 20 and 30N .What is the mass of the ball, and where is it on the tablerelative to the leftmost leg (the 10N leg)?

Chapter 9

Angular Momentum

Conservation of angular momentum is one of the most important of all conservation laws in nature. In any systemwith no torques exerted by outside agents we find that

×m

v=(v,0,0)

r=(−x,y,0)

L=mr × v=(0,0, −mvy)

x

y

d

dtL =

∑i

τi,ext ni = 0 (9.1)

and soL = L(t) = L(0) (9.2)

and so both magnitude and direction of theangular momentum are conserved. In theabsence of externally applied torquesabout the point, angular momentum isalways conserved with respect to thepoint. Angular momentum is computed withrespect to a reference point regarded as apoint about which the system is in instanta-neous rotation. For example in the figure below

the particle is instantaneous rotation about the origin. The magnitude of the angular momentum is the area of theparallelepiped whose sides are the velocity and position vector. This turns out to be v times the component of r thatis perpendicular to v, in other words y.

165

166 CHAPTER 9. ANGULAR MOMENTUM

x

y

v0

ω n × r’

rcom

r’

This is called the lever arm of the momen-tum about the origin. The angular momentumvector becomes

L = mr× v = m(−x i+ y j)× (v i) = −myv k(9.3)

and points into the paper by the right handrule.Compute the angular momentum of a rollingdisk about the origin; the fragment of massdm at body-relative coordinate r′ has positionvector r+ r′ and total velocity

v = v0 + ωn× r′ (9.4)

and angular momentum with respect to the origin

dL =(r+ r′

)× dm

(v0 + ωn× r′

)(9.5)

and the entire body has angular momentum

L =

∫dm(r× v0 + ωr× (n× r′) + r′ × v0 + ωr′ × (n× r′)

)(9.6)

Using ∫dm r′ = 0 (9.7)

for the position vectors in the body (relative to center of mass) and

r′ × (n× r′) = |r′|2 n, we get L = m r× v0 + Icomωn (9.8)

which is the angular momentum of the center of mass motion plus the angular momentum about the center of mass.

×

P r

ω

v

Example 1. A tin can of moment of inertiaIcom about its center of mass axis and massm, height ℓ sits on a post. It is dealt a blownear its top, delivering momentum P to it.Compute the speed of its center of mass androtation rate after being struck.

The blow carried momentum P = P i and angular momentum

L = r×P = (− ℓ

2i+

ℓ

2j)× (P i) = −Pℓ

2k (9.9)

about the center of mass of the can. Both of these are delivered to the can, which results in center of mass motion v;

P i = mv = mv i (9.10)

and spin about the center of mass at rate and direction

L = −Pℓ2k = Icomωn = Icomω (−k) (9.11)

167

Example 2. A stick of length ℓ and moment of inertia Icom about its center of mass is nailed to the wall with asingle nail through its center, allowing it to pivot and freely rotate about its center. A ball of mass m, speed v0collides with it and sticks to the end. Compute the rotation rate of the whole mess after the collision.The ball has angular momentum

L = Ln = mr× v = m(−xi+ ℓ

2j)× (v0i) = mv0

ℓ

2(−k) (9.12)

x

y

x

ym

v0

with respect to the pivot point (where we put the origin) before the collision. Since the collision isinelastic, only momenta, angular momenta, can be conserved, and so the angular momentum before the collisionequals that after the collision. Linear momentum is not conserved; the nail holding the stick in place exertsforces on the system constraining the center of the stick to remain fixed. The presence of external forces foils theconservation of momentum.Angular momentum contains contributions from the stick and the ball stuck to its end

mv0ℓ

2(−k) =

(Icom +m(

ℓ

2)2)ω n =

(Icom +m(

ℓ

2)2)ω (−k) (9.13)

which we can solve for ω. We find that the axis of rotation is n = −k, so rotation is clockwise, and

ω =mv0

ℓ2

Icom +m( ℓ2 )2

(9.14)

Example 3. A slightly more challenging example is the following. Suppose that the collision with the stick is elastic,but still the stick must rotate about its stationary center of mass.

x

y

x

ym

v0 v1


Conservation of energy says1

2mv20 =

1

2mv21 +

1

2Icomω

2 (9.15)

let angular momenta into the paper be positive, then conservation of angular momentum about the pivot point says

m(−xi+ ℓ

2j)× (v0i) = m(−x′i+ ℓ

2j)× (−v1i) + Icomω(−k) (9.16)

or

mv0ℓ

2= Icomω −mv1

ℓ

2(9.17)

which gives us

ω =4v0

ℓ+ 4Icommℓ

, v1 =Icom −m ℓ2

4

Icom +m ℓ2

4

v0 (9.18)

RP

h

Example 4. A billiard ball is to be struckwith a cue stick such that it rolls withoutslipping instantly after the blow is struck.How high above the table must the cue strikethe ball?

Let the stick impart all of its momentumP = P i to the ball. If the ball rolls withoutslipping then no forces act on it after thecollision (except for gravity and the normalforce,

the presence of static friction is not required, and so the solution is the same whether it is present or not providedthe table is horizontal!) and so momentum is conserved.Let the cue stick deliver impulse (change in momentum) P to the ball, resulting in center of mass motion at speed v

P = P i =Mv =Mvi, v =P

M(9.19)

The impulse delivered had angular momentum-magnitude P (h−R) about the center of mass of the ball before thecollision. Afterwards the ball rolls with angular speed ω and so has angular momentum Iω about its center. Therolling condition states v = Rω and so using Icom = 2

5MR2 for the ball we find

L = r×P = (−xi+ (h−R)j)× (P i) =2

5MR2ω n

= Mv(h−R) (−k) =2

5MR2ω n (9.20)

we see that n = −k;

L =2

5MR2 v

R(−k), eliminating R we find that H =

7

5R (9.21)

is the required height above the table of the blow to be delivered by the cue, and this does not depend on how hardthe ball is struck, or whether or not friction is present.Why? Lets model the force applied by the cue as some F(t) = F (t) i. The actual time δt for which the cue is incontact with the ball is infinitismal, nevertheless in that amount of time the cue delivers a finite impulse (momentumchange) into the ball. ∫ δt

0

F (t) dt = P (9.22)

Write the equations of motion for the ball including friction if there is any

F (t)i+ Ff (t)(−i) =Ma(t) i, (h−R)j× F (t)i+ (−Rj)× (−Ff (t)i) = Iα(t)(−k) (9.23)

169

solve;

F (t) =Ma(t) + Ff (t), F (t) =2MR2

5(h−R)α(t) +

R

h−RFf (t) (9.24)

integrate over time ∫ δt

0

F (t) dt = P = M

∫ δt

0

a(t) dt+

∫ δt

0

Ff (t) dt∫ δt

0

F (t) dt = P =2MR2

5(h−R)

∫ δt

0

α(t) dt+R

h−R

∫ δt

0

Ff (t) dt (9.25)

We obtain

P =M(vf − 0) +

∫ δt

0

Ff (t) dt, P =2MR2

5(h−R)(ωf − 0) +

R

h−R

∫ δt

0

Ff (t) dt, set ωf =vfR

(9.26)

but friction has a maximal magnitude µsN , therefore∫ δt

0

Ff (t) dt ≤∫ δt

0

µsN dt = µsNδt→ 0 (9.27)

and so provided the impulse is delivered instantaneously and the ball does not slip momentum is con-served with friction present or not, and we can solve for h as we did above! The presence of the word “instantly”in the wording of the problem implies that the impulse is so delivered. Compare this problem with example 78, inwhich friction does work and momentum is not conserved.

Example 5. A stick of length ℓ, mass M , moment of inertia I floats freely in space. It is struck elastically by aball of mass m, moving perpendicular to the stick at speed v0. Compute the resulting spin rate and velocity of thestick after the collision.

x

y

x

ym

v2

v0 v1

Labeling velocities before and after as in the figure, with momenta to the right positive, and angular momenta intothe paper positive, the momentum, energy and angular momentum about the center of mass if the stick are allconserved

mv0 i = mv1 (−i) +Mv2 (i), mv0ℓ

2(−k) = mv1

ℓ

2(k) + Icomω (−k) (9.28)

and1

2mv20 =

1

2mv21 +

1

2Mv22 +

1

2Icomω

2 (9.29)

since the stick has two types of kinetic energy after the collision, center of mass rectilinear and angular. We solvethese three equations for v1, v2, and ω.


x

y

R

m2

xcom

ycom

×b

a

v1

v0

Example 6. A mass m2 on a disk of mass m1 moves in such a way that it travels in a circle at speed v0 around thecenter of mass, which we know is a fixed-point. The body m2 is a distance R from the center of the disk.The center of mass is a distance a from the disk-center, with

a+ b = R, m1 a = m2 b (9.30)

or

a =m2

m1 +m2R, b =

m1

m1 +m2R (9.31)

The disk behaves like a point-mass m1 and also circles the center of mass at speed v1 such that the center of massis fixed. This requires that the two masses move “equal angles in equal times” on their respective circles, bothcompleting an “orbit” in the same amount of time

T =2πa

v1=

2πb

v0, v1 =

a

bv0 =

m2

m1v0 (9.32)

Computation of angular momentum with respect to the fixed origin of the C.O.M. system leads to

m2bv0k+m1av1k+ Icomωn = 0 (9.33)

requiring that the disk rotate about its center clockwise at rate

ω =m2bv0 +m1av1

Icom=m2(a+ b)v0

Icom=m2Rv0Icom

(9.34)

Example 7. Consider two disks on parallel axes. One has an initial spin rate ω0, the other is stationary. They arebrought into contact and friction between the two causes them to slip until eventually they spin at such a rate thatno mutual slipping occurs, they roll on one another. Compute this final spin rate.

171

R1

R2

Ff

Ff

× ×

ω0 ω1

ω2

This problem is very deceptive. If we try to use angular momentum conservation, we must pick a single point aboutwhich to compute angular momenta for each disk. Instead we apply τ = Iα. For the large disk we begin with

ω1(t)n = −ω1(t)k, ω1(t = 0)n = −ω0 k (9.35)

τn = (−R1j)× (Ff i) = R1Ffk = I1,comα1 n = I1,com

( ddtω1

)n (9.36)

integrating, we find

− ω1(t)k = −ω0k+FfR1t

I1,comk, ω1(t) = ω0 −

FfR1

I1,comt (9.37)

and

ω2(t)n = ω2(t)k, ω2(t = 0)n = 0 (9.38)

τn = (R2j)× (−Ff i) = FfR2k = I2α2 k = I2,com

( ddtω2,com(t)

)k (9.39)

integrate

ω2(t)k = 0 +FfR2

I2,comtk, ω2(t) =

FfR2

I2,comt (9.40)

The rolling condition is

ω1,fR1 = ω2,fR2 (9.41)

for the final non-slipping motions, eliminating Ff and t we obtain

ω1 =ω0

1 +R2

1I2R2

2I1

(9.42)

This problem illustrates the basic principles behind the automobile clutch and power transmission systems. It is(slightly) complicated by not knowing the value of Ff , since the normal force has not been specified. You can seethat it is not necessary to know N or Ff if we resort to basic principles.


9.1 Problems

x

y

x

ym

v0 v1

1∗. A ball of mass m = 0.5 kg traveling at v0 = 2.0 ms

collides elastically with a stick of mass M = 1.0 kgof length ℓ = 1.0m, which is free to rotate and movehorizontally and vertically. How far d above the stickcenter must the ball hit the stick in order to be stoppeddead by the collision? Find the final rotation rate ω ofthe stick and the speed v of its center of mass.A classic exam problem.

m v0

x

ω n2. A stick of mass m1 and length ℓhangs from a perfect hinge at oneend. It is struck elastically by aball of mass m moving horizontallyat v0 = v0i. Determine how far xbelow the hinge that the ball muststrike in order to be stopped deadby the collision. Find the rotationrate ω of the stick instantly afterthe collision.

ω1 n

ω2 n

3. Two sticks of mass m1, m2,length ℓ hang from a commonperfect hinge at one end. One(m1) is displaced to horizontal,and released. It collides elasticallywith the second stick. Computeboth rotation rates ω1, ω2 of thetwo sticks instantly after thecollision.

×

v1 v2

ω1 ω2

4. Two sticks of mass m1, m2, length ℓ float freely inempty space (no gravity). A powerful spring of forceconstant k is compressed by a very small∗ amountx, placed between the top ends, and released. Thesticks are flung apart. Find the velocities v1, v2 ofthe centers of mass and both rotation rates ω1, ω2 ofthe two sticks.∗This lets you assume F is ⊥ to the stick while thespring is acting on it.

m v0

x

ω n

θ 5. A rod of mass m and length 2ℓ hangs belowa fixed end. It is struck inelastically at a pointx below the pivot point by a ball of mass mtraveling at v0. Compute the final angle θ towhich the rod swings (under the influence ofgravity).

9.1. PROBLEMS 173

x

y

x

ym

v0

v1

ω

6. A rod of mass M and length 2ℓ floats freelyin space. It is struck inelastically at its endby a ball of mass m traveling at v0. After thecollision the combined object rotates about itscenter of mass at ω, which moves with velocityv1. Find ω and v1.

×v0

h

7. A disk of radius R = 0.25m, massm = 10.0 kg rolls towards a step of heighth < R at speed v0. Determine the maximalheight h step that the disk can “hop” bysnagging on the corner and pivoting (withoutslipping) onto the top of the step.Hint; angular momentum with respect tothe corner of the step is conserved, computeit instantly before and instantly after thecollision.

v0

vf

ω0

ωf

8∗. Two discs (I = 12mR

2) of radii R = 0.1m and massm = 1.0kg, with one initially moving at v0 = 2.0ms andthe other spinning at ω0 = 20 rad/s are lined with Velcroand are on a collision course. They stick together at apoint on the rims. Find the final spin rate ωf about thenew center of mass, and the speed vf of the center of massafter the collision.

This occurs in space, there are no forces present.


v0

vf

ω0

ωf

9∗. Two discs (I = 12mR

2) of radii R = 0.1m and massm = 1.0kg, with one initially moving at v0 = 2.0ms andthe other spinning at ω0 = 10 rad/s are lined with Velcroand are on a collision course. They stick together at apoint on the rims. Find the final spin rate ωf about thenew center of mass, and the speed vf of the center of massafter the collision.

This occurs in space, there are no forces present. Fullcredit for numerical answer.

y

x

10∗. A person of mass m = 50 kg standsat the rim of a freely rotating disk ofmass M = 200 kg, radius R = 5m,spinning at ω0 = 1 rad

s j. She walksinwards to the center of the disk. Findthe final spin rate when she arrives atthe center.

11. In the preceding problem, there were no torques applied to the system (why?) yet there was angular accelerationof the disk. From

d

dtL =

d

dt

(Iω)

show that if our person riding the disk walks towards the center at rate v, then at the instant that she leaves therim the angular acceleration of the disk is

α =( 2mRv

12MR2 +mR2

)ω0

x

y

x

ym

v0

v1

ω

12. Two balls of mass m each connected bya massless string of length 2ℓ floats freely inspace. It is struck inelastically at its end bya ball of mass 3m traveling at v0. After thecollision the combined object rotates about itscenter of mass at ω, which moves with velocityv1. Find ω and v1.

9.1. PROBLEMS 175

R

x

y

z

Mg

13. A disk of moment of inertia I forms agyroscope with a ball-joint (red) a distanceR from the disk. The disk rotates at angularspeed ω (fast) so that its angular momentumvector at t = 0 is L = Iω j. The torque ofgravity acting on the disk causes the angularmomentum vector to rotate at rate Ω, soits tip travels in the dotted circle. Using

L(t) = Iω(sin(Ωt) j + cos(Ωt) j

), compute Ω

in terms of M, g, I, ω.

14. Suppose that the gyroscope is nearly vertical, with its axis tipped at angle θ with respect to perfect verticality.Let

L =1

2Ma2ω

(sin θ sin(Ωt) j+ sin θ cos(Ωt) j+ cos θ k

)+(14Ma2 +MR2

)ωk

and find the precession rate Ω caused by the torque of gravity. What angle maximizes the precession rate?

15. A mass M is made totravel in a circle of radiusr0 at speed v0 on a friction-less horizontal plane witha string under tension Tthrough a hole as illustrated.If the string is pulled downuntil the radius of the pathis reduced to r0/2, computethe work done on the system.

Chapter 10

Fluids1

Different parts of a fluid can move with different velocities, there can be currents in the fluid, and so let v(x(t), t)be the velocity of the fluid element at x. The flux of a fluid through a surface of area dA and normal n is

dF = v · n dA (10.1)

The main complications with fluids come from the fact that a fluid could be compressible, and flow at different ratesin different places, so this means that v = v(t, r). The velocity will then have two parts to its rate of change, sincethe position r of a block of fluid also changes r = r(t). Compute the acceleration of a block of fluid

d

dtv(t, x(t), y(t), z(t)) =

∂v

∂t+∂v

∂x

dx

dt+∂v

∂y

dy

dt+∂v

∂z

dz

dt

=∂v

∂t+dx

dt·(∇v)=∂v

∂t+ v ·

(∇v)

(10.2)

This is the convective derivative. The last idea that we need is pressure, force exerted normally on a surfaceper unit area. The gadget ∇v is altogether new, it is an example of a tensor. No holds are barred in what follows,since much of this material is needed for the physics GRE (for physics students), but the consolation prize is thatonly Archimedes principle and Bernoulli’s law are needed by everyone else.

10.1 Velocity fields of fluids

A fluid can be compressible, flowing out of a “source” suchas

v =(x i+ y j+ zz

)/s (10.3)

Flow lines (lines tangent to v) are straight lines throughthe origin

dy

dx=

vyvx

=y

xdy

y=dx

xln y = lnx+ c

y = C ′ x, c′ = ec (10.4)

This type of field corresponds to fluid enteringspace at a regular rate, measure the flux or flow-ratethrough a sphere of radius R around the origin, which isthe point where the fluid is entering space

1This chapter is skipped in 201, it is included for MCAT and GRE preparation, which requires only section 10.2.1. Problems 1-10are covered by the GRE and MCAT. This treatment requires calculus 222 and 223.

177

178 CHAPTER 10. FLUIDS

Note v is perpendicular or normal to any such sphere of radius R, so on such a sphere S v = R/sn

∮S

v · n dS = 4πR3/s (10.5)

Now multiply by the density ρ, and you get the amount of mass per second of fluid entering space at the origin;

ρ

∮S

v · n dS = 4πρR3/s =dm

dt(10.6)

Lets calculate the “source strength” of this field

∇ ·(ρv)=∂ρvx∂x

+∂ρvy∂y

+∂ρvz∂z

= ρ(1 + 1 + 1)/s = 3ρ/s (10.7)

Note that this is exactly the correct mass per second per volume entering our sphere; multiply it by thesphere-volume and check for yourself, you get the result of Eq. 10.6. In other words

±∇ ·(ρv)=dρ

dt(10.8)

with a plus sign if fluid enters, a minus if it leaves (source versus sink).

On the other hand a fluid field might contain vortices or whirlpools. This property is measured by ∇ × v, the“swirliness” of a field, such fields have closed field-lines.

So what is ∇v for this particular flow/velocity function? This quantity is a tensor, which is a generalization of thevector concept. Where a vector v is a 3× 1 array of components (vx, vy, vz) a tensor such as ∇v is a 3× 3 array ofcomponents

∇v =

∂vx∂x

∂vx∂y

∂vx∂z

∂vy∂x

∂vy∂y

∂vy∂z

∂vz∂x

∂vz∂y

∂vz∂z

(10.9)

which for our flow v = (αx, αy, αz) with α = 1s looks like

∇v =

α 0 00 α 00 0 α

(10.10)

What would the acceleration of a block of fluid be? Eq. 10.2 says

axayaz

=

αdxdtαdydtαdzdt

+

α 0 00 α 00 0 α

·

αxαyαz

=

αdxdt + α2x

αdydt + α2yαdzdt + α2z

(10.11)

using the normal rules of matrix multiplication (from linear algebra).

10.2. NEWTON’S LAWS APPLIED TO FLUIDS 179

Example 1.

v = (y i− x j)/s (10.12)

is a vortex field. Field lines are circles

dy

dx=

vyvx

= −xy

y dy = −x dy x2 + y2 = 2c(10.13)

Lets measure its “swirliness” (the vortexstrength)

ω = ∇× v = −2/sk (10.14)

The −k says it swirls clockwise (the −) aroundthe z-axis.

10.2 Newton’s laws applied to fluids

x x+dx

P(x) P(x+dx)

Fluid elements accelerate because there are dif-ferent pressures on the different walls of the el-ement (pressure gradients), consider a volumeof fluid in a pipe of cross-section A between xand x + dx. Its momentum changes becauseforces (pressures times areas) are exerted on it

d

dt

(ρAdxv

)= A

(P (x)−P (x+dx)

)i (10.15)

so the fluid equation of motion is

ρd

dtv = −dP

dxi = −∇P (10.16)

Re-write this using our convective derivative;

ρ(∂v∂t

+ v ·(∇v))

= −∇P, Euler’s equation (10.17)

In addition we have the mass transport equation Eq. 10.6; fluid can leave a volume V0 by flowing out in a current.Let m be the fluid mass within V0, a volume bounded by surface S or Eq. 10.8

∂ρ

∂t= −∇ · (ρv), mass conservation (10.18)

Momentum flow and stress. The equation for the ith component of the fluid equation of motion is

ρ(∂vi∂t

+∑j

(vj

∂

∂xj

)vi

)= − ∂P

∂xi, x1 = x, x2 = y, x3 = z (10.19)

take the mass conservation equation

∂ρ

∂t= −

∑j

∂

∂xj(ρ vj) = −

∑j

∂ρ

∂xjvj −

∑j

ρ∂vj∂xj

(10.20)

multiply by vi

vi∂ρ

∂t= −

∑j

∂ρ

∂xjvi vj −

∑j

ρ vi∂vj∂xj

(10.21)


and add∂

∂t

(ρ vi

)= − ∂P

∂xi−∑j

∂

∂xj

(ρ vi vj

)= −

∑j

∂

∂xj

(P δij + ρ vi vj

)= −

∑j

∂

∂xjTij (10.22)

We have invented the stress-momentum tensor T

Tij = P δij + ρvi vj ,∂

∂t

(ρv)= −∇ ·T, with δij =

1 i = j0 i = j

(10.23)

which in explicit component form looks like

T =

Txx Tyx TzxTxy Tyy TzyTxz Tyz Tzz

=

P + ρv2x ρvxvy ρvxvzρvxvy P + ρv2y ρvyvzρvxvz ρvyvz P + ρv2z

(10.24)

The left side of this equation is the rate of change of momentum density (momentum per unit volume), the rightside is the divergence of the stress momentum, so integrate over volume and apply the divergence theorem2

∂

∂t

∫V0

ρv d3x = −∮S

T · n dA (10.25)

the objectdp

dt= −T · n dA, dpi

dt= Tij nj dA (10.26)

is the flow of momentum through the area dA (force exerted on the area) whose normal is n (in other words T isthe momentum flux density).

For an incompressible fluid with flow in the u|u| = n direction.

This breaks up forces into various types that are separately important in many applications,

j, Tyy

Tyx

Tyz i, Txx

Txz

Txy

Txx (Normal) force in the x-direction on a surface with normal i

Txy (Shear) force in the y-direction on a surface with normal i

Txz (Shear) force in the z-direction on a surface with normal i

The second of Eq. 10.23 applies to all sorts of continuous media, not just fluids, but the explicit form of Tij =Pδij+ρvivj is for fluids only. Elastic media will typically have the components of the tensor as variables to be solvedfor given specified deformations of the material, or the components of T will be known applied forces and one willsolve for the material deformations.

10.2.1 Incompressible, irrotational fluids

Incompressible, irrotational fluids (constant ρ) have ∇ · v = 0, ∇ × v = ω = 0, and so the velocity field v is thegradient of some “velocity potential” Φ

v = ∇Φ (10.27)

2see the appendix of this chapter

10.2. NEWTON’S LAWS APPLIED TO FLUIDS 181

This means that the fluid flowing in a pipe would flow without any vortices (“swirliness”) and the fluid speeds upthrough a constriction in the pipe so that the flux stays constant

Av · n = A′ v′ · n (10.28)

Put this into the fluid equation of motion

∂v

∂t+(∇v)· v =

∂v

∂t+

1

2∇(v · v)− v × (∇× v) = −1

ρ∇P (10.29)

∂v

∂t+

1

2∇(v · v)− v × (∇× v) =

∂

∂t(∇Φ) +

1

2∇(v · v)− v × (∇× v) = −1

ρ∇P (10.30)

both sides are gradients

∇(∂Φ∂t

+1

2v2 +

P

ρ

)= 0,

(∂Φ∂t

+1

2v2 +

P

ρ

)= f(t) (10.31)

(a function of time only, not x), and so if the flow is steady (constant in time) we obtain Bernoulli’s law

1

2ρ v2 + P = constant (10.32)

to which we could add external potential energy densities dϕdV (such as gravitational dϕ

dV = dmghdV = ρgh)

1

2ρ v2 +

dϕ

dV+ P = constant (10.33)

Example 2. Find the speed with which water is ejected from atiny hole a distance h below the waterline of a very large tub ofwater open to the atmosphere.Bernoulli’s law says that the left side of Eq. 10.33 is constantthroughout the fluid; apply it at a molecule on the surface (“A”)(open to the atmosphere) and at the end of the water jet (“B”);the waterline barely moves

1

2ρ 02 + ρ gh+ Patm =

1

2ρ v2 + 0 + Patm

v =√

2gh (10.34)

A

BC

Example 3. What is the pressure a distance h below the surface of a body of water? This is the hydrostaticpressure. Use the previous example; plug the hole and apply Bernoulli to points “A” and “C”;

1

2ρ 02 + ρ gh+ Patm =

1

2ρ 02 + 0 + P

P = Patm + ρ gh (10.35)

Archimedes principle; the pressure at depth h depends only on h at the pressure on the surface.

Example 4. Find the stress-momentum tensor of a still, pres-surized fluid, and the pressure exerted by the fluid on the foursides of a submerged still cube of side area A.From Eq. 10.22

Tij = Pδij , T =

P 0 00 P 00 0 P

Fright = −T · iA =

−PA00

(10.36)


Ftop = −T · jA =

0−PA0

Fleft = −T · (−i)A =

PA00

(10.37)

and so you see that all of these forces are directed inwards on the cube walls. The cubic “bubble” would collapse ifthese forces were not countered by internal pressure. A fluid without viscosity cannot exert shear forces.

Example 5. Find the stress-momentum tensor of a moving (v = vi), unpressurized fluid, such as a tenuous molec-ular cloud in space, and the pressure exerted by the fluid on the four sides of a submerged still cube of side area A.

Tij = ρ vi vj , T =

ρv2 0 00 0 00 0 0

, Fleft = −T · (−i)A =

ρv2A00

, Ftop = 0 (10.38)

Notice that there are no shear forces exerted by the fluid (again).

A

dx

Example 6. This is a common GRE problem; Ablock of fluid of width dx, area A carries momentumdp = dmv = (ρAdx) v. Of it is stopped bythe wall placed in its way (and made to turn asillustrated), within time dt the momentum impartedto the wall per unit time (force on the wall) isdpdt = (ρA dx

dt ) v = ρ v2A.

A

B

h1 h2

C

D

Example 7. A Pitot tube is asimple analog device for measuringair-speed of an aircraft. Air (fluid#1) at point “A” at Patm rushesthrough the pipe at vair, but is stillat point “B”, so for the air

1

2ρairv

2air + Patm = PB

For the mercury column (fluid #2),the pressure on top at “D” is Pcabinand comparing two formulas for thehydrostatic pressure at “C”

PC = PB + ρmg h1

= Pcabin + ρmg h2

we simplify things by lettingPcabin = Patm

and eliminate PC and PB to get

vair =

√ρmρag(h2 − h1) (10.39)

10.3. ADVANCED TOPICS 183

10.3 Advanced topics3

Lets take Euler’s equation and separate off the pressure term

ρdv

dt= −∇P +∇ ·T′ (10.40)

from T, calling what is left over T′. What follows is a very typical model-building approach for describing fluidflow in which we could have different layers moving at different velocities because of inter-layer friction (viscosity).This is probably the only example of the scientific method performed at the very highest levels that you will find inthese (freshman physics) notes.

Lets assume that T′ depends on the fluid velocity and other scalar properties possessed by the matter that the fluidis composed of, and make a constitutive assumption that every component of T′ is linear in components of v.This we hope will result in simple equations.

We begin by looking for candidates for second-rank tensors (a vector is a first-rank tensor, components have a singleindex label) that involve the components of v. We can come up with three

T ′ij = ρ

(bi vj + cj vi

)T ′ij = ρα vi vj

T ′ij = ρµ

∂vi∂xj

(10.41)

in which α, µ are scalars and b and c are vectors that may depend on x, y, z.

We reject the second candidate; it is quadratic, too complicated! We reject the first candidate; we are hoping todescribe flow that has different velocities for different layers, so there should be velocity gradients, spatial deriva-tives of v involved. This leaves the third choice as the most suitable.

Suppose that we wish our fluid to be irrotational, possess no vortices. Then ∇× v = 0, or(∂vz∂y

− ∂vy∂z

,∂vz∂x

− ∂vx∂z

,∂vy∂x

− ∂vx∂y

)= (0, 0, 0) (10.42)

Lets write T′ as

T′ =

ρµ∂vx∂x ρµ∂vx∂y ρµ∂vx∂zρµ

∂vy∂x ρµ

∂vy∂y ρµ

∂vy∂z

ρµ∂vz∂x ρµ∂vz∂y ρµ∂vz∂z

=1

2

ρµ(∂vx∂x

)ρµ(∂vx∂y +

∂vy∂x

)ρµ(∂vx∂z + ∂vz

∂x

)ρµ(∂vx∂y +

∂vy∂x

)ρµ

∂vy∂y ρµ

(∂vy∂z + ∂vz

∂y

)ρµ(∂vx∂z + ∂vz

∂x

)ρµ(∂vy∂z + ∂vz

∂y

)ρµ∂vz∂z

+1

2

0 ρµ

(∂vx∂y − ∂vy

∂x

)ρµ(∂vx∂z − ∂vz

∂x

)ρµ(− ∂vx

∂y +∂vy∂x

)0 ρµ

(∂vy∂z − ∂vz

∂y

)ρµ(− ∂vx

∂z + ∂vz∂x

)ρµ(− ∂vy

∂z + ∂vz∂y

)0

(10.43)

Notice that for an irrotational fluid the last part is zero. We have refined our constitutive description of anirrotational, viscous fluid to

Tij = 2ρµDij , Dij =1

2

( ∂vi∂xj

+∂vj∂xi

), ρ

dv

dt= −∇P +∇ ·T′ (10.44)

3This is for the benefit of physics and hydrology students. It is not part of the 201 curriculum, it was part of our 401 course.


which is called the linearized Navier-Stokes equation. Do you need some spare cash? The Clay Institute isoffering a $1, 000, 000.00 for a general solution to the fully-developed equations containing the convective term v ·∇v.

Example 8. Couette flow. Consider such a viscous, irro-tational and incompressible fluid trapped between two planar,parallel surface, to top surface moving at speed u to the right.Find the velocity of the fluid between the plates.

x

h

The fluid layer touching the bottom plate must be stationary, the fluid touching the top plate moves with it becauseof the viscosity (friction) with the plate

v(y = 0) = 0, v(h) = ui, v = v(y) i (10.45)

Assume that the flow is steady; dvdt = 0, and that the pressure in the fluid is uniform P0; for example the fluid

layer height h is so small that height variations can be neglected. This is an excellent model of how an automatictransmission in a car works. Then

T′ =

0 ρµdvdy 0

ρµdvdy 0 0

0 0 0

∇ ·T′ = (

∂

∂x,∂

∂y,∂

∂z) ·

0 ρµdvdy 0

ρµdvdy 0 0

0 0 0

=

0

ρµd2vdy2

0

(10.46)

and the Navier-Stokes equation gives us (∇P0 = 0)

0 = ρµd2v

dy2, v = a+ by, v =

u

hy (10.47)

using our boundary conditions. For my next trick, I find the viscous drag on the moving plate.

Example 9. What force does this flow exert on the upper (moving) plate? It will exert a shear force (drag) on theplate. This example tells us very clearly what µ represents; viscosity is basically friction

T′ = −

0 ρµuh 0ρµuh 0 00 0 0

, F = −T′ · nA =

0 ρµuh 0ρµuh 0 00 0 0

·

010

A =

−ρµAuh

00

(10.48)


x

h

−h

Example 10. Poisseville or capillary flow. In this case wewill have a fluid trapped between two surfaces at y = ±h, butlet it be forced with a pressure gradient

∇P = −Ci, P = −Cx (10.49)

(higher pressure to the left) and again try for a solution

v = v(y) i, v(±h) = 0 (10.50)

We can re-use our T′ tensor from the previous example to get(for steady flow again)

− 1

ρ

∂P

∂x+ µ

d2v

dy2= 0,

d2v

dy2= − C

ρµ(10.51)

which has solution

v = − C

2ρµy2 +Ay +B (10.52)

which reduces to

v = − C

2ρµ

(y2 − h2

)(10.53)

after enforcing our boundary conditions. There must be quite a high pressure at the left end of the “pipe” to pushthe flow through at this rate. Lets say the “pipe” has length ℓ in the x-direction and square cross section; the flowrate is

dm

dt= −

∫ h

−hdz

∫ h

−hdy ρ v(y) =

4Ch4

3µ=CA2

12µ(10.54)

in which A = (2h)2 is the cross-sectional area of the “pipe”. Lets suppose that this flow rate is constant, since

C =Pleft−Pright

ℓ , we find that there can be a huge pressure drop in the pipe

Pleft = Pright +12µℓ

A2

dm

dt(10.55)

This has obvious biomedical ramifications; the pressure drop in an artery is proportional to its length, the bloodviscosity, is smaller for larger cross-section arteries, and increases with flow-rate.

The Fully Developed Navier-Stokes equations for steady flow are nonlinear equations (keeping the vorticity, wartsand all)

ρ∂v

∂t+ ρv ·∇v = ρv ·∇v = −∇P + µρ∇2v (10.56)

Let u, ℓ, p be typical order of magnitude velocity, length and pressure scales for the fluid, so that we can createdimensionless quantities w, ξ, π

v = uw, x = ℓξ, P = pπ (10.57)

Substituting all of this stuff into the F.D.N-S. equations we end up with dimensionless equations

w ·∇w = −( p

ρu2

)∇π +

( µuℓ

)∇2w (10.58)

The number

R =uℓ

µ(10.59)

is called the Reynolds number, it is a ratio of inertial to viscous forces. There can be layers in a fluid nearstationary surfaces where viscous forces dominate the fluid motion, even though elsewhere in the fluid viscosity my


play little or no role at all in the fluid motion.

x=L

δ(L)

For example if an otherwise constant, uninterrupted fluid hits astationary planar edge, like a wing, within a layer of thicknessδ ≈ ℓ√

Rviscous forces are too large to neglect, whereas beyond

this distance we can completely neglect them. This meansthat you can solve the F.D.N.-S. equations in the two regionsseparately, and blend the solutions along the boundary of thetwo regions.

Outside of this envelope, vx = u(x) and vy = 0, within the envelope

vx∂vx∂x

+ vy∂vx∂y

= −1

ρ

∂P

∂x+ µ

(∂2vx∂x2

+∂2vx∂y2

)vx∂vy∂x

+ vy∂vy∂y

= −1

ρ

∂P

∂y+ µ

(∂2vy∂x2

+∂2vy∂y2

)(10.60)

These can’t be solved exactly (welcome to my world), we can try a sequence of successive approximations, startingwith Bernoulli to get rid of the pressure, assuming that it is the same in both regions

P +1

2ρu2x = constant,

∂P

∂x= 0 (10.61)

How do we use the Reynolds number? We use it to simplify the equations by deciding which terms can be dropped;within the boundary layer

∂2vx∂x2

scales likeu

ℓ2(10.62)

∂2vx∂y2

scales likeu

δ2(10.63)

which means that the second term is the largest, so we discard the first term, and to lowest approximation, for largey we have a uniform flow, so ux = u0, we should have little or no pressure variations in the fluid as well, reducingthings down to

vx∂vx∂x

+ vy∂vx∂y

≈ µ∂2vx∂y2

∂vx∂x

+∂vy∂y

= 0 (incompressible flow) (10.64)

with vx(x, y = 0) = 0, vy(x, y = 0) = 0 and vx(x, y → ∞) = u0, which are the Prandtl flow equations; nonlinearbut somewhat simpler.

Incompressibility can be enforced by assuming that v can be gotten from a stream function ψ, which is much likea potential function

vx =∂ψ

∂y, vy = −∂ψ

∂x(10.65)

which gives us a third-order non-linear equation to solve(∂ψ∂y

)( ∂2ψ∂x∂y

)−(∂ψ∂x

)(∂2ψ∂y2

)= µ

(∂3ψ∂y3

)(10.66)

You sure do have to break a lot of eggs to do science (this is aeronautical and hydrodynamical engineering) at anylevel above the most superficial. This is why mathematics is your best educational investment.

The high-concept technique at this final stage is to look for scaling solutions; construct a new variable

ξ = y

√u0µx, ψ =

√µxu0 Ψ(ξ) (10.67)


to get the vastly simpler but still most heinous equation

Ψd2Ψ

dξ2+ 2

d3Ψ

dξ3= 0, Ψ(0) = 0,

dΨ

dξ

∣∣∣0= 0,

dΨ

dξ

∣∣∣∞

= 1 (10.68)

which can be solved numerically. This is a real physics problem, its solution is necessary for the design of supersonicwing and turbine-blade cross sections.

10.3.1 Incompressible flow around obstacles

Lets begin with the Navier-Stokes equation, but toss out the convective, non-linear term, and look at steady flow

0 = −∇P + µ∇2v (10.69)

We can simplify this tremendously if we can get rid of the pressure, and there is any easy way to do that;

0 = ∇×(−∇P + µ∇2v

)0 − µ∇2

(∇× v

)= µ∇2ω, ω = ∇× v (10.70)

For two-dimensional flow ωz =∂vy∂x − ∂vx

∂y , and if we do our familiar trick of using a stream function ψ

vx =∂ψ

∂y, vy = −∂ψ

∂x, ω = −∂

2ψ

∂x2− ∂2ψ

∂y2= −∇2ψ (10.71)

and our fluid velocity obeys the biharmonic equation

∇2∇2ψ = 0 (10.72)

which is fourth order, but a suffi-cient condition that ψ obeys thisis the much simpler harmonic orLaplace equation

∇2ψ = 0 (10.73)

(this is irrotational, if we wantconstant rotation ωz = c we lookfor solutions ∇2ψ = c).

As an example, lets consider flowaround a cylinder. Far from thecylinder the flow is undisturbed

limr→∞

v = −v0i

and so

limr→∞

ψ = −v0 y

= −v0 r sin θ


Because our obstacle is a cylinder, it might be a good idea to switch to cylindrical coordinates, in which our equationbecomes

∇2ψ =(1r

∂

∂r

(r∂ψ

∂r

)+

1

r2∂2ψ

∂θ2

)= 0 (10.74)

It is easy to verify that the following stream function is a solution, and it obeys Eq. 10.74

ψ = −v0(r +

α

r

)sin θ (10.75)

We can’t have fluid flow through the obstacle, it must go around it; this is our second boundary condition that weuse to get the unknown constant α

vr(r = a) = 0, vr = vx cos θ + vy sin θ (10.76)

Fromr =

√x2 + y2, θ = tan−1 y

x(10.77)

and∂r

∂x=x

r= cos θ,

∂r

∂y=y

r= sin θ,

∂θ

∂y=

x

x2 + y2=

cos θ

r,

∂θ

∂x= − sin θ

r(10.78)

together with the chain rule

vr = vx cos θ + vy sin θ

=∂ψ

∂ycos θ − ∂ψ

∂xsin θ

=(∂ψ∂r

∂r

∂y+∂ψ

∂θ

∂θ

∂y

)cos θ −

(∂ψ∂r

∂r

∂x+∂ψ

∂θ

∂θ

∂x

)sin θ (10.79)

you can show that

vr =1

r

∂ψ

∂θ(10.80)

and deduce that the stream function is

ψ = −v0(r − a2

r

)sin θ

= −v0(y − a2 y

x2 + y2

)(10.81)

and from it find the velocity components that I have illustrated on the previous page, and compute the drag on thecylinder or the pressure in the fluid around it. Don’t be surprised if the drag is zero, we threw out the baby whenwe went from the full biharmonic to the harmonic equation!

My hope, and one reason for including material in this book that we don’t cover in 201, is that it might pique yourinterest. If it does, come and see me, we have something to talk about. If it does not, well, we just don’t haveanything to talk about.

10.3.2 The divergence theorem

The divergence theorem lets us translate area integrals into volume integrals. For any vector field v, the flux of vthrough a closed surface S can be expressed as an integral of the source strength of v over the volume enclosed by S.

FS =

∮S

v · n dA =

∫V

∇ · v d3x (10.82)

It makes some sense when stated in English; the flux counts the fluid flow outward through S, and this is related tohow many sources of fluid there are within S.

10.4. APPENDIX 189

10.4 Appendix

REDUCE code for div and curl

procedure divergence(vect);

begin

retval:=df(part(vect,1),x)+df(part(vect,2),y)+df(part(vect,3),z);

return(retval);

end;

procedure curl(vect);

begin

retval1:=df(part(vect,3),y)-df(part(vect,2),z);

retval2:=-df(part(vect,3),x)+df(part(vect,1),z);

retval3:=df(part(vect,2),x)-df(part(vect,1),y);

return(retval1,retval2,retval3);

end;

% try it

Evect:=2*x-3*y,2*y+x,2*x*y;

divergence(Evect);

curl(Evect);


10.5 Problems

1. The density of ice is 0.92 gcm3 and of liquid water 1.0 g

cm3 . What height of a floating cubic block of ice of sidea = 10.0m will be above the surface of the water?

2. A balloon payload of total mass M = 200 kg is attached to aspherical balloon to be filled with helium. The density of helium is0.1786 kg

m3 , and of air 1.2 kgm3 . What radius R balloon will be needed to“float” the payload (no accelerations, just vertical static equilibrium)?

3. Find the total force on the wall of a dam of wall areahℓ holding back water of density ρ and height h. Hint,divide the wall into strips of height dy, depth y below thesurface, area ℓ dy and compute the force on them.

4. An incompressible fluid flows at speed v through a pipe of cross-sectional area A. Find the flow speed at a constric-tion in the pipe where the cross-sectional area drops to A/4. Suppose that both regions of the pipe are at the sameheight, and the pressure at the wide (cross-sectional area A) portion is P . Find the pressure in the constricted portion.

5. Hydraulics is a nice application of static fluidmechanics. A vessel of fluid of density ρ connectstwo pistons of areas A1 and A2 sealed at the top withmovable piston heads. Show that if forces F1 and F2,presumably one of these being the weight of a heavyobject that you wish to elevate, are applied to thepiston-heads, the condition for static equilibrium is

F1

A1+ ρg h1 =

F2

A2+ ρg h2

so that a small F1 can be used to support or raise alarge F2. This is how the lifts in an auto repair shopwork.

F1

A1

h1

F2

A2

h2

10.5. PROBLEMS 191

h

v0, R0

vf, Rf

6. Water leaks from a faucet of radius R0 = 1.0 cm at the rate of1.0 gs . Find the speed v0 with which it exits the faucet. As waterfalls, the stream narrows as illustrated. By the time it has descendedh = 5.0 cm, find the radius of the stream Rf and the velocity vfof the fluid. At some point the fluid stream will break up into droplets.

h1h2

7∗. Here is a crude “density-meter” for immiscible liquids. It isinitially filled with a fluid of density ρ1 (light) presumably known. Adarker fluid of density ρ2 is poured into one arm, and the fluids areallowed to settle down. the heights h1, h2 are measured. Supposeρ1 = 1.0 g

cm3 , h1 = 5 cm and h2 = 8 cm. Find ρ2.

h

P1P0

8. The manometer can be used to measure pressures. Typically oneend (say the right) is sealed at the top after a fluid of density ρ isput into the tube. This makes P0 = 0. The pressure P1. is gotten bymeasuring the height of the column h. If P1 = 1 atm = 101, 000 N

m ,what would h be if the fluid were water? What would h be if thefluid were mercury? You will need to look up the density of mercury.I suggest google.

9∗. Show that on a windy day the force pushing out on the windows is

|F| = 1

2ρairv

2windA


P, v

Patm

10. Show that on a windy day you must exert a force

|F| = 1

4ρairv

2air A

on a doorknob at the edge of the door of area A inorder to keep it closed.Hint; compute the torque on the door.

11. Bernoulli’s law is responsible for aerodynamiclift. A wing cross section is designed to cause air toflow faster over the top surface than it does over thebottom surface.

When the airplane has airspeed v, suppose that the speed of air-flow over the top surface is v′ = α v where α isslightly larger than one. If the surface area of the wing is A for both top and bottom surface (approximately), showthat the upward-directed force on the wing (the lift) is

|F | = 1

2ρair(α

2 − 1) v2A

12. Show that a stream function

vx =∂ψ

∂y, vy = −∂ψ

∂x

will ensure that the fluid is incompressible ∇ · v = 0.

13. Show that a for Prandtl flow

vx =∂ψ

∂y= u0

∂Ψ

∂ξ

Use ode to generate a solution to the Prandtl flow equation Eq. 10.66

0 1 2 3 4 5 60.0

0.2

0.4

0.6

0.8

1.0

1.2

ξ

v x

# Our equations (ode only solves first order!)

# so break it into three first order equations

# Here I use u_0=1

v’=-v*psi

psi’=p

p’=v

# our initial data

psi=0

p=0

v=0.47 # this makes v_x/u_0-->1 for large y

print x,p

step 0,6

You must fool around with the initial value of v = d2Ψdξ2 until you get a solution whose vx → 1 for large ξ (large

y). Once you get there, the pay-off is some real science; you discover that the edge of the boundary layer where

vx ≈ 0.99u0 is at δ ≈ 5√

xµu0

, a famous and non-trivial result.

10.5. PROBLEMS 193

14. Scaling ideas can be applied to solve some of the linear multi-variable equations in chemistry, such as the diffusionequation

∂f

∂t= J

∂2f

∂x2

First define τ = J t to get

∂f

∂τ=∂2f

∂x2

Now note that

f(x, τ) =1√τg(

x√τ) =

1√τg(ξ), ξ =

x√τ

Show thatd2g

dξ2+ξ

2

dg

dξ+

1

2g = 0

and the solution is

g(ξ) = e−ξ2

4

According to this solution, the depth to which for diffusion occurs in time t is

x ∝√4Jt

a result which you probably encountered in your chemistry classes.

15. Why does a rising bubble flatten out transverselyto the direction it travels in? For a circular-crosssection obstacle of radius a, fluid flowing (far away)at velocity v0i has velocity components

vx = −v0(1− a2

r2cos(2θ)

)vy = v0

a2

r2sin(2θ)

In the fluid and bubble the pressure is P0 at restand at equilibrium.

Use Bernoulli’s law to compute the pressure on itssurface, explaining why it will deform as I haveillustrated.

16. Use Eq. 10.81 to verify that the velocity of fluid flow around a cylinder of radius a are given by Eq. 10.83.

17. A sphere of radius a and mass m moving at speed v through a viscous fluid of density ρ experiences drag

|F | = 6πµ v a

If released from rest, find its terminal velocity as it falls through the fluid.


18. Consider Poisseville flow in a long cylinder of radius a with a pressure gradient ∇P = ck. Look for a solutionv = v(r)k that vanishes on the wall. Show that v(r) = c

4µ (r2 − a2), and find the mass-flow rate

dm

dt=

∫ 2π

0

dθ

∫ a

0

(ρ v(r)

)r dr

and use it to compute the pressure drop ∆P in a pipe of length ℓ as a function of mass-flow rate. Hint; note c = ∆Pℓ .

19∗. Consider a rotating cylindrical vessel of fluid, with rotation vector ω = ω k. If you co-rotate with it, a bit ofmatter in the fluid a distance r from the center would move radially outward were it not for the forces exertedby surrounding matter forcing it to travel in a circle of radius r. This means that in the rotating frame there is anoutward directed force of magnitude Fr = mω2r on it, which has a centrifugal potential

Vc(r) = −1

2mω2 r2

Use Bernoulli’s law to find the shape of the fluid surface in the rotating frame under the combined influences of thegravitational and centrifugal potentials.

Chapter 11

Oscillations

Oscillatory motion is periodic motion; the pattern of displacements of a body under the influence of forces is periodicof period T if

r(t+ T ) = r(t) (11.1)

Newton’s laws in one dimension can easily be reduced from a second order equation for the case of an external forcedepending only on position

md2x

dt2= Fext(x) (11.2)

to a first order equation by a simple transformation equivalent to constructing a conserved energy; multiply bothsides by dx

dt

m(d2xdt2

)(dxdt

)= F (x)

dx

dt=

d

dt

(12m(dxdt

)2)(11.3)

and if there exists a function V (x) such that

F (x) = −dVdx

(11.4)

then the chain rule allows us to write the differential equation as

d

dt

(12m(dxdt

)2)+d

dtV (x) = 0 (11.5)

since ddtV (x) = dV

dxdxdt , and we can integrate directly by introducing a constant of integration C = E. This is a

conserved quantity, also called a first integral since it was obtained by an integration;

1

2m(dxdt

)2+ V (x) = E (11.6)

and invert this to get

t− 0 =

∫ x(t)

x(0)

dx′√2m (E − V (x′))

(11.7)

which we perform and again invert to obtain x(t).

For the case of a force exerted by a spring;

Fx = F (x) = −kx, V (x) =1

2kx2 (11.8)

we can perform this integral, and the result is a periodic function. Starting with

t =

∫ x(t)

x(0)

dx′√2m (E − 1

2kx′2)

(11.9)

195

196 CHAPTER 11. OSCILLATIONS

make the variable substitution

x′ =

√2E

ksin θ, dx′ =

√2E

kcos θ dθ (11.10)

and the integral becomes

t =

∫ θ(t)

θ0

√2Ek cos θ dθ√

2m (E − E sin2 θ)

=

√m

k

∫ θ(t)

θ0

dθ =

√m

k(θ(t)− θ0) (11.11)

We solve this equation;

θ(t) = θ0 +

√k

mt (11.12)

and insert it into the equation for x;

x(t) =

√2E

ksin θ =

√2E

ksin(θ0 +

√k

mt)

(11.13)

The result is periodic motion, since the sine function is periodic. Assuming that the period is T , then

x(t+ T ) = x(t) (11.14)

implies that √2E

ksin(θ0 +

√k

mt)=

√2E

ksin(θ0 +

√k

m(t+ T )

)(11.15)

and we can use the trig identity

sin(θ0 +

√k

m(t+ T )

)= sin

(θ0 +

√k

mt)cos(√ k

mT)+ cos

(θ0 +

√k

mt)sin(√ k

mT)

(11.16)

Periodicity will be met if

cos(√ k

mT)= 1, sin

(√ k

mT)= 0 (11.17)

and this is true if √k

mT = 2π, T = 2π

√m

k(11.18)

We often say that the motion repeats with frequency

f =1

T=

ω

2π, ω =

√k

m(11.19)

This type of motion appears all over nature. A simple mass suspended from a spring will exhibit this periodic motion.

The generic form of periodic motion is any motion whose displacement function is of the type

x(t) = C sin(θ0 + ωt), T =2π

ω(11.20)

The constants C and θ0 are determined from initial data, and are referred to as amplitude and phase respectively.

Example 1. Find the amplitude and phase for a mass m connected to a spring k such that at t = 0 the displacementof the mass is zero but its velocity is v0.From our work above we start with

x(t) = C sin(θ0 +

√k

mt) (11.21)

and first enforcex(0) = 0 = C sin(θ0) (11.22)

197

which gives us θ0 = 0. Next we compute

v(t) =dx

dt= C

√k

mcos(

√k

mt), v(0) = v0 = C

√k

mcos(0) = C

√k

m(11.23)

and we arrive at the solution

x(t) =v0√km

sin(

√k

mt) (11.24)

Example 2.Find the amplitude and phase for a mass m connected to a spring k such that at t = 0 the displacementof the mass is x0 but its velocity is 0.Again we start with

x(t) = C sin(θ0 +

√k

mt) (11.25)

and first enforcex(0) = x0 = C sin(θ0), θ0 = sin−1 x0

C(11.26)

then

v(t) =dx

dt= C

√k

mcos(θ0 +

√k

mt), v(0) = 0 = C

√k

mcos(sin−1 x0

C) (11.27)

this indicates thatsin−1 x0

C=π

2, x0 = C sin

π

2= C (11.28)

and we are finished

x(t) = x0 sin(π

2+

√k

mt) = x0 cos(

√k

mt) (11.29)

Example 3. Lets create a master formula; suppose that we have two conditions at t = 0;

x(0) = x0, v(0) = x(0) = v0 (11.30)

then x(t) = C sin(θ0 + ω t) reduces to

x0 = C sin θ0, v0 = C ω cos θ0 (11.31)

dividingx0ω

v0= tan θ0, θ0 = tan−1 x0ω

v0(11.32)

instead, square and add

x20 +v20ω2

= C2 sin2 θ0 + C2 cos2 θ0 = C2, C =

√x20 +

v20ω2

(11.33)

and the master formula is

x(t) =

√x20 +

v20ω2

sin(ωt+ tan−1 x0ω

v0) (11.34)

Example 4. How are these factors (amplitude and phase) related to the energy of the oscillating mass? The totalmechanical energy is the sum of kinetic and potential, and is conserved;

Emech =1

2mv2(t) +

1

2kx2(t) =

1

2mv2(0) +

1

2kx2(0) (11.35)

Multiply by two, divide by k

2E

k=v2(t)km

+ x2(t) =v2(t)

ω2+ x2(t) =

v2(0)

ω2+ x2(0) = C2 (11.36)

and we discover that the oscillator energy depends only on its amplitude

x(t) =

√x20 +

v20ω2

sin(ωt+ tan−1 x0ω

v0) =

√2E

ksin(ωt+ tan−1 x0ω

v0) (11.37)


Example 5. A masssuspended from aspring is pictured be-low in an equilibriumconfiguration, with thehanging mass m atrest. The un-stretchedspring has a length L.Under these condi-tions all vertical forces(x−direction) on themass are in balance

Fx = 0 = −k x0 +mg, and so the spring is stretched by amount x0 =mg

k(11.38)

If the mass is now pulled down by an additional distance A and released, the position x relative to the equilibriumposition will satisfy

Fx = max = md2x

dt2= −k (x0 + x) +mg = −k x (11.39)

when the mass is displaced by x. This equation has general solution

x(t) = C1 cos(√ k

mt)+ C2 sin

(√ k

mt)=√C2

1 + C22 sin

(√ k

mt+ ϕ

)(11.40)

with

C1

C2= tanϕ (11.41)

For our particular problem, with displacement x equal to A at t = 0 when the mass is released from rest vx = dxdt = 0

as in the figure below.We find that

A = x(0) = C1, 0 =(dx(t)

dt

)t=0

= C2

√k

m(11.42)

and so

x(t) = A sin(√ k

mt)

(11.43)

describes the motion of the mass. This is called simply periodic motion of frequency

f =ω

2π=

1

2π

√k

m(11.44)

and of period T = 1f , which is the time needed to complete one full cycle of the motion.

199

Any object that executes such periodic motionis called a simple harmonic oscillator.A more detailed analysis shows that the massof the spring ms has a significant affect on theperiod of the motion, and if properly taken intoaccount, the entire system executes periodicmotion with

T = 2π

√(m+ ms

3 )

k(11.45)

T 2 =4π2

km+

4π2ms

3k(11.46)

and so a plot of T 2 versus the hanging mass

m should be a straight line of slope 4π2

k .

This together with an independent measurement of k with which to make a comparison provides us with a means oftesting that the motion of a real mass and spring is actually simple harmonic motion.

x

y

x

y

−k2 x2

k2 x2

−k1 x1

Example 6. The frequency of oscillations canbe very simply found by reducing equations ofmotion to the standard form for oscillation

x = −ω2x (11.47)

Consider a pair of springs k1 and k2 joined ata massless junction (the red ball m1 = 0) con-nected to a massm. The procedure to computeω is to displace the mass by x and find its New-tonian equation of motion. Displace m by x,then spring k1 stretches by x1 and k2 by x2,with

x1 + x2 = x (11.48)

Each spring pulls in opposite directions on m1 = 0

m1a1 = 0 = −k1x1 + k2x2, k1x1 = k2x2 (11.49)

Solving we find that

x1 =k2

k1 + k2x, x2 =

k1k1 + k2

x (11.50)


Only k2 is attached to m;

ma = md2x

dt2= mx = −k2x2 = − k1k2

k1 + k2x (11.51)

Compare this to the oscillator equation and extract the frequency of oscillation

x = −ω2x, ω2 =k1k2

m(k1 + k2)(11.52)

x

y

x

y

−k x

k x

M1 M2

x1 x2

Example 7. These two masses slide on a fric-tionless surface, and are attached to nothingbut one another, and so there are no appliedexternal forces. Therefore momentum is con-served. Displace m1 to the right by x1 andm2 to the left by x2, stretching the spring byx = x1+x2. Apply conservation of momentum;

0 = m1dx1dt

−m2dx2dt

, m1x1 = m2x2

(11.53)Solving

x1 =m2

m1 +m2x, x2 =

m1

m1 +m2x (11.54)

Pick one (say m1) and write its equation ofmotion

m1a1,x = −m1d2x1dt2

= F1,x = kx (11.55)

We obtain

−m1m2

m1 +m2x = kx, ω2 =

(m1 +m2)k

m1m2(11.56)

x

y

x

y

−k1 2x

k1 2x

M M

k2 x −k2 x

Example 8. Momentum is not conservedhere; the walls exert external forces throughthe springs. This system (many masses) canoscillate in two modes each with a differentfrequency. Displace each mass symmetricallyby x, stretching the middle spring k1 by 2x andcompressing each outer spring k2 by x. Theequation of motion for the right mass is

md2x

dt2= −2k1x− k2x, ω2 =

k2 + 2k1m(11.57)

Chemical applications. Oscillatory modes of molecules are stimulated by collisions and by the absorption of light.For a CO2 molecule, the mode of oscillation illustrated below is stimulated by the absorption of infrared light, making

201

CO2 a potent contributor to the greenhouse effect.

Example 9. A triatomic linear molecule (twoatoms of mass m, one atom of mass 2m canhave two linear oscillatory modes. No exter-nal forces are applied, so all motions preservethe center of mass position. Displace eachlight mass to the left by x and the heavy oneto the right by x; this conserves momentum.The right spring is compressed by 2x, the leftstretched by 2x Pick any mass, say the far rightone, and write its equation of motion;

max = −md2x

dt2= k(2x), ω2 =

2k

m(11.58)

Example 10. The other oscillatory mode ofthis linear molecule must also involve displace-ments that conserve momentum (leave the cen-ter of mass fixed). For the m, 2m,m moleculethe only other such mode has the middle atomstationary, the outer atoms displace away fromit by x;for the rightmost atom the equation of motionis

md2x

dt2= −kx, ω2 =

k

m(11.59)


v

Ff

−k x

Example 11. This disk of mass m and ra-dius R rolls without slipping. It is drawn withspring stretched by x, and so m is displacedfrom equilibrium by x to the right. We writethe force, torque and rolling conditions

− kxi− Ff i = ma = md2

dt2i (11.60)

FfR(−k) = Icomα(−k) (11.61)

and

a =d2x

dt2= Rα (11.62)

and eliminate the force of friction, and install the rolling condition. Compare the result to x = −ω2x;

− kx− IcomR2

d2x

dt2= m

d2x

dt2, ω2 =

k

m+ IcomR2

(11.63)

θ

x

−k x

mg

Fx

Fy

Example 12. This is an example of a physical pen-dulum, a rigid by pivoted about a fixed point, allowedto execute periodic oscillations. It is illustrated with therod (mass m length 2ℓ) displaced by some small amountx from equilibrium (hanging vertically with the spring un-stretched). The force equations are complex, momentumis not conserved, and the hinge exerts unknown forces thatwe could in principle compute. To find the frequency of os-cillations we hammer the torque equation about the pivotpoint into the oscillator generic form x = −ω2x. Assumingthat x << ℓ we can say that

θ ≈ x

ℓ, α = θ ≈ x

ℓ(11.64)

then

(xi−√ℓ2 − x2 j)× (−kxi) + (xi−

√ℓ2 − x2 j)× (−mgj) = Iendαn (11.65)

− kx√ℓ2 − x2k−mgxk ≈ Iend

x

ℓk (11.66)

approximating the square root (neglect the x2) we arrive at

ω2 ≈ kℓ2 +mgℓ

Iend(11.67)

In this example we must have αn = αk for periodic motion. In addition the results for the frequency are onlyapproximate; pendula do not exhibit perfect harmonic motion. The frequency of a pendulum is only approximatelyharmonic for very small displacements. In general the frequency of a pendulum depends on the amplitude of the swing.

It is possible for a particle subjected to any simple binding potential to experience simple harmonic motion, providedits equilibrium point is stable, and the deviations from equilibrium are small.

203

For any potential V (x) we can locate the equilibrium position (where Fx = 0) by solving for

Fx(xeq) = 0 = − d

dxV (x)

∣∣∣xeq

= −V ′(xeq) (11.68)

and by expanding V (x) in a power series about that point

V (x) = V (xeq) +(x− xeq

)V ′(xeq) +

1

2!

(x− xeq

)2V ′′(xeq) + · · ·

= V (xeq) +1

2!

(x− xeq

)2V ′′(xeq) + · · · (11.69)

Using this formula for the potential the equation of motion becomes

max = md2x

dt2= − d

dxV (x) = −

(x− xeq

)V ′′(xeq) + · · · (11.70)

Now let X = x− xeq, and we obtain a harmonic oscillator equation

md2X

dt2= −V ′′(xeq)X, X(t) = A sin

(√V ′′(xeq)

mt− ϕ

), x(t) = xeq +A sin

(√V ′′(xeq)

mt− ϕ

)(11.71)

and so any potential with an equilibrium point and a non-zero second derivative will possess simple harmonicmotions of small amplitude.. If you have trouble, use REDUCE.Example 13. For example consider

V (x) =L2

2mx2− mM1G

x(11.72)

V:=(L^2/(2*m*x^2))-m*M1*G/x;

solve(df(V,x),x);

% this gives me x=L^2/(m^2 M1 G)

xeq:=L^2/(m^2*M1*G);


taylor(V,x,xeq,4);

REDUCE tells me that

V ′′(xeq) =G4m7M4

1

L6(11.73)

and our small-amplitude oscillations about equilibrium for this (planetary orbit) potential have frequency

ω =G2m3M2

1

L3(11.74)


11.1 The simple pendulum

The pendulum motion is very similar to that of a harmonic oscillator, butfirst appearances are deceptive. The pendulum is more complex; the periodof its motion is dependent on the amplitude (maximal displacement) of itsmotion.

There is of course an elementary (but only approximate) treatment. Weagain compute torques about the string fixed point by letting the horizontaldisplacement of the pendulum be x with x << ℓ. If the pendulum length isℓ we have

(−xi−√ℓ2 − x2j)× (−mgj) = Iendαn (11.75)

or−mgxk = Iendαk = mℓ2αk (11.76)

and once again we use the approximation

θ =x

ℓ, θ = α ≈ x

ℓ(11.77)

and we get

−mgxk = mℓxk, ω2 ≈ g

ℓ(11.78)

and the period

T =2π

ω≈ 2π

√ℓ

g(11.79)

The exact treatment of the pendulum is much more complex, but some aspects are understandable at this level ofmathematical sophistication. Given an initial angular displacement θ0, the motion of the pendulum is given by theenergy conservation formula

1

2I(dθdt

)2−mgℓ cos θ = −mgℓ cos θ0, I = mℓ2 (11.80)

We can rearrange this to

mgℓ(cos θ − cos θ0) =1

2I(dθdt

)2,

√2mgℓ

Idt =

dθ√cos θ − cos θ0

(11.81)

The period is gotten by noting that in period T , the pendulum begins at θ0, goes to −θ0 and returns to θ0;∫ T

0

√2mgℓ

Idt =

√2mgℓ

IT = 4

∫ θ0

0

dθ√cos θ − cos θ0

= 4

∫ θ0

0

dθ√2 sin2 θ02 − 2 sin2 θ2

(11.82)

and change variables, let cosϕ =sin θ

2

sinθ02

,

dθ = −2sin θ0

2

cos θ2sinϕdϕ (11.83)

and we arrive at

T = 4

√I

mgℓ

∫ π2

0

dϕ√1− sin2 θ02 cos2 ϕ

(11.84)

11.1. THE SIMPLE PENDULUM 205

This is a very non-trivial integral, it results in an elliptic function, but as long as the value of sin2 θ02 is small, wecan expand this in a series and integrate term by term. First lets assume that

sin2θ02<< 1 (11.85)

so that

sin2θ02

cos2 ϕ << 1 (11.86)

and then we can use the binomial theorem

(1 + δ)n = 1 + nδ +nn− 1

2δ2 +

n(n− 1)(n− 2)

6δ3 + · · · (11.87)

This is easy to prove by just multiplying the thing out, but it is also true even if n is not an integer, in particular

(1 + δ)−12 = 1− 1

2δ +

3

8δ2 + · · · (11.88)

Apply this to our pendulum;

T = 4

√mℓ2

mgℓ

∫ π2

0

dϕ(1 +

1

2sin2

θ

2cos2 ϕ+

3

8sin4

θ

2cos4 ϕ+ · · ·

)(11.89)

and do the integrals. This results in our final formula

T = 2π

√ℓ

g

(1 +

1

4sin2

θ02

+9

64sin4

θ02

+ · · ·)

(11.90)

Notice that the smallest that the period can be is Tmin = 2π√

ℓg , which is the period of very small amplitude

oscillations. For larger initial deflections θ0, the period increases.

L

x

Example 14. A bead slides on a horizontalbar, and is connected by a spring of un-stretched length ℓ to a fixed point a distance ℓabove the bar. If the mass is displaced by anamount x = A very small and released it willexecute motion about its equilibrium pointwith position x between −A and A. If theamplitude A of the motion is doubled, whathappens to the period of this motion?

Find the potential energy stored in the spring;

V (x) =1

2

(√ℓ2 + x2 − ℓ

)2≈ k

4ℓ2x4 (11.91)

The energy of small amplitude motion is conserved; begin at x = A and x = 0;

1

2mx2 +

k

4ℓ2x4 = 0 +

k

4ℓ2A4 (11.92)

or

x =dx

dt=

√k

2mℓ2(A4 − x4), 2

∫ A

−A

1

A2

dx√1− kx4

2mℓ2A4

=

∫ T

0

dt = T (11.93)

Let x = az;

2

∫ A

−A

1

A2

dx√1− kx4

2mℓ2A4

= 2

∫ 1

−1

1

A

dz√1− kz4

2mℓ2

= T (11.94)


and so its a nasty integral, but the amplitude dependence can be extracted

T =ξ

A, ξ = 2

∫ 1

−1

1

A

dz√1− kz4

2mℓ2

(11.95)

so if you double the amplitude, the period is cut in half.

11.2 Oscillations in general1

Oscillations occur in nature whenever the solution to the Newtonian equations of motion can be written in terms ofsines and cosines. The most important properties of sines and cosines can be stated in this way, let

f(x) = sinx, g(x) = cosx (11.96)

then

f2(x) + g2(x) = 1, anddf

dx= g(x) (11.97)

There are plenty of non-oscillatory functions that obey one equation or the other, such as

f(x) = x, g(x) =√1− x2, f2(x) + g2(x) = 1 (11.98)

butdf

dx= g(x) (11.99)

so systems whose Newtonian laws of motion are these functions do not oscillate.Lets try some other possible combinations; exponential functions are very nice to work with because they are so easyto differentiate

d

dxe±λx = ±λ e±x (11.100)

The combination

f(x) =ex − e−x

2, g(x) =

ex + e−x

2(11.101)

does obeydf

dx= g(x) (11.102)

but not f2(x) + g2(x) = 1; ( (e2x − 2exe−x + e−2x)

4+

(e2x + 2exe−x + e−2x)

4

)= 1 (11.103)

but what if there was a number (call it i) such that i2 = −1, then

f(x) =eix − e−ix

2i, g(x) =

eix + e−ix

2(11.104)

would obeydf

dx= g(x) (11.105)

and f2(x) + g2(x) = 1; ( (e2ix − 2eixe−ix + e−2ix)

4i2+

(e2ix + 2eixe−ix + e−2ix)

4

)= 1 (11.106)

and so it must be that

sinx =eix − e−ix

2i, cosx =

eix + e−ix

2(11.107)

1This section is optional. It is here for physics and math majors

11.2. OSCILLATIONS IN GENERAL 207

This is a remarkable fact discovered by Leonardo Euler over 300 years ago, and allows us to switch sines andcosines for exponentials.

This neat pair of formulas allow us to construct a method of solving practically any differential equation that atypical undergraduate science or engineering student would need to solve; any equation of the form

ad2x

dt2+ b

dx

dt+ cx = 0 (11.108)

has a solutionx = Aeλt, λ = λ1 + iλ2 (11.109)

The proof is almost trivial; just compute derivatives and stick them into the equation

x = Aλeλt, x = Aλ2eλt (11.110)

aAλ2eλt + bAλeλt + cAλeλt = 0 (11.111)

Nearly everything cancels out except

aλ2 + bλ+ c = 0, λ =−b±

√b2 − 4ac

2a(11.112)

If 4ac > b2, these two roots are complex

λ =−b2a

± i

√4ac− b2

2a= λ1 + iλ2, 4ac > b2 (11.113)

Example 15. A harmonic oscillator such as a mass connected to a spring, that experiences air resistance thatincreases with speed will have equation of motion

ma = −kx− 2mγv, md2x

dt2+ 2mγ

dx

dt+ kx = 0 (11.114)

where γ is a measure of how severe the air resistance is. Lets assume that it is small, then

λ± =−2mγ

2m± i

√4km− 4m2γ2

2m= λ1 + iλ2 (11.115)

= −γ ± i

√k

m− γ2 (11.116)

Since the equation has two roots, we need two terms in the solution

x(t) = A+e(−γ+i

√km−γ2) t +A−e

(−γ−i√

km−γ2) t (11.117)

This is called a damped oscillator, note that A± are gotten from initial conditions, and if A− = Ap =A2 then

x(t) = Ae−γt cos(

√k

m− γ2 t) (11.118)

the mass does oscillate, bit the amplitude of the motion dies out slowly over time.

Example 16. Without solving the equations of motion, show that the energy of a damped oscillator monotonicallydecays to zero.

Solution The equation of motion ismx+mω2x = −2mγx, γ ≥ 0 (11.119)

Multiply both sides by x;mx x+mω2x x = −2mγx2 (11.120)

ord

dt

(12mx2 +

1

2mω2x2

)=

d

dtE = −2mγx2 (11.121)

We see that E ≥ 0 since it is the sum of squares, and the right-hand side is manifestly negative, therefore the energywill monotonically drop to zero.


11.3 Classical elastica. Stress and strain

F

∆x

A

L

When forces are applied to the surfaces of a solid block of elastic material, they canbe applied normally or tangentially to the surface (or at any angle in between). Webegin with the most rudimentary summarized descriptions, for those who need onlyan equation, for example people studying for MCATs. This material is a normal partof 201 if time permits.

Imagine a rod of elastic material of cross-sectional area A, length L with a force |F|applied normally to its top surface. This will result in a strain or fractional deformationof the rod (as well as bulging of its sides, which we will ignore). If the rod changes lengthby

∆x = L− Lnew (11.122)

we define the strain e to be

e =∆x

L=L− Lnew

L(11.123)

and the stress Tn to be

txx = Tn =|F|A

(11.124)

Since this is applied normally, this equals the pressure P put on the rod.In linear elastica we assume that there is a linear relationship between Tn and e, theproportionality constant is the Young’s modulus E or Y in some texts;

Tn = E e (11.125)

γ

F∆x

A

L

If the force F is applied tangentially to a cross sectional area of the rod,we call the resulting stress a shear stress, and the rod is assumedto bend like a stack of cards, making an angle γ that results in adeformation ∆x of the end where the strain is applied a distance Lfrom the end of the rod that is clamped in place and cannot move.

Again we assume a linear relationship between applied stress Ts =|F|A

and deformation γ, with proportionality constant G called the shearmodulus

txy = Ts = Gγ = G∆x

L(11.126)

Liquids simply cannot resist shear forces, and materials like plasticsand rubber have very low shear moduli (making them easy to cutwith shears or scissors).

Material E, Nm2 G, N

m2 Material E, Nm2 G, N

m2

Rubber 1.0× 107 − 1.0× 108 3.0× 105 White Oak 1.5∗ × 1010 0.94× 109

1.1† × 1010

Aluminum 6.9× 1010 2.6× 1010 Titanium 1.1× 1011 4.1× 1010

Steel 2.1× 1011 7.9× 1010 Brass/Copper 1.1× 1011 6.3× 1010

Nylon 5× 109 2.0× 109 Carbon nanotube 1.0× 1012

∗ Parallel to grain. † Across the grain.

As you can see, most metals are harder to compress with a given stress applied normally, than they are to shear withthe same stress applied tangentially. The ratio is about three to one (Poisson’s ratio ν)

ν = |γϵ| ≈ 0.3 (11.127)

11.3. CLASSICAL ELASTICA. STRESS AND STRAIN 209

11.3.1 Torsional strain

Rr

F(r)

Suppose that you were to grab a circular cross-section bolt orsolid cylinder with a wrench that distributes the force uniformlyaround the perimeter, and apply a torque τ = |τ |. This willresult in a shear stress on the cross section which at a distance rfrom the center depends on r. Certainly there is no stress rightat the center. Lets assume that

Ts(r) = Ts(R)r

R(11.128)

The total torque applied to the ring illustrated, of radius r andwidth dr is

dτ = r dF (r) = r(Ts(r) 2πr dr

)(11.129)

and the total torque applied to the bolt is

τ =

∫ R

0

r(Ts(r) 2πr dr

)=

∫ R

0

(Ts(R) r

r

R2πr dr

)=Ts(R)

R

πR4

2(11.130)

Suppose that we take a rod or bolt of length h and radius R, clamp the lower end, and apply a torque τ to the upperend. By what angle θ will the cross-section of the torqued end rotate?

h

R

γ

θs

According to our definition of shear strain, we see that

Tn(R) = γ G (11.131)

and the arc of length s is situated such that

s = Rθ = h γ (11.132)

and thereforeR

hθ = γ =

Tn(R)

G=

Rτ

G πR4

2

(11.133)

or

θ =h τ

G πR4

2

(11.134)

With this torque τ applied, the bolt is twisted but is static, thereis an equal but opposite internal torque. If we deform the bolt bytwist θ, the internal torque opposes this, resulting in a restoringtorque

τ = −κ θ, κ = GπR4

2h(11.135)

We have a torsion spring of force-constant κ. A twisted rod or bolt tries to restore itself with an internal torque.Suppose that we clamp a metal bar of length L and cross-sectional area A in a vise, displace the free end by x, andlet go. There will be an internal restoring force Fx = −TsA = −GA

L x, and so the released bar will oscillate with a

frequency that is proportional to√G.


11.3.2 Complete treatment of elasticity2

The physics of elastic media is virtually identical to the physics of fluids, many equations will be the same or similar,only the interpretations will be different. This treatment is a simple as I can make it, but it still requires the use ofmatrices (linear algebra). By now you must have realized that real science can’t be done without mathematics, andhave come to terms with that truth. Hopefully you will be moved to take more mathematics.

We will make use of two coordinate systems

X = (X1, X2, X3) material point in an un-deformed body

x = (x1, x2, x3) coordinates of X after deformation (11.136)

Consider a particle whose coordinate xi = xi(Xj , t) depends on some initial location in an un-displaced medium,such as a fluid or elastic medium. The components Xj are called Lagrangian or material coordinates, the xi are thespatial or Eulerian coordinates. We will assume a non-degeneracy; mathematically

J = det∣∣∣ ∂xi∂Xj

∣∣∣ > 0

and colloquially; no two X’s end up with same x, so that the equations xi(Xj , t) could potentially be inverted togive Xj = Xj(xi, t). We limit our attention to linear elasticity for which xi is a linear function of Xj .

The coordinates xi represent the actual coordinates of a particle of matter in a fluid or elastic medium after adisplacement, and so

ui = xi −Xi

is the actual displacement, and

vi =dxi

dt=∂xi(Xj , t)

∂t=dui

dt=

∂

∂tui(Xj , t)

is the particle velocity. The equation Xj = Xj(xi, t) can be used to eliminate Xj , so that the velocity is a functionof time and actual position

vi = vi(xj , t)

If we hold Xj fixed, then

ai =d

dtvii(xj , t) =

∂vi

∂t+ vj

∂vi

∂xj

We see our familiar friend the time-derivative operator (convective derivative)

d

dt=

∂

∂t+ vj

∂

∂xj

that will be used in conditions under which Xj are fixed.

Let Xi, X∗i be adjacent points in space at t = 0 before any displacement. Then

x∗i = xi +∂xi

∂Xj

(X∗j −Xj

)+ · · ·

We define the deformation gradients

dxi = F i jdXj , F i j =

∂xi

∂Xj(11.137)

with initial (pre-deformation) and final (post-deformation) displacements

dXj = X∗j −Xj , dxj = x∗j − xj

2This is not covered in 201, it appears for the benefit of physics majors and engineering or architecture students (this section isspecialized and was formerly covered in 401).


We are going to be most interested in three types of material displacements; rotations, stretches or dilations, andshears. Only the latter two are actual deformations (our primary objects of interest), a rotated object is not deformed;

X1 X1

X2 X2

(x1(X1 , X2),x2(X1 , X2))(X1 , X2)

Example A rotation(dx1

dx2

)=

(cos θ sin θ− sin θ cos θ

)(dX1

dX2

)dx = F · dX (11.138)

has detF = 1. Note that F is anorthogonal matrix.

X1 X1

X2 X2

(x1(X1 , X2),x2(X1 , X2))(X1 , X2)

Example. A (non-isotropic)stretch(dx1

dx2

)=

(1 + ϵ 00 1− ϵ

)(dX1

dX2

)dx = F · dX (11.139)

has detF = 1− ϵ2. Note that F isan upper-triangular matrix (diago-nal qualifies as upper triangular).

X1 X1

X2 X2

(x1(X1 , X2),x2(X1 , X2))(X1 , X2)

Example. A shear (our first illus-tration was of a shear)(dx1

dx2

)=

(1 2γ0 1

)(dX1

dX2

)dx = F · dX (11.140)

has detF = 1. Note that F is anupper-triangular matrix.

QR-decomposition theorem. In physics 307 we prove the QR decomposition theorem; any matrix F can bedecomposed into a product of an upper-triangular matrix R and an orthogonal matrix Q. This is sometimes called apolar decomposition. The punch-line is that any deformation can be thought of as being accomplished by two simpledeformations in succession, one being a rotation.

We will use this theorem to “filter out” the part of a solid’s displacement that is just a rotation, the idea being thatsince a rotation is not a deformation, a rotated body looks exactly the same as the unrotated if we co-rotate withit. This is called in technical parlance “passing to the principle strain coordinate system or axes”.

Small deformations. The study of small deformations gives us a “proof” of the QR-decomposition theorem. We


break up Fij into two parts, one symmetric and one antisymmetric

F i j = δi j +∂ui

∂Xj

= δi j +1

2

( ∂ui∂Xj

+∂uj

∂Xi

)+

1

2

( ∂ui∂Xj

− ∂uj

∂Xi

)= δi j + ei j + wi j

=∑k

(δi k + ei k

)(δk j + wk j

)dxi = dXi +

(ei j + wij

)dXj (11.141)

since ei k wkj = 0. The part representing a deformation is eij , we will discard the rotational part wij .

How do we describe forces and stresses on the blocks of elastic matter? Exactly the same way in which we did thisfor fluids, with Euler’s equation and a stress-momentum tensor t;

0 = ρdv

dt= ∇ · t+ dF

dV(11.142)

for a static deformed body, with applied forces per unit volume (such as gravity).

In order to relate stresses t to the strains or deformations e that they presumably cause, we hypothesize thatthere is a strain energy U with the basic form of Hooke’s law;

tij =∂U

∂eij

U = U0 +∑i,j

αij

∣∣∣0eij +

∑i,j,k,ℓ

αijkℓ

∣∣∣0eijekℓ + · · ·

tij = 0 + αij

∣∣∣0+∑k,ℓ

∂

∂eij

(αijkℓ

∣∣∣0eijekℓ

)+ · · · (11.143)

then to lowest order in deformations

tij = αij

∣∣∣0+∑k,ℓ

(αijkℓ

∣∣∣0ekℓ + αkℓij

∣∣∣0ekℓ

)We introduce a new (fourth-rank!) tensor of material elasticity coefficients

Cijkℓ = αijkℓ + αkℓij (11.144)

we see that it is symmetric

Cijkℓ = Ckℓij

and

U =1

2

∑i,j,k,ℓ

Cijkℓ eij ekℓ (11.145)

is a sort of generalized Hooke’s law.

Lets study isotropic materials only (for which these really are constants, no position dependence). Let Qij be anorthogonal matrix (a rotation), then our material coefficient tensor is rotationally isotropic if rotations don’t changethe elastic coefficients (it is as elastic in the x-direction as it is in the y...)

Cijkℓ =∑a,b,c,d

QiaQjbQkcQℓdCabcd (11.146)


and by picking out a collection of simple rotations to apply, we can drastically cut down the number of elasticitycoefficients that we need, and the complexity of the equations an be reduced.

Case 1. Let

Q =

−1 0 00 −1 00 0 0

which just represents a 180o degree rotation about the z-axis. Then

C1311 = Q11Q33Q11Q11C1311 = −C1311, C1311 = 0

C1333 = Q11Q33Q33Q33C1333 = −C1333, C1333 = 0

C1322 = Q11Q33Q22Q22C1322 = −C1322, C1322 = 0

C2311 = Q22Q33Q11Q11C1311 = −C2311, C2311 = 0 (11.147)

so in fact any Cijkℓ with an odd number of indices being 1 or 2 will be zero. This really cuts down the number ofnon-zero coefficients!

t11 = C1111 e11 + C1122 e22 + C1133 e33

t22 = C2211 e11 + C2222 e22 + C2233 e33

t33 = C3311 e11 + C3322 e22 + C3333 e33

t12 = C1212 e12

t13 = C1313 e13

t23 = C2323 e23 (11.148)

Case 2. Let

Q =

1 0 00 0 10 −1 0

which just represents a 90o degree rotation about the x-axis. Then

C1313 = Q11Q32Q11Q32C1212 = C1212

C2323 = Q23Q32Q23Q32C3232 = C3232

C2323 = Q21Q32Q21Q32C1212 = C1212 = C1313 (11.149)

Case 3. Let

Q =

1√2

1√2

0

− 1√2

1√2

0

0 0 1

then

C1212 = Q11Q21Q11Q21C1111 +Q12Q22Q12Q22C2222

+ Q11Q21Q12Q22C1122 +Q12Q22Q11Q21C2211

=1

2

(C1111 − C1122

)(11.150)

The convention is to callC1212 = 2µ, C1122 = λ, C1111 = 2µ+ λ

(introducing the Lame’ constants) and we have thefundamental relations between stresses and strainsfor elastic materials, together with Euler’s equation

∇ · t = 0 (11.151)

t11 = (2µ+ λ) e11 + λ(e22 + e33

)t22 = (2µ+ λ) e22 + λ

(e11 + e33

)t33 = = (2µ+ λ) e33 + λ

(e11 + e22

)t12 = 2µ e12

t13 = 2µ e13

t23 = 2µ e23 (11.152)


Example 17. The stretching of a bar. We know that if a band or bar is stretched, its cross section must shrinkin order for its volume to stay constant if the matter it is made of is incompressible. Consider a tension loading of abar such that t33 is a specified external stress, we solve for the resulting deformations;

t33

t33

0 = (2µ+ λ) e11 + λ(e22 + e33

)0 = (2µ+ λ) e22 + λ

(e11 + e33

)t33 = = (2µ+ λ) e33 + λ

(e11 + e22

)0 = 2µ e12

0 = 2µ e13

0 = 2µ e23 (11.153)

then e12 = e13 = e23 = 0 and we can solve these simple linear equations

e11 = − 2µλt33(λ+ 2µ)3 + 2λ3 − 3λ2(λ+ 2µ)

= − λt332µ(3λ+ 2µ)

e22 = − λ

2(λ+ µ)e33

e11 = − λ

2(λ+ µ)e33

t33 = µ(3λ+ 2µ)

λ+ µe33 (11.154)

Notice that e33 > 0 since the bar stretches bute11, e22 < 0 so the cross-section shrinks, andwe have recovered (better yet, obtained fromfirst principles) the summarized version of thefirst page of section 11.3;

ν ≡ e11e33

=λ

2(λ+ µ), Poisson’s ratio

E ≡ t33e33

= µ(3λ+ 2µ)

λ+ µYoung’s modulus (11.155)

Poisson’s ratio tells us how much transverse dimensions contract when we stretch or squeeze an object (the shrink orbulging of the sides and cross-section), and Young’s modulus is like a spring-constant for Hooke’s law, the relationbetween applied forces/stresses and the resulting deformations/strains.

x1

x2

Example 18. Elastic longitudinal waves (a 202 example). Considera semi-infinite elastic medium bounded by the plane x = 0 that suffers anormal displacement

ui =

f(t) i = 10 i = 2, 3

then u1 = u(X1, t), uj = 0 for j = 2, 3 in ρdvdt = ∇ · t;

ρ∂2u1

∂2t= (λ+ µ)

∂2uj

∂X1∂Xj+ µ

∂2u1

∂Xj∂Xj

= (λ+ µ)∂

∂X1

( ∂u1∂X1

+∂u2

∂X2+∂u3

∂X3

)+ µ

( ∂2u1

∂(X1)2+

∂2u1

∂(X2)2+

∂2u1

∂(X3)2

)= (λ+ 2µ)

∂2u1

∂(X1)2(11.156)


which has wave-like solutions

u1 = f(t−X1/c) + g(t+X1/c), c2 =λ+ 2µ

ρ

x1

x2

Lets consider instead a sudden shear displacement of the surface x = 0

ui =

f(t) i = 20 i = 1, 3

ρ∂2u2

∂2t= (λ+ µ)

∂2uj

∂X2∂Xj+ µ

∂2u2

∂Xj∂Xj

= (λ+ µ)∂

∂X2

( ∂u1∂X1

+∂u2

∂X2+∂u3

∂X3

)+ µ

( ∂2u2

∂(X1)2+

∂2u2

∂(X2)2+

∂2u2

∂(X3)2

)= µ

∂2u2

∂(X1)2(11.157)

which has wave-like solutions

u2 = f(t−X1/c′) + g(t+X1/c′), c′2 =µ

ρ< c2

which is very interesting, these waves are slower than the previous group of waves (in an earthquake there are bothtypes of waves, one is much faster than the other. The shear wave cannot pass through the earth’s fluid core).

For future reference, lets invert Eq. 11.152

e11 =1

2µ(3λ+ 2µ)

((2λ+ 2µ)t11 − λ(t22 + t33)

)e22 =

1

2µ(3λ+ 2µ)

((2λ+ 2µ)t22 − λ(t11 + t33)

)e33 =

1

2µ(3λ+ 2µ)

((2λ+ 2µ)t33 − λ(t11 + t22)

)e12 =

3λ+ 2µ

2µ(3λ+ 2µ)t12

e23 =3λ+ 2µ

2µ(3λ+ 2µ)t23

e13 =3λ+ 2µ

2µ(3λ+ 2µ)t13 (11.158)

Example 19. The twisting of a bar. Let

X1 = x, X2 = y, X3 = z, u1 = u, u2 = v, u3 = w

When the bar is twisted, its cross-section along the length undergoes a rotation about the central axis by an anglethat depends on how far the cross-section is from the end, and the cross-sections themselves warp out of plane;(

uv

)=

(cos θ − 1 sin θ− sin θ cos θ − 1

)(X1

X2

)≈(

0 αz−αz 0

)(xy

)using small angle approximations; let the beam twist at a fixed rate along its length, so at the end of the bar it istwisted by (see the figure of section 11.3)

Rθ(ℓ)

ℓ= tan γ ≈ γ, θ(ℓ) =

ℓ

Rγ (11.159)


and at some point z from the top

θ(z) =z

Rγ ≈ αz, α =

γ

R(11.160)

Lets look for cross sections with

u = −αzy, v = αzx, w = αψ(x, y)

For static warping

ρ∂2ui

∂t2= 0 = (λ+ µ)

∂2uj

∂Xi∂Xj+ µ

∂2ui

∂Xi∂Xj

ρ∂2u3

∂t2= 0 = (λ+ µ)

( ∂∂z

( ∂∂x

(− αzy

)+

∂

∂y

(αzx

)+

∂

∂z

(αψ))

+ µα(∂2ψ∂x2

+∂2ψ

∂y2+∂2ψ

∂z2

)0 =

∂2ψ

∂x2+∂2ψ

∂y2, (∇2ψ = 0) (11.161)

the last being a Laplace equation, which is part of the description of electric and magnetic fields, chemical diffusion,the motion of waves, the spread of diseases, and the flow of heat.What do the strains look like?

e =

0 0 1

2

(− αy + α∂ψ∂x

)0 0 1

2

(αx+ α∂ψ∂y

)12

(− αy + α∂ψ∂x

)12

(αx+ α∂ψ∂y

)0

(11.162)

and for this problem t = 2µe gives

t =

0 0 µ

(− αy + α∂ψ∂x

)0 0 µ

(αx+ α∂ψ∂y

)µ(− αy + α∂ψ∂x

)µ(αx+ α∂ψ∂y

)0

(11.163)

which we will use to establish boundary conditions for the sides of the bar.On the lateral sides of the bar, which are free of forces

tx = 0 = nxtxx + ny txy

ty = 0 = nxtyx + ny tyy

tz = 0 = nxtzx + ny tzy =(∂ψ∂x

− y)nx +

(∂ψ∂y

+ x)ny (11.164)

For a circular cross-section rod the entire set

txz = µα(∂ψ∂x

− y)

tyz = µα(∂ψ∂y

+ x)

Fx =

∫A

µα(∂ψ∂x

− y)dA = 0

Fy = 0

∇2ψ = 0 (11.165)

can be satisfied with ψ = 0! (no warping), since y = r sin θ and∫ R

0

r dr

∫ 2π

0

dθ(−µα r sin θ) = 0


and the same for x. The cross-sections remain planar circles, with end-torque

M = µα

∫A

(x2 + y2) dA

= µα

∫ R

0

r2 r dr

∫ 2π

0

dθ =πR4

2µα

= µαJ (11.166)

The stresses are then simply

txz = −yµα = −MJy, tyz = µαa =

M

Jx (11.167)

which maxes out on the surface. These are the equations gotten in section 11.3.1;

Eq. 11.135 τ =M = GπR4

2ℓθ(ℓ)

= GπR4

2ℓ

ℓγ

R

= GπR4

2ℓ

ℓ

RαR =

πR4

2Gα =

πR4

2µα (11.168)

and we discover what µ is; µ = G the so-called shear modulus.

We have recovered all of the simplified, summarized versions of elasticity at this stage, and can move on.

11.3.3 Architectural examples

I will only show you a small number of cases, all in the category of planar stresses; the materials are assumed tobe extremely rigid in one or more directions. This will be sufficient to cover the only really important elementaryarchitectural elements, such as beams and columns.

This is a special case with tiz = 0 for i = x, y, z and the body in question is infinite in extent in the z-direction(in/out of the page), or z-variation is irrelevant. Then

exz = eyz = 0, tzz = 0 = λ(exx + eyy + ezz) + 2µezz

so

ezz = − λ

λ+ 2µ

(exx + eyy

)Our set of strains Eq. 11.158 collapses to

exx =1

Etxx −

ν

Etyy

eyy =1

Etyy −

ν

Etxx

ezz = − ν

E

(txx + tyy

)exy =

1 + ν

Etxy (11.169)

The lack of all body forces results in ∇ · t = 0 or

∂txx∂x

+∂txy∂y

= 0,∂txy∂x

+∂tyy∂y

= 0 (11.170)

Introduce the Airy stress function

txx =∂2ϕ

∂y2, tyy =

∂2ϕ

∂x2, txy = − ∂2ϕ

∂x∂y(11.171)


which automatically satisfies ∇ · t = 0, however it makes a big assumption about the stress components, and weshould question whether or not they are compatible with boundary conditions. In other words Eq. 11.171 is sufficientto satisfy Eq. 11.170 but may not be compatible with typical boundary conditions.

Let X1 = x and X2 = y, and recall that

∂ui∂Xj

= eij + wij

∂2ui∂Xj∂Xk

=∂eij∂Xk

+∂wij∂Xk

=∂eik∂Xj

+∂wik∂Xj

(11.172)

since the order of the derivatives on the left side is irrelevant

∂

∂Xk

∂ui∂Xj

=∂

∂Xj

∂ui∂Xk

therefore

∂eij∂Xk

− ∂eik∂Xj

=∂wik∂Xj

− ∂wij∂Xk

=∂

∂Xj

(12

( ∂ui∂Xk

− ∂uk∂Xi

))− ∂

∂Xk

(12

( ∂ui∂Xj

− ∂uj∂Xi

)=

∂wjk∂Xi

(11.173)

Then

∂

∂Xℓ

( ∂eij∂Xk

− ∂eik∂Xj

)=

∂

∂Xℓ

(∂wjk∂Xi

)=

∂

∂Xi

(∂wjk∂Xℓ

)=

∂

∂Xi

( ∂eℓj∂Xk

− ∂eℓk∂Xj

)(11.174)

which is the compatibility relation that we are after; if we specify stresses, the solution of the equation Eq. 11.171gives rise to a consistent strain tensor if these are true. In the present case this means that

2∂2exy∂x∂y

=∂2exx∂2y

+∂2eyy∂2x

2(1 + ν)

E

∂2txy∂x∂y

=∂2

∂y2

( txx − νtyyE

)+

∂2

∂x2

( tyy − νtxxE

)−2

∂4ϕ

∂x2∂y2=

∂4ϕ

∂x4+∂4ϕ

∂y4

Commonly written as ∇2∇2ϕ = 0 (11.175)

the biharmonic equation, probably the highest order equation that you will see in undergraduate physics. Thisequation is one of the foundations of architectural engineering (I apologize, that was the worst pun ever).

To use the Airy stress function, you solve ∇2∇2ϕ = 0 subject to your problem’s boundary conditions, which shouldbe the applied loads on surfaces. The resulting ϕ gives t everywhere within the block of elastic matter, and fromEq. 11.169 you obtain the stresses.


Typical boundary conditions. Construct torques and load forces;

Nxx =

∫A

txx dA

Nxy =

∫A

txy dA

Mxx =

∫A

y txx dA (11.176)

Free end. At a free end of a beam, Nxx, Nxy,Mxx all vanish. This is the case of a cantilevered beam.

Simply supported end. No axial forces Nxx = 0.

Built-in end. No horizontal displacements at the end.

These forces and torques are related to one another through ∇ · t = 0; for example examine a loaded beam of length

ℓ top area 2bℓ loaded such that txy = 0 and tyy = q(x)2b on y = h, txy = tyy = 0 on y = −h∫ b

−bdz

∫ h

−hdy(∂txx∂x

+∂txy∂y

)= 0 (11.177)

∂

∂x

∫ b

−bdz

∫ h

−hdy txx = −

∫ b

−bdz(txy(h)− txy(−h)

)= 0

∂Nxx∂x

= 0∫ b

−bdz

∫ h

−hdy(∂txy∂x

+∂tyy∂y

)= 0 (11.178)

∂

∂y

∫ b

−bdz

∫ h

−hdy txy = −

∫ b

−bdz(tyy(h)− tyy(−h)

)= −q(x)

∂Nxy∂y

= −q(x) (11.179)

and ∫ b

−bdz

∫ h

−hdy y

(∂txx∂x

+∂txy∂y

)= 0 (11.180)

∂

∂x

∫ b

−bdz

∫ h

−hdy ytxx = −

∫ b

−bdz(ytxy

∣∣∣y=hy=−h

−∫ h

−hdy txy

)∂Mxx

∂x= Nxy (11.181)

x

y

Example 20. Pressure loading of a beam. We load the topsurface of the beam with a normal load force, the bottom of the beamis not loaded;

tx = txy = 0, ty = tyy = − P

2bfor y = h

tx = txx = ty = tyy = 0 for y = −h

attempt a polynomial solution

ϕ =∑m=0

∑n=0

am,n xn ym = ϕ2(y) + xϕ3(y) + x2ϕ1(y)

because we should have a symmetric solution ϕ(x, y) = ϕ(−x, y) for a uniform beam supported at both ends as Ihave illustrated, we reduce this to

ϕ = ϕ2(y) + x2ϕ1(y) (11.182)


This is the usual strategy taken in solving a complicated equation, we guess a solution containing coeffi-cients and parameters, substitute it into the equation, solve for the coefficients and parameters. It is low-tech, andrequires little or no advanced training in mathematical techniques.

Put it into ∇2∇2ϕ = 0 and we obtain

4∂2ϕ1∂y2

+ x2∂4ϕ1∂y4

+∂4ϕ2∂y4

= 0,∂4ϕ1∂y4

= 0 (11.183)

so thatϕ1 = a+ c y + b y2 + e y3

which in turn specifies

8b+ 24 ey = −∂4ϕ2∂y4

Now employ the surface conditions

txy

∣∣∣y=±h

= 0 =∂ϕ

∂y(±h) (11.184)

so ∂ϕ∂y must be even in y, forcing b = 0, then

24 ey = −∂4ϕ2∂y4

, ϕ2 = −e5y5 +B y4 + C y3 +Dy2 +Gy +H (11.185)

but ∂ϕ2

∂y must be even, forcing B = D = 0 (H is not important since it does not affect the stresses specified by the

Airy stress formulas). We have arrived at

ϕ2 = −e5y5 +

f

6y3 +

g

2y, ϕ1 = a+ c y + e y3 (11.186)

after renaming two constants. The stresses are

txx = 2e y(3x2 − y2) + f y + g

tyy = 2(a+ c y + e y3)

txy = −2x(c+ 3e y2) (11.187)

Boundary conditions on y = ±h make

a = − P

8b, c = − 3P

16bh, e =

P

16bh3(11.188)

with stresses

txx = y(3P x2 − 2P y2 + 48bh3 f)

8bh3

tyy =P

8bh3(y3 − 3h2y − 2h3)

txy =3P

8bh3x(h2 − y2) (11.189)

but both f and g are unspecified (and g is irrelevant) unless end-conditions at x = ± ℓ2 are specified. For example if it

is simply supported (as I have illustrated it in the figure), we have displacement conditions v(±ℓ/2) = d2vdx2 (±ℓ/2) = 0,

and u(±ℓ/2) = 0.

Example 21. A cantilevered beam. Consider an asymmetrical (left/right) solution;

ϕ = ϕ1(y) + xϕ2(y) (11.190)

which is automatically biharmonic if d4ϕ2

dy4 = 0,

ϕ2 = A+By + Cy2 +Dy3 (11.191)


and we can try to discard ϕ1 altogether and see if we can solve some meaningful problem with just

ϕ = x(A+By + Cy2 +Dy3) (11.192)

alone.

txy = −(B + 2Cy + 3Dy2)

txx = x(2C + 6Dy)

tyy = 0

Nxx = (2b)

∫ h

−htxx dy = 8bhC x

Nxy = (2b)

∫ h

−htxy dy = −4bh(B +Dh2)

Mxx = (2b)

∫ h

−hy txx dy = 4bh2 x (C + 2Dh) (11.193)

The first thing that we see is that with tyy = 0, there are no normal loads on the top (or bottom) surfaces of thebeam. We can place no shear loads on these surfaces if txy(±h) = 0. There is no x-directed force on the left end atx = 0, with the choice C = 0 we have no net x-directed forces at the right end either. Then no shear loads on thetop and bottom surface will set B + 3Dh2 = 0. This leaves us with

txy = −B(1− y2

h2)

txx = −2B

h2xy

tyy = 0

Nxx = (2b)

∫ h

−htxx dy = 0

Nxy = (2b)

∫ h

−htxy dy = −8bh

3B

Mxx = (2b)

∫ h

−hy txx dy =

8bh2

3B x (11.194)

Lets choose the coefficients remaining to create an applied downward force F on the left end,∫ b

−bdz

∫ h

−hdy t · (−i) = 2b

∫ h

−hdy

(0 −B(1− y2

h2 )

−B(1− y2

h2 ) 0

)(−10

)= 2b

∫ h

−hdy

(0

B(1− y2

h2 )

)=

(0

8bhB3

)=

(0

−F

)(11.195)

so we have a beam with no top/bottom forces applied to it, just an end-applied downward shear at the left end.

ϕ = − 3F

8bhxy +

F

8bh3xy3 (11.196)

What is going on at the right end? Compute strains and displacements

exx =txx − νtyy

E=

3F

4bh3Exy

eyy =tyy − νtxx

E= − 3Fν

4bh3Exy

exy =1 + ν

Etxy = (1 + ν)

3F

8bhE(1− y2

h2) (11.197)


Integrate to get displacements u and v;

u =

∫exx dx =

3F

8bh3Ex2y + F (y)

v =

∫eyy dy = − 3F

8bh3Exy2 +G(x) (11.198)

from which we will guess that F and G are no higher than cubic, and let REDUCE do all of the work;

% REDUCE code for cantilever

phi:=x*(A+B*y+C*y^2+D*y^3);

txx:=df(phi,y,2);

tyy:=df(phi,x,2);

txy:=-df(df(phi,x),y);

Nxx:=int(txx,y,-h,h);

Nxy:=int(txy,y,-h,h);

% Examination of the output indicates that...

c:=0;

a:=0;

b:=-3*d*h^2;

d:=F/(4*h^3);

% Lets get displacements, expand in powers up to cubic

operator cu,cv;

u:=cu(1)*x^2*y+cu(2)*y^3+cu(3)*y^2+cu(4)*y+cu(5);

v:=cv(1)*x*y^2+cv(2)*x^3+cv(3)*x^2+cv(4)*x+cv(5);

exx:=df(u,x)-txx/EE;

eyy:=df(v,y)+nu*txx/EE;

exy:=(df(u,y)+df(v,x))/2-(1+nu)*txy/EE;

% now examine the output, and match coefficients

% of like powers of x, y

let solve(exx,cu(1));

let solve(eyy,cv(1));

let solve(coeffn(exy,y,2),cu(2));

let solve(coeffn(exy,x,2),cv(2));

let solve(coeffn(exy,y,1),cu(3));

let solve(coeffn(exy,x,1),cv(3));

% write out the results

off div;

u;

v;

% which are

3 3 3 2 3

4*cu(5)*ee*h + 4*cu(4)*ee*h *y - f*nu*y + 3*f*x *y - 2*f*y

---------------------------------------------------------------

3

4*ee*h

36:

3 3 2 3

4*cv(5)*ee*h + 4*cv(4)*ee*h *x - 3*f*nu*x*y - f*x

------------------------------------------------------

3

4*ee*h


We have (let b = 1/2, I did this to simplify things)

u =F

4Eh3

(3x2y − (2 + ν)y3 + C1y + C2

)v =

F

4Eh3

(− 3ν xy2 − x3 +D1x+D2

)(11.199)

after renaming some constants to absorb numerical factors.

If the right end of the beam is rigidly mounted to a wall, we would need zero displacement at the x = ℓ end, so bothu, ∂v∂x and v should vanish there, but you can see this can’t be done for all y, The best we can do is to demandthat u(x = ℓ, y = 0) = 0 and v(x = ℓ, y = 0) = 0 (no displacements at the fixed end at its median plane) by taking(for example)

0 =F

4Eh3

(3 ℓ2y − (2 + ν)y3 + C1y + C2

)0 =

F

4Eh3

(− 3ν y2 − 3ℓ2 +D1

)0 =

F

4Eh3

(− 3ν ℓy2 − ℓ3 +D1ℓ+D2

)(11.200)

all for y = 0 give D1 = 3ℓ2, C2 = 0, D2 = −2ℓ3. You have to conclude that our original Airy function is onlyan approximate solution to this problem, but it is still rather good because it is simple (a polynomial). Our finaldisplacements are

u =F

4Eh3

(3x2y − (2 + ν)y3 + C1y

)v =

F

4Eh3

(− 3ν xy2 − x3 + 3ℓ2x− 2ℓ3

)(11.201)

and the moral of the story is that simple polynomial Airy functions are good approximate solutions to a variety ofbeam and support problems, in general most but not all boundary conditions can be satisfied by them.


11.3.4 Problems

1. Show that the speed of a harmonic oscillator of energy E and amplitude A and frequency ω when its displacementis x is

v = ω√A2 − x2

2. The displacement of a mass m = 0.01 kg is x(t) = 0.25m sin(

62.83s t − 0.785398

). Find its amplitude, its total

energy, and its speed at t = 0.

3. A mass m = 0.01 kg attached to a spring k = 100Nm initially at rest at zero displacement is given an impulsecausing its energy to jump up to 10 J . Find the amplitude of the resulting oscillations and the phase ϕ in the formula

x = A sin(ωt− ϕ

)

Solid; V(x) = 2a e−2ax−a e −ax

Dotted; V(x) = V(x0)+1/2 k (x−x0)2 + ...

4. Very small amplitude deviations from equi-librium (zero-force condition) for any potentialwill result in simple harmonic motion. For ex-ample a commonly used inter-molecular poten-tial (Morse) is

V (x) = 2ae−2ax − ae−ax

Show that a mass subjected to this potentialhas no force acting on it if it is located atx0 = 1

a ln 4, and that if it is allowed to devi-ate slightly from this position by amount x, itundergoes oscillations of frequency

ω =

√8a3e−2ax0 − a3e−ax0

m=

√k

m, k =

d2

dx2V (x)

∣∣∣x=x0

x

y

R xy

5. A very small mass m, to be thoughtof as a point mass, slides without frictionunder the influence of gravity in a bowl ofradius R as shown. The equilibrium point isat the bottom of the bowl. For very smallamplitude displacements x from equilibrium,find the frequency of the periodic motion thatit executes.Solve the problem if the mass is a sphere ofradius r that rolls without slipping.


x

θ

6. Rocking motion. A solid disk of mass Mand radius R has a mass m embedded in it at adistance ℓ from its center. It is placed on-edgeon a smooth surface. The dotted outline showsits initial position and orientation. It is rolledslightly (by amount x) to the left and released.It rolls back and forth about the equilibriuminitial position. Find the frequency of thisrocking motion.

x

y

7. Find the center of mass of a half-ring ofradius R and mass m. If this is placed on aflat surface and allowed to rock (roll) back andforth, find the frequency of the motion.

See the previous problem.

m1 m2

v0

8. A mass m1 connected to a spring of force constant kis at rest at equilibrium at the origin. It is struck inelas-tically by a mass m2 moving at speed v0 at t = 0. Findthe frequency, amplitude and phase of the resulting simpleharmonic motion

x(t) = A sin(ωt− ϕ)

Solve the problem again if the collision is elastic.


m1

m2

k

9. A mass m1 rests on a frictionless surface, and a massm2 rests upon m1. There is friction with coefficientsµk, µs between the blocks. The top block m2, shown atrest at equilibrium, is connected to a wall by a spring offorce constant k. Determine the maximal amplitude ofsimple harmonic motion such that both blocks co-movewith no slipping of m2 on m1.

10. Find the amplitude and phase of the simple harmonic motion executed by a mass m = 0.05 kg connected to aspring of force constant k = 100Nm with initial displacement x(0) = 0.1m and initial velocity v(0) = 5.0 m

s .

11. Find the maximal speed and acceleration of the simple harmonic motion executed by a mass m = 0.05 kg con-nected to a spring of force constant k = 100Nm with initial displacement x(0) = 0.1m and initial velocity v(0) = 5.0 m

s .

12. Find the equilibrium value of displacement x, and the frequency of small-amplitude motion for a particle of mass

m in potential V (x) =r120x12 − 4r60

x6 . Hints;

V:=(r0/x)^(12)-4(r0/x)^4;

solve(df(V,x),x); % will give xeq


taylor(V,x,xeq,4); % will expand around xeq

13. A mass m connected to a spring executes simple harmonic motion with velocity v(t) = 6.0 ms cos( 120s t −

π12 )

when connected to a spring of force-constant 100kgs2 . Compute m and the total mechanical energy possessed by thesystem.

14. A mass m = 0.05 kg connected to a fixed support with a spring of force constant k = 100 Nm has initial velocity

v0 = 0.4 ms and initial displacement x0 = 0.05m. Determine the amplitude of the simple harmonic motion that it

executes.


ρB ρWL−d

d

x

FB

FG

15. Compute the length d of an iceberg of volumeL3 that is out of the water when in static equilibrium(ρB = 0.92 g

cm3 , ρw = 1.0 gcm3 ). If the iceberg is pushed

down by an amount x and released, find the frequencywith which it bobs up and down in the water.

16. A SiO2 molecule has a linear oscillatory modethat is illustrated in the figure. The atomic massesare mO = 15.999Amu, mSi = 28.086Amu and thechemical bond between Si and O has force constantk = 120Amu

(ps2) , where 1 ps = 1.0 × 10−9 s. Find the fre-

quency of the oscillatory mode. 1Amu = 1.661×10−27 kg.

x

y

x

y

M M

17. Two blocks of mass M = 0.2 kg are connected asillustrated with springs. The far left and far right springshave k = 200 N

m , the middle spring has k = 300 Nm . The

masses are each displaced 2.0 cm away from equilibrium(top figure) in opposite directions and released. Find thefrequency of the resulting periodic motion. Find theamplitude A and total energy E of the oscillations.


ω n

θ18. This is a physical pendulum; an object allowed topivot about a fixed point under the influence of gravity.It consists of a stick of mass M of length ℓ with a slidingmass m attached a distance d below its center. Find thefrequency of its oscillations.

A

B

C

D

E

F

19∗. A disk of mass M and radius R is fixed to the wall with a nailthrough it at the rim. If displaced (center moved away from a verticalline through the nail) and released, it acts like a physical pendulum.Find its frequency of oscillation.

20∗. A hoop of mass M and radius R is fixed to the wall with a nailthrough it at the rim. If displaced (center moved away from a verticalline through the nail) and released, it acts like a physical pendulum.Find its frequency of oscillation.

21. To your left is a torsional pendulum, a steel disk (A) of radiusR = 15 cm and mass M = 4.0 kg suspended from a steel circular crosssection wire (B) of radius r = 0.1 cm and length ℓ = 1.0m. The diskis rotated slightly about its axis and released. Find the frequency ofthe torsional oscillations.

By how much is the steel wire stretched by the disk?

22. Compact bone has approximately the same Young’s modulusas aluminum. By how much does the lower leg-bone (radius 2.0cm,length 1.0m) of a 1200 pound thoroughbred racehorse compresssupporting its weight at rest?

M1 M2

k23. The spring is stretched by some smallamount and released. The disks roll back andforth at what frequency?


m1

m2k

24. A mass m1 rests on a frictionless surface, and a massm2 rests upon m1. There is friction with coefficientsµk, µs between the blocks. The bottom block m1, shownat rest at equilibrium, is connected to a wall by a springof force constant k. Determine the maximal amplitude ofsimple harmonic motion such that both blocks co-movewith no slipping of m2 on m1.

25. Determine the shearing force Ts needed to cut through a 1.0 cm radiuscircular cross-section steel bolt if the “kerf” is no more than 0.1 cm in length.

26. By how much does the length of a column of material of height h,cross-sectional area A and Young’s modulus Y and density ρ sag underits own weight? Think of any cross-section as supporting the weight ofall of the matter above it, and sub-divide it into thin slices. Add up the

deformations of all of the slices. Show that you get ρgh2

2E .

27. By how much does the end of a cantilevered beam of material of length ℓ, cross-sectional area A and shearmodulus G and density ρ deflect under its own weigh? Think of any cross-section as supporting the shear of all ofthe matter to the right of it, with the left end attached firmly to a wall, and sub-divide it into thin slices. Add up

the deformations of all of the slices. Show that the right end is ρgh2

2G lower than the left end. This assumes that thebeam deflects like a stack of cards, cross-sections remain vertical.

28. For most of the metals studied in section 11.3, the Poisson ratio is ν ≈ 0.3. From the table of moduli E,Gcompute the Lame’ coefficients for steel, aluminum, brass and titanium.

29. Show that if the components of the planar stress t are given by an Airy stress function Eq. 11.171, then ∇ ·t = 0is automatically satisfied. Does this remind you of the potential function from which all force components are gottenby taking derivatives?

30. Show that the free end of a cantilevered beam with end-load F is vertically deflected by Fℓ2

8bh3E from the Airysolution.

31. Consider a uniform pressure loading of a block of matter of volume V = h3;

txx = tyy = tzz = −P, txy = tyz = txz = 0

Show that the volume change of the block is

∆V = exx + eyy + ezz

The bulk modulus κ of a material is defined by

−P = κ∆V

V, show that κ = λ+

2

3µ


32. A traveling wave of wavelength λ moving at speed c on a string is a displacement of a point x on the string attime t

∆x = A sin(2π

λ(x− ct)

)Show that every point x on the string executes simple harmonic motion. Find its amplitude and frequency.

33. Consider Eq. 11.142 for a block of matter at equilibrium acted upon by gravity

∇ · t = −ρg k

Show that

tij = 0, i, j = 3, t33 = −ρg z

Suppose that the block of matter is supported at its lower end at z = 0, has height ℓ and width/depth 2h (afree-standing square cross-section column). Show that

e12 = e13 = e23 = 0, e11 = e22 =ν

Eρg z, e33 = −ρg

Ez

Consider deformations u3 = w, u1 = u, u2 = v, and find the deformations that account for these stresses and strains.What is the maximal lateral “bulge” of the column? By how much does it sag under its own weight?

34. Consider the Airy function ϕ = (A+By+5Cy3)x2+Dy3−Cy5, which we have shown is biharmonic, and applyit to the problem of a square cross-section beam (A = (2h)2) of length ℓ that is torque-free at its left end x = 0.Use REDUCE to find all stresses;

% Torque-free left end

phi:=(A+B*y+5*C*y^3)*x^2+D*y^3-C*y^5;

txx:=df(phi,y,2);

tyy:=df(phi,x,2);

txy:=-df(df(phi,x),y);

% the equations

M:=int(y*txx,y,-h,h);

Nxx:=int(txx,y,-h,h);

EQ1:=sub(y=h,tyy)+P/(2*h);

Eq2:=sub(y=-h,tyy);

Eq3:=sub(y=-h,txy);

Eq4:=sub(x=0,M);

% solve for A,B,C,D

sols:=solve(Eq1,Eq2,Eq3,Eq4,A,B,C,D);

let sols;

txy;

tyy;

txx;

Can you find the displacement functions, and satisfy the additional condition that the right end at x = ℓ is lockedinto place (no displacements) at least on the median plane y = 0?

35. Find the strains and stresses on an h× h block of matter with moduli E, ν subjected to a pure planar shear(dx1

dx2

)=

(1 2γ0 1

)(dX1

dX2

)36. Find the strains and stresses on an h× h block of matter with moduli E, ν subjected to a pure planar isotropicdilation (

dx1

dx2

)=

(1 + ϵ 01 1 + ϵ

)(dX1

dX2

)


37. Most types of wood cannot suffer through a maximal compressionof more than 1.5% of the length of the piece without rupturing on thecompressed side, or maximal elongation of more than 2.0% of the lengthof the piece without rupturing on the stretched side. Consider a piece ofwood h × h × ℓ as illustrated. Bend it into an arc of a circle subtendingtotal angle θ, so that its median line (dotted) remains ℓ in length. Findthe largest θ for which the plank does not rupture on its “belly” (concave)side.

38. Find the deformations within the plank in the previous problem for small θ, let x start at the center of theplank on the median line and follow the median, and y be ⊥ to the median, with origin at the plank center. Findthe stresses in the plank.

39. The energy method. Suppose that you can write the energy of a system as

E = D(dxdt

)2+Bx2 + C

then you can solve for x ∫dx√

(E−C)D − B

Dx2

=

√D

Bsin−1

(√ B

E − Cx)=

∫dt = t− t0

inverting this to get

x(t) =

√E − C

Bsin(√B

D(t− t0)

)from which you can read off both amplitude A =

√E−CB and frequency ω =

√BD . Rework 5, 8, 17, and 23 by this

method.Example. Revisit example 11; when the spring is stretched by x and disk rolls at v = ω/R (note we use symbol ωfor rolling spin-rate and for frequency of oscillation, which are two different things)

E =1

2mv2 +

1

2I( vR

)2+

1

2kx2, ω =

√k

m+ IR2

x y

Example. A diatomic molecule in space moves freely with astationary center of mass. This requires that when the atomsdisplace,

−m1 x+m2 y = 0

The energy is

E =1

2m1x

2 +1

2m2y

2 +1

2k(x+ y)2

or

E =1

2

(m1 +m2

(m1

m2

)2)x2 +

1

2k(1 +

m1

m2

)2x2, ω2 =

k(m1 +m2)

m1m2

40. Re-examine the physical pendulum (problem 11.18). Suppose that a planar body of mass m has moment ofinertial Icom about an axis through its center of mass. You pass a nail through it a distance x from the COM aboutwhich it will swing as a physical pendulum under gravity. Find its frequency ω as a function of x. Find the value ofx that maximizes the frequency of oscillation.

41. A mass m is subjected to potential energy function V = b2 x

3 + c4x

4 with b, c > 0. Find the equilibrium positionxeq where the body experiences no force, and the frequency of small-amplitude oscillations about this equilibriumpoint.


42. In analogy with a spring, which stores potential energy

U =1

2k (ℓ− ℓ0)

2 =1

2k (∆ℓ)2

when stretched or compressed, show that an elastic bar of length ℓ0 and cross-sectional area A0 stores energy

U =1

2

EA0

ℓ0(∆ℓ)2

if stretched or compressed in such a way that its cross-sectional area does not change.

Chapter 12

Gravitation and Planetary Motion

12.1 Newton’s universal law

Isaac Newton was the first person that we know of to have made the direct cause and effect link between the motion ofplanets around the sun, and the force of gravity. By physical and purely mental experimentation Newton determinedthat the force of gravity exerted on mass M1 by mass M2 satisfies

x

y

M1

r1

F1,2

M2

r2

F2,1

F12 = −M1M2G(r1 − r2)

|r1 − r2|3

G = 6.67× 10−11Nm2

kg2(12.1)

This was an amazing deduction that can easily betaken for granted today, but at the time at which itwas made, it ranked as the most important physicaldiscovery of the century. The magnitude of theforce between two masses varies with the inversesquare of the distance between them, and is alwaysattractive, pulling one mass towards the other. Thisforce acts vectorially just like any other vector forces.

Example 1. Four identical masses m are held in place at the corners (0, 0), (0, d), (d, 0), and (d, d) of a square.Compute the force on the mass placed at the origin.

F = −m2G( (0, 0)− (0, d)

d3+

(0, 0)− (d, 0)

d3+

(0, 0)− (d, d)

(√2d)3

)(12.2)

which can be simplified a bit to

F =m2G

d2

((1 +

1√8), (1 +

1√8))

(12.3)

Example 2. Compute the period of a binary star system of masses M1 and M2.Such a star system could for example consist of the stars in circular orbits about their center of mass. If the centerof mass is at the origin then the planets travel in circles of radii r1 and r2 that satisfy

0 =M1r1 −M2r2 (12.4)

and are separated by a distancer = r1 + r2 (12.5)

233

234 CHAPTER 12. GRAVITATION AND PLANETARY MOTION

x

y

M1

v1

r1

M2v2

r2

Examine planet number one. It is in a circularorbit and therefore has centripetal accelerationac

ac =v21r1

(12.6)

and the centripetal force must be supplied bythe gravitational attraction to mass M2

M1v21r1

=M1M2G

(r1 + r2)2(12.7)

The period is the time needed to orbit the cen-ter of mass once

T =2πr1v1

(12.8)

we find that

r1 =M2r

M1 +M2, v1 =

√M2

2G

r(M1 +M2)(12.9)

which leads to

T =2πr

32√

G(M1 +M2)(12.10)

which we have written in terms of the interstellar distance r.

x

y

Mv

RF1F2

Example 3. Compute the period of a triplestar system of masses M and equal radii R.Such a star system again consists of the starsin circular orbits about their center of mass. Ifthe center of mass is at the origin then the plan-ets travel in circles of radii R. The interstellardistance is R12 = R

√3. The acceleration vec-

tor of the top star is

ac = −v2

Rj (12.11)

and this is produced by the forces exerted bythe two lower stars;

F1,2 + F1,3 =M2G

(√3R)2

(sin 30oi− cos 30oj

)+

M2G

(√3R)2

(− sin 30oi− cos 30oj

)= − M2G√

3R2j (12.12)

12.2. GRAVITATIONAL POTENTIAL ENERGY 235

and so

M(− v2

Rj)=M

(− 22π2R2

T 2Rj)= − M2G√

3R2j (12.13)

The period T of the motion is therefore

T 2 =4π2

√3R3

MG(12.14)

12.2 Gravitational potential energy

We can construct a potential energy function for the gravitational field by equating the difference in potential energybetween points at ri and rf with the work done against the gravitational field in moving from the first to the secondpoint.

WFg = −(V (rf )− V (ri)

)=

∫ rf

ri

−mMGr

r3· dr (12.15)

where the mass M at the origin exerts gravitational forces on the test mass m that we are moving through the forcefield of stationary mass M . We can rewrite this by introducing the vector derivative or gradient

∇ =( ∂∂x,∂

∂y,∂

∂z

)= i

∂

∂x+ j

∂

∂y+ k

∂

∂z(12.16)

and by noting thatx dx+ y dy + z dz

(√x2 + y2 + z2)3

=r · drr3

= −(∇ 1√x2 + y2 + z2

) · dr = −(∇1

r) · dr (12.17)

and the integral becomes an integral of an exact differential

−(V (rf )− V (ri)

)=

∫ rf

ri

∇(− MmG

r

)· dr

= −MmG

r

∣∣∣rfri

= −mMG

|rf |+MmG

|ri|(12.18)

we discover that the point mass M placed at the origin produces a gravitational potential energy field whose valuedepends only on how far one is from the origin

V (r) = V (r) = −MmG

|r|(12.19)

This formula completely replaces Vg = mgh in any calculation in which the altitude h is a significant distancecompared to the radius of the planet.

Example 4. A satellite falls from a radius of two earth radii to the ground. With what speed does it hit the earth?Conservation of energy says

− MmG

2Re=

1

2mv2 − mMG

Re(12.20)

so the impact speed will be

v =

√MG

Re(12.21)

where M is the mass of the earth.

Example 5. Two planets of mass M1,M2 and radii R1, R2 are separated by distance R >> Ri and are initially atrest. They fall towards one another. How fast are they each moving when they collide?They collide when the centers are separated by distance R1 +R2. Energy and momentum are conserved;

M1v1 =M2v2,1

2M1v1 +

1

2M2v2 −

M1M2G

(R1 +R2)= 0 + 0− M1M2G

R(12.22)


Solving we get

v21 =2M2

2G

(M1 +M2)

( 1

R1 +R2− 1

R

)(12.23)

Example 6. At what speed must a projectile be launched from the earth’s surface at in order to barely escape theearth’s gravitational field?This is the escape velocity, and is found from the condition that v = 0 when R→ ∞, since the gravitational fieldreaches to infinity. Conserve energy;

1

mv2e −

mMeG

Re=

1

2m02 − mMeG

“∞′′ = 0 (12.24)

and we obtain

ve =

√2MeG

Re(12.25)

Example 7. A black hole is a star that has collapsed under its own gravitational field. This happens when energyproduction in the stellar core is too slow to provide enough heat and pressure to support the star from within. Theevent-horizon is a roughly spherical shell around the hole upon which the escape velocity is the speed of light. Fora one-solar mass black hole, find its radius.We have

c =

√2M⊙G

Reh, Reh =

2M⊙G

c2=

(2)(2× 1030kg)(6.67× 10−11Nm2

kg2 )

(3× 108ms )2

= 3.0 km (12.26)

This is the physical size of a black hole of the same mass as the sun. Black holes do not form from stars whose initialmass is below three solar masses, although in the process some mass is blown off in the explosion and collapse.

M1

M1

M2

M2

a b

b a

M3

M3

Example 8. Find the gravitational force exerted on M3,and find the work needed to move it from the top positionto the lower.For the top figure

F = −M1M3G

a2i+

M2M3G

b2i (12.27)

and

Wf = −(Vf − Vi))

= −(−(M1M3G

b+M2M3G

a

)+

(M1M3G

a+M2M3G

b

))(12.28)

Example 9. Two stationary masses (held in place) M are located at points (x, y, z) = (0, b, 0), (0,−b, 0). A thirdmass m is located at (a, 0, 0). Find the force on m and the work needed to move it to (“∞′′, 0, 0).The work is easy;

W = −(Vf − Vi) = −(0 + 0− −mMG√a2 + b2

− −mMG√a2 + b2

) (12.29)

(you add up the potentials between m and each other mass). The force is

F =mMG

(√a2 + b2)2

(− a√

a2 + b2i+

b√a2 + b2

j)+

mMG

(√a2 + b2)2

(− a√

a2 + b2i+

−b√a2 + b2

j)

= 2mMG

a2 + b2−a√a2 + b2

i (12.30)


x

y

M

dm=ρ dy

y

x

Example 10. Find the gravitational force exertedby a stick of mass m length 2ℓ on a point mass Mheld a distance x from the stick center.All we have to go on is the form of the force be-tween two point-masses, so divide the stick into pointmasses of length dy and mass

dm = ρ dy =m

2ℓdy (12.31)

The fragment illustrated exerts force

dF =dmMG

(√x2 + y2)3

(− xi+ yj

)(12.32)

on the point-mass M , and integration gives a totalforce

F =

∫ ℓ

−ℓ

mdy

2ℓ

MG

(√x2 + y2)3

(− xi+ yj

)(12.33)

x

y

Mdm=ρ dy y x

Example 11. Find the gravitational force exertedby a stick of mass m length 2ℓ on a point mass Mheld a distance x from the stick end.divide the stick into point masses of length dy andmass

dm = ρ dy =m

2ℓdy (12.34)

The fragment illustrated exerts force

dF =dmMG

((x+ y)3

(− (x+ y)i

)(12.35)

and integration gives

F =

∫ 2ℓ

0

mdyMG

2ℓ((x+ y)3

(− (x+ y)i

)=

mMG

2ℓ

( 1

x+ 2ℓ− 1

x

)i (12.36)


z

M

θ

Rdm=ρ R dθ

Example 12. Find the gravitational force exertedby a ring of mass m radius R on a point mass Mheld a distance z from the center.Divide the ring into point masses of length Rdθ andmass ρR dθ, ρ = m

2πR . The fragment at location

r1 =(R cos θi+R sin θj

)(12.37)

is a distance√z2 +R2 from our mass, and so exerts

dF =dmMG

(√z2 +R2)3

(R cos θi+R sin θj−zk

)(12.38)

on it. Two of three integrals are zero;

F =

∫ 2π

0

ρRdθMG

(√z2 +R2)3

(R cos θi+R sin θj− zk

)= − ρ2πRMGz

(√z2 +R2)3

k = − mMGz

(√z2 +R2)3

k (12.39)

Example 13. Compute the gravitational potential energy function for a ring of mass m radius R, at a distance zfrom its center.The potential is much easier to compute than the force by integration. In the setup of the previous example, the

fragment d at r1 =(R cos θi+R sin θj

)produces

dV (z) = − dmMG√z2 +R2

(12.40)

at the location of the mass M , and so this is the potential energy of the pair of charges M and dm. Add up thepotential energies of M with all of the ring fragments;

V (z) = −∫

dmMG√z2 +R2

= − mMG√z2 +R2

(12.41)

for the total. Furthermore we can compute Fz from the fact that the gravitational force is conservative;

Fz = −∂V (z)

∂z= − mMGz

(√z2 +R2)3

(12.42)

in agreement with the previous example.

z

M

r dr

dm=ρ 2π r dr

Example 14. Compute the gravitational force on amass M exerted by a flat disk of matter of mass ma distance z away.Divide the disk into concentric rings of thickness dr,radius r and mass dm = m

πR2 2πr dr. Use the expres-sion for the force exerted on M by a ring;

F =

∫ R

0

− MGz

(√z2 + r2)3

m

πR22πr dr k

= −2mMG

R2

( z|z|

− z√z2 +R2

)k (12.43)


z

M

R

θ

R sinθ

dm

Example 15. Compute the gravitational potentialenergy of a mass M inside of a thin-skinned hollowspherical shell of radius R and mass m.Subdivide the shell into a stack of rings, one seenedge-on in the figure. Each ring has radius r =R sin θ, is a distance z + R cos θ from M , and ifcut and laid flat would be a strip of area dA =(Rdθ)(2πR sin θ) and mass

dm = ρ dA =m

4πR22π R2 sin θ dθ =

m sin θ dθ

2(12.44)

The potential energy of M in the field of the stackof rings is

V (z) =

∫ π

0

− dmMG√(z +R cos θ)2 +R2 sin2 θ

(12.45)

=

∫ π

0

− MG√(z +R cos θ)2 +R2 sin2 θ

m sin θ dθ

2=mMG

2Rz

(|R− z| − |R+ z|

)(12.46)

If z < R, this results in

V (z) =mMG

2Rz

((R− z)− (R+ z)

)= −mMG

R(12.47)

a result completely independent of z, therefore inside of this hollow sphere there is no gravitational field and Mexperiences no force

Fz = −∂V∂z

= 0 (12.48)

If z > R we obtain

V (z) =mMG

2Rz

((z −R)− (R+ z)

)= −mMG

z(12.49)

and m experiences a force

Fz = −∂V∂z

= −mMG

z2(12.50)

which is exactly the same as if all of the mass M comprising the shell were concentrated at is center.


M

ζ

z−ζ

dζ√r2−ζ2

Example 16. Compute the force exerted bya ball of radius r, mass m on a point mass Ma distance z > R away (measured center tocenter).Subdivide the sphere into a stack of disks, theone shown is z − ζ away from M , has radius√r2 − ζ2 and has mass

dm =m

4π3 r

32π(√r2 − ζ2)2 dζ (12.51)

It exerts a force

dF = −2 dmMG

r2

(1− z − ζ√

(z − ζ)2 + r2

)k

(12.52)on M , and integration results in

F =

∫ r

−r−2MG

r2

(1− z − ζ√

(z − ζ)2 + r2

) m4π3 r

32π(√r2 − ζ2)2 dζ k = −mMG

z2k (12.53)

This is the same result that we would get by replacing the entire ball with a point mass m and placingit at the center of the ball. We can see this at work in the previous problem as well.

Example 17. Find the force of gravity exerted by a solid ball of massM radius R on a point mass m located insideof it a distance r < R from the center.The previous examples suggest dividing the ball into concentric shells of radii ρ, thickness dρ and mass dm(ρ) =

(4πρ2)(dρ) m4πR3

3

= 3mρ2 dρR3 , each acting like a point mass concentrated at the center of the sphere and exerting

dFr =−mdm(ρ)G

r2if ρ < r (12.54)

on m. Adding up the forces exerted by these shells we arrive at

Fr =

∫ r

0

−mG

r23ρ2 dρ

R3= −mMGr

R3(12.55)

which vanishes at the center of the ball (r = 0).

12.3. PLANETARY ORBITS 241

12.3 Planetary orbits

x

y

a

b

aε

r1

r2

The ellipse is a conic section consisting ofall points whose summed distances fromthe two focal points is a constant 2a Oneof the great achievements of Newtoniangravitational theory was the correctcomputation of all of the laws of plan-etary motion, previously established byobservation and computation by Braheand Kepler, as consequences of the lawof gravitation. By Newton’s time itwas known that planets orbited the sunon elliptical paths with the sun at a focus

r1 + r2 = 2a (12.56)

where a is called the semi-major axis.The degree of deviation away from a per-fect circle is measured by the eccentricityϵ, such that the distance between the fociis 2aϵ.

x

y

θ

r

ra

va

rpvp

The semi-minor axis is then the shortaxis of the ellipse and satisfies

b2 + (aϵ)2 = a2, b = a√1− ϵ2

(12.57)The point of closest approach of a planetto the sun, which resides at a focal pointof the orbit, is called perihelion and is thedistance

rp = a(1− ϵ) (12.58)

The point of greatest departure from thesun is aphelion, the distance being

ra = a(1 + ϵ) (12.59)

The angular momentum of the planet is conserved, since the torque on it is

τ =d

dtL = r× F = r× (

−MmG

r3r) = 0 (12.60)

and so the orbit will be planar, and angular momentum magnitude and direction will be constant. If at perihelionthe velocity is vp and at aphelion is va we see that conservation of angular momentum says

mvaa(1 + ϵ) = mvpa(1− ϵ) (12.61)

since at both of these points the velocity being tangential to the orbit is perpendicular to the radius vector from thesun. The consequence is that as the planet nears perihelion it speeds up, and slows as it reaches aphelion. This can


be seen in a different way. The velocity vector can be expressed in polar coordinates as

v = rdr

dt+ θr

dθ

dt(12.62)

and sincer × r = 0, r × θ = k (12.63)

the angular momentum is

L = mr× v = mr2dθ

dtz (12.64)

but of course the area of the wedge spanned by r and the vector r dθ θ is dA = 12r× (r dθ θ) and so

|L| = 2mdA

dt(12.65)

x

y

θr

since the magnitude of L is constant, thissays that the line connecting the planetto the sun will sweep out equal areas inequal times. This means that the planetmust speed up as r shrinks in order toalways sweep area at the same rate.Integrate this expression over one full pe-riod, the time for a complete orbit to oc-cur. The area swept out will be the areaof the entire ellipse

TL = 2mA

T =2m

Lπab (12.66)

=2m

Lπa2

√1− ϵ2

We will now establish the relation between (a, ϵ) and (E,L) so that the orbital geometry and period can be expressedas functions of the energy and angular momentum of the orbit.

First of all we write the velocity squared in polar form starting with

v =dx

dti+

dy

dtj (12.67)

andx = r cos θ, y = r sin θ (12.68)

Differentiatedx

dt=dr

dtcos θ − r

dθ

dtsin θ,

dy

dt=dr

dtsin θ + r

dθ

dtcos θ (12.69)

Now square these and add them together

v2 =(dxdt

)2+(dydt

)2=(drdt

)2+(rdθ

dt

)2(12.70)

and the energy, which is constant, is (in polar coordinates)

E =1

2m((dr

dt))2

+(rdθ

dt

)2)− mMG

r(12.71)

Why do we do this? We want to make use of the fact that at perihelion and aphelion that the velocity is purelytangential so that

dr

dt= 0 when r = a(1± ϵ) (12.72)


In addition we can eliminate the theta derivative via

L

mr2=dθ

dt(12.73)

and so

E =1

2

L2

mr2− mMG

r, r = ra, rp (12.74)

at aphelion or perihelion. This we can solve for ra,p which will be two formulas relating E and L to a and ϵ.

1

ra,p=mMG±

√(mMG)2 + 2EL2

m

L2

m

(12.75)

Set each up separately

1

a(1− ϵ)=mMG+

√(mMG)2 + 2EL2

m

L2

m

(12.76)

and

1

a(1 + ϵ)=mMG−

√(mMG)2 + 2EL2

m

L2

m

(12.77)

eliminate ϵ between these to obtain

E = −mMG

2a(12.78)

The energy depends only on the semi-major axis of the orbit and nothing else. Substitute this back into the equationto obtain

ϵ =

√1 +

2L2E

m3M2G2(12.79)

and so the angular momentum determines the eccentricity of the orbit. Notice that the energy is negative, meaningthat the planet is bound, it will not escape the sun’s gravitational field. These results can be inserted back into theperiod expression

L =

√m2MGa

2(1− ϵ2), T =

2ma2√1− ϵ2

L(12.80)

to arrive at Kepler’s third law of planetary motion

T 2 =4π2a3

MG(12.81)

The period of a planets orbit depends on the suns mass and the orbital semi major axis only.

Lets wrap it all up by showing that the orbital energy equation Eq. 12.74 really does imply elliptical orbits. We usethe fact that angular momentum is conserved, since a radial force exerts no torque

dL

dt= 0 =

d

dt

(mr2

dθ

dt

)(12.82)

This we express as

dt =mr2

Ldθ (12.83)

and substitute this into Eq. 12.74 to get

E =1

2m( L

mr2dr

dθ

)2+

1

2m( Lr

mr2

)− mMG

r

=L2

2m

( 1

r2dr

dθ

)2+

L2

2mr2− mMG

r

=L2

2m

(dudθ

)2+L2

2mu2 −mMGu, u =

1

r(12.84)


Twist this around, solving for dudθ and integrate∫

du√2mEL2 − u2 + 2m2MG

L2 u=

∫dθ = θ − θ0 (12.85)

This is an integral that you hopefully have seen in your calculus courses, it is just an inverse-trig function∫du√

α2 − u2 + 2βu= − sin−1

( β − u√α2 + β2

)= θ − θ0

u = β +√α2 + β2 sin(θ − θ0)

=m2MG

L2−√

2mE

L2+ (

m2MG

L2)2 cos θ, θ0 = π/2

=m2MG

L2

(1−

√1 +

2EL2

m3M2G2cos θ

)=

( 1

a(1− ϵ2)

)(1− ϵ cos θ

)r =

a(1− ϵ2)

1− ϵ cos θ(12.86)

which is the equation of an ellipse of semi-major axis a and eccentricity ϵ, oriented so that aphelion occurs atθ = 0 as in all of our pictures illustrating elliptical orbits.

Example 18. At what points on the orbit are you at r = a?Two locations, solutions to

cos θa = ϵ, θ = cos−1 ϵ (12.87)

Example 19. Where are you on your orbit? If you know E, ϵ, ℓ and r, your distance from the sun, can the angle θbetween r and true aphelion be found?Well of course it ca, that is the purpose of this equation;

cos θ =(r − a(1− ϵ2)

ϵ r

)(12.88)

(note the θ = 0 if you are at the aphelion point).

12.3.1 Orbital corrections

Suppose that you are in an orbit, and you want to alter it (increase/decrease a or ϵ, move inward or outward).You will do this by applying thrust, which will instantaneously alter your velocity but not your position.Instantaneous velocity changes always alter E, therefore a as well, but need not alter L (if they are radial).

Suppose that you make an orbital alteration that leaves L fixed (a radial speed boost) but changes E, a, ϵ, then notethat since

m2MG

L2=

1

a(1− ϵ2)(12.89)

you do not change the numerator of the r versus θ equation, so the new semi-magor axis and eccentricity obey

a′(1− ϵ′2) = a(1− ϵ2) (12.90)

and your aphelion point must shift, it can’t remain at θ = 0 since

a(1− ϵ2)

1− ϵ cos θ= r =

a′(1− ϵ′2)

1− ϵ′ cos θ′ϵ cos θ = ϵ′ cos θ′ (12.91)

cannot be satisfied by θ = θ′ and the orbit equation gives your position relative to aphelion. For exampleif you are initially in a circular orbit, ϵ = 0, then cos θ′ = 0 so θ′ = π

2 ! A purely radial rocket burst will have the


following outcomes;

a

In this picture we begin at “a” with a purely verticalvelocity and go into a circular orbit (black) for oneperiod. Passing through “a” we give the satellite anoutward sudden push. We begin moving outward ina new orbit (red) with aphelion/perihelion shiftedby ±π/2.

a

In this picture we begin at “a” with a purelyvertical velocity and go into a circular orbit forone period. Passing through “a” we give thesatellite an inward sudden push. In both casesthe angle between our position vector at the timeof the push, and the point of closest approach is π/2.

These figures are the output of numerical integration of the equations of motion starting with the initial data at “a”,using MG = 2.0, a = 1.0, vy = 1.414, and vx = 0.0. The outward radial velocity boost is by 0.4 units delivered atT = 4.442 time units (one period) and the inward boost is 0.2 velocity units.

a

Begin at “a” with a purely vertical velocity and gointo a circular orbit for one period. Passing through“a” we give the satellite an upwards (tangential)push, increasing its orbital speed. We begin movingoutward in the orbit, but the boost-point becomesperihelion.

a

Begin at “a” with a purely vertical velocity and gointo a circular orbit for one period. Passing through“a” we give the satellite a sudden “de-boost”, aretrograde tangential rocket burst, slowing it inits orbit. Now the boost-point where the rocketdeployed becomes aphelion.


12.3.2 High-altitude projectiles

The equation of an ellipse in Cartesian coordinates, with the origin at the center is

x2

a2+y2

b2= 1 (12.92)

With the origin at (−aϵ, 0) it becomes(x+ aϵ)2

a2+y2

b2= 1 (12.93)

and in polar coordinates(r cos θ + aϵ)2

a2+r2 sin2 θ

b2= 1 (12.94)

Think of this as a quadratic equation for r and solve it for our more familiar form up above;

r =a(1− ϵ2)

1− ϵ cos θ(12.95)

which places aphelion ra = a(1 + ϵ) at θ = 0 and perihelion at θ = π. This coincides with our earlier pictures. Ifyou know the energy and angular momentum of the orbit, one can use them to compute a and ϵ and this expressiongives the equation for the orbit itself as a curve in space.

φ

θ Re

v0

sIn this example we will compute therange of a high-altitude projectile,launched from earth-radius Re at veloc-ity v0 at angle ϕ elevated above the sur-face of the earth. The energy and angularmomentum will then be

E =1

2mv20 −

mMeG

Re|L| = mv0Re cosϕ (12.96)

The range of the projectile over theearth’s surface is then the arc-lengths between the two places where the“orbit” intersects the earth’s surface,

Range = s = 2Reθ = 2Re cos−1(1− a(1−ϵ2)

Re

ϵ

), Re =

a(1− ϵ2)

1− ϵ cos θ(12.97)

and this angle is computed from the a and ϵ that we get from the energy and angular momentum.

12.4. PROBLEMS 247

12.4 Problems

1. Suppose that you are able to launch yourself at initial speed v0 = 0.45 ms from any surface. Stoney asteroids are

made of silicates of density ρ = 3000 kgm3 , and have masses m = 43πR

3ρ where R is the radius. What is the largestradius stoney asteroid that you could leap off of in an escape trajectory (and drift off into the cold, dark and scaryreaches of outer space)?

y

x

F1 F3

F2

2. Consider the quadruple star system illustrated below.Each star is a distance R = 1.0 × 1011m from the origin,and have masses m = 2.0× 1031 kg.A. Find the magnitude and direction of the net gravita-tional force acting on the top star.B. Find the orbital speed v of the top star.C. Find the gravitational potential energy of the top star.

3. A mass m placed a distance z << R above the center of a ring of mass M and radius R on the symmetry axis ofthe ring will execute simple harmonic motion about the equilibrium point at the center of the ring. Find its frequency.

θ

M

m

m

4. Find the force acting on the upper-right star in this triple-starsystem, withM = 7m. Determine the period of the orbit as a functionof G, R, and m.If you want numbers, use m = 1.0 × 1031 kg and R = 3.0 × 1010m,θ = 60o.

M

m

µ

v

w

R

5. A planet of mass m has a circular orbit arround a very massivestar M >> m of radius R. Determine the speed v of its orbit. It isstruck inelastically by a fast-moving comet of mass µ << m travelingat w = 2v. Find the new semimajor axis and eccentricity of theplanets orbit.

6. A planet of mass m moves in a circular orbit of radius R around a star. A satellite of mass µ << m attempts a


gravitational sling-shot by making a close-pass around the planet, swinging by in a path of radius r << R. Afterthe encounter, find the new speed w′ of the satellite, and v′ of the planet.

M

m

µvw

r

R

M

m

µv’

w’

r

R

R

M

md

7. Find the force exerted on a point-mass m by aplanet of mass M and radius R that has a hollowwithin it in the shape of a sphere of radius R

2 asillustrated. The point-mass is a distance d from thecenter of the larger sphere.

8. Imagine that the earth (mass M , radius R) hada tunnel that passed through its center, connectingtwo diametrically opposed points on the surface(geographical, not political). Find the total gravita-tional force exerted by the earth on a body of massm that is in the tunnel, a distance r < R from thecenter. If r << R, this mass could execute simpleharmonic motion about the earth’s cnter if released.Find the frequency.

12.4. PROBLEMS 249

θ

dm

R

Mx

y

9. Find the gravitational potential energy of a mass M inthe field of a half-ring of mass m, radius R, held at thecenter of the ring.Find the force on M .

10. Show that the time spent in the two “halves” of an elliptic orbit on either side of the semi-minor axis are

Tℓ/r =(12± ϵ

π

)T

in which T is the orbital period (ℓ/r stands for left/right, in which the sun is in the left half of the orbit as in ourpicture of an elliptic orbit).

11. Show that the planetary speeds at perihelion/aphelion for an elliptical orbit of semi-major axis a, eccentricity ϵare

v2peri =MG

a

(1 + ϵ

1− ϵ

), v2aph =

MG

a

(1− ϵ

1 + ϵ

)12. Suppose that you must bring down a satellite from a circular orbit at an altitude h above the earth. You do thisby making an orbital correction that puts it into a new orbit with a perigee distance Re. How must the thrust bedirected? What must the energy change given to the satellite be? What must the angular momentum change givento the satellite be? Where will it land?

13. Suppose that a satellite has gone into an elliptical orbit of semimajor axis a, eccentricity ϵ, and as it passesthrough apogee you must make an orbital correction that puts it into a circular orbit through that point. Find the∆E and ∆L for the correction.

14. Find the smallest radius stoney-asteroid (ρ = 3000 kgm3 )radius R that you could tee-off on and hit a golf ballwith initial velocity 60ms directed horizontally, and haveit pass all the way around the asteroid on a circular orbitof radius R.

M1

M2

x R−x

15. Dynamic equilibrium is a state of having no net forceacting on you. Suppose that there twp planets M1 andM2 a distance R apart. Find a place between them adistance x from M1 where a rock m can be placed suchthat there is no force acting on it (find x).Equilibrium is stable if a slight displacement results in anet force trying to pull the object back to its equilibriumposition. Is the equilibrium point in this problem stable?


16. One astronomical unit is equal to 149, 597, 900 km, the radius of the Earth’s very nearly circular orbit aroundthe sun. The aphelion of the orbit of Mars is 1.666AU and perihelion is 1.382AU . Find its semimajor axis andeccentricity.

17. Find the orbital period of Mars in years from your results of the previous problem and the mass of the Sun1.989× 1030 kg.

18. Compute the orbital speed of the earth from the radius 149, 597, 900 km of the Earth’s orbit around the sun. Ifa rocket is to be launched on an escape orbit from the solar system, launched from the Earth tangentially to Earth’sorbit, with what velocity relative to the earth must it be launched?

19. How fast would the earth have to be spinning on its axis in order that you would “float away” at the equator(the normal force exerted by the Earth on you would vanish)? If you are worried, you could move to one of the poles.

20. What is the ratio of your weight as measured at the north pole compared to the equator? Your weight is thenormal force exerted by the Earth on you.

21. Show that the equilibrium position for the potential

V (r) =L2

2mr2− mMG

r

describes a circular orbit. Examine the small-amplitude radial oscillations of Example 11.14. What are the eccentric-ities of such orbits? Show that the equation for the frequencies of such small-amplitude oscillations about a circularorbit is in fact Kepler’s third law.

r

22. Find the force exerted upon you (your massis m) by a planet of total mass M and radius R(you are r from its center) which has been partiallyhollowed out by space-miners extracting dark matterto be used as starship fuel (end gratuitous Futuramareference).

r

R

23. You discover an underground pocket of naturalgas (density 0) a distance r = 8000m below theearth’s surface, by discovering at on the surfaceabove it the acceleration of gravity differs from g by1× 10−6m/s2. Find its radius R. Let the density ofthe earth be 5500 kgm3 .

24. Consider a spherical galaxy of stars, rotating at rate ω on its axis so that stars a distanc r from the axis travelin circles at speed v(r) = ω r. If the galaxy has a uniform density ρ, compute M(r), the mass within a radius of r

12.4. PROBLEMS 251

from the origin, and use it to show that stars at that radius should have

v(r) =

√4π

3ρG r

Spectroscopic data (Doppler shifts) can be used to measure galaxy rotation rates. Although the visual appearanceof spherical clusters and galaxies indicate a constant density ρ(r) = ρ of stars, spectroscopic data reveals that theorbital speeds of stars v(r) is very nearly constant

v(r) = Vc

throughout the galaxy, indicating that there must be vast amounts of matter present that we cannot see (DarkMatter). Using this v(r) compute the density versus radius ρ(r) of matter in the galaxy, and the amount of matterM(r) between 0 and r.

Chapter 13

Heat and temperature

Our study of thermodynamics starts with the Joule-Thompson experiment, in which it was established that heatis a form of energy, and that mechanical energy can be readily turned into heat. We already know that friction tendsto dissipate mechanical energy and create heat. What we will demonstrate is that the reverse process, transformingheat into mechanical energy, can never be performed with 100% efficiency in a cyclic or repeatable process.Thermodynamics is the study of the bulk physical properties of systems composed of large numbers of atoms ormolecules, which we will refer to as subsystems. Bulk properties are quantities such as total energy or internalpressure. In particular we will study how a system in quasi-equilibrium absorbs or dumps heat into another systemwith which it is in thermal contact.In thermodynamics we deal with a non-mechanical form of energy (heat energy) in that we will have no means ofcomputing it from particle positions and momenta. In fact our systems will have such large numbers of particles inthem that any such computation would be practically impossible. The fact that thermodynamical concepts predateknowledge of the existence of the atom indicates to us that the whole concept of heat energy can be understoodwithout any reference to particle positions or momenta.

13.1 Early experiments

Historically early experiments with heat and temperature were able to establish some simple properties of heat andtemperature without shedding much light on exactly what these quantities are. Simple calorimetry experimentsindicated that heat could flow from one block of matter to another as if it were a fluid of some sort, and whenheat entered a material, its empirical temperature (indicated with some sort of thermometer, literally a heat-meter)would rise. When an amount of heat dQ is put into an object, there is an resulting change in the objects empir-ical temperature. The amount of heat needed to cause a empirical temperature change dT is proportional to theamount of matter present (the mass) and the temperature change. The proportionality constant is called the heatcapacity. What “true” temperature actually is was not understood at this point. Empirical temperature was ananalog measurement of the sensation of “hotness”. Empirical temperature is a measure of a certain qualitypossessed by the objects heat content, and is not a measure of the quantity of heat possessed. Clearlythere is a big difference between heat and temperature; a bath-tub full of warm water has a lot of heat but a lowtemperature, a thimble-full of boiling water has relatively little heat, but has a high temperature.

13.1.1 Joule-Thompson experiment

This crucial experiment established heat as a form of energy. The experiment is very simple; a calorimeter consistingof a sealed metal container filled with water was subjected to frictional work; in our version of the experiment a beltunder tension is wrapped around the cylinder. The cylinder is then rotated against the frictional force between beltand cylinder. It is easy to compute the work done in N revolutions in terms of the belt tension T and cylinder radius

253

254 CHAPTER 13. HEAT AND TEMPERATURE

dW = 2πR T N (13.1)

As work is done, the temperature is monitored. There is a linear relationship between the work and calorimetertemperature

dW ∝ dT (13.2)

Calorimetry relates the heat required to raise the temperature of an object by a given amount, dQ ∝ dT andmathematical transitivity show that

dW ∝ dQ, dW = J dQ, J = 4.814J

cal(13.3)

equating one Joule, the unit of work, to its equivalent value in units of heat, the calorie. All of our formulas willuse the unit of Kelvin or centigrade for temperature. Before this experiment was done, there was no establishedrelationship between these two quantities in the minds of scientists.

13.1.2 Calorimetry

The amount of heat dQ needed to raise the temperature of a mass m by amount dT at constant pressure and atconstant volume (respectively) are

dQ = mCpdT, dQ = mCvdT (13.4)

The two heat capacities are slightly different. We will study the reasons for this for gases very soon. For experimentscarried out in the open atmosphere it is appropriate to use the constant pressure version.It is important to realize that at this point in history, nobody knew what temperature actually was, there was onlyan empirical relation; temperature as measured by some analog device would rise when heat flowed into a body.Water has a rather high heat capacity of 1 cal

g oC . This means that water is an excellent coolant, capable of extractinga great deal of heat from a heat source without changing its temperature so much that it boils away.

The most common temperature scales are centigrade and Fahrenheit. The conversion between the two is accom-plished with

T (C) =5

9

(T (F )− 32

), T (F ) = 32 +

9

5T (C) (13.5)

Basic benchmark temperatures are the freezing point of water at one atmosphere of pressure which is at 0C and32F . The boiling point of water at one atmosphere e pressure is 100C or 212F . The entire world except for theUnited States uses centigrade colloquially, and centigrade is universally used for scientific work.

The empirical temperature scale used in all scientific work is the Kelvin scale which is given a concrete definition bykinetic theory; it starts at absolute zero which the temperature at which all random molecular motion ceases; atabsolute zero the average speed of gas atoms is zero according to kinetic theory. This is −273.15 degrees centigrade.The Kelvin temperature scale uses this as it’s starting point but uses the centigrade degree size. The degree size iscalibrated by making there be one hundred Kelvin or centigrade degrees between the melting and boilingtemperatures of water at sea-level pressures.

13.1. EARLY EXPERIMENTS 255

Temperature scales do have a lower absolute limit, but nature seems to recognize no upper bound on temperature.Stellar cores, such as that of the sun, are typically 15− 20 million degrees centigrade, and very hot stellar cores arein the billions.

Example 1. The simplest empirical thermometers exploit the observation that as heat flows into a material, itexpands dimensionally. The expansion of each dimension is proportional to that dimension, the “temperature”change, and a constant intrinsic to the material called the thermal expansion coefficient α;

∆ℓ = ℓ− ℓ0 = ℓα∆T (13.6)

A pendulum clock makes one swing per second at a temperature of 20C. The pendulum rod is brass. How long willone swing take at 30C?The pendulum could be used as a simple thermometer.A pendulum has a period of

τ0 = 2π

√ℓ0g

(13.7)

at temperature T0. At a higher value of T

τ = 2π

√ℓ0(1 + α∆T )

g(13.8)

taking the ratio

τ = τ0√(1 + α∆T ) (13.9)

will be the new period in terms of the old. Since more time will be required per swing at high temperature, the clockwill make fewer complete swings in a given time period and therefore will run slow.Using the binomial theorem, for small temperature changes

τ

τ0=√(1 + α∆T ) ≈ 1 +

1

2α∆T (13.10)

and by rearranging this we can establish an empirical “temperature” Θ(T ) which is a function of the “true” temper-ature T , without even knowing the thermal expansion coefficient of brass

T = T0 +2

α

( ττ0

− 1), Θ = Θ0 +

( ττ0

− 1)

(13.11)

by simply measuring the period τ and comparing it to the period τ0, the period when the “temperature” is Θ0. Thefact that Θ(T ) = α

2 T is irrelevant, since technically the “true” temperature scale T in centigrade or Kelvin units isalso empirical, it is simply established from observations of different phenomena.

Example 2. A copper cup of volume V0 is filled to the brim with mercury at temperature T0 . How much mercurywill spill out if the temperature is raised to T?The new volume of the cup will be

V0 = πR20h0 at T0, V = πR2h = πR0(1 + αcu∆T )

2h0(1 + αcu∆T ) at T (13.12)

or

V ≈ V0(1 + 3αcu∆T ) (13.13)

but the mercury will require a volume

V ′ = V0(1 + 3αHg∆T ) (13.14)

to contain it. The difference

∆V = V03(αHg − αcu)∆T (13.15)

will be the overflow.


Example 3. The ubiquitous glass-mercury thermometer is a simple analog device for measuring empirical temper-ature changes. Glass has a very low α, and mercury has a large α. A hollow glass tube of circular cross section willhave a radius

R = R0(1 + αgl∆T ) (13.16)

and a mass m of mercury will occupy a volume

V0(1 + αHg∆T )3 ≈ πR2

0h0(1 + 3αHg∆T ) (13.17)

If it is confined to the glass tube, as the temperature rises, so does the height of the column of mercury in the tube

πR20h0(1 + 3αHg∆T ) = πR2

0(1 + 2αgl∆T )h(T ) (13.18)

dividing and using some approximation

h(T ) = h0

(1 + (3αHg − 2αgl)∆T

)(13.19)

and the height h(T ) can be used as an empirical temperature Θ(T ). The “thermometer” is standardized by calibrat-ing it against the more common Kelvin empirical temperature scale by choosing h0 carefully and cutting graduationsand scales into the glass so that the mercury rises one unit when its Kelvin temperature is increased by one degree.

Example 4. A glass cup of massmc at temperature T1 is filled withmw grams of water at T2 > T1. After a sufficientlylong time, heat exchange ceases. Assuming no loss of heat with the environment, find the final temperature of thesystem.What is temperature? A good question, but very early on it was discovered that heat transfer ceasesbetween two bodies when they reach the same temperature. In fact, this equilibrium condition will becomethe tool with which we will define temperature. In this problem we let both objects reach Tf , and equate heat loss bythe water to heat gain by the cup, realizing that heat is a form of energy and exploiting energy conservation;

∆Qw +∆Qc = 0, mcCc(Tf − T1) +mwCw(Tf − T2) = 0 (13.20)

and solve;

Tf =mwCwT2 +mcCcT1mwCw +mcCc

(13.21)

and we note that this indicatesT1 ≤ Tf ≤ T2 (13.22)

Kinetic theory will give us a better understanding of what thermal expansion is all about; at higher temperaturesrandom molecular motions make the average distances between atoms and molecules in matter increase.

13.2 Phase Changes

Matter comes in three material phases, gas, solid and liquid. When a material changes phase it requires the absorptionor emission of heat energy. To pass from solid to liquid, heat must be absorbed to raise the level of molecular motion tothe point that atoms in a solid bonded structure have sufficient kinetic energy to escape the chemical bond potentialsthat lock it in place with its neighbors.In the liquid state there are chemical long range bonding interactions (VanDer Waals) forces that must be overcome with addition of heat before the liquid can become a gas. In addition forcesof surface tension must be overcome for the gas atoms to escape the liquid. For these reasons, when heat is addedto a material, this heat is used to break the bonds and cause the phase transition and none is used to raise thetemperature until the transition is complete. That is why a kettle of water placed on the stove will continueto rise in temperature as it is heated until boiling begins. At that point all of the added heat is going into the phasechange to steam, and the temperature of the water remains at 212F as it boils.Of course none of this molecular explanation was available to early experimenters. They simply measuredthe amount of heat that was required to affect a phase change.The heat needed to effect a phase change of mass m from solid to liquid is called the latent heat of fusion and

dQ = mLf (13.23)

13.3. HEAT FLOW 257

For transition from liquid to gas the heat per gram is called the latent heat of vaporization and the heat requiredsatisfies

dQ = mLv (13.24)

For water these are

Lf = 80cal

gand Lv = 540

cal

g(13.25)

Example 5. An ice cube of mass mi and temperature 273K is dropped into mass Mw of water at temperature Th.After the ice melts, compute the final temperature of the mixture.Since energy is conserved the heat lost by the water equals the heat gained by the ice. If everything ends up at finaltemperature Tf we have

MwCw(Th − Tf ) = miLf +miCw(Tf − 273) (13.26)

which we proceed to solve for Tf . What if we get an answer that is less than 273K? We know that the finaltemperature must be between 273 and Th, so we must conclude that all of the ice did not melt. We set up theproblem differently. The final temperature will be 273 and instead we solve for the mass of ice that does melt

MwCw(Th − 273) = mmeltLf (13.27)

13.3 Heat flow

Heat flow was studied by Newton, long before it was fully appreciated that heat was a form of energy. At thattime heat was simply regarded as a “fluid” that could flow into matter, increasing its sensation of “hotness”. Thereare three mechanisms for heat flow through various materials, conduction, convection and radiation. In most casesall three are at work when an object cools in an atmosphere. We will study only conduction and radiation, sinceconvection is quit a bit more work.Conduction is the mechanism of heat flow through solid material originally studied by Newton. The basic microscopicmechanism is that random kinetic energy can be exchanged between neighboring atoms in a solid by collisions, or bytransmission through inter-atomic bonds as wave motion. Newton was the first to study conduction quantitatively.He noted that the rate of cooling of an object by conduction depended on the area through which heat can flow,and the temperature gradient dT

dx between the hot and cool surfaces of the object. The proportionality constantis called the thermal conductivity

dQ

dt= κA

dT

dx(13.28)

Example 6. A pane of glass has thermal conductivity 1.0 WmK (see the table in the chapter appendix) and an area

of 1m2and a thickness of 4mm. On a day in which the outside temperature is −10C and the inside temperature is23o, compute the rate at which heat is being lost through the window.The rate is

dQ

dt= (1

W

mK)(1m2)(

33K

0.004m) = 8250W = 8250

J

s(13.29)

which is really staggering if you think about it, like leaving 80 one-hundred Watt light bulbs on continuously. This iswhy we use thermo-pane windows, which trap a layer of air between two glass panes. Air has a very low thermalconductivity.


T2 T1

L L L

airTc Th

dQ/dTdQ/dTdQ/dT

Example 7. Perform the same computation for a thermo-pane madeof 2 sheets of glass with the same dimensions as in the previous exam-ple, but with an interposed 4mm air layer.At equilibrium, no heat can build up on any of the two intermediatesurfaces and so heat flow rates must be the same through all threelayers

κgATh − T1

L= κaA

T1 − T2L

= κgAT2 − TcL

(13.30)

Calling

x =κgκa

(13.31)

we find

xTc = (1 + x)T2 − T1, x Th = (1 + x)T1 − T2 (13.32)

which

we can solve for the first junction temperature

T1 =xTc + x(1 + x)Th

2x+ x2(13.33)

and substitute this intodQ

dt= κgA

Th − T1L

=2κgκaA

L(2κa + κg)(Th − Tc) (13.34)

evaluate this using κa = 0.026 WmK and we find that

dQ

dt= 429W (13.35)

which is only one twentieth the heat loss rate through a normal window!

13.4 Radiative cooling and heating

The derivation of the rate of cooling for a body by radiation is beyond the scope of this course. The actual result isknown as the Stefan-Boltzmann Law and goes as follows for a perfect radiator of heat

dQ

dt= ℘ = σ AT 4

s (13.36)

where Ts is the surface temperature of the radiator, A is its surface area and σ = 5.67 × 10−8 Wm2K4 is the Stefan-

Boltzmann constant.This law is the result of basic quantum theory (the Planck hypothesis) applied to the problem of determining whichfrequencies of light, and in what abundance, will be found inside of and at thermal equilibrium with a hot container.It is outside of the scope of the course to derive, but not to use.

Example 8. The earth receives a total power flux of 1200 Wm2 from the sun. Compute the surface temperature ofthe earth when it is at equilibrium with this incoming radiation.Equilibrium means the rate of absorbing heat equals the rate of heat loss. The earth has radius Re and presents asurface of area πR2

e to the sun. It radiates over its entire surface 4πR2e and so

1200W

m2π R2

e = 4π R2e σT

4s (13.37)

13.4. RADIATIVE COOLING AND HEATING 259

and so

Ts = (1200 Wm2

4 · 5.67× 10−8 Wm2K4

)14 = 269K (13.38)

which is pretty close to the actual average surface temperature of the earth. We are a bit warmer than this becausethe atmosphere traps some of the heat.The most interesting applications of these topics involve several of the concepts in any given problem.

Example 9. Matter in the core of the sun is very hot and dense, one gram of it would fit into a cube 1.0mm on aside. Its temperature would be around 2× 107K. How far from it must you be in order for the radiation emitted byone gram of stellar core matter to be below the flux that would be fatal?Lets figure that if you were exposed to ten times the flux of light falling on the earth from the sun at noon, thatyou would buy the proverbial farm (fatal sunburn, not a pretty way to go). This would be around 12, 000 W

m2 . Thepower emitted by this fragment of matter is

℘ = 6 · (1× 10−3m)2 · 5.67× 10−8 W

m2K4· (2× 107K)4 = 5.44× 1016W (13.39)

Surround the fragment with a sphere of radius R, all of this power is intercepted by the sphere, and upon it we wantthe flux to be

12, 000W

m2=

℘

4πR2=

5.44× 1016W

4πR2, R = 6× 105m (13.40)

Think about that, if you stood 600 km away from one gram of stellar core matter, the light emanatingfrom it would kill you instantly.

Example 10. Compute the rate at which ice forms on the surface of a lake with water temperature Tw and airtemperature Tair assuming Tair < 273K.The amount of heat we must extract from a layer of water of thickness dx in order to freeze it if its temperature isTw = 273K is

dQ = ρwAdxLf (13.41)

The rate at which heat flows through an ice layer of thickness x is

dQ

dt=κiA (273− Tair)

x(13.42)

and so ice will form at rate dxdt if

ρwAdx

dtLf =

κiA (273− Tair)

x, or

dx2

dt=

2κi (273− Tair)

Lfρw(13.43)

solving this with initial condition x(0) = 0 we find

x(t) =

√2κi (273− Tair) t

Lf ρw(13.44)

Example 11. An excellent way to cool water under primitive conditions is to store it in a breezy place in a per-meable bag, one that will sweat or leak. Water on the surface will evaporate drawing heat from inside of the bag.Aboriginal Americans used rawhide bags or animal bladders for water storage. At what rate must water evaporatefrom the surface of a water-bag of area 0.5m2 area, thickness ℓ = 2mm and thermal conductivity k = 0.01 W

mK inorder that the water inside be kept at 10 degrees cooler than the surrounding air.

Lfdm

dt=kA(Ta − Tw)

ℓ(13.45)

where dmdt is the rate of mass lose through evaporation. We find

dm

dt= 0.064

g

s= 0.234

Kg

hr(13.46)


13.5 Statistical mechanics. Kinetic theory

Kinetic theory is an application of the laws of statistics to the problem of determining what heat and temperatureactually are at the molecular level. The main idea behind kinetic theory is that it may be impossible to describeexactly what every gas atom is actually doing, but very quantitatively accurate assessments of the thermodynamicproperties of the gas can be made by replacing exact details of individual molecular motions with reasonablyconstructed average values.In such a statistical model, exact details of particle positions and moment will be replaced by a distribution function

that estimates the probability f(v) = d℘(v)dv that any one particle will have speed between v and v + dv for example.

To compute the average speed of a molecule in the gas,

⟨v⟩ =∫vd℘(v)

dvdv (13.47)

If there are N molecules in the gas sample, the average energy will be

⟨E⟩ = N1

2m⟨v2⟩ (13.48)

which will be related to the gas temperature once the distribution is properly calibrated.

13.5.1 Kronig-Clausius kinetic theory

In this first and simplest statistical model of a gas we treat a gas as being composed of hard spheres. Collisionsbetween atoms are totally ignored.It is virtually impossible for us to know at any time where the individual particles are located, which would re-quire knowledge of 3N numbers r1, r2, · · · , rN, or the individual particle velocities, comprising another 3N datav1, v2, · · · , vN.The earliest statistical approach to the problem of computing an equation of state P = P (U, V ) for this gas wastaken by Kronig-Clausius in 1856, and consisted of the following statistical assumptions

1. The density of the gas is approximately uniform,2. there are no collisions take place between atoms, and3. molecules in a box are divided into 6 classes with speed v in directions ±x,±y,±z.

This amounts to replacing the actual molecular velocities with an average value.

Just how drastic these assumptions are is illustrated by the figure above, showing gas molecules with a random dis-tribution of velocities, and the next figure in which only four cardinal velocity directions are taken in two dimensions,

13.5. STATISTICAL MECHANICS. KINETIC THEORY 261

and each speed is replaced by an average.

The pressure on a wall will be the momentum transfer rate divided by the wall area

P = (N

6

v dt L2

L3)(2mv

dt)(

1

L2=

1

3

Nmv2

L3=

2

3(1

2Nmv2)(

1

V) =

2

3

E

V(13.49)

from which we extract Boyle’s Law

PV =2

3E (13.50)

This law is true (except at very low temperatures) despite the absurdity of the assumptions used to establish it.The probability distribution function in this case is very simple, let δ(v,v0) = 1 if v = v0, and zero if they are notequal, then

f(v) =1

6

(δ(v, vi) + δ(v,−vi) + δ(v, vj) + δ(v,−vj) + δ(v, vk) + δ(v,−vk)

)(13.51)

13.5.2 Boltzmann’s theory

Boltzmann developed a much more reasonable description of gas kinetics by considering the factors responsible forthe time evolution of the distribution of particle velocities in each differential volume comprising the gas,f (1)(v,x, t), using the following assumptions

1. Only pairs of molecules collide. Triple simultaneous collisions are so rare at low density that they can be ignored.2. f (1)(v,x, t) d3v is the number of atoms per volume in a cell located at x with velocity within vx ± dvx, vy ± dvy,vz ± dvz at time t. This could be a local function for a spatially inhomogeneous gas. This must be normalized to∫

f (1)(v,x, t) d3v =dN(x)

dV= ρ(x) (13.52)

in which the right hand side is the local particle density.3. Consider now a uniform gas such that the velocity distribution does not depend on x. Let f (2)(v1,v2, t) be thedensity of pairs of atoms simultaneously having velocities v1, v2 at time t. This function is much more complex thanf (1)(v,x, t), but clearly should depend on it somehow.

Boltzmann developed an extremely complex formula for the evolution of the distribution function in response tointer-molecular collisions. Solution of this equation focuses on the very deep and crucial assumption of molecularchaos or Stosszahlansatz

f (2)(v1,v2, t) = f (1)(v1, t) f(1)(v2, t) (13.53)

This is a statistical assumption and turns an otherwise deterministic mechanical model of a gas into a statisticalmodel, the meaning of which will require very careful interpretation. This assumption takes the place of theClausius-Kronig assumption that one-sixth of the molecules travel in any cardinal direction at theaverage molecular speed at any given time.

Boltzmann solved this difficult evolution equation, subject to the assumption of molecular chaos, by proving thatat as equilibrium is approached, and the function f1(v,x, t) stops changing and takes on its spatially uniformequilibrium value f1eq(v), two-particle collisions such as

v1, v2 → v1′, v2

′ (13.54)

preserve the two-particle distribution function, and so

f1eq(v1) f1eq(v2) = f1eq(v1

′) f1eq(v2′) (13.55)

If we take logs, this is a conservation law

ln f1eq(v1) + ln f1eq(v2) = ln f1eq(v1′) + ln f1eq(v2

′) (13.56)


and since elastic collisions conserve kinetic energy, we are left with

ln f1eq(v) = C − β1

2mv2 (13.57)

The two constants C and β are gotten by requiring that this distribution function give the correct number of particlesin the system, and the correct total energy of the gas;

N

V=

∫f1eq(v) d

3v =

∫ ∞

−∞dvx

∫ ∞

−∞dvy

∫ ∞

−∞dvz C e

− βm2 (v2x+v

2y+v

2z)

= C(√ 2π

βm

)3EtotalV

=

∫1

2mv2 f1eq(v) d

3v

=

∫ ∞

−∞dvx

∫ ∞

−∞dvy

∫ ∞

−∞dvz C

1

2m(v2x + v2y + v2z) e

− βm2 (v2x+v

2y+v

2z)

= C3m

4

(√ 2π

βm

)2√ 8π

β3m3(13.58)

I am making use of the integral

I =

∫ ∞

−∞e−ax

2

dx =

√π

a(13.59)

which is easy to prove;

I2 =

∫ ∞

−∞

∫ ∞

−∞e−a(x

2+y2) dx dy

=

∫ ∞

0

r dr

∫ 2π

0

e−ar2

= π

∫ ∞

0

e−a(r2) d(r2) =

π

a(13.60)

Divide these two results;

EtotalN

=3

2β, C =

N

V

(√mβ

2π

)3(13.61)

We can now calculate this energy by heating a gas of atoms (monatomic and inert such as Helium) from absolutezero to temperature T

Etotal = Cv(T − 0) = n3

2RT =

3

2N k T (13.62)

where Cv =32R is the specific heat of such a gas (determined from calorimetry), n is the number of moles, N is the

number of molecules, and k is called Boltzmann’s constant. The final distribution of molecular velocities is

f1eq(v) =N

V

(√ m

2πkT

)3e−

mv2

2kT (13.63)

13.5.3 The meaning of temperature

Kinetic theory provides an answer to that difficult nagging question “what is temperature?” We compute the averagemolecular speed of a single molecule;

⟨|v|⟩ =∫ √

v2x + v2y + v2z

(√ m

2πkT

)3e−

mv2

2kT dvx dvy dvz (13.64)

Let

vx = v sin θ cosϕ, vy = v sin θ sinϕ, vz = v cos θ (13.65)

13.5. STATISTICAL MECHANICS. KINETIC THEORY 263

thendvx dvy dvz = dϕ sin θ dθ v2 dv (13.66)

and

⟨|v|⟩ =∫ 2π

0

dϕ

∫ π

0

sin θdθ

∫ ∞

0

v(√ m

2πkT

)3e−

mv2

2kT v2 dv =

√8kT

πm(13.67)

And there we have it; temperature is a measure of the average molecular speed of a gas. The distributionof molecular speeds is not zero for any speed, its just that very fast molecules are rare, as are very slow ones. Atany given temperature most of the molecules have speeds clustered around the most likely molecular speed, foundby locating the speed vmp at which the speed distribution is maximal

d

dv

(4π(√ m

2πkT

)3e−

mv2

2kT v2)vmp

= 0, vmp =

√2kT

m(13.68)

0 1 2 3 4 5 60.0

0.2

0.4

0.6

0.8

speed v

(v2 /

T3/

2 ) e

−m

v2 /2k

T

A gas feels “hot” when the average molecular speedis high, and “cold” when the average speed is low.How fast are the average oxygen molecules that youbreath traveling?

⟨|v|⟩ =

√8 · 1.38× 10−23 J

K · 300Kπ · 32AMU · 1.6606× 10−27 kg

AMU

= 445m

s(13.69)

13.5.4 Boltzmann entropy

In the process of proving that the equilibrium distribution of molecular velocities obeys Eq. 13.55, Boltzmann wasable to show that as the system of particles evolved towards an equilibrium state,

H(t) =

∫f1(v, t) ln

(f1(v, t)

)d2v obeyed

dH

dt≤ 0 (13.70)

and that equilibrium is attained when this derivative is actually zero; equilibium is the state of maximal H. This iscalled the Boltzmann entropy, and the theorem states that equilibrium is the state of maximal entropy. Thisis one of the deepest principles in physics.

Using Boltzmann’s ideas as a foundation, Gibbs was able to prove that the distribution of energy values that a systemcan be observed to have when in contact with a large reservoir of heat (a heat sink/source) held at temperature T is

℘(Ei) =S(Ei) e

−Ei/kT∑i S(Ei) e

−Ei/kT=S(Ei) e

−Ei/kT

Z(T )(13.71)

where S(E) is the number of states that the system can be found to be in if its energy is E. ℘(Ei) gives us theprobability that if you were to measure the energy of such a system, you would get Ei, and Boltzmann called S(E)the number of complexions of the system. Z(T ) is called the partition function.


Example 12∗. Suppose that a system in thermal equilibrium with a heat sink at T can be in two states, one ofenergy 0, one of energy ϵ, each with S(E) = 1. What is the probability of it being in the zero-energy state?

℘(0) =1

1 + e−ϵ/kT(13.72)

Example 13∗. A system consists of a gas of N atoms, each can be in two states, one of energy 0, one of energy ϵ,each with S(E) = 1. How many atoms on average are in the state E = ϵ when the system in thermal equilibriumwith a heat sink at T?

N(ϵ) = N℘(ϵ) =Ne−ϵ/kT

1 + e−ϵ/kT(13.73)

Example 14∗. A system consists of a gas of N atoms, each can be in two states, one of energy 0, one of energy ϵ,each with S(E) = 1. What is the average energy that the system can be found in?

E =∑

states i

Ei ℘(Ei) =0 +Nϵ e−ϵ/kT

1 + e−ϵ/kT(13.74)

(these three examples are standard chemistry and physics GRE problems).

Since ℘(E) is the state distribution function, the Boltzmann maximal entropy principle states that the equilibriumenergy Ee that a system will be in when in thermal equilibrium with a heat sink at T obeys

d℘(E)

dE

∣∣∣E−Ee

= 0 (13.75)

This formula is so powerful that it can be used to derive most if not all of the thermodynamics of undergraduatechemistry. It is the secret weapon of physical chemistry.

13.6 Convection 1

Heat flow is governed by the full, not watered-down version of Newton’s law, involving a heat flux vector for the heatflow normal to area dA

JQ =d2Q

dt dAn = κ∇T (13.76)

Our version is gotten from estimating ∇T ·n = Th−Tc

ℓ . Compute the total heat leaving volume V enclosed by surfaceS

dQ

dt=

∫S

JQ · n dA =

∫S

κ∇T · n dA = κ

∫V

∇2T d3x

by the divergence theorem. Now use the fact that the total heat within V is

Q =

∫ρCvT d

3x

and we obtain the diffusion equation∂T

∂t=

κ

ρCv∇2T = D∇2T

(the new constant is the diffusion coefficient D = κρCv

. The ρ and Cv factors are often simply absorbed into the

definition of κ). You see this equation everywhere in the sciences.

Example 15. Suppose that a spherical animal of radius R is able to maintain its body temperature at Tb in coldwater Te. At what rate does it loose heat?

1This requires that you read the fluid mechanics section (Chapter 10). It appears for physics and environmental science students. Itis not covered as part of physics 201. Meteorology and oceanography isn’t all balloon rides and beach parties, there is some math!

13.6. CONVECTION 265

The water near the animal will be heated by its body, but far away from it the water will be cold, so we haveboundary conditions

T (r = R, t) = Tb, T (r → ∞, t) = Te

and at equilibrium the water temperature every where stabilizes;

0 = κ∇2T,1

r2d

dr

(r2dT

dr

)= 0

Solving we get

T =a

r+ b

and deploying the boundary conditions

T =R(Tb − Te)

r+ Te

The rate of heat flow from the critter is

JQ = −κ∇T∣∣∣r=R

= −κ ddrT∣∣∣r=R

= κ(Tb − Te)

R2

and it loses heat at ratedQ

dt=

∫JQ dA = 4πR2κ

(Tb − Te)

R2= 4πκ(Tb − Te) (13.77)

If the animal swims, it continually replaces the warmed water around it with cooler water, and so loses heat muchfaster.

Convective heat flow moves heat by moving heated matter; hot matter rises under gravity since it becomes lessdense, cooler, denser matter falls under gravity to take its place. Convection dominates the motion of our atmo-sphere, and is the best means of heat flow for moderate temperatures.

For a moving fluid material, we know that equilibrium under forces results in

ρ(∂v∂t

+ v ·∇v)

= −∇P +∇ · t′ + dF

dV∂ρ

∂t+∇

(ρv)

= = 0

dT

dt=(∂T∂t

+ v ·∇T)

= D∇2T (13.78)

and we will limit our attention to gases, which possess something that we begin studying in detail in the next chapter,an equation of state

P = RρT

We will study it under a system of approximations (the Boussinesq approximations), imagining that thermal con-ductivity and viscosity don’t change much with temperature, and that density decreases with temperature. Callingα the thermal expansion coefficient of the gas;

dV = αV dT

How does this affect density?

ρ0 =m

V0, ρ =

m

V,

ρ

ρ0=V0V,

dρ

ρ0= −V0 dV

V 2≈ −αV0 V dT

V 2= −αdT

and we obtain

ρ = ρ0(1− α(T − T0))


It is conventional to use centigrade in these calculations, so we let T0 = 0.Using dF

dV = −ρgk we have (∂v∂t

+ v ·∇v)

= −1

ρ∇P + µ∇2v + g(1− αT )k

∂ρ

∂t+∇

(ρv)

= = 0(∂T∂t

+ v ·∇T)

= D∇2T

P = ρRT (13.79)

for a viscous, heated, gaseous atmosphere.

We simplify the whole geometrical layout by imagining that the atmosphere is trapped between two surfaces, one aty = 0 held at Th, one at y = h held at a cooler temperature Tc, and we study first an equilibrium with no moving

air, so v = 0, and dT (y,t)dt = 0. This results in

0 = −1

ρ∇P + g(1− αT (y))k

0 = D∇2T = Dd2T

dy2(13.80)

which is easy to solve for the static equilibrium state

Teq(y) = Th +(Tc − Th)

hy,

dP

dy= ρg

(1− αT (y)

), Peq = P0 + ρg y − ρgα

∫ y

0

T (y) dy (13.81)

The next step is to consider small disturbances of this static equilibrium, that are stable in the sense thattheir amplitudes do not grow over time. These disturbances will involve slowly moving air, so v = 0, andsmall temperature disturbances

T (y, t) = θ(y, t) + Teq(y, t), P (y, t) = ℘(y, t) + Peq(y, t)

Then our fourth equation will become (∂T∂t

+ v ·∇T)

= D∇2T

dθ

dt+ v ·∇

(θ(y, t) + Teq(y, t)

)= D∇2

(θ(y, t) + Teq(y, t)

)dθ

dt+ v ·∇(Teq(y, t)) ≈ D∇2θ(y, t)

dθ

dt+

(Tc − Th)

hvy ≈ D∇2θ(y, t) (13.82)

in which I have dropped all terms quadratic in small quantities.

Trial solutions involving only exponential functions (all derivatives of exponential functions are proportional to thefunctions themselves) will transform all of the linearized equations into simple algebraic equations, from which thestable disturbances can be easily extracted. These trial solutions are vx

vyvz

= eiℓx+nt

A(y)B(y)0

δP = eiℓx+nt ℘(y), (13.83)

θ = eiℓx+ntθ(y) (13.84)

and we most certainly want the real part of n to be zero or positive, so that as time progresses, these distur-bances do not grow.


Lets use incompressibility ∇ · v = 0 together with the boundary conditions vy(y = 0) = vy(y = h) = 0. Using someforsight, we will try

B(y) = B sin(sy), s =mπ

h, m = 1, 2, 3, · · · (0) (13.85)

since it obeys our boundary conditions vy(y = 0, t) = 0, vy(y = h, t) = 0. Then (call γ = gα

∇ · v = 0, iℓA+dB

dy= iℓA+ sB = 0 (1)

dv

dt= −1

ρ∇P + (g(1− αT )k+ µ∇2v

nvx = −i ℓρ℘− µ(ℓ2 + s2) vx (2)

nvy = −i 1ρd℘dy + γθ − µ(ℓ2 + s2) vy (3)

nθ +(Tc − Th)

hvy ≈ −D(ℓ2 + s2) θ(y, t) (4) (13.86)

What can we deduce from these?

Equations (0), (1) force A(y) =i

ℓ

d

dyB sin(sy) =

isBℓ

cos(sy) = A cos(sy)

equation (4) says θ(y) = − (Tc − Th)

h(n+D(ℓ2 + s2))B sin(sy) ≡ − β

n+Davy ≡ Θsin(sy)

and equation (3) says ℘ = iρ

∫ ((n+ µ(ℓ2 + s2)) vy − γ θ

)dy = η cos(sy) (13.87)

Since we know that s = mπh already (satisfies vy(0) = vy(h) = 0), all we need to do now is to get allowed n and ℓ values.

Call a = (ℓ2 + s2), and n′ = n+ µa, and note I abbreviate β = Tc−Th

h , γ = gα, then we have(n+ µ(ℓ2 + s2)

)( isℓ

)B = n′

( isℓ

)B = −i ℓ

ρη (2’)(

n+ µ(ℓ2 + s2))B = n′ B =

s

ρη + γΘ (3’)

(n+Da)Θ = −β B (4’) (13.88)

Eliminate η between the first two(n+ η(ℓ2 + s2)

)(s2 + ℓ2)B = γ ℓ2 Θ

(n+Da)Θ = −β B (4’) (13.89)

Stand back now, I’m going to do some science; these must be compatible with one another, dividing them wefind that this requires

− n′a

β=

γ ℓ2

(n+Da)(13.90)

Solving this for n

n =a(D + µ)

2

(− 1±

√1− 4

a2(D + µ)2

(Dµa2 +

βγ

aℓ2))

(13.91)

If this is positive, the disturbances grow, and our equilibrium degenerates into chaos. The condition that thedisturbances remain small is that the radical must be less than 1 in magnitude, requiring that

Dµa2 +βγℓ2

a≥ 0 (13.92)

β = (Tc−Th)h(n+D(ℓ2+s2)) < 0, so the whole solution will become unstable if

Dµa3 < −γ ℓ2 β, Dµa3 < γ ℓ2 |β| = γ (a− s2) |β| (13.93)

so we letf(a, s) = |β|γ(a− s2)−Dµa3 (13.94)

and we graph it, looking for where the solution becomes unstable;


-1.2

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0 0.5 1 1.5 2

f(a)

a

f(x,0.6)f(x,1)

f(x,0.88)0

The onset of instability occurs when any part of curve crosses the a-axis and changes sign, the critical curve isthe one for which it turns back and does not cross (the middle curve, the Rayleigh-Benard instability criteria)

f(ac, sc) = 0,df

da(ac, sc) = 0 (13.95)

which gives us two equations|β|γ = 3Dµa2c , |β|γ(ac − s2c) = Dµa3c (13.96)

from which we deduce that

ac = 3(ac − s2c), (ℓ2c + s2c) = 3ℓ2c , ℓ2c =1

2s2c , ℓc =

sc√2

(13.97)

which in turn gives us the integer m in the formula for s

s4c =4

27

|β|γDµ

=4

27

|Th − Tc|gαhDµ

, m4c =

4

27π4

|Th − Tc|gαDµ

h3 =4

27π4RL s =

mπ

h(13.98)

RL is called the Rayleigh number, in order to get a stable convection cellmc = 1 we must have RL = 274 π

4 ≈ 657.5.

Taking the real parts of our trial solutions, we find that the motion of the air when mc is chosen as small as possible(mc = 1, the cells just fill the height of the fluid layer) is

vx =(√

2 v0 ent)cos(

πy

h) cos(ℓx), vy =

(v0 e

nt)sin(

πy

h) sin(ℓx), with n = 0, ℓ = ℓc (13.99)

and the velocity field pattern is periodic, repeating itself horizontally over a distance

2π

ℓc=

2ππ√2h

= 2√2h (13.100)


when the convection or Benard cells are of maximal height h

2√2 hTh at y=0

Tc at y=h

What happens if Eq. 13.92 is violated? The magnitudes of the velocity components Eq. 13.99 will grow since thenn > 0, and the whole premise that this motion of the fluid is a small fluctuation imposed on the equilibrium Eq. 13.81becomes false, the motion will degenerate into chaos. The cells are stable and persistent when n = 0.

You can see alternating zones of strong up-drafts carying heated air upwards, and down-drafts carrying cooler airdown to the lower heated surface. These convection cells give the photosphere of the sun a granulated appearance,and you can see them forming in any fluid heated on the surface elements of a stove or oven. Convection plays animportant part in our understanding of stars, the oceans, the atmosphere and meteorology.

This section is presented for the benefit of those of you who want or need to understand this topic, but we will notdiscuss it in class. What background do you need to go into oceanography or meteorology? You can see that youneed the full battery of calculus courses, and differential equations, and the two semesters of physics plus a coursein dynamics, a course in thermodynamics, and another in fluid dynamics or transport phenomena. Oceanographerswill also need several chemistry courses, such as physical chemistry, and analytical or a specialized water chemistrycourse. Generally speaking these very challenging fields are not concentrations within other programs, but are multi-disciplinary majors with their own departments most closely affiliated with physics and engineering.


13.7 Appendix. Some numerical values of material properties

Material Thermal Cond. κ ( Jm·s·K ) Specific heat Cp (

Jg·K ) Density ρ ( kgm3 = g

cm3 )

Air 0.024 1.0035 1.26Aluminum 250 0.897 2712Asbestos 2.07 · · · 320 (shredded), 1600 (rock)Brick 1.31 0.84 1900-2500Copper 401 0.385 8930Cork 0.07 · · · 240Cotton wool 0 029 · · · · · ·Dry earth 1.5 0.8 1522 (dry)Fiber glass 0.04 · · · · · ·Glass 0.96 0.84 2600Stone (Granite) 1.7-4.0 0.79 2515Hardwood 0.15 1.2-2.3 400-900Iron 80 0.45 7850Marble 2.5 0.88 2563Rock 2.0-7.0 0.79 2300-3000Snow 0.05-0.25 · · · 160 (fresh), 480 (packed)Water 0.58 4.18 1000

Material α (10−6m/m ·K)Aluminum 22.2Brick 5.5Concrete 14.5Copper 16.6Plate glass 9.0Granite 7.9Ice 51.0Iron 12.0Lead 28.0Mercury 61.0Nylon 72.0Steel 13.0Water 69Wood (oak) 4.9-5.4 (lower=parallel to grain)Wood (oak) 4.9-5.4 (higher=across grain)

13.8. PROBLEMS 271

13.8 Problems

1. Suppose that you own a poorly insulated small house, with outside walls ℓ = 0.25m thick, total outer wall plus roofsurface area 210m2 (about 1992 ft2) and your walls have a thermal conductivity of 1.3 W

m·K (limestone). Calculatehow much power your air conditioner will require to keep your house at 293K when the outside temperature is 308K.

2. The Jovian moon Europa is the smoothest body in the solar system, indicating thatits surface is probably ice, of thickness 2.0 km (water ice in fact) blanketing a deep oceanof fluid. Ice has thermal conductivity κ = 1.67 W

mK . Compute the surface temperature

of the moon if Europa gets only 90 Wm2 from the sun. Assuming that it is liquid wa-

ter at 273K below the ice, compute the rate with which the ice layer grows in thickness.

3. Schmidt’s lesser Neptunian penguin is perfectly spherical, of radius R = 1m andbody temperature T = 60K. How many Joules of heat does a flock of 1000 of thesebirds lose in one hour (the environment is at 0K)?

These birds are capable of merging together into a giant super-penguin-ball to keepwarm. If 1000 of these birds merge into a ball, how much heat will they lose in anhour? This is why arctic animals have large volume-to-surface area ratios.

4. The planet Mars is 1.52 times as far from the Sun as the Earth is. Mars is irradiated by the Sun at 520 Wm2 .Assuming that the space around Mars is at 0K, determine the equilibrium average surface temperature of the planet.

5. On a very cold winter night a puddle of water of mass 1.0 kg and surface area 1m3 at 0 degrees centigrade will loseheat almost exclusively by radiation. Assuming that it absorbs no heat from its surroundings, and radiates perfectly,how long will it take to freeze into ice? Let the air be at −20C.

6. A perfectly insulating cup contains 100 gm of water at 30 degrees centigrade. An ice cube of mass 50 gm at 0degrees centigrade is dropped into it. Determine the the equilibrium temperature after the ice has melted. Howmuch ice melts?

7. A 100 gm sample of water at 50 degrees centigrade, specific heat CH20 = 1 calgmOC

is mixed with 100 grams of

ethyl alcohol at 10 degrees centigrade, with specific heat Ceth = 0.6 calgmOC

. Find the final temperature.

8. A concrete slab of length L0 = 2.0m at T = 0oC andcross-sectional area A = (0.2m)2 perfectly bridges a gap atT = 0oC. If the temperature increases to 100C, the gap is stillL0 = 2.0m but the slab has expanded, and has buckled, itscenter rising by amount x. Compute x.

L0

Cross−sectional area at T=00 C, a02

9. A steel bar of length L0 = 2.0m and cross-sectional areaA = a20 = (0.1m)2 at T = 0o is wedged between two rigidsupports L0 apart. If the temperature is increases to T = 30oc,compute the force that the rod exerts on the supports, assumingthat they are still 2.0m apart.


L0

a02

Th=100o CTc=0o C

10. A copper bar of length L0 = 0.2m andcross-sectional area A = 0.05m2 is used as aheat pipe to transfer heat between a steambath at Th = 100oC and a beaker of waterat Tc = 0oC containing ice cubes. Determinethe equilibrium rate with which the ice cubesmelt, on kg

s .

11. Suppose that molecular velocities have the following experimentallymeasured distribution.Find the average molecular speed, and rms speed.

velocity (m/s) ℘(v)−2.0 1/13−1.0 3/130.0 5/131.0 3/132.0 1/13

12. The amount of heat that you lose due to radiation alone when you are in a cold environment is a small fractionof the total heat loss by the body. Your body temperature is 37oC = 310K, and if you have 2.0m2 of surface area,your net radiative rate of heat loss when exposed to a 5oC = 278K environment is

dQ

dt= (5.58× 10−8 w

m2K4) · (2.0m2) ·

(310)4 − (278)4

)= 364W

Conduction and convection of heat from the body account for much more. All told, you probably lose around 3500Wof heat in a 5oC = 41F environment.Compute the rate of heat loss through the dome of an igloo made of ice with κ = 0.25 W

mK , inner radius of 2.0m andwall thickness of 0.8m when the inside temperature is 5o C and the outer is −40o C. Two people and one candlewould be very cozy inside.Hint; modify the solution of example 12 so that T (r = 2.0) = 278K and T (r = 2.5) = 233K.

13. Determine the (classical) specific heat of a linear triatomic molecule AB2 with mA > mB using the equipartitiontheorem.

14. Consider a simple “kinetic theory” in which the number of particles with energies between E and E + dE perunit volume in a homogeneous system in thermal equilibrium with a source of heat at temperature T is

dN

V= C E3 e−

EkT

For a system of N particles, find C and the average energy of the system.What is the probability that a randomly selected particle will have energy between E and E + dE?

15. Consider a simple “kinetic theory” of γ-rays produced in a cosmic ray event. Suppose a cosmic ray of energy Ecollides with an atom, and let n (a non-negative integer) be the number of γ-rays that are produced. Suppose thatthe probability that n γ-rays are created is

℘(n) =( EEc

)n

n!e−

EEc

where Ec is some constant, for example Ec = 1× 106 eV .Find the average number of γ-rays produced in a 4×106 eV cosmic ray collision. What is the most probable numberof γ-rays produced?

16. Suppose that your apartment has a total of 6.0m2 of windows, each a single pane of glass of thickness ℓ = 4.0mm,κglass = 1.05 W

mK . You heat the apartment with an electric furnace that can produce a maximal heat output of10, 000W . On a day in which the outside temperature is 0o C = 273K, determine the equilibrium temperatureinside of the apartment when the furnace is running at maximal output. You may think that this furnace poweroutput is high, but most furnaces produce many times this amount of heat.

Close the storm windows! This creates a layer of dead air, κair = 0.024 WmK , ℓ′ = 2.0 cm thick between the inner and

outer pane. The only way for this layer to exchange heat with the room or outside environment is through conduction.

13.8. PROBLEMS 273

Neglecting the contribution to conduction through the glass, compute the equilibrium room temperature with yourfurnace at maximal output. What heat output could you run the furnace at in order to have a room temperature of20o C = 68o F?

17. Find the smallest mass earth-like planet (ρ = 5.5 gcm3 ) for which the average molecular speed for H2 at 300K is

less than the escape velocity of the planet. This is why we can’t have nice things (like H2 and He in our atmosphere).

18. How long would it take 1.0 kg of water at 273K to freeze through heat loss by radiation alone on a very coldwinter night Te = 253K? Let its radiating area be 1m2.

19. Find the temperature at an altitude y in an atmosphere of height h undergoing convection with air velocityEq. 13.99. Note that heat is flowing upwards because of conduction and convection.

20. Note that our spherical sea-creature loses heat at a rate that does not depend on its surface area. Wouldnon-spherical animal of the same body mass and density lose heat at a higher or lower rate than this? Explain.

21. Suppose that you have a circular cross-section hot water pipe of radius a, length ℓ transporting hot water at Thfrom your hot water heater. If it is surrounded with thermal insulation of conductivity κ of inner radius a and outerradius b, the outside temperature of which is maintained at Tc, at what rate do you lose heat through the insulator?

22. Estimate the core temperature Th of a planet whose radius is R = 1, 000 km, surface temperature is Ts =20K << Th, which loses heat into the surrounding space (take Tspace = 0) by radiation only. Assume that the heatis not replenished by a star or sun. Suppose that it is made of granite-like rock.

23. Use Eq. 13.75, the Boltzmann maximal entropy principle, to find the equilibrium temperature of a so-calledthermodynamically normal system; one for which S(E) = C E

3N2 where N is the number of particles in the

system in three dimensions.

24. Systems for which the number of complexions S(E) does not increase as E increases pose real problems. Supposethat for a magnetic system S(E) = C E(Em−E) where C and Em are constants. Graph this versus E, so you knowthe range over which S(E) increases. At what absolute temperature would such a system in thermal equilibriumwith a resevoir at T be found to be in an equilibrium energy state with E > Em/2?

25∗. How fas is the average H2 molecule in the earth’s atmosphere (T = 300K) moving? How fast is the average H2

molecule in the sun’s photosphere (T = 5800K) moving? The photosphere is actually too hot for molecules, theywould dissociate to atoms or ions at that temperature.

26∗. The density of the earth’s atmosphere is about ρ = 1.26 kgm3 . Approximately how far apart are the nitrogen andoxygen molecules in the atmosphere? How long will it take the average atmospheric molecule to travel this distance?Estimate how many collisions with other molecules the typical O2 in our atmosphere will collide with each second.

27. A mass 1.0 kg of water is contained inside of a box of volume 1000 cm3 with area 726 cm2 and thickness 1.0 cmmade of material with thermal conductivity κ = 0.3 W

m·K . Outside of the box the room temperature is 300K. Thewater in the box is at 273K. How long will it take the water to reach 290K?

Chapter 14

Equilibrium thermodynamics

Thermodynamics deals with the physical properties of complex systems in equilibrium states, or systems under-going quasi-static evolution from one quasi-static state to another, as energy is moved into or out of the system intwo forms, as mechanical energy (work) or as heat. This sentence involves all of the major keywords and conceptsof which we will begin a detailed study.

14.1 Core definitions

1. System; typically a gas or some other simple object with a well-defined physical behavior.

2. State; for a gas this is a specification of P and V . The state of a system is a specification of all of the bulkvariables needed to determine a complete set of independent thermodynamic properties. By state we will alwaysmean a global state, one that prevails throughout the entire system. This means that the system will have a uniformdensity, and no position and time dependent fluctuations in its thermodynamic variables. If the system does possessa position and time dependent matter or energy density, it will not even be possible to define certain thermodynamicvariables.

3. Thermal equilibrium; the condition of the system being in a state that does not change over time. We willlearn later on that the tendency to equilibrium is a non-mechanical concept.

4. Equation of state; a relationship or constraint linking the internal energy of a system with its thermodynamicvariables, an example being the ideal gas law PV = NkT . In the early years of thermodynamics, equations of statewere established experimentally. Later on they were constructed or proposed in such a way as to explicitly exhibitcertain behaviors that were seen experimentally.

5. Temperature; our first concept of temperature will be arrived at through the concept of equilibrium; Contemplatetwo systems, A, and B, associate with A, and B a function fAB(P1, V1;P2, V2) of the states (P1, V1), (P2, V2)of A and B respectively. If the systems are in equilibrium, the net heat exchange becomes zero,and the states of the two systems cannot be independently and arbitrarily specified; the are closelycorrelated

fAB(P1, V1;P2, V2) = 0 (14.1)

Equilibrium between the two systems is to be understood as a constraint between the states of the systems. This isoften called the Zeroth law of thermodynamics.

From this we make a very simple statement of one of the most important laws of thermodynamics; If A is inequilibrium with B and B is in equilibrium with C, then A is in equilibrium with C. There is transitivityof equilibrium. In other words

fAB(P1, V1;P2, V2) = 0 and fBC(P2, V2;P3, V3) = 0, implied fAC(P1, V1;P3, V3) = 0 (14.2)

275

276 CHAPTER 14. EQUILIBRIUM THERMODYNAMICS

This permits us to define an empirical temperature. The simplest way to do this is to think of system B as thethermometer. Solve the relation fAB = 0 and fBC = 0 for P2 and equate the expressions

TA(P1, V1, V2) = TC(P3, V3, V2) (14.3)

is equivalent to saying A and C are in equilibrium. We say that two systems are in equilibrium if they’re empiricaltemperatures are equal. What relation this bears to the actual system temperature is yet to be determined.

The introduction of these functions f(P, V ) has allowed us to show that a necessary and sufficient condition forA and C to be in equilibrium is the equality of their empirical temperatures. Under these circumstanceA and C will stop exchanging heat.We still do not have a precise relationship between temperature and some dynamical quantity, onlythe mathematical connection between the states of two systems in equilibrium being the statement that their tem-peratures are the same.

6. We now invert T = T (P, V ) to get the equation of state

P = P (V, T ) (14.4)

and so all of the machinery of thermodynamics stems from the fundamental postulate and the Zeroth law.

The collection of states (P, V ) for which T is a constant is called an isotherm. In an adiabatic or isentropicprocess a change of state occurs during which the system is thermally isolated, with no contact with another system.Adiabatic processes are those for which no heat is exchanged.

7. A quasi-static process is a change of state that occurs so slowly that the system remains in equilibrium duringthe process. A process is reversible if it can be carried out in reverse order; the initial state gotten from the finalby reversing all steps. This requires time reversal insensitivity of the system in the succession of states along theprocess path. We will see that the entropy concept is related to time-direction sensitivity of the system, and that areversible process is one that takes the system quasi-statically from one equilibrium state to another isentropically.Is the propagation of a sound wave through a gas a quasi-static process? Absolutely not, since the pressure of thegas is not a global variable, it has spatial variations P = P (x, t). This brings us to an important juncture, wherewe ask the question “what kinds of processes and systems will thermodynamics deal with”? We will study onlyquasi-static processes, such that the entire system possesses global pressures and temperatures, in other words wewill exclude (for now) any such processes for which T = T (x, t) and or P = P (x, t). This will allow us to map outour processes of study on a PV diagram (or BI diagram for magnetic systems).

For a gas, a finite volume expansion accomplished in finite time is irreversible; during the expansion large localvariations in the pressure and temperature will be created, and these will not disappear until a final equilibriumstate is reached. More importantly, they cannot be perfectly replicated if we try to undo the expansion.The only finite-volume change processes that can be regarded as reversible are those that take an infinite time toaccomplish, so that during the process we can regard the gas at any time as being in a static or quasistatic state with a constant P and T throughout.

14.2. THE LAWS OF THERMODYNAMICS 277

8. The work done by a gas during a change of stateis the integral

∆W =

∫Γ

P dV (14.5)

This is easy to understand from the definition of pres-sure; the force exerted per unit area by a piston-headconfining the gas to a piston of cross sectional areaA and normal direction n;

P =F · nA

(14.6)

The work done in moving the piston-head by amountdx in the direction n is

dW = F · (dxn)

=(F · n

A

)(Adx

)= P dV (14.7)

where dV is the accompanying volume change of the gas.Work is one of only a few means of getting energy into or out of a system. In all forms of dynamics, energy isconserved if all forms are included in the balance. We now state the principle of conservation of energy in the contextof thermodynamics;

14.2 The laws of thermodynamics

14.2.1 First Law of thermodynamics

In thermodynamics we recognize heat as a form of energy, and so there are only two ways to change the energy of asystem; we do work on it or give it heat;1. A system A has a unique function UA(P, V ) the internal energy, such that during an adiabatic process from(P1, V1) to (P2, V2)

∆W = UA(P1, V1)− UA(P2, V2) = −∆UA (14.8)

is the work done by the system ( a convention).2. The heat absorbed by A during a state change is

QA = ∆UA −∆W (14.9)

and for an infinitismal quasi-static change

δQ = dU + PdV, and U(P2, V2)− U(P1, V1) =

∫Γ

(dQ− PdV ) (14.10)

This is all merely energy conservation, identifying heat as an energy form. We will actually use this as our definitionof heat energy.The Joule experiment established heat as a form of energy. In this experiment a sealed, insulated container ofmercury was vigorously stirred, and its temperature monitored. It’s temperature rose in direct proportion to theamount of work done on it, indicating a transformation of work energy into heat energy.

For a quasi-static adiabatic process

dQ = 0 = dU + PdV = (∂U

∂PdP +

∂U

∂VdV ) + PdV, or 0 =

∂U

∂PdP + (P +

∂U

∂V)dV (14.11)


The solutions to this equation are functions of P an V called adiabatic curves.

A heat bath will be a system so large that heat can be taken from it or given to it without altering its temperature.This is an integral part of gas thermodynamics, which is the best laboratory for learning the ropes of this branch ofphysics. In gas thermodynamics we will study the behavior of a confined gas as it is subjected to the two means ofchanging its energy; we do work on a confined gas, or bring it into thermal contact with another system at a differenttemperature so heat can be exchanged.One of the most important problems that thermodynamics was able to solve during its inception was the problem ofhow a machine can be used to convert heat energy into usable work. Mechanical energy can be converted from oneform (kinetic, potential) to another with perfect efficiency in the absence of dissipative or non-conservative forces,but heat energy cannot be perfectly converted to work.

A machine that is used to convert heat into work is called a heat engine, and it requires a working substancesuch as a gas. One of the most important heat engines is called the Carnot cycle.A heat engine consists of three objects, a primary system, in the case of the Carnot engine this is a thermal reser-voir, a heat source at high temperature, a secondary system, which is a reversible heat sink at lower temperature,and a working substance trapped in an auxiliary system that can be restored to its original state.

A Carnot cycle is a reversible cyclic process consist-ing of two isotherms and two adiabatic processesundergone by the auxiliary system.

From (P1, V1) to (P3, V3) the auxiliary system(presumably a gas) absorbs heat Q1 at empiricaltemperature T1 from a heat bath (primary system)at this temperature.

From (P2, V2) to (P4, V4) the system dumps heat Q2

into a heat bath (secondary) at T2. Let all subpro-cesses be quasi-static. Compute the heat in and out,

Q1 =

∫ V3

V1

P (V, T1)dV + U(V3, T1)− U(V1, T1)

Q2 =

∫ V2

V4

P (V, T2)dV + U(V2, T2)− U(V4, T2) (14.12)

and on the two adiabatic lines the heat in is zero

0 =

∫ V2

V3

PdV + U(V2, T2)− U(V3, T1)

0 =

∫ V4

V1

PdV + U(V4, T2)− U(V1, T1) (14.13)

Subtract the first two equations and add in the last two

Q1 −Q2 =

∫ V3

V1

PdV +

∫ V2

V3

PdV +

∫ V4

V2

PdV +

∫ V1

V4

PdV =

∮c

PdV (14.14)


or in other words

Q1 −Q2 = ∆W (14.15)

This implies that over the entire cycle

∆U = 0 (14.16)

but this is just a consequence of the first law, U depends on P and V and the gas returns to its original values aftera cycle.Let the efficiency of the cycle be the ratio of work extracted to input heat

ϵ =W

Q1= 1− Q2

Q1(14.17)

14.2.2 Kelvin’s Second Law

No cycle exists whose efficiency is 1.

Basically this says that it is impossible to convert heat to work withno discarded waste heat.

The symbolic representation of a Carnot cycle is illustrated below,showing flow of heat to and from the primary/secondary systemsinto/out of the auxiliary, along with work flow in/out of the auxiliaryinto a reversible work source.

Contemplate two cycles, run one in reverse, such as that illustratedbelow.This is equivalent to one perfectly efficient cycle, which is in violationof the second law, and so

W =W ′ and Q1 = Q′1 (14.18)

this means that a cycle run between 2 baths T1 and T2 cannot depend on the system (A,A′, B) but only on thetemperatures T1, T2 because the two machines, clearly different in the figure, operating between the same empirical


temperatures, are importing/exporting the same heats and exporting the same works. Therefore

Q2

Q1= f(T1, T2) (14.19)

must be a universal function, applying to all systems running between the two temperatures. We can in fact do evenbetter and demonstrate that

Q2

Q1=ϕ(T2)

ϕ(T1)(14.20)

For the two cycles shown

Q2

Q1= f(T1, T2)

Q3

Q2= f(T2, T3) (14.21)

for the combined cycle

Q3

Q1= f(T1, T3) (14.22)

multiply the first two equations together, andwe find using the third that

f(T1, T2) · f(T2, T3)

= f(T1, T3) (14.23)

which implies that

Q2

Q1=ϕ(T2)

ϕ(T1)(14.24)

Let us now define the absolute temperature

T = αϕ(T ) (14.25)

then

ϵ = 1− Q2

Q1= 1− T2

T1, or

Q1

T1+−Q2

T2= 0 (14.26)

We use the Kelvin temperature scale to measure absolute temperature.

14.2.3 Carnot’s Theorem

Let CR be a closed reversible cycle, then ∮CR

dQ

T= 0 (14.27)

This is trivially proven by covering CR with a mesh of isotherms and adiabatic processes and building up the cycleas a chain of infinitismal adiabatics and isotherms


since we have proven it for a Carnot process. If we compute this line integral around each little Carnot-cell thatcovers an arbitrary cycle, the integrations over the isothermal and adiabatic segments shared by two cells cancel,leaving only the integrations over the segments forming the arbitrary cycle.

Define a new function of state (a function that depends only on the state of the material, not how it got into thatstate), called the entropy

S = S(V, T ), dS =dQ

T(14.28)

Then the entropy change during a process that begins at point 1 in the P − V plane and ends at point 2 is processindependent and can be expressed as

S(V2, T2)− S(V1, T1) =

∫ 2

1

dQ

T(14.29)

with ∮dQ

T= 0 (14.30)

14.2.4 Non-reversible Process

Let C be a non-reversible process absorbing heat Q1 from a heat bath at T1 and absorbing −Q2 from a bath attemperature T2. This is the machine at the bottom the the left-hand figure below. We can make no claim suchas Q1

T1= Q2

T2for it.


Then for the left Carnot engine

Q0

Q1=T0T1

(14.31)

and for the right Carnot engine

Q′0

−Q2=T0T2

(14.32)

What does this say about the irreversible device in the bottom of the figure? The entire apparatus violates thesecond law of course, unless Q ≤ 0 since in that case some of the work done on the cycle is discarded back into thebath at T0 as waste heat. This requires

Q0 +Q′0 = Q = Q1

T0T1

+ (−Q2)T0T2

≤ 0 (14.33)

but since T is absolute temperature, T0 is positive and so its division does not change the sign of the inequality

Q1

T1− Q2

T2≤ 0 (14.34)

Now look at the bottom engine and compute∮cdQT for, noting the reversal of the arrows representing heat flow;

∮c

dQ

T=Q1

T1− Q2

T2≤ 0 (14.35)

14.3. THERMODYNAMIC POTENTIALS 283

14.2.5 Clausius’ Theorem

For C an irreversible cyclic process∮c

dQ

T≤ 0 (14.36)

This applies to any cycle containing anirreversible part.

A corollary of this theorem is that for anyprocess ∆S ≥ 0 for an isolated system,equality is for reversible processes, in whichthe entropy change is for the universe; the sys-tem together with anything that it exchangesheat and work with.

The implications of this theorem are as fol-lows, consider a cycle that has an irreversiblesub-process c → a; Since a → b and b → c arereversible, we can say that

∫ b

a

dQ

T+

∫ c

b

dQ

T+

∫ a

c

dQ

T=

∮C

dQ

T≤ 0 (14.37)

however for the reversible part;

∆S =(S(b)− S(a)

)+(S(c)− S(b)

)=

∫ b

a

dQ

T+

∫ c

b

dQ

T(14.38)

and so we obtain

S(c)− S(a) ≤ −∫ a

c

dQ

T(14.39)

Multiply by (−1) to reverse the inequality; for the irreversible process c→ a we find then that

S(a)− S(c) ≥∫ a

c

dQ

T, c→ a irreversible (14.40)

The significance of this is quite great, we have illustrated that in any process the entropy of an isolated systemdoes not decrease since dQ = 0 for an isolated system. Reversible processes are at the threshold, ∆Srev = 0.Consider a simple example of two vast reservoirs of heat, one at T1, the other at T2 > T1. Place them in thermalcontact, and let dQ flow from the hot to the cold. Then for the whole irreversible process

dQ

T1+

−dQT2

≥ 0 since T1 < T2 (14.41)

if the two objects together cannot exchange heat or work with the environment.

14.3 Thermodynamic Potentials

We now define several state functions; all are expressions of the energy of the system, and each one is an appropriateform of the energy useful for computation when a specific set of conditions are met.1. Helmholtz Free Energy

F = U − TS (14.42)


2. Gibbs Free Energy

G = U − TS + PV (14.43)

3. Enthalpy

H = G+ TS (14.44)

14.3.1 The meaning of F ; the F-Theorem;

In a mechanically isolated system at constant temperature, F never increases and thermal equilibriumis a state of extremal F .

Proof;dQ

T≤ ∆S (14.45)

at constant temperature, and so∆W ≤ −∆U + T∆S (14.46)

but−∆F = −∆U + T∆S (14.47)

at fixed T . In a mechanically isolated system no work can be done on the system and so

∆W = 0 ≤ −∆U + T∆S = −∆F (14.48)

Therefore under these conditions any deviation from equilibrium lowers F

∆F ≤ 0 (14.49)

The significance of the Helmholtz free energy is that the work delivered in a reversible process with contactwith a heat reservoir at constant volume is the decrease in F . Chemists tend to use the Gibbs free energysince all of their experiments are conducted in the open atmosphere, at constant P . Physicists use the Helmholtzfree energy because they tend to study confined systems (fixed V ).

Consider a sealed container with a movable wall held fixed by a magnetic force. On one side is one mole of idealmonatomic gas at volume 10 ℓ, on the other side is one mole of the same gas at volume 2 ℓ. The entire apparatusis immersed in a heat bath that maintains both gases at 273K. Compute the Helmholtz free energy change as thepartition is moved to the center of the piston, leaving each gas at volume 6 ℓ.We will show in the examples that for an ideal gas

F

nRT=

F0

nRT0− ln

((T

T0)

32V

V0

)(14.50)

and so∆F = −nRT (ln 6− ln 10 + ln 6− ln 2) = −∆W (14.51)

The work delivered in a reversible process at constant temperature for a system in contact with a heat bath is thedecrease in Helmholtz free energy. This work is delivered to the field, or whatever agent maintains it.

14.3.2 The meaning of G; the G-Theorem;

For a system at constant T and P , G never increases and equilibrium is a state of extremal G.

Proof;∆W = P∆V (14.52)

at constant pressure, and∆G = P∆V +∆U − T∆S (14.53)


at constant T . From the first law, for a constant T process

∆Q

T≤ ∆S (14.54)

substituting this we find thatP∆V = ∆W ≤ −∆U + T∆S (14.55)

and so0 ≤ −P∆V −∆U + T∆S = −∆G (14.56)

proving that under these conditions, any deviation from equilibrium lowers G

∆G ≤ 0 (14.57)

The functional dependence of these variables are

F = F (V, T ) G = G(P, T ) (14.58)

taking differentialsdF = dU − T dS − S dT = dQ− P dV − T dS − S dT (14.59)

but for a reversible processdQ

T= dS (14.60)

which causes two terms to cancel, leaving only

dF = −P dV − S dT (14.61)

This tells us how the derivatives of F are related to the properties of the system

P = −(∂F∂V

)T, S = −

(∂F∂T

)V

(14.62)

These are extremely general expressions that apply to any physical system.

The decrease in the Gibbs free energy is the non-PV work done by a system in a reversible, constant P constant Tprocess, but any exploration of this must wait until we discuss non-PV work, such as electrical, magnetic or chemicalwork.

14.3.3 The meaning of entropy

Clausius said that the universe has fixed energy and its entropy tends to a maximum. The exact meaning of thisstatement can be unclear since people are inclined to think that entropy, like energy, is a material property of matter.It is not, entropy is simply a mathematical function that measures a system’s capacity for further spon-taneous change. We will discover that in a spontaneous or irreversible process, the entropy always increases. Somechemistry people say that entropy is a measure of exhaustion, in that it increases as a systems ability to undergofurther spontaneous change decreases.

The entropy function and second law gives us a powerful tool for the computation of equilibrium temperatures in thecase of two or more systems brought into thermal contact. Consider two identical systems of completely unknowncharacters, with constant specific heats and respective initial temperatures T1 and T2. If they are brought intothermal contact and allowed to equilibrate by reversible heat exchange, they will reach a temperature Tf determinedby the second law;

Case 1; Both volumes held fixed by rigid walls.No work can be done on either system, of any form, so ∆U1 = ∆Q1, ∆U2 = −∆Q1, so that energy is conserved;

∆U1 +∆U2 = 0 =

∫ Tf

T1

C dT +

∫ Tf

T2

C dT (14.63)


and the final temperature is found to be

Tf =T1 + T2

2(14.64)

and entropy change

∆S =

∫ Tf

T1

CdT

T+

∫ Tf

T2

CdT

T> 0 (14.65)

indicating that the heat transfer is spontaneous and irreversible.

Case 2; Volumes held fixed by reversible work source.For example; the systems are gases held in cylinders, with a movable piston head, but the head is held fixed, in place,by a powerful electromagnet. As energy is transfered, the power supplied to maintain the field needed to hold thevolumes fixed will increase or decrease.The final temperature is determined by the second law

∆S = 0 =

∫ Tf

T1

CdT

T+

∫ Tf

T2

CdT

T(14.66)

giving, in the case of identical specific heats

Tf =√T1 T2 (14.67)

and and a computation using the first law shows

∆U1 = ∆Q1 −∆W1, ∆U2 = −∆Q1 −∆W2 (14.68)

and so the amount of work supplied to the electromagnets is

∆W = −∆W1 −∆W2 = ∆U1 +∆U2 (14.69)

Notice that this is not PV work, since the volumes are fixed.If the two systems are simple gases, the only way to work on them is by volume change, so that

∆U1 = ∆Q1 + P1∆V1 = ∆Q1 =

∫ Tf

T1

C dT (14.70)

∆U2 = ∆Q2 + P2∆V2 = ∆Q2 =

∫ Tf

T2

C dT (14.71)

and so in this case we learn that the amount of work supplied to the magnetic fields is

W = C(T1 + T2 − 2

√T1 T2

)(14.72)

At this point you should ask, “if the volumes are fixed in each case, what difference does it make how they are fixed,so why are these two cases different?” The difference is that in the second case the two systems are in contact witha work reservoir, and so there is energy flow into or out of the system..You can see then that the overwhelming urge to increase the entropy places constraints on how much work can beextracted from a change of state of a system. Overall energy must be conserved, taking all forms into account, andentropy can at best be constant, and in any but a reversible process, must increase. The most work that a statechange could deliver then will be delivered when the state change is carried out reversibly, so dS = 0 and heattransfer is minimized. This is a principle of maximal work, a consequence of the increase of entropy. Dependingon exactly how an irreversible process is carried out, the entropy changes and heat transfers can vary, and thereforeso can the work delivered to a work source. When the heat transfer is zero though, so is ∆S and the delivered workis maximal.

The connection between entropy and exhaustion of spontaneity is illustrated very well by the following example;consider a sample of one mole of ideal gas atoms trapped in a perfectly insulated cylinder of volume V0 = 5.0 ℓ.


p1 p2 p3 p4 p5 p6V0 = 5.0 liters

There are six partitions, between each is an additional 1 ℓ of empty space. We can compute the entropy change aseach partition is removed, and we know that once removed, the molecules will not ignore the new volume opened upto them, they will spontaneously expand into it.Entropy calculations require equations of state if you wish to carry them out. For an ideal gas

PV = nRT, U = nCV T (14.73)

You can’t increase the gas energy by removing a partition, so

dU = 0 = nCV dT, ⇒ dT = 0

dU = 0 = T dS − P dV, ⇒ dS =P

TdV = nR

dV

V

Remove p1; ∆S =

∫ 6 ℓ

5 ℓ

RdV

V= R ln(6/5) = 0.182R

Remove p2; ∆S =

∫ 7 ℓ

6 ℓ

RdV

V= R ln(7/6) = 0.154R

Remove p3; ∆S =

∫ 8 ℓ

7 ℓ

RdV

V= R ln(8/7) = 0.134R (14.74)

In each case ∆S > 0, indicating that the change is spontaneous, but in each case ∆S is smaller than in the previousstep, indicating a loss of enthusiasm for additional expansions.

14.3.4 The meaning of enthalpy

Chemistry is generally done under constant atmospheric pressure, and bond energies can be determined by measuringheats of reaction; the amount of heat needed to initiate a reaction or the amount liberated by a reaction. If theamount of heat absorbed in a chemical reaction depended upon the thermodynamic path, rather than just initialand final states, experiments would have to be conducted accordingly. In 1840 Hess demonstrated that the heatabsorbed in a chemical reaction was independent of the number of steps in the reaction; different reaction pathsgave the same total heat absorbed. The mathematical statement of Hess’s law is that since

dU = T dS − P dV, U = U(S, V ) (14.75)

then the quantityH = H(S, P ), H = U + PV, dH = T dS + V dP (14.76)

can be used to measure the heat of reaction under constant atmospheric pressure since in such circumstances

dU = dQ− δW → dQ− P dV, dH = dU + P dV + V dP → dQ (14.77)

since dP = 0. A nice chemical application of Hess’s law is the computation of the heat of reaction for

Ca+O2 +H2 Ca(OH)2 (14.78)


from

CaO +H2O Ca(OH)2, ∆H18o C = −15260cal

mole

H2 +1

2O2 H2O, ∆H18o C = −68315

cal

mole

Ca+1

2O2 CaO, ∆H18o C = −151800

cal

mole

∆H18o C = −235375cal

mole(14.79)

adding all of it together we obtain ∆H18o C = −235375 calmole for the net reaction.

The most important aspect and application of enthalpy in physics is its interpretation as being a potential for heat,in the sense that at constant pressure an equilibrium condition is the minimization of H.

14.3.5 Material properties of matter

The isochoric (constant volume) heat capacity is

Cv =(∂Q∂T

)V=(∂U∂T

)V

(14.80)

isobaric heat capacity is

CP =(∂Q∂T

)P= P

(∂V∂T

)P+(∂U∂T

)P

(14.81)

These quantities tell us how much the energy must change by with an appropriate variable held fixed, in order forthe temperature to change by a specific amount. If CP is high, like it is for water, a small temperature change couldrequire quite a bit of energy.The coefficient of thermal expansion is

3α =1

V

(∂V∂T

)P

(14.82)

which measures an objects size changes at constant pressure as its temperature is changed.

14.4 Ideal gases

The ideal gas is one of the few systems for which one can very easily construct a detailed and exact picture ofthermodynamical processes. This is because the two equations of state

PV = NkT = nNAkT = nRT, U =3

2NkT =

3

2nNAkT = nCV T (14.83)

are extremely simple, and so are the calculations that extract information from them.

The tools of the trade are;1. The equations of state of the matter to be studied, and their differential relations. For example

PV = nRT, U = nCV T

P dV + V dP = nRdT dU = nCv dT (14.84)

2. The first law dU = dQ − dW specialized to reversible (equilibrium) processes without any form of work otherthan PV , so we have only one chemical species present

U = U(S, V ), dU = T dS − P dV

and the second law for equilibrium processes

dS =dQ

T

14.4. IDEAL GASES 289

3. The differential relations for the potentials F = F (T, V ), G = G(T, P ). These are universal (true for all materials)

dU = T dS − P dV

dF = −S dT − P dV

dG = −S dT + V dP

dH = T dS + V dP

From this we begin examining particular processes that the gas can be put through. One of the original applicationsof thermodynamics was to converet heat energy into mechanical work. A machine that does this is a heat engine,and most use an ideal gas as a working substance.

Lets look at a collection of same calculations for the most important type of processes that a gas can be subjectedto.

Example 1. Compute the increase in entropy of an ideal gas that is heated from T1 to T2 at constant volume.

Use the first law with dV = 0 to get dQ;

dS = nCVdT

T,

∫ S2

S1

dS = S2 − S1 =

∫ T2

T1

nCVdT

T= nCV ln

T2T1

(14.85)

Example 2. Compute the increase in internal energy of an ideal gas that is allowed to expand from V1 to V2 atconstant temperature.

There is no change in internal energy, since U depends only on T . If T does not change, neither does U

dU = Cv dT = 0 (14.86)

Example 3. Compute the increase in entropy of an ideal gas that is allowed to expand from V1 to V2 at constanttemperature.Use the first law with dT = 0, note that the second equation of state implies dU = 0;

dS = nRdV

V,

∫ S2

S1

dS = S2 − S1 =

∫ V2

V1

nRdV

V= nR ln

V2V1

(14.87)

Example 4. Compute the increase in internal energy of an ideal gas that is heated from T1 to T2 at constantvolume.Use

ndU = nCV dT,

∫ U2

U1

dU = U2 − U1 =

∫ T2

T1

nCV dT = CV (T2 − T1) (14.88)

An adiabatic process is one for vwhich there is no heat transfer, dQ = 0.

Example 5. Compute the increase in internal energy of an ideal gas that is allowed to expand reversibly andadiabatically from V1 to V2.Start with

dU = T dS − P dV = −P dV (dS = 0) (14.89)

and recognize that we simply need to compute the temperature change, since ∆U = nCV ∆T . Insert the ideal gaslaw

dU = nCV dT = −nRTV

dV,dT

T= − nR

nCV

dV

V= −2

3

dV

V(14.90)

and integrate ∫ T2

T1

dT

T= ln

T2T1

= −2

3

∫ V2

V1

dV

V= −2

3lnV2V1

(14.91)

rearrange (remember the second of these, it is very handy)

lnT 32

T 31

= lnV 21

V 22

, T2V232 = T1V

231 for adiabatic processes (14.92)


and we get

∆U = nCV (T2 − T1) = nCV T1

((V1V2

) 23 − 1

)(14.93)

If you use the ideal gas law to eliminate the temperature in the so-called adiabatic relation, you get a second adiabaticrelation

T2V232 = T1V

231 , P2V

532 = P1V

531 for adiabatic processes (14.94)

for a monatomic ideal gas.

Example 6. Compute the change in Helmholtz free energy if an ideal gas is allowed to isothermally expand fromV1 to V2.Start with

dU = δQ− P dV = T dS − P dV (14.95)

and useF = U − TS, dF = dU − T dS − S dT = −S dT − P dV (14.96)

For isothermal expansion dT = 0, and insert the ideal gas law

dF = −P dV = nRTdV

V, F2 − F1 = −nRT

∫ V2

V1

dV

V= −nRT ln

V2V1

(14.97)

Example 7. Compute the change in Gibbs free energy if an ideal gas is isothermally re-pressurized from P1 to P2.Start with

dU = δQ− P dV = T dS − P dV (14.98)

and useG = U − TS + PV, dG = dU − T dS − S dT + P dV + V dP = −S dT + V dP (14.99)

For isothermal expansion dT = 0, and insert the ideal gas law

G2 −G1 =

∫ P2

P1

nRT

PdP = nRT ln

P2

P1(14.100)

Example 8. Compute the heat that flows into a gas during a constant pressure heating process in which thetemperature rises from T1 to T2.Start with

dU = T dS − P dV = δQ− P dV, δQ = dU + P dV (14.101)

in any constant pressure ideal gas process

dP = 0 = d(nRT

V

)=nRdT

V− nRT dV

V 2,

dT

T=dV

V(14.102)

therefore inserting this and the ideal gas law;

δQ = dU + P dV = dU + PV

TdT = dU + nRdT =

3

2nRdT + nRdT (14.103)

This leads us to

∆Q =

∫δQ =

5

2nR(T2 − T1) = CP (T2 − T1) (14.104)

Example 9. There is a master entropy formula for the ideal gas, from which you can get ∆S for any state-to-stateprocess. Find it!

Rearrange the first law

dS = nCVdT

T+ nR

dV

V(14.105)

can be integrated to a final entropy formula for the ideal gas. We first note that it can be written as

dS = nCVdT

T+ nR

dV

V=(∂S∂T

)VdT +

( ∂S∂V

)TdV (14.106)

14.5. PV -DIAGRAMS 291

Divide by dσdS

dσ=(∂S∂T

)V

dT

dσ+( ∂S∂V

)T

dV

dσ(14.107)

and integrate ∫dS

dσdσ =

∫ (∂S∂T

)V

dT

dσdσ +

∫ ( ∂S∂V

)T

dV

dσdσ (14.108)

Suppose that when σ = σ0, we have S = S0, T = T0 and V = V0, and when σ = σ1, we have S = S1, T = T1 andV = V1. We can change variables in each of the three integrals and perform each one over a separate, appropriatevariable; ∫ σ1

σ0

dS

dσdσ =

∫ S1

S0

dS = S1 − S0,

∫ σ1

σ0

(∂S∂T

)V

dT

dσdσ =

∫ T1

T0

(∂S∂T

)VdT (14.109)

and ∫ σ1

σ0

( ∂S∂V

)T

dV

dσdσ =

∫ V1

V0

( ∂S∂V

)TdV (14.110)

Performing these last two explicitly for the ideal gas;∫ T1

T0

(∂S∂T

)VdT =

∫ T1

T0

nCVdT

T= nCV ln

T1T0

(14.111)

Performing these last two explicitly for the ideal gas;∫ V1

V0

( ∂S∂V

)TdV =

∫ V1

V0

nRdV

V= nR ln

V1V0

(14.112)

and we arrive at

S1 − S0 = nCV lnT1T0

+ nR lnV1V0

= nR ln((T1

T0

) 32(V1V0

))(14.113)

Example 10. There is a master free energy formula for the ideal gas, from which you can get ∆F for any state-to-state process. Find it!A master Helmholtz free energy equation can be obtained by exactly the same means. Starting with the definition

F = U − TS,F

T=U

T− S (14.114)

we obtainF (T0, V0)

T0=U(T0, V0)

T0− S(T0, V0),

F (T1, V1)

T1=U(T1, V1)

T1− S(T1, V1) (14.115)

subtract these and use UT = CV = 3

2nR for an ideal gas

F (T1, V1)

T1− F (T0, V0)

T0= S(T1, V1)− S(T0, V0) = nR ln

((T1T0

) 32(V1V0

))(14.116)

14.5 PV -diagrams

Because the first equation of state is a constraint f(V, T, P ) = 0, the state of the ideal gas is specified by twoquantities, usually we use P and V . For an equilibrium process in which the gas is always in a well-defined global(homogeneous T , P ) state, we can plot the progression of equilibrium states that a gas passes through during aprocess on a PV -diagram.

The most technically important processes are cycles, in which the gas is subjected to a sequence of sub-processesthat ultimately return it to its original state. These sub-processes could be arbitrary, but we will study only constantV , constant P , isothermal and adiabatic


Process ∆U ∆Q ∆W

Constant V nCV (Tf − Ti) nCV (Tf − Ti) 0

Constant T 0 nRT lnVf

VinRT ln

Vf

Vi

Adiabatic nCV (Tf − Ti) 0 −nCv(Tf − Ti)Constant P nCV (Tf − Ti) n(CV +R)(Tf − Ti) nR(Tf − Ti)

(14.117)

from which you can see that the secret of working out these problems is to get Tf and Ti, using the diagramand the equations of state.

One very important property of a cyclic process is that for any state variable such as U, S, F,G,H, the sum of allchanges over the cycle sub-processes must be zero, because the cycle returns the gas to its original state

∆X =∑i

∆Xi = 0, X = U, S, F,G,H, for any cycle (14.118)

Example 11. Consider the thermodynamic cycle carriedout on one mole of monatomic ideal gas illustrated to theleft, a → b is constant P , b → c is constant T , c → dis constant P , d → a is constant T . Fill in the table ofthermodynamic quantities below. Pa = 1.0 × 104 N

m2 ,

Va = 1.0m3, Vb = 2.0m3, Pc = 0.5 × 104 Nm2 . Note that

Vd = Vb.

The first thing you do is get the temperatures;

Ta =PaVaR

=(1.0× 104)(1.0)

8.314= 1203K

Tb =PbVbR

=(2.0× 104)(1.0)

8.314= 2406K

Tc = Tb = 2406K

Td = Ta = 1203K (14.119)

Now its all pretty easy; for example

∆Ua−>b = 1 · 32(8.314)(2406− 1203) = 15, 000J

∆Wa−>b = Pa(Vb − Va) = (1× 104)(2.0− 1.0)

= 10, 000J

∆Wb−>c = 1 ·RTb lnVcVb

= (8.314)(2406) ln(2)

= 13863 J (14.120)

Process ∆U ∆Q ∆W

a→ b 15000.0 J 25000.0 J 10000.0 J

b→ c 0.0 13862.94 J 13862.94 J

c→ d -15000.0 J -25000.0 J -10000.0 J

d→ a 0.0 -6931.47 J -6931.47 J

Ta = 1203.4C Td = 1203.4C Vc = 4m3


Example 12. Consider the thermodynamiccycle carried out on one mole of monatomicideal gas illustrated to the left, a → b isadiabatic, b → c is constant T , c → a isconstant V . Fill in the table of thermody-namic quantities below. Pa = 1.0 × 104 N

m2 ,

Va = 1.0m3, Pb = 0.25× 104 Nm2 .

Get the temperatures

Ta =PaVaR

= 1203K

Vb = Va ·(PaPb

) 35

= 2.3m3

Tb = Ta

(VaVb

) 23

= 690K

Tc = Tb = 690K (14.121)

Once you have the temperatures, its very easy to wrap it all up.

∆Ua−>b =3

2(8.314)(690− 1203)

= −6397J

∆Sa−>b = 0 (adiabatic)

∆Wb−>c = RTb ln(VcVb

)= −4778J

∆Sb−>c = R ln(VcVb

)= −6.92J/K

(14.122)

Process ∆U ∆Q ∆W ∆S

a→ b -6397 J 0.0 6397 J 0.0

b→ c 0.0 -4778 J -4778 J -6.92 J/K

c→ a 6397 J 6397 J 0.0 6.92 J/K

Always use the first law ∆U = ∆Q −∆W ; apply it to a row in a table, you need only two out of three entries perrow, the third is for free from the first law. Remember that U, S are state variables, their columns must add up to zero.

14.5.1 The Carnot cycle

In an abstract sense the Carnot cycle is incredibly simple; there is an energy balance

∆U = 0, Uin = Uout, Qh =W +Qc (14.123)

and the heat in/out steps are isothermal at Th/Tc respectively, and the only other processes are adiabatic, so

∆S = 0 =QhTh

− QcTc

(14.124)

The figure below details a four step cycle for a heat engine in which two isothermal processes perfectly turn heat intowork, and cool the gas by requiring work input to expel unwanted residual heat in the gas. The cycle is completedwith adiabatic steps. Heat in/out is with thermal (conductive) contact with heat sources/sinks


It is a trivial matter to compute the work, heatand energy changes for each step of the cycle,since

dUa→b = 0, dQb→c = 0, dUc→d = 0

dQd→a = 0

We can fill in a table for all relevant data. No-tice first

PaVa = nRTh = PbVb, PcVc = nRTc = PdVd

∆Qa→b = ∆Wa→b

∆Ub→c = −∆Wb→c

and so forth, using the first law of thermodynamics.

process ∆U ∆Q ∆W ∆S

a→ b 0 nRTh ln(Vb

Va) nRTh ln(

Vb

Va) nR ln( Vb

Va)

b→ c 32nR(Tc − Th) 0 −3

2nR(Tc − Th) 0

c→ d 0 nRTc ln(Vd

Vc) nRTc ln(

Vd

Vc) nR ln(Vd

Vc)

d→ a 32nR(Th − Tc) 0 −3

2nR(Th − Tc) 0

Notice that the sum of all of the energy changes for the entire cycle is zero. This is because the gas returns to itsoriginal state, and therefore original temperature, at the end of the cycle. Since the energy of an ideal gas dependsonly on the temperature, the energy change per cycle is zero. So what does the cycle do? It extracts heat from thehigh temperature heat bath, which could be burning fuel or any heat source, converts some of it into net out-flowingwork (notice that the net area under the curve=area within the cycle is positive) and discards unused heat as waste


into the coolant bath. This is what your car engine does.

Take note of the fact that the total cycle ∆U = 0, and the total cycle ∆S = 0 since both are state variables. Theentropy calculation is not immediately obvious, but

Ta = Tb, a→ b isothermal

Td = Tc, c→ d isothermal

TbV23

b = TcV23c , b→ c adiabatic

TdV23

d = TaV23a , d→ a adiabatic

divideTcTd

(VcVd

) 23

=TbTa

(VbVa

) 23

or(VcVd

)=

(VbVa

)Therefore ∆S = nR ln

(VbVa

)+ nR ln

(VdVc

)= 0 (14.125)

S

T

1 2

34

Example 13. The Carnot cycle is the simplest in theS − T plane, since the four sub-processes will leave eitherT or S unchanged.

One mole of a monatomic ideal gas is made to undergothe reversible Carnot cycle in the figure.

T1 = 200K, T2 = 500K, S1 = 5.0 JK and S4 = 3.0 JK .

All of the calculations are particularly simple becauseevery row has a zero in it.

Line ∆U ∆Q ∆W ∆S

1 → 2 (3/2)(1)(8.31)(500-200) 0 -3740 J 0=3740 J

2 → 3 0 (500)(3-5)=-1000J (500)(3-5)=-1000J (3-5)J/K

3 → 4 (3/2)(1)(8.31)(200-500) 0 3740 J 0=-3740 J

4 → 1 0 (200)(5-3)=400 J (200)(5-3)=400J (5-3)J/K

The Carnot cycle is a heat engine, a machine that takes in heat at Th on the a→ b isothermal sub-process, caus-ing the gas trapped in the cylinder to expand, doing work that can be exported or harnessed to perform a mechanicaltask. This is the power-stroke part of the engine cycle. Afterwards the gas must be brought back to its original


state, so that the process can be repeated. This is done by thermally isolating the gas from b → c and allowing itto expand, which causes a temperature drop. Now heat is extracted and discarded into a heat sink and thegas contracts isothermally to a much lower volume, after which it is again thermally isolated from d→ a while it iscompressed mechanically to its original pressure, volume and temperature. Now the process can be repeated endlessly.

Heat is one of the most readily created but degenerate form of energy, and harnessing it by using a machine to turnit into work was a giant technical achievement that made the Industrial Revolution possible.

The two really important parts of the cycle are a → b; a perfect transformation of heat into work, since dT = 0means dQ = dW ; input heat becomes output work. But this can’t go on forever; the spontaneous expansion willlead to exhaustion. The adiabatic expansion from b → c that reduces the gas temperature is equally important,because a smaller amount of mechanical work can now be done to restore the gas to its original state. Thisresults in a net positive outflow of work, and a net positive inflow of heat, but some heat must be lost as waste,so the net engine efficiency is less than one, all to avoid exhaustion of the working gas. The efficiency is

ϵ =Wout

Qin

=Qin −Qout

Qin= 1− Qout

Qin

but ∆S = 0 =QinTh

− QoutTc

therefore ϵ = 1− TcTh

(14.126)

The Carnot engine is as perfect as you can get, no other heat engine has an efficiency this high.

If you run a Carnot cycle in reverse, you input more work than you extract, and you output more heat (into ahigh-T reservoir) as waste than you input as fuel (from a low-T reservoir); this is a refrigerator. A measure of howwell a reversed Carnot cycle is at extracting heat is

K =QinWin

but Qin +Win = Qout

K =Qin

Qout −Qin

=Tc

Th − Tc(14.127)

By dividing by time, we can compute the power it takes to remove heat from an already chilled object at rate dQc

dt

Win =(Th − Tc

Tc

)Qin, ℘ =

(Th − TcTc

) dQindt

(14.128)

Example 14. Suppose that you own a poorly insulated small house, with outside walls ℓ = 0.25m thick, total outerwall plus roof surface area 210m2 (about 1992 ft2) and your walls have a thermal conductivity of 1.3 W

m·K (limestone).Calculate how much power your (Carnot) air conditioner will require to keep your house at 293K when the outsidetemperature is 308K.

You must remove the heat that gets in through the walls

dQcdt

=(1.3)(210)(308− 293)

0.25= 16, 380J,

dW

dt= ℘ =

(308293

− 1)16, 380J = 838W


14.5.2 Arbitrary systems

There are a few simple systems other than ideal gases that one can paint a complete picture of thermodynamically,because all we need in order to do this is two equations of state and the basic thermodynamical principles.

P

V

1 2

34

Light (electromagnetic radiation) trapped in a cavity ofvolume V in thermal equilibrium with hot conducting wallsat temperature T has equations of state

U = aV T 4, a = 7.56 × 10−16 J

m3K4

U = 3PV

Suppose that it is made to undergo the reversible processillustrated, with P1 = 100 N

m2 , V1 = 0.1m3, V2 = 0.3m3

and P4 = 25 Nm2 . Compute ∆U, ∆Q, ∆W for each process

line on the cycle.

Everything can be calculated with just dU = 3P dV ordU = 3V dP since all processes are constant P or V , alongwith ∆W = P ∆V or ∆W = 0.

Line ∆U ∆Q ∆W

1 → 2 3(100)(0.3− 0.1) 4(100)(0.3− 0.1) (100)(0.3− 0.1)

2 → 3 3(0.3)(25− 100) 3(0.3)(25− 100) 0

3 → 4 3(25)(0.1− 0.3) 4(25)(0.1− 0.3) (25)(0.1− 0.3)

4 → 1 3(0.1)(100− 25) 3(0.1)(100− 25) 0

By combining the equations of state we discover that for this material P = a3T

4, so constant P processes like1 → 2 and 3 → 4 are isothermal.

Entropy problems just go back to basics as well; suppose the “photon gas” undergoes isothermal expansion from(V1, P1) = (0.1m3, 100 N

m2 ) to (3V1, P1). Find ∆S.

Since P = a3T

4, constant T means constant P ; so dU = 3P dV

dS =dU

T+P

TdV

= 3P

TdV +

P

TdV

= 4P

TdV

∆S = 4P

T∆V = 4

(100)

4

√3(100)

5.67×10−8

(0.3− 0.1) = 0.297J

K(14.129)


What about ∆S2→3? Once again we go back to basics. Note that dV = 0 for 2 → 3, and dU = 3d(PV ) = 3V dP

dS =dU

T+P

TdV

= 3V

TdP

= 3V

∫ P4

P3

dP4√3P/a

=4V

4√3/a

(P

343 − P

342

)(14.130)

so it is just a matter of going back to basics; the first law-definition of dS for a reversible process, into which thecorrect equations of state are substituted.

The Van der Waals equations of state(P +

n2a

V 2

)(V − nb) = nRT, U =

3

2nRT − n2a

V(14.131)

is a better description of the behavior of a gas at low temperatures than the ideal equations of state (which describegasses accurately at high-T ).In an isothermal expansion T1 = T2 from V1 to V2, take note that the internal energy of a VDW gas does change,unlike the ideal gas

∆U =(32nRT2 −

n2a

V2

)−(32nRT1 −

n2a

V1

)=n2a

V1− n2a

V2= 0

so isothermal ∆U = 0 is an ideal gas result not necessarily exhibited by other materials.

The adiabatic relations (some at least) for an ideal gas and VDW gas are very similar

dU = dQ− PdV = −PdV3

2nRdT +

n2a

V 2dV = −

( nRT

V − nb− n2a

V 2

)dV

3

2

dT

T= − dV

V − nb(TfTi

) 32

=( Vi − nb

Vf − nb

)(14.132)

14.6. NON-PV WORK. CHEMICAL AND ELECTRICAL 299

14.6 Non-PV work. Chemical and electrical1

Work in general has the formdW = (intensive) · d(extensive) (14.133)

Intensive quantities (like P and T ) do not depend on the size of the system (the number of particles) but extensivequantities (like V ) do. For very simple systems such as an ideal gas, work is easy to compute (its just P dV ). Keepin mind that the introduction of a new form of work will require the introduction of a new extensive variable.

What about non-P dV work? Perhaps the simplest such examples are chemical work, and the work done when chargedc is transported through an electro-chemical battery of voltage E . This electro-chemical work is

dWE = E dc (14.134)

Unfortunately at this point we have not yet discussed electric or magnetic fields, and so we are very limited in ourabilities to incorporate non-P dV work into problems.

The only other candidate would be chemical work, in which we study the work done when chemical recombinationof molecules occur. The chemical potential µi is the work done in adding another particle of species i tothe system. In a system with multiple chemical species, the first law reads

dU = dQ− P dV +∑i

µi dNi + E dc (14.135)

The total number of particles of any element must be conserved in a chemical process, let N =∑iNi where we sum

over the species, and if we write Ni in moles, we can work with mole-fractions

xi =NiN

(14.136)

For example in an ideal gas (in these formulas k is Boltzmann’s constant, related to the ideal gas constant byAvagadro’s number R = kNA)

P =(∑

i

Ni

)kTV

=∑i

xiNkT

V=∑i

Pi (14.137)

which is the law of partial pressures; the pressure exerted by species i alone is Pi = xiNkTV .

Example 15. For example in a gas of H-atoms we could allow for bonding to H2 to occur and include the workdone by adding a chemical work term such as

dWc = µHdNH + µH2 dNH2 (14.138)

to the first law.In a chemical reaction the total number of H atoms will be conserved, and so there is a conservation law

NH + 2NH2 = constant, dNH + 2 dNH2 = 0 (14.139)

Lets consider only reversible chemical processes, these are processes for which dQ = T dS where S is the entropy.Then the master thermodynamic relation that we begin with is

dU = T dS − P dV +∑i

µi dNi + E dc (14.140)

which we invert and write as

dS =( 1

T

)dU +

(PV

)dV −

∑i

(µiT

)dNi −

ETdc (14.141)

1It is not likely that we will get this far in physics 201. Those of you that must take P-chem, or the MCAT, or the GRE, read on;everything beyond this point is for you.


In order to study chemical thermodynamics, we will need to figure out what µ depends on. To do this, we limitourselves of course to reversible, equilubrium processes, and employ the maximal entropy principle, which is oneof the most powerful and insightful concepts in physics; the equilibrium state is the most probable state.

S, v

E, V

Consider a universe U of volume V divided into a system S to bestudied, of volume v, and an environment E of volume V − v. If thereare a total of n particles to be shared by U = E

∪S, how are they

distributed between E and S? The probability that any one particlewill be found in S is ℘S = v

V , and in E ℘E = 1− vV , so the probability

that m of the n particles will be found is S is

℘ =n!

m!(n−m)!

( vV

)m(1− v

V

)n−m(14.142)

There is a value m∗ of m that maximizes this, lets us the fact thatfor large N ; lnN ! ≈ N lnN is a very good approximation

d℘

dm

∣∣∣m∗

= 0 =d

dm

(n lnn−n−m lnm+m−(n−m) ln(n−m)+(n−m)+m ln(v/V )+(n−m) ln(1−v/V )

∣∣∣m∗

(14.143)

and we get

− lnm∗ + ln(n−m∗) + ln v − ln(V − v) = 0 (14.144)

There are two useful and insightful way to write this.Firstly we write it as

lnm∗

v= ln

n−m∗

V − vor

m∗

v=n−m∗

V − v(14.145)

which says that the equilibrium state is one in which E and S have equal particle densities. This issomething that we all feel is correct at an intuitive level.Secondly we leave it as

lnm∗

v= ln

n−m∗

V − v, ln cs = ln cE (14.146)

introducing particle concentrations or densities, and return to the First Law of Thermodynamics, eq. 14.141,

dS =( 1

T

)dU +

(PV

)dV −

(µST

)dm−

(µET

)d(n−m) (14.147)

which says that if we maximize S with respect to particle distribution, entropy is maximal when

dS

dm= 0 =

µST

=µET

(14.148)

recalling the Boltzmann definition of entropy

S = k ln℘ (14.149)

we obtain our formula for the chemical potential

µS = µ0 + kT ln cS (14.150)

which is literally the foundation of basic chemistry.

Example 16. Find the equilibrium constant for the chemical reaction A+B AB at constant T and P .

Chemists love the Gibbs free energy G = U − TS + PV for the following reason. Differentiate this thing

dG = dU − TdS − SdT + PdV + V dP (14.151)


and put eq. 14.140 into it, several things cancel

dG = −SdT + V dP + µA dNA + µB dNB + µAB dNAB (14.152)

Suppose that you study your chemical reaction at constant T and P , then dT = dP = 0 and you get

dG = µA dNA + µB dNB + µAB dNAB (14.153)

and at chemical equilibrium G stops changing, so we get the condition for equilibrium

0 = µA dNA + µB dNB + µAB dNAB (14.154)

into which we insert conservation of total species subject to stoichiometric constraints; if we lose an A we must losea B and gain an AB

dNA = dNB = −dNAB (14.155)

This makes all of the dN ’s cancel0 = µA + µB − µAB (14.156)

or by putting in Eq.14.150

− µ0A − µ0

B + µ0AB = kT ln(

cAcBcAB

),cABcAcB

= e−(−µ0

A−µ0B+µ0

AB)

kT = Keq (14.157)

in which each gas has a different µ0i , which depends on things like its mass, specific heat, binding energy, and so on.

This quantity is basically the energy that it takes to create a member of the species at a standard temperature andpressure. Chemistry convention is that K concentration ratios of product over reactant.

14.6.1 Chemical potential for ideal gases

The fundamental equation of a thermodynamic system is the formula for its absolute entropy, which for an idealgas is Eq. 14.113

S1 − S0 = nCV lnT1T0

+ nR lnV1V0

= nR ln((T1

T0

)CvR(V1V0

))(14.158)

Lets carefully seperate out the dependence on N,V and T ; let V0 = N ν0 and V = N ν where ν, ν0 are vol-ume/reference volume per particle, and pick ν0 and T0 so that S is zero when T = T0 and ν = ν0

S = Nk ln(( T

T0

)CvR( νν0

))(14.159)

and into this, for gaseous chemical species i, we insert P = NkTV = kT

ν , and P0 = NkT0

V0= kT0

ν0;

S = Nk ln(( T

T0

)CvR +1(P0

P

))(14.160)

We know that both T and P are not dependent on N , so this formula contains all of the N -dependence of the entropyout front as the coefficient, which is very nice.Our next step is to start with the first law;

dU = TdS − PdV + µdN

dS =dU

T+P

TdV − µ

TdN( ∂S

∂N

)U,V

= −µ

T(14.161)

from which we get an explicit formula for the chemical potential of an ideal gas, suitable for doing gas chemistry

µ = −kT ln(( T

T0

)CvR +1(P0

P

))(14.162)


Since both Cv and P for a gas of species i in a mixture could depend on species (P is the pressure exerted by thespecies, a partial pressure), for a mixture of gases we have

µi = −kT ln(( T

T0

)Cv,iR +1(P0

Pi

))= −kT ln

(( TT0

)Cv,iR +1( P0

xiP

))= µ0

i + kT lnxi (14.163)

from which we get an actual formula displaying the temperature and pressure dependence of µ0i that you see in

Eq. 14.157This formula is extremely useful for understanding principles layed out (without proof) in freshman chemistry, suchas Lechatelier’s principle. For example, in the reaction A+B AB, our equilibrium constant would be

lnKeq = −

(µ0A + µ0

B − µ0AB

)kT

= ln(( T

T0

)Cv,AR +1(P0

P

))+ ln

(( TT0

)Cv,BR +1(P0

P

))− ln

(( TT0

)Cv,ABR +1(P0

P

))= ln

(( TT0

)Cv,A+Cv,B−Cv,ABR +1(P0

P

))(14.164)

so thatxA xBxAB

=( TT0

)Cv,A+Cv,B−Cv,ABR +1(P0

P

)(14.165)

If you increase P , the right-side gets smaller, meaning that xAB gets bigger, in other words increasing the pressurepushes A+B AB to the right. Thermodynamics gives you formulas for Keq!

14.6.2 Electrochemical work

Zn (anode)

Pt (cat.)

Zn++, 1 M H3O+, 1 M

H2 gas

When a known amount of charge dc passes through aknown voltage difference E such as that of a battery,a calculable amount of work is done dW = E dc.A battery is simply a device that gives anelectrical charge dc some potential energy

−(Vf − Vi

)= dW = E dc (14.166)

The cell pictured has a zinc strip undergoing an ox-idation reaction (loss of electrons) on its anode

Zn (s) → Zn++(aq) + 2e−

E0anode = +0.76V

The other electrode (cathode, where chemical re-duction occurs) is a solution of 1M H+ created bybubbling H2 gas over a platinum (inert) strip. Thishydrogen half-cell reaction is the zero-point againstwhich all other half-cell reactions are measured,being assigned a cell-voltage of zero

H2 (g) → 2H+(aq) + 2e−, E0

cat = 0.0V

The entire cell reaction is the sum of the half-cellreactions

2H+(aq) + Zn (s) → Zn++

(aq) +H2 (g)

What do these voltages E0cathode, E0

anode mean? How can this device actually supply voltage?


Put this device together with chemical concentrations nowhere near the equilibrium values. Then the chemical re-action will run spontaneously and electrons will be driven one way or the other through the connecting wire. Ittakes voltage to drive the electrons along, the voltage of the battery. So how much voltage is this? Put your ownexternal battery into the circuit where the meter is, and tune it up to a sufficient value to stop the flow of charge.At that value it is exactly negating the battery voltage. These are the half-cell voltages E0

anode and E0cathode, each

is separately measured by setting up a half-cell with an external variable battery and tuning the external voltageuntil no current flows. This must be done by momentarily completing the circuit, so no major chemical changesoccur in the cells while under measurement.

In a typical chemical process such as (note the sign convention)

ν1X1 + ν2X2 ν3X3 + ν4X4, more generally written∑i

νiXi = 0, νprod > 0, νreact < 0 (14.167)

the stoichiometric coefficients play a big role in the thermodynamics because the total number of atoms must beconserved (they simply rearrange in different molecular combinations), so that there exists a number ξ such that thenumber of particles of species i changes as

Ni = Ni,0 + νiξ (14.168)

as the reaction progresses (Ni,0 is the initial number when the reaction is started), so

dNi = νi dξ (14.169)

If the cell is assembled with any concentrations of reactants, the chemical reaction will run spontaneously in whateverdirection minimizes the Gibbs free energy (so ∆G = Gf − Gi < 0). This is the natural tendency. So instead weset it up so that the chemical concentrations are at or near equilibrium values. What are those values? Ifwe work at constant T and P

2H+(aq) + Zn (s) Zn++

(aq) +H2 (g) (14.170)

dG = −S dT + V dP +∑i

νi

(µ0i + kT ln ci

)dξ → 0 =

∑i

νi

(µ0i + kT ln ci

)(14.171)

or

− 1

kT

∑i

νiµ0i = ln

(∏i

cνii

),

∏i

cνii = Keq = e−1

kT

∑i νiµ

0i (14.172)

Solids are generally taken to have zero total chemical potential in such reactions, they are treated simply as particlereservoirs, not as chemical reactants.Using the stoichiometric numbers for our reaction eq. 14.170 we can find these concentrations ci. Under equilibriumoperation there will be no net flow of charge through the wire connecting the half-cells.

Example 17. Lets describe how the cell can be used to measure equilibrium constants.

Set the cell up with concentrations ci = 1 (molar) for all reactants, add in an external battery, and tune the batteryuntil equilibrium is restored, and no current flows. The battery is doing reversible work on the system, and sincewe are at equilibrium, if an infinitismal shift dξ occurs in the participating chemical species numbers, 2dξ electronswill be transferred; let F = NAe = 96500C be the charge on a mole of electrons

dG = −S dT + V dP + Eext2e dξ +∑i

νi

(µ0i + kT ln ci

)dξ = 0

0 =(Eext2e+

∑i

νi

(µ0i + kT ln ci

))dξ

or − 2eEext =∑i

νiµi0 = −kT lnKeq

Keq = e2eEext

kT = enFEext

RT (14.173)

(if n electrons are transferred) so by reading the external battery voltage needed to restore reversibility, we obtainthe equilibrium constant.


We generally write eq. 14.173 asdGuniv = dGcell + dGbattery (14.174)

from which we obtain the useful result that when the external battery restores equilibrium to the system

dGcell = −dGbattery = −Ee dnelectron (14.175)

when dnelectron electrons would flow out of the battery into the cell to halt the reaction.

Electrode E0 ReactionLi+;Li -3.045 V Li+ + e− → LiK+;K -2.925 V K+ + e− → K

Al+++;Al -1.66 V 13Al

+++ + e− → 13Al

Cd++;Cd -0.403 V 12Cd

++ + e− → 12Cd

Cl−;AgCl(s), Ag 0.2224 V AgCl + e− → Ag + Cl−1

Cu+;Cu 0.521 V Cu+ + e− → CuCu++;Cu 0.337 V 1

2Cu++ + e− → 1

2Cu

Chemists use some simple conventions to take the guessing out of computing some of these quantities. First

Ecell = E0right − E0

left = E0red(cat)− E0

red(an) =

> 0 Spontaneous (∆G < 0)< 0 non-spontaneous

(14.176)

and second;the reaction on the left electrode is written as an oxidation, X → X+ + e−, and on the right electrode we have areduction Y ++ e− → Y . The net cell reaction is then the sum of the two half-cell reactions (the electrons “cancel”).

14.7 More applications of µ

The chemical potential can be used to describe any kind of mass-motion or concentration gradient, not just chemicaltransformation.

The process of sedimentation (this requires some force/potential) will make the concentration of a chemical speciesvary with position. The simplest sedimentation is caused by gravity.

z

z

z+dz

The chemical potential of the ideal solute will be of the form

µ(z) = µ0 + kT ln c(z) (14.177)

where c(z) is the concentration. Chemical potential is Gibbs free energy perparticle, and we have seen several times that Gibbs free energy is a measure ofwork, in particular −µdN is the work done in moving another dN particles intothe system. Consider two thin “cells” of solution in the cylinder, one at height z,one at height z + dz, and compute the work done in moving a solute molecule ofmass m from the lower cell into the upper (dN = ±1)

−(µ(z + dz)− µ(z)

)· 1 = mg dz = −kT

(ln c(z + dz)− ln c(z)

)(14.178)

Expand

− mg dz

kT= ln

c(z + dz)

c(z)= ln(

c(z) + dc(z)dz dz + · · ·c(z)

) ≈ 1

c(z)

dc(z)

dzdz (14.179)

14.7. MORE APPLICATIONS OF µ 305

and sodc(z)

c(z)= −mg dz

kT,

∫ c(z)

c(0)

dc(z)

c(z)=

∫ z

0

−mg dzkT

(14.180)

orc(z) = c(0) e−

mg zkT (14.181)

and the concentration is greatest at the bottom of the cylinder. The work done against the gravitational potentialis non-PV work.

14.7.1 The centrifuge

Example 18. The centrifuge works by the same principle. Macro and bio-molecules can be seperated from solventsby application of a much greater force than gravity.

rdr

Consider two thin “cells” of solution in a test-tube in the machine, one at radius r, the otherat r+dr, and compute the work done in movinga solute molecule of mass m from the inner cellinto the outer (dN = ±1)

−(µ(r + dr)− µ(r)

)· 1 = dW (14.182)

= −kT(ln c(r + dr)− ln c(r)

)(14.183)

Examine a molecule of mass m within the testtube, relative to the stationary xy-coordinatesystem outside of the tube,

for which x = r cosωt, y = r sinωt where ω is the spin-rate of the centrifuge. The energy of such a molecule is (seeproblem 10.19)

Ustat. =1

2m(x2 + y2) =

1

2m(r2 + ω2r2) (14.184)

Now think about what the molecule experiences with respect to the coordinate system inside of the tube. Themolecule only moves radially within the tube, relative to a co-rotating set of coordinate axes, and so has energy

Urot. =1

2mr2 = Ustat. −

1

2mω2r2 (14.185)

so even if it is not moving radially, it has “potential energy”

Vcent.(r) = −1

2mω2r2 (14.186)

which acts like a simulated force of gravity within the tube. Equilibrium (with respect to particle exchange) betweenthe adjacent cells illustrated is then

dG = 0 = 1 ·(kT ln c(r + dr)− 1

2m(r + dr)2ω2

)− 1 ·

(kT ln c(r)− 1

2mr2ω2

)(14.187)

or

kT ln(c(r) + dc(r)

dr dr + . . .

c(r)

)= mrω2dr (14.188)


and finishing the problem by the methods of the previous example results in

c(r) = c(0) emω2r2

2kT (14.189)

indicating that the concentrations will be enriched at the end of the test-tube where r is largest.

14.7.2 Osmosis

Example 19. Consider a solution occupying the left sideof a two-part cell, with pure solvent in the right side anda solvent-permeable membrane seperating the two sides.Solvent molecules will flow through the membrane frompure-solvent to solution side, because the chemical poten-tial of the solvent is lower on the left side (“matter flowsfrom high chemical potential to low”). This is osmosis,and the flow can be stopped by pressurizing the solution(left) side with osmotic pressure π. If the solvent is pure,its mole-fraction is one. We know

µsolv = µ0solv + kT ln c (14.190)

Let the mole fraction of solute be X, then that of the solvent is 1 − X, and in the solution-side the solvent haschemical potential

µ = µ0solv + kT ln(1−X) (14.191)

Let the solute mole fraction change by dX, and the pressure change by dP , then

dµsolv =∂µsolv∂X

dX +∂µsolv∂P

dP (14.192)

but the first part is easy for an ideal solvent∂µsolv∂X

= − kT

1−X(14.193)

The chemical potential is the Gibbs free energy per particle, so

∂µ

∂P=∂ GN∂P

= v (14.194)

and for a mixture, this will be the molal volume per particle, so that if the particle exchange and pressure changeare performed in such a way that equilibrium is maintained,

µsolvent, left = µsolvent, right, dµsolvent = 0 (14.195)

we find that

0 = v dP − kTdX

1−X(14.196)

and we integrate ∫ Patm+π

Patm

v dP = kT

∫ Xf

0

dX

1−X(14.197)

14.7. MORE APPLICATIONS OF µ 307

and we arrive at the following for a dilute solution (X ≈ 0)

vπ = −kT ln(1−Xf ) ≈ kT Xf (14.198)

This can be written as

π ≈ kTXf

v= RT

Xf

Nv= RT [c] =

RT

Msolute[C] (14.199)

in which [c] is the particle per volume concentration and [C] is the gram per volume concentration of the solute.Osmotic pressure is easy to measure, and its measurement in one of the simplest ways to determine the molecularweight of dissolved macromolecules.

14.7.3 Boiling point elevation

The condition for chemical equilibrium applies equally well to phase equilibroium. The work needed to liberate agas atom from a liquid equals the work needed to condense a liquid atom from the gas at equilibrium

µg = µℓ (14.200)

If the liquid is a solvent in a solution, its mole fraction will play a role in this story. Suppose that a mole fraction Xof solvent os present in the solution, then for an ideal solution

µℓ = µ0ℓ + kT ln(1−X) (14.201)

where again µ0ℓ is the chemical potential of pure solvent. The gas/liquid equilibrium condition becomes

µg − µ0ℓ

T= R ln(1−X) (14.202)

Differentiate with respect to temperature Keeping pressure fixed

∂µg

T

∂T

∣∣∣P−∂µ0ℓ

T

∂T

∣∣∣P= R

∂ ln(1−X)

∂T

∣∣∣P

(14.203)

Now use the fact that the chemical potential is the Gibbs free energy per particle, and

G = G(T, P ) = U − TS + PV = H − TS (14.204)

dG = −S dT + V dP,∂G

∂T

∣∣∣P= −S, ∂G

∂P

∣∣∣T= V (14.205)

andG

T=H

T− S =

H

T+∂G

∂T

∣∣∣P

(14.206)

so∂GT∂T

∣∣∣P=

1

T

∂G

∂T

∣∣∣P+G

∂ 1T

∂T=

1

T

∂G

∂T

∣∣∣P− G

T 2(14.207)

and this means that∂GT∂T

∣∣∣P= −S

T− G

T 2= −H

T 2(14.208)

and we obtain

− Hv

T 2+H0ℓ

T 2= R

∂ ln(1−X)

∂T

∣∣∣P

(14.209)

but the difference in the heat of formation of the pure liquid and the heat of formation of the gas is the heat ofvaporization

− ∆Hvap

T 2= R

∂ ln(1−X)

∂T

∣∣∣P

(14.210)

This can be integrated if ∆Hvap is a known function of T . Lets pretend that it is constant

−∆Hvap

∫ T

T0

dT

T 2= −R

∫ X

X0

dX

1−X(14.211)


−∆Hvap

( 1

T− 1

T0

)= −R ln

(1−X)

(1−X0)(14.212)

Let X0 = 0, and let X be small (dilute solution), this leads to

∆Hvap

( 1

T0− 1

T

)≈ RX (14.213)

giving us the change in the temperature at which gas/liquid equilibrium occurs as a function of changes in theconcentration of a solute present in the liquid.

14.8 Non-equilibrium applications. Diffusion, Graham’s and Fick’slaws

Consider a tube of cross-sectional area A, filled with matter. Let JN (x) be the flux or rate per unit area with whichmatter crosses cross-sections of the tube, moving to the right.

x x+dx

Consider how the number of particles dN in themiddle cell could change, by a current of matter intothe cell from the left, our out into the cell to the right;

d

dt

(dN)= AJN (x)−AJN (x+ dx),

d

dt

( 1

A

dN

dx

)=JN (x)− JN (x+ dx)

dx= − d

dxJN (x) (14.214)

This is a matter conservation law, call n(x) = 1AdNdx , which is the concentration or number of particles per unit

volume. We have a matter conservation lawdn

dt= − d

dxJN (14.215)

Now lets figure out what potential is responsible for our matter flow. Examine the energy formula for candidates

dF = −S dT + µdN + E dc (14.216)

This exhibits the couplings (Onsager’s theory) between quantities that flow (T,N, c) and the potential responsiblefor its flow. It is clearly the chemical potential that is responsible for matter flow, and so in the simplest linearrelationship between current and potential gradient we would suppose that

JN = −ν ddxµ (14.217)

where ν is some constant (that can actually be determined from kinetic theory). Lets suppose that the matter inthe cells, liquids or gases, are ideal, so that we can use our formula for the chemical potential of a normal system

µ(x) = kT lnn(x) + f(T ),dµ

dx= kT

1

n

dn

dx(14.218)

then we obtain Fick’s law

JN = −νkTn

dn

dx(14.219)

anddn

dt= νkT

( 1n

d2n

dx2− 1

n2(dn

dx)2)

(14.220)

If we let n(x) = n0 + c(x) where c(x) is a small variation we get the diffusion equation upon which volumes ofchemistry and biology depends

dc(x)

dt≈ νkT

n0

d2c(x)

dx2+ · · · (14.221)

14.9. FRESHMEN CHEMISTRY CRAMMED INTO THE CHEMICAL POTENTIAL NUTSHELL 309

We see derived what chemists take for granted; matter flow is driven by concentration gradients. Graham’s law ofdiffusion is obtained by illustrating that the solution to the diffusion equation, subject to all of the particles beingat the origin at t = 0, is

c(x) =1√

4π νkTn0te− x2

4 νkTn0

t(14.222)

and so the distance d that molecules diffuse through in time t is

d =

√4νkT

n0t (14.223)

Within equal times t, changing the temperature from T0 to T results in a ratio of average diffusion distances of

d

d0=

√T

T0(14.224)

14.9 Freshmen chemistry crammed into the chemical potential nutshell

We like our physics majors to know something about chemistry (thermodynamically ideal systems), and so we pro-vide a very short review of the most basic concepts of freshman chemistry.

A typical chemical reaction such asA+ 1

2B2 AB (14.225)

can be written in terms of stoichiometric coefficients νi as∑i νiXi = 0, for example in this case νA = 1, νB2 = 1

2and νAB = −1. The reaction formula looks like a mathematical equation, and can be thought of one.

It is a constraint on the changes in thenumbers of each reactant or product as thereaction proceeds. Suppose that you putN0A moles of A, N0

B2moles of B2 and N0

AB

moles of AB into a reaction vessel, andallow the reaction to occur. As time andthe reaction progresses, we see the num-bers of molecules of each species change inaccordance with the chemical reaction formula.

Time step NA NB2 NAB0 N0

A N0B2

N0AB

1 N0A − ξ1 N0

B2− 1

2ξ1 N0AB + ξ1

2 N0A − ξ2 N0

B2− 1

2ξ2 N0AB + ξ2

3 N0A − ξ3 N0

B2− 1

2ξ3 N0AB + ξ3

Consider another example, 3A + 2B2 + C CA2B2 + AB2. Suppose that we begin with N0A, N

0B2

, N0C , N

0CA2B2

and N0AB2

moles of each species, and from there the reaction proceeds to create an additional ξ moles of AB2. Thechemical reaction formula tells us that for every mole of AB2 created, a mole of CA2B2 is created as a biproduct,three moles of A are consumed, one mole of C and two moles of B2 are consumed. If we use the number ξ ofmoles of AB2 created as a reaction progress indicator, we find that the number of molecules of each speciespresent in the reaction chamber have changed in proportion to the stoichiometric coefficients;

N0A −NA = 3ξ, N0

B2−NB2 = 2ξ, N0

C −NC = ξ, N0CA2B2

−NCA2B2 = −ξ, N0AB2

−NAB2 = −ξ (14.226)

Therefore when we are given a chemical reaction’s stoichiometric equation we can write down the changes in themole numbers very easily in terms of some reaction progress variable ξ of our choosing∑

i

νiXi = 0, N0i −Ni = νi ξ, −dNi = νi dξ (14.227)

At constant temperature (dT = 0) and pressure (dP = 0), the condition for chemical equilibrium is

dG = −S dT + V dP +∑i

dNi µi =∑i

dNi µi = 0 (14.228)


which becomes0 = −dξ

(∑i

νi µi

), 0 =

(∑i

νi µi

)(14.229)

and for ideal reactants and products, µi = µ0i + kT lnxi, which leads to the equilibrium formula

−∑i

νiµ0i =

∑i

kT lnxνii = kT ln∏i

(xνii

)(14.230)

which chemists write ase−

1kT

∑i µ

0i (T,P ) = xν11 xν22 · · ·xνNN (14.231)

This allows one to compute the mole-fractions of all reactants and products at chemical equilibrium.One can go a bit further and seperate the P and T dependence using

µi = −kT ln(( P0

xiP

)( TT0

)CV +k

k))

= ϕ0i (T ) + kT lnP + kT lnxi (14.232)

to get

P−∑

i νi e−1

kT

∑i ϕ

0i (T ) = xν11 xν22 · · ·xνNN (14.233)

Example 20. Study the gas reaction N2 + 3H2 2NH3. The equilibrium equation is

P−1−3+2 e−1

kT (ϕ0N2

+3ϕ0H2

−2ϕ0NH3

) =xN2 x

3H2

x2NH3

(14.234)

or in terms of partial pressures pi = xiP (remember that K is product/reactant);

e−1

kT (ϕ0N2

(T )+3ϕ0H2

(T )−2ϕ0NH3

(T )) = K−1(T ) =pN2 p

3H2

p2NH3

(14.235)

A chemistry student conducting the experiment in a sealed vessel of volume V = 5 × 10−3m3 at T = 723K findsthat at equilibrium, with an internal pressure of 3.35 × 107 Pa, that the partial pressures of the constituents arepH2 = pNH3 = 1.44× 107 Pa, pN2 = 0.48× 107 Pa. Determine the equilibrium constant K(T ).In terms of Pa this is simple

K−1(723) =(0.48× 107 Pa)(1.44× 107 Pa)3

(1.44× 107 Pa)2= 6.912× 1013 Pa2 (14.236)

Example 21. The same wacky chemist puts a mole of N2 and three moles of H2 in a bomb and maintains itstemperature at 723K. Find the mole fractions of all reactants at a pressure of 100 atm = 1.01× 107 Pa.Suppose ξ moles of ammonia form. That means that there are 1 − 1

2ξ moles of nitrogen, and 3 − 3 12ξ moles of

hydrogen in the mix. Therefore the mole fractions are

xN2=

1− 12ξ

(ξ) + (1− 12ξ) + (3− 3 1

2ξ)=

1− 12ξ

4− ξ, xNH3

=ξ

4− ξ, xH2

=3(1− 1

2ξ)

4− ξ(14.237)

6.912× 1013 Pa2 · (1.01× 107 Pa)−2 = 27(1− 1

2ξ)4

ξ2(4− ξ)2(14.238)

which should be solved numerically. You can see that since none of the molecules can exist in a negative molefraction, the larges ξ can be is ξmax = 2. One finds that ξ ≈ 0.744.

Example 22. Our heroic chemist notes that all of the chemistry books insist on computing equilibrium constantsin terms of molar concentrations, and so re-embarks on his quest to understand the reaction N2 + 3H2 2NH3.Returning to the original equilibrium mixture in a V = 5 × 10−3m3 vessel at T = 723K with an internal pressureof 3.35× 107 Pa, and partial pressures of the constituents are pH2 = pNH3 = 1.44× 107 Pa, pN2 = 0.48× 107 Pa, hecomputes actual numbers of moles using

nN2 =pN2V

RT=

(0.48× 107 Pa)(5× 10−3m3)

(8.314J/mole ·K)(723K)= 4.0moles, nH2 = nNH3 = 12moles (14.239)


and then computes the concentrations in moles per liter

[N2] =(4.0moles)

(5.0ℓ)= 0.8

moles

ℓ, [H2] =

(12.0moles)

(5.0ℓ)= 2.4

moles

ℓ(14.240)

[NH3] =(12.0moles)

(5.0ℓ)= 2.4

moles

ℓ(14.241)

and therefore

K−1 =[H2]

3[N2]

[NH3]2= 1.92

moles2

ℓ2(14.242)

Example 23. In molarity units the equilibrium constant for

CO +H2O CO2 +H2, is 0.25 =[CO][H2O]

[H2][CO2](14.243)

At equilibrium one finds [CO] = 0.6 moleℓ , [H2O] = 0.2 mole

ℓ and [CO2] = 0.5 moleℓ . Our intrepid experimenter deduces

that the molarity of H2 in the equilibrium mixture is [H2] = 4 (0.6)(0.2)(0.5) = 0.96 moles

ℓ .

Example 24. One wonders what the precise relationship is between the equilibrium constants expressed in molaritiesto those written in terms of partial pressures. Since

[Ni] =NiV

=piRT

=xiP

RT(14.244)

we obtain ∏i

[Ni]νi =

( P

RT

)∑i νi ∏

i

xνii = (RT )−∑

i νi∏i

pνii (14.245)

In our first example withN2+3H2 2NH3 we found the equilibrium constant in pressure units to be 6.912×1013 Pa2

at 723K. Therefore it should be (6.912×1013 Pa2)(8.314·723 J/mole)2 = 1.913× 108mole

2

m3 = 1.912 mole2

ℓ2 , which is exactly what we found

in a subsequent example.

Example 25. Our erstwhile scientist wishes to measure the volume of a bomb calorimeter. He injects 0.2 moles ofCH4 and 0.1 mole of H2S into the calorimeter, and maintains it at 1000K while the reaction

CH4 + 2H2S 4H2 + CS2 (14.246)

equilibrates at a total pressure of 9.2× 105Pa. The partial pressure of the hydrogen is found to be 2.0× 105 Pa.Assuming that ξ moles of hydrogen and 1

4ξ moles of CS2 have formed, leaving 0.2− 14ξ moles of methane and 0.1− 1

2ξmoles of H2S, Ilya computes the mole fraction of of hydrogen

xH2 =pH2

P=

2.0

9.2=

ξ

ξ + 14ξ + 0.1− 1

2ξ + 0.2− 14ξ, ξ = 0.0732 (14.247)

and therefore the vessel contains 0.3 + 12ξ = 0.3366moles of ideal gas, and its volume is V = (0.3366)(8.314)(1000)

9.2×105 =

0.003m3, which he proudly announces.During this experiment, a young physicist devises a simpler approach; she fills an identical calorimeter with water,pours it out into a graduated cylinder, and records the calorimeter volume to be V = 3.0 ℓ.

14.9.1 Ionic equilibria

Aqueous solutions are ideal in the sense that the chemical potentials of the various species in solution decomposeas µi = µ0

i + kT ln[i]. It is conventional for chemists to refer to the formality of a solute as the number of molesof solute used per liter of resulting solution to prepare the solution. A 1.0F solution of XY may not contain 1.0M(1.0 moles per liter) of anything, the actual molarities of the ions in solution could be quite different. The aqueous


ionization process XY X−+Y + is a chemical process whose equilibrium at constant T and P is again determinedby minimization of G, resulting in an equilibrium or solubility equation

K(T ) =[X−][Y +]

[XY ](14.248)

The constant K could be called the ionization constant. Suppose that a 0.01F solution of XY is 5% ionized atequilibrium. What is K(T )?

This means that the actual molarity of XY is [XY ] =(0.01 − (0.01)(0.05)

)= 9.5 × 10−3M , and [X−] = [Y +] =

5.0× 10−4, so K = 2.63× 10−5M .

Example 26. What formality of XY solution will be 0.10% ionized?Let the solution be ξ F . The amount of X− and Y + that will form will be [X−] = [Y +] = 0.001 ξ;

(0.001ξ)2

ξ − 0.001ξ= K = 2.63× 10−5M (14.249)

and we solve for the formality ξ = 26.27F

Example 27. The common ion effect is fancy-talk for adding more [X−] or [Y +] to push the reaction to the left.For example, suppose we toss in some XZ into the mix, which completely ionizes to X− + Z+.What is the molarity of X− in a 0.1F solution of XY and 0.2F XZ?Letting ξ = [Y +] at equilibrium, the two reactions

XY X− + Y +, K = 2.63× 10−5M, XZ → X− + Z+ (14.250)

will have [Z+] = 0.2M , [X−] = (0.2 + ξ)M , [XY ] = 0.1− ξ so that

ξ(0.2 + ξ)

0.1− ξ= 2.63× 10−5M, ξ ≈ 1.32× 10−5M (14.251)

Example 28. Complex equilibrium calculations involve multiple reactions, for example a formality f solution ofB2C undergoes

B2C B+ +BC−,[B+][BC−]

[B2C]= K1, BC− B+ + C−−,

[B+][C−−]

[BC−]= K2 (14.252)

It would be imprudent to contemplate a net reaction, since the equilibrium constants are radically different.Let x moles of B2C ionize, leaving [B2C] = f − x. Then in the second stage y moles of BC− ionize, so the solutioncontains

[B2C] = f − x, [B+] = x+ y, [BC−] = x− y, [C−−] = y (14.253)

then(x+ y)(x− y)

f − x= K1,

y(x+ y)

x− y= K2 (14.254)

Eliminate y between the two equations to get

x2 +K1x− fK1 =(√(x+K2)2 + 4xK2 − (x+K2)

2

)2(14.255)

This we solve graphically by plotting each side versus x, over the range 0 ≤ x ≤ f . What if K2 << K1, and K1 isvery small. Then y << x and we find that y ≈ K2 and x ≈

√f K1.

14.9.2 Acids and bases. pH

Suppose that these reactions involve hydrogen liberation

XH2 H+ +XH−,[H+][XH−]

[XH2]= K1, XH− H+ +X−−,

[H+][X−−]

[XH−]= K2 (14.256)


Calculate the concentrations of the various species of an f -formality solution of XH2 in hydrochloric acid, with apH= 1.The pH of an acid is defined to be pH = log10

1[H+] , so this solution has [H+] = 0.1M . The common ion effect comes

into play. Using the same equations as before, but pinning [H+] = 0.1M we obtain

(0.1)(x− y)

f − x= K1,

y(0.1)

x− y= K2 (14.257)

The results are linear equations

x(1 + 10K1)− y = 10K1 f, y(1 + 10K2)− 10K2 x = 0 (14.258)

x =10K1f (1 + 10K2)

(1 + 10K1)(1 + 10K2)− 10K2≈ 10 f K1, y ≈ 100K1K2 f (14.259)

if K1,K2 << 1.

Example 29. Strong acids have high levels of ionization, meaning large ionization constants

XH X− +H+, Ki =[X−][H+]

[XH](14.260)

Really strong acids like HCl completely ionize, a f -formal solution of HCl will have [H+] = f . Keep in mind thatwater also dissociates, H2O H+ +OH− with K = 1.0× 10−7M . This means that the concentration [H+] = x inpure water will be (a liter of water contains 55.6 moles of water)

x2

55.6− x= 1.799× 10−16 (14.261)

but clearly x is small, so this becomes

[H+][OH−] = 1× 10−14 = Kw, x = [H+] = 1.0× 10−7, at T = 298K (14.262)

and so the pH of pure water will be 7, which we take as neither basic or acidic. Suppose we make an f -formalitysolution of hydrochloric acid. The hydrogen ion concentration is then [H+] = f + 1× 10−7 ≈ f .In any aqueous solution we always have [OH−][H+] = Kw = 1 × 10−14M2. Notice that the constancy of theconcentration of water in dilute solutions means that the concentration of solvent does not appear inthe equilibrium equations.

14.9.3 Buffering

The salt of a weak acid added to a strong acid will drop the concentration of hydrogen ion down dramatically,effectively neutralizing its acidity by binding up the hydrogen ion. A nice example is the addition of the completelysoluble salt NaCN to HCl. Add 0.1M of NaCN to a 0.01F HCl solution. The equilibrium

CN− +H+ HCN,[CN−][H+]

[HCN ]= Ki = 4.0× 10−10M (14.263)

What happens? The acid dissociates completely, we have [H+] = 0.01M before the salt is added. Now add the salt,making [CN−] = [Na+] = 0.1M . Supose x molecules of HCN form

(0.1− x)(0.01− x)

x= 4.0× 10−10M, x ≈ 0.01M (14.264)

which has gobbled up all of the hydrogen ion. The solution is nearly neutral. Why do they call it buffering?Bubble in another 0.01M of hydrogen chloride. Again check for the amount of HCN formed

(0.1− x)(0.02− x)

x= 4.0× 10−10M, x ≈ 0.02M (14.265)

again, all of the hydrogen ion is consumed. This is classic buffering action; the addition of acid (up to a certainamount) has no effect on the pH.


14.9.4 Hydrolysis

is more fancy-talk for a reaction of the anion of the salt of a weak acid with water. Consider weak acid HX,

HX H+ +X− with [H+][X−][HX] = Ki, Ki << 1 and a salt such as NaX. The hydrolysis reaction is

X− +H2O HX +OH−,[HX]

[X−][H+]= K−1

i (14.266)

by the concentration of water for dilute (and it is dilute) solutions remains close to 55.6molesℓ , and so it can belumped with the other constants

K · 55.6molesℓ

= Kh =[HX][OH−]

[X−](14.267)

and we discover rather simply that because of the constraint on the product [OH−][H+] in aqueous solutions,

Kh =[HX][OH−]

[X−]=( [HX]

[X−][H+]

)([H+][OH−]

)=Kw

Ki(14.268)

Example 30. Determine [OH−] for 0.1F KCN . We know already that [CN−][H+][HCN ] = Ki = 4.0 × 10−10M , so let

[OH−] = x. The salt should dissolve completely to [K+] = 0.1M and [CN−] = 0.1, but hydrolysis alters this to[CN−] = 0.1− x by

CN− +H2O HCN +OH− (14.269)

and so ( [HCN ][OH−]

[CN−]

)=

x2

0.1− x= 2.5× 10−5M, [OH−] = 1.59× 10−3 (14.270)

Example 31. Calculate the pH of 0.1F KCN . We have

[H+] =1× 10−14M2

1.59× 10−3M, pH = −log10[H+] = 11.2 (14.271)

The salt of a weak acid forms a basic solution in water via hydrolysis.

Example 32. Hydrolysis can also occur with a salt derived from a weak acid and a weak base. Consider NH4OH NH+

4 +OH− with Ki,NH4OH and HQ H+ +Q− with Ki,HQ, both ionization constants small.There is hydrolysis of both cation and anion

NH−4 +H2O NH4OH +H+, Q− +H2O HQ+OH− (14.272)

with practically all of the liberated H+ and OH− recombining to form water in a net reaction

Q− +NH+4 +H2O NH4OH +HQ, Kh =

[NH4OH][HQ]

[NH+4 ][Q−]

=Kw

Ki,HQKi,NH4OH(14.273)

I’ve got blisters on me fingers. Thats all for now.

14.9.5 Chemistry glossary

For those of you preparing for the MCAT, this is a short list of terminologies that may prove to be helpful.

Lewis acid is an electron-pair receptor. In the process a covalent bond is formed. Examples AlCl3.

Lewis base forms covalent bond by donating electron pairs. This can be anything with lone-pair electrons (since itcan attach to any cation Z+)Bronsted acid donates H+.

NH3 +H2O → NH+4 +OH−, ammonia is base


Bronsted base accepts H+

H3O+ + CH3COO

− → CH3COOH +H2O, H3O− is an acid

Conjugate pairs HX → H+ +X−, HX is the conjugate acid and X− the conjugate base.

Arrhenius acid when dissolved in water yields a hydronium H3O+ plus complementary negative ion.

Arrhenius base when dissolved in water yields a hydroxide HO− plus complementary negative ion.

Groups are vertical columns, periods are horizontal rows of the periodic table.

Atomic radii increase down groups (new shells added) and decrease left to right as Z increases.

Electronegativity varies from 0 to 4. Flourine is 3.98, Lithium is about 1.0. Electronegativity generally decreasesdown the group, and increases left to right.Electronegativity measures ability to attract electrons to form bonds.

Ionization energy (energy neaded to remove an electron) increases left to right. In kJ/mole;

Na Mg Al Si P S Cl Ar496 738 577 786 1060 999.6 1256 1520

Spontaneous reactions diminish G (so ∆G < 0).

Exothermic reactions liberate heat (so ∆H < 0).

Synthesis or direct combination N2 + 3H2 → 2NH3

Decomposition or analysis 2H2O → 2H2 +O2

Substitution 2Na+ 2HCl → 2NaCl +H2

Combustion (oxydation) CH2S + 6F2 → CF4 + 2HF + SF6

Oxydation is loss of electrons by atom or molecule.

Reduction is the gain of electrons.

Lewis acid is an electron-pair receptor. In the process a covalent bond is formed. Examples AlCl3.

Lewis base forms covalent bond by donating electron pairs. This can be anything with lone-pair electrons (since itcan attach to any cation Z+)

14.10 Advanced topics2

14.10.1 Maxwell relations

Start with the first law for a reversible process

dU = T dS − P dV =

(∂U

∂S

)V

dS +

(∂U

∂V

)S

dV (14.274)

2Presented for your amusement only. We will not cover these topics in class, but students facing P-Chem might wish to read this.


We see that U is a function of S and V , U = U (S, V ) and so we use this thermodynamic function in problems inwhich these quantities are available. The Legendre transformation

F = U − TS (14.275)

is used to express a potential such as U in terms of other variables; take differentials

dF = dU − T dS − S dT = −S dT − P dV

=

(∂F

∂T

)V

dT +

(∂F

∂V

)T

dV (14.276)

and so F = F (T, V ) and is the appropriate function to use in cases where the values of these variables are accessible.

The symbol(∂F∂T

)Vmeans “differentiate F while holding V constant”. This is done by setting dV = 0 in Eq. 14.276

and dividing by dT (to get −S).The transformation

G = F + PV (14.277)

results in the Gibbs free energy

dG = dF + P dV + V dP = −S dT + V dP

=

(∂G

∂T

)P

dT +

(∂G

∂P

)T

dP (14.278)

so that G = G (T, P ). Finally transform to the enthalpy

H = U + PV (14.279)

taking differentials

dH = dU + V dP + P dV = T dS + V dP

=

(∂H

∂S

)P

dS +

(∂H

∂P

)S

dP (14.280)

which makes the enthalpy a function of S and P ; H = H (S, P ). From these expressions we obtain the following setof fundamental identities (

∂U

∂S

)V

= T,

(∂U

∂V

)S

= −P(∂H

∂S

)P

= T,

(∂H

∂P

)S

= V(∂F

∂T

)V

= −S,(∂F

∂V

)T

= −P(∂G

∂T

)P

= −S,(∂G

∂P

)T

= V (14.281)

One of the most important classes of thermodynamic identities are the Maxwell Relations. These are resultsderived from the following identity obeyed by the partial derivatives (independence of the order of differentiation);(

∂

∂y

(∂U

∂x

)y

)x

=

(∂

∂x

(∂U

∂y

)x

)y

(14.282)

Example 33. using the fact thatU = U (S, V ) (14.283)

we have

dU = TdS − PdV, so that T =

(∂U

∂S

)V

, P = −(∂U

∂V

)S

(14.284)


Differentiate the first with S fixed and the second with V fixed and we establish the Maxwell relation(∂T

∂V

)S

= −(∂P

∂S

)V

(14.285)

which is universally true for all materials.

Keep in mind the following differential identities from elementary calculus(∂x

∂y

)z

=1(∂y∂x

)z

, and

(∂x

∂y

)z

(∂y

∂w

)z

=

(∂x

∂w

)z

(14.286)

A far from obvious differential relation that is in fact the most useful of them all is that if you have a three-variableconstraint such as f(x, y, z) = 0 then (

∂x

∂y

)z

(∂y

∂z

)x

(∂z

∂x

)y

= −1 (14.287)

The proof is just elementary calculus, and is a good test of whether or not you know how to compute derivatives;simply choose one of the variables to be dependent (say z) and write

dz =

(∂z

∂x

)y

dx+

(∂z

∂y

)x

dy

and let z be held fixed

0 =

(∂z

∂x

)y

dxz +

(∂z

∂y

)x

dyz

and divide

0 =

(∂z

∂x

)y

(∂x

∂y

)z

+

(∂z

∂y

)x

The remaining Maxwell’s relations can be computed by applying these identities to derivative relations obtainedfrom the differentials of F ,G, and H.

14.10.2 The Gibbs-Duhem relation

The Gibbs-Duhem relation gives us a very important constraint on the thermodynamical variables that is dueto the nature of particle-number dependences; we develop this first in a one-species environment in the entropyrepresentation, in which we think of the entropy as being a function of energy and volume

S = S(U, V ) (14.288)

What happens if we multiply the number of particles in the system by λ? Then everything in this formula mustincrease by this factor, since S,U, V all are proportional to N ;

S(λU, λV, λN) = λS(U, V,N) (14.289)

Differentiate both sides with respect to λ;

∂

∂λS(λU, λV, λN) =

∂

∂λλS(U, V,N) = S(U, V,N)

=∂

∂(λU)S(λU, λV, λN)

∂(λU)

∂λ+

∂

∂(λV )S(λU, λV λN)

∂(λV )

∂λ+

∂

∂(λN)S(λU, λV λN)

∂(λN)

∂λ

=∂

∂(λU)S(λU, λV, λN)U +

∂

∂(λV )S(λU, λV, λN)V +

∂

∂(λN)S(λU, λV, λN)N (14.290)

This statement is true for all λ, including λ = 1, and therefore

S(U, V,N) =( ∂S∂U

)V,N

U +( ∂S∂V

)U,N

V +( ∂S∂N

)U,V

N =( 1

T

)U +

(PT

)V −

(µT

)N (14.291)


This is true for any one-species thermodynamic system, including the ideal gas. It encapsulates the particle-numberdependence of these thermodynamic quantities.How do we use it? It is most important in chemical processes. We exploit this formula by using it in tandem withthe First Law;

dU = dQ− dW = T dS − P dV + µdN, or dS =( 1

T

)dU +

(PT

)dV −

(µT

)dN (14.292)

which relates particular differentials of the same quantities. This allowed us to figure out what the derivativeswere in the Gibbs-Duhem relation, since

dS(U, V,N) =( ∂S∂U

)V,N

dU +( ∂S∂V

)U,N

dV +( ∂S∂N

)U,V

dN =( 1

T

)dU +

(PT

)dV −

(µT

)dN (14.293)

Take differentials of the Gibbs-Duhem equation and compare it to the first law;

dS = d( 1

T

)U +

( 1

T

)dU + d

(PT

)V +

(PT

)dV − d

(µT

)N −

(µT

)dN (14.294)

Direct comparison shows that the chemical potential (the third equation of state) can be obtained from these tworelations

0 = d( 1

T

)U + d

(PT

)V − d

(µT

)N (14.295)

if we know the other two equations of state (which are substituted into this formula). This opens the doors to ahost of chemical applications; computations of equilibrium constants and concentrations, chemical entropy/Gibbsfree energy changes, and electro-chemistry.

14.10.3 The Van der Waals gas

The equation of state of a gas contains all of the thermodynamic data for the gas, it is merely a question of extractingit. The most important equations of state are the ideal gas and the Van der Waals gas(

P +n2a

V 2

)(V − nb) = nRT (14.296)

The ideal gas molecules do not interact with each other, and have no size, they are point particles. The VDWmolecules have a finite size, and do weakly interact with each other. Typically for real gases described by the Vander Waals equation of state, a and b are very small numbers, and in real life problems very good accuracy can begotten by expanding actual thermodynamic functions in power series in a and b and dropping quadratic or higherterms. Some actual numbers are

Gas a ( liter2 atm

mole2 ) b ( litermole )H2 0.2444 0.02661He 0.03412 0.02370O2 1.360 0.03183

The number a is a measure of the weak interatomic forces, and b represents the excluded volume, or volumeoccupied by a gas molecule, within which no other molecule can infringe.

Example 34. Compute the change in U for an isothermal expansion.We do this for both gases in one shot, since the ideal gas is a sub-case of the Van Der Waals with a and b parametersset to zero.First

P =nRT

V − nb− n2a

V 2(14.297)

and (∂P

∂T

)V

=nR

V − nb(14.298)


but F = U − TS keep T fixed and differentiate this(∂F

∂V

)T

=

(∂U

∂V

)T

− T

(∂S

∂V

)T

(14.299)

use Maxwell identities to rewrite this as (∂U

∂V

)T

= T

(∂P

∂T

)V

− P (14.300)

and insert the derivatives computed directly from the Van Der Waals equation(∂U

∂V

)T

=n2a

V 2(14.301)

integrate both sides

∆U =

∫ V2

V1

n2a

V 2dV = n2a

V2 − V1V2V1

(14.302)

Note that this will be zero for the ideal gas, we have discovered that the energy of an ideal gas depends only on thetemperature, not the volume!The second equation of state for the Van der Waals gas is

U =3

2nRT − n2a

V(14.303)

Example 35. The work done in the isothermal expansion of the VdW gas is

W =

∫ V2

V1

(RT

V − b− a

V 2

)dV = RT ln

V2 − b

V1 − b+

a

V2− a

V1(14.304)

and the change in internal energy is

∆U =a

V1− a

V2(14.305)

This means the change in heat is

∆Q = RT lnV2 − b

V1 − b(14.306)

Then since

∆H = ∆U +∆(PV ) (14.307)

we find

(PV )2 − (PV )1 = bRT

(1

V2 − b− 1

V1 − b

)− a

V2+

a

V1(14.308)

and so putting all of it together

∆H = bRT

(1

V2 − b− 1

V1 − b

)− 2a

V2+

2a

V1(14.309)

14.10.4 An application of enthalpy; refrigeration

One of the most effective means of refrigeration uses the Joule-Thompson effect , an isenthalpic process by whicha gas is forced under high pressure P1 through a permeable plug into a region of lower pressure P2. The gas is cooledin the process if its initial temperature T1 is below the so called inversion temperature Tinv, and is heated in theprocess if T1 > Tinv.

Suppose a single mole of gas is driven under these circumstances through the plug, from an initial volume (on theleft) of V1 to final volume (on right) of V2.


Cylinder 1; P1 Cylinder 2; P2

Plug

Valve 1 Valve 2

Holding

Tank

The process can be cycled; in the figuregas from a holding tank is pulled intocylinder 1 by opening both valves, andpulling piston 1 to the left while pushingpiston 2 to the left. Close both valvesand push piston 1 to the right withhigh pressure P1 while pulling piston 2to the right with pressure P2. The gasis drawn from cylinder one to cylindertwo through the porous plug (whichneeds to dissipate heat with an attachedheat sink and fan). The gas is cooledin the process, which can be repeatedby opening the valves and sending bothpistons left again. The gas is drawn intoa holding tank and back into cylinderone. Objects placed in good thermalcontact with the holding tank can becooled by the cold gas.

The change in heat of the gas (which is kept insulated) is

∆Q = ∆U +∆W = 0 = U1 − U2 =

∫ V1

0

P1 dV −∫ V2

0

P2 dV =(U1 + P1 V1

)−(U2 + P2V2

)(14.310)

and a rearrangement shows that the process is indeed isenthalpic

Uf + P2 V2 = Hf = Ui + P1 V1 = H1 (14.311)

however during the actual transfer, the gas is not in an equilibrium state, and so H cannot even be defined for thegas during the process of passage through the plug, however for the initial and final states it can be.We can compute the actual temperature change with each stroke of the pistons from the derivative

dT =(∂T∂P

)HdP = −

(∂H∂P

)T(

∂H∂T

)P

dP (14.312)

which is just Eq. 14.287 (∂x

∂y

)z

(∂y

∂z

)x

(∂z

∂x

)y

= −1 (14.313)

with x = T, z = H, y = P , and the definition

dH = T dS + V dP,(∂H∂P

)T= T

( ∂S∂P

)T+ V (14.314)

and (∂H∂T

)P= T

(∂S∂T

)P=(∂Q∂T

)P= Cp (14.315)

We now invoke the Maxwell relations Eq. 14.285 and the definition of the thermal expansion coefficient Eq. 14.82( ∂S∂P

)T= −

(∂V∂T

)P= −V 3α (14.316)

and substitution gives us the interesting result

dT =V

Cp(3Tα− 1) dP (14.317)

for the Joule-Thompson effect. The right-hand side is zero for an ideal gas, so if all gases actually obeyed the idealgas law at all T -ranges, such refrigeration would be impossible. All gases obey the ideal gas law remarkably well


at high-T , and all obey the VDW gas law very well at low-T . The two laws are barely distinguishable at hightemperatures, they argee with each other in that domain.

I hope that this example points out that the mathematics of thermodynamics is not just formal, there is a point toit, it is necessary to invent and quantify technological innovations such as simple refrigeration.


14.11 Problems

1. Consider one mole of ideal monatomic gas undergoingthe cyclic process illustrated. Point 1 is at (Va, Pa) andthe cycle proceeds clockwise.In this figure Pa = 83100.0 N

m2 , Va = 0.1m3. Pa =2Pd, Vb = 0.8m3. One full atmosphere of pressure is101, 000 N

m2 . In this cycle a → b and c → d are adia-batic, b→ c and d→ a are constant volume.


a→ b

b→ c

c→ d

d→ a

2. Find the entropy change when 100 g of water (C = 1.0 calg C ) at 50C is mixed with 80 g of ethyl alcohol (C = 0.62 cal

g C )at 20C.

3. A Carnot heat engine has an efficiency of ϵ = 0.7 and produces work at a rate of 1000watts. A watt is one Jouleper second. If the engine is run in reverse as a refrigerator, at what rate can it remove heat energy from its lowtemperature reserviour? Express your answer in watts.

4. The molar heats of reaction for two chemical processes are

C +O2 CO2, ∆H = −94.05 kCal

CO +1

2O2 CO2, ∆H = −26.42 kCal

Use the fact that enthalpy is a state function to determine the heat liberated (these combustions are all exothermic)in the reaction

C +1

2O2 CO

5. A 25W light-bulb is left on inside of the ultimate dorm refrigerator; no heat can get in or out through its perfectlyinsulated walls. At what rate must its motor, a Carnot heat engine, do work to pull the heat created by the bulbout of the 0o C compartment and dump it into the 27o C room?

14.11. PROBLEMS 323

P

V

1

2

3

6. One mole of a monatomic ideal gas is made to undergothe reversible cycle in the figure, with P1 = 1.0 × 105 N

m2 ,

V1 = 0.1m3, and P3 = 2.667 × 104Nm . Process 1 → 2 isadiabatic, 2 → 3 is isothermal. Find T1, T2, T3, V2 andcompute ∆U, ∆Q, ∆W, ∆S for each process line on thecycle.


1 → 2

2 → 3

3 → 1

T1 T2 T3 V2


S

T

1 2

34

7∗. One mole of a monatomic ideal gas is made to undergothe reversible Carnot cycle in the figure, which is illus-trated in the S/T or entropy /temperature plane ratherthan the more conventional P/V plane. T1 = 200K,ϵ = 0.6, S1 = 5.0 JK and S4 = 3.0 JK . Find ∆U, ∆Q,∆W, ∆S for each process line on the cycle. Note that

U =3

2nRT, PV = nRT

dS =dU

T+P

TdV


1 → 2

2 → 3

3 → 4

4 → 1

8∗. Find the entropy change when 100 g of water at 273K is solidified into ice at 273K.

9. Find the entropy change when 100 g of water at 300K is chilled to 273K.

10. A refrigerator that runs in a room at T = 300K and has an icebox maintained at 273K draws 1000W of power.How long will it take the refrigerator to freeze a kilogram of chilled water into ice?

11. One mole of ideal gas is made to undergo an adiabatic process that takes it from (Vi, Ti) = (0.2m3, 600K) to(Vf , Tf ) = (Vf , 300K). Compute Vf .

12. One mole of ideal gas is made to undergo an irreversible process that takes it from (Vi, Pi) = (0.2m3, 2×104 Nm2 )

to (Vf , Pf ) = (0.3m3, 1.333× 104 Nm2 ). Compute the entropy change of the gas.

14.11. PROBLEMS 325

P

V

1 2

3

13. One mole of a monatomic ideal gas ismade to undergo the reversible cycle in thefigure, with P1 = 81340 N

m2 , V1 = 0.2m3, andV2 = 0.4m3. Process 2 → 3 is adiabatic,3 → 1 is isothermal. Find T1, T2, T3, V3 andcompute ∆U, ∆Q, ∆W, ∆S for each processline on the cycle.

P

V

1 2

34

14. A material with equations of state

U = aV T 4, a = 7.56 × 10−16 J

m3K4, U = 3PV

is made to undergo the reversible process illustrated, withP1 = 100 N

m2 , V1 = 0.1m3, V2 = 0.3m3 and P4 = 25 Nm2 .

Compute ∆U, ∆Q, ∆W for each process line on the cycle.

Line ∆U ∆Q ∆W

1 → 2

2 → 3

3 → 4

4 → 1

15. A material with equations of state U = aV T 2, a = 2.0 ×10−6 Jm3K2 , PV = U undergoes a constant pressure

volume tripling from (V, P ) = (1m3, 1× 104 Nm2 ) to (3m3, 1× 104 N

m2 ). Find its entropy change.

16. A material with equations of state U = aV T 2, a = 2.0 × 10−6 Jm3K2 , PV = U undergoes isothermal expan-

sion from V1, P1 to 4V1, P2. FindP2

P1.

17. A material with equations of state U = a V T 4, a = 2.0 × 10−6 Jm3K4 , 3PV = U undergoes adiabatic expan-

sion from V1, T1 to 3V1, T2. FindT2

T1.


P

V

1 2

34

18∗. One mole of ideal gas is made to undergo the cyclicprocess illustrated. P1 = 1.0 × 104Pa, P3 = 5.0 × 103Pa,V1 = 1m3, V2 = 3m3.Compute ∆U, ∆Q, ∆W for each process line on the cycle.

Line ∆U ∆Q ∆W

1 → 2

2 → 3

3 → 4

4 → 1

19∗. In the thermodynamic heat enginecycle illustrated, let a→ b be isothermal,compute all thermodynamic quantitiesfor each part of the cycle.

20. Consider an electrolytic cell with the two electrode reactions

Zn(s) → Zn++(aq) + 2e−, E0 = 0.76V

Cu(s) → Cu++(aq) + 2e−, E0 = −0.34V

Compute the ratiocZn++

cCu++

=[Zn++]

[Cu++]

at 25o C. At equilibrium do you expect to see any copper in solution? This illustrates a simple way to copper-platea metal surface.

14.11. PROBLEMS 327

21. Show that for a chemical reaction of ideal gases with stoichiometric coefficients νi∑i

νiXi = 0

that in terms of the partial pressures

pi = xiP =Ni∑j Nj

P

we have the equationpν11 pν22 · · · pνNN = K(T )

for the equilibrium constant.

22. The “gas of photons” or light quanta trapped in a cavity of volume V in thermal equilibrium with a heat sourceat temperature T has total internal energy and pressure

U = b V T 4, P =U

3V, b a constant

Show that during an adiabatic expansionT 31 V1 = T 3

2 V2

You should find that

S =4

3

(bU3V

) 14

=4b

3V T 3

Show that for reversible processes (note dQ = T dS)

U =3

4ST

dU =3

4

(S dT + T dS

)dW =

1

4

(T dS − 3S dT

)so that the description of this system’s thermodynamics is simplest in the ST -plane.

P

V

1

2

3

4

23 One mole of a monatomic ideal gasis made to undergo the reversible cyclein the figure, with P1 = 1.0 × 105 N

m2 ,

V1 = 0.1m3, V2 = 0.3m3 and P4 = P1

2 .Process 1 → 2 and 3 → 4 are adiabatic.Find T1, T2, T3, T4, P2, P3 and compute∆U, ∆Q, ∆W, ∆S for each process lineon the cycle.


P

V

1 2

34

24 One mole of a monatomicideal gas is made to undergo thereversible cycle in the figure, withP1 = 1.0 × 105 N

m2 , V1 = 0.1m3,

V2 = 0.2m3 and P3 = P1

4 . Process2 → 3 and 4 → 1 are adiabatic.Find T1, T2, T3, T4, V3, V4 andcompute ∆U, ∆Q, ∆W, ∆S foreach process line on the cycle.

P

V

1

23

25 One mole of a monatomicideal gas is made to undergo thereversible cycle in the figure, withP1 = 1.0 × 105 N

m2 , V1 = 0.1m3,

and P3 = 0.25 × 105 Nm2 . Process

3 → 1 is adiabatic, 1 → 2 isisothermal. Find T1, T2, T3, V3and compute ∆U, ∆Q, ∆W, ∆Sfor each process line on the cycle.

26. Compute the change in the Gibbs free energy for the zinc/copper voltaic cell in the first example of this section,if two moles of electrons are transferred. Is the reaction spontaneous? One mole of electrons has charge magnitude

1.6× 10−19C · 6.023× 1023 = 96, 368C (14.318)

27. Look up the oxidation/reduction potentials for the lead/acid battery described in this section and determinewhat voltage the battery should produce.

28. Find the cell voltage for the cellCd;Cd++ ||Cu++;Cu

14.11. PROBLEMS 329

(left side is cadmium half-cell, right side is copper) and the change in Gibbs free energy when two moles of electronsare transferred.

z z+dz

z=0 z=L

29. Electrophoresis can be usedto enrich one end of a gel-columncontaining macromolecules suchas DNA, or colloidally suspendedparticles, because the moleculesor particles in solution will havesurface charges.

Suppose that you impose a constant electric field E = Ek with potential (voltage function) E(z) = −E z on thecolumn, in which particles of charge q > 0 (assumed to be chemically ideal µ(z) = µ0 + kT ln c(z)) are dispersed.Show that the equilibrium concentration of suspended molecules varies exponentially with z along the tube. Hint.Let an infinitismal number of molecules migrate out of the little volume at z into the volume at z + dz.

30. By considering a column of gas above the earth’s surface, use µ to compute the pressure and density of theatmosphere as a function on altitude z, with P (z = 0) = Patm, the sea-level value.

31∗. Show that mixing is a spontaneous process. Suppose that you have a vessel partitioned into two volumes V1,V2 initially occupied by different species of gases, with N1 particles of type 1 in V1, and N2 particles of type 2segregated in V2. Remove the divider. Compute ∆S.

32∗. Suppose that you have a vessel partitioned into two volumes V1, V2 initially occupied by a single species ofgases, withN1 particles of type 1 in V1, andN2 particles of type 1 segregated in V2. Remove the divider. Compute ∆S.

33∗. Suppose that you use a large mirror or lens to focus a large amount of solar radiation (the sun’s surfaceT = 5800K) onto a block of tungsten. Can you heat the block up to 6000K this way?

34. A material with equations of state U = aV T 4, a = 7.56 × 10−16 Jm3K4 , 3PV = U undergoes isothermal

expansion from (V1, P1) = (0.1m3, 100 Nm2 ) to (3V1, P1). Find ∆S.

35. Suppose that you own a poorly insulated small house, with outside walls ℓ = 0.25m thick, total outer wallplus roof surface area 210m2 (about 1992 ft2) and your walls have a thermal conductivity of 1.3 W

m·K (limestone).Calculate how much power your (Carnot) air conditioner will require to keep your house at 293K when the outsidetemperature is 308K.

36. Derive the Maxwell relations(∂T∂P

)S=(∂V∂S

)P,

( ∂S∂V

)T=(∂P∂T

)V,

( ∂S∂P

)T= −

(∂V∂T

)P

37. Use the two Van der Waals equations of state to show that

∆S = R ln((V − b)(U +

a

V)c)

for one mole of VDW gas, and find c.

38. Show that the thermal expansion coefficient Eq. 14.82 of an ideal gas is

3α =1

V

(∂V∂T

)P=

1

T

39. Show that the isothermal compressibility, defined to be

KT = − 1

V

(∂V∂P

)T

of an ideal gas is KT = 1P .


40. Show that the constant volume specific heat of an ideal gas is

CV =(∂Q∂T

)V=

3

2R

per mole, and the constant pressure specific heat of an ideal gas is

Cp =(∂Q∂T

)P=(∂H∂T

)P= T

(∂S∂T

)P=

5

2R

41. Show that the thermal expansion coefficient Eq. 14.82 of a Van der Waals gas is

3α =1

V

(∂V∂T

)P≈ 1

T

(1−

(b− 2aRT )

V

)(remember a and b are small).

42. Show that the constant volume specific heat of a Van der Waals gas is

CV =(∂Q∂T

)V=

3

2R

per mole, and the constant pressure specific heat of a Van der Waals gas is

Cp =(∂Q∂T

)P=(∂H∂T

)P≈ 5

2R+

2a

TV

43. For a material with equations of state

U = aV T 4, PV =1

3U

show that in an adiabatic processV1T

31 = V2T

32

44. For a gas in a reversible Carnot cycle undergoing isothermal expansion ∆S = nR ln(Vf/Vi) > 0. But I thoughtthat for a reversible process ∆S = 0! Is there a problem here? Resolve this apparent paradox.

45. Show that for Van der Waals gas refrigerator, each compressor stroke will lower the reservoir temperature by

dT = −(b− 2a

RT )52R+ 2a

V T

dP

as long as T < 2aRb (the inversion temperature Tinv). Compute this number for O2, and for freon, a common refrig-

erant with a = 10.78 ℓ2·atmmole2 , b = 0.0998 ℓ

mole . Perhaps this explains what constitutes a good refrigerant.

46. Another adiabatic relation for a Van der Waals gas is(Pf +

n2a

V 2f

)(Vf − nb

) 53

=(Pi +

n2a

V 2i

)(Vi − nb

) 53

Derive this (hint; start with the VDW equation of state and Eq. 14.132).

47. You must supply the binding energy ϵ0 to transform a diatomic molecule AB into seperated A and B molecules,this is a modification of chemical potential

µdiat = ϵ0 + µ0 + kT lnx

Find the changes that this creates in the formula Eq. 14.165 for the equilibrium constant for the gas reactionA+B AB.

14.11. PROBLEMS 331

48. Lets call Eq. 14.165xA xBxAB

=(P0

P

)K(T )

Mix n moles of A with n moles of B in a reaction vessel. Suppose that n′ moles of AB form at P, T . ComputexA, xB, xAB in terms of n, n′.

Let(P0

P

)K(T ) = 10. Find n′/n.

49. Lets call Eq. 14.165xA xBxAB

=(P0

P

)K(T )

Mix n moles of A with n moles of B in a reaction vessel. Suppose that n′ moles of AB form at P, T . ComputexA, xB, xAB .Let K(T ) = 10. Find the pressure ratio P/P0 for which n′/n = 0.5.

50. A biochemist makes a graph of osmotic pressure versus concentration of some macromolecule at T = 300K.The slope of the graph is 1.64× 10−4 ℓ·atm

g . Find the molecular weight of the molecule.

51. The salinity of the oceans is about 35g of NaCl per liter of water. Compute its molar concentration.

52. Show that for small boiling-point changes, the boiling-point elevation formula Eq. 14.213 can be written as

∆T = T − T0 ≈ RT 20X

∆Hvap

Find the boiling point of seawater.

53. In practice biochemists must measure osmotic pressure of solutions with rather substantial concentrations. Showthat if you keep more than one term in

vπ = −kT ln(1−Xf )

that it is the intercept of a graph of π versus Xf that gives you the formula for computing molecular weightEq. 14.199.

54. For the material with equations of state

U = PV = aV T 2, a constant

show that integration of the first law

dS =dU

T+P

TdV

leads to the fundamental relationS = 2aV T

55. For the material with equations of state

U = PV = aV T 2, a constant

show that for a reversible process

U =ST

2, dU =

1

2

(S dT + T dS

)dW =

1

2

(T dS − S dT

)and since dQ = T dS, this material has a very simple thermodynamics in the ST -plane.

56. A room has volume V0 and atmospheric pressure T0, and contains an ideal gas at T0. You are a bit chilly soyou turn on the heat. Compute the internal energy of the gas in the room before and after heating in terms of thepressure and volume of the gas. So where did the energy produced by the furnace go? Is the gas in the room at ahigher T or not after you turn on the furnace?

Chapter 15

Answers to the problems

Chapter 1

5. v = v0 + a (t2+t1)2

6. v = v0 +12a

(t22−t21)

(t2−t1) = v0 + a (t2+t1)2 .

12. dxdt =

√d2+h2

d u13. Yes.16. c = a× b = (−6,−2, 10)17. c/|c| = (−6/

√40,−2/

√40, 0)

18. cos θ = 0, θ = 90o

19. |a|2 = 1 = a · a, d1dt = 0 = d

dt

(a · a

)26.

f1(x)=(x<=1) ? 1+2*x : 0

f2(x)=(x>1) ? 3+4*(x-1) : f1(x)

f3(x)=(x>2) ? 7+3*(x-2) : f2(x)

plot[0:3] f3(x)

Chapter 2

1. R =2v20g

sin 20o

cos2 20o , v = v0√1 + 4 tan2 20o

2. v0 = 29.9ms . ϕ = 0.313 rad, v0 = 46.7ms , descending.

3. v0 = 20√

9.820

ms

4. v = 35.9ms , tanϕ = 0.66.

5. R =2v20g cos2 θ

(tan θ−tanα

cosα

)7. s =

√v20 + (g xv0 )

2

12. ymax = 95.7m, R = 191.32m20. v0 = 50.34ms26. It looks like θ = 0.7 rad, about 40.1o gives the longest range.

Chapter 3

1. (0.3ms2 , 0.4ms2 )

2. (27N, 26N)3. (6 m

s , 9ms ) at t = 3 s

4. N = mg − F sin θ = 93N, a = 0.866ms2

333

334 CHAPTER 15. ANSWERS TO THE PROBLEMS

5. Ff = −3.0N i, Ff = −4.0N i

7. a = M2gM1+M2

, T = M1M2gM1+M2

8. Run the following through REDUCE; EQS:=M1*g-T-M1*a1,M2*g-2*T-M2*a2,M3*g-T-M3*a3,a1+2*a2+a3;vars:=a1,a2,a3,T; solve(EQS,vars);9. Run the following through REDUCE; EQS:=M1*g-3*T-M1*a1,M2*g-T-M2*a2,3*a1+2*a2;vars:=a1,a2,T;solve(EQS,vars);10. Run the following through REDUCE; EQS:=M1*g-5*T-M1*a1,M2*g-T-M2*a2,5*a1+2*a2;vars:=a1,a2,T;solve(EQS,vars);12. T ′ = mg = T ′′ cos 60, T ′ = (20)(9.8) = 196N, T ′′ = 392N , Ff = T = T ′′ sin 60 = 339.5N13. Run the following through REDUCE; EQS:=120*a1-T, 20*a2-20*9.8+2*T,a1+2*a2;vars:=a1,a2,T;solve(EQS,vars);14. M2 =M1 sin 60 = M1

2 = 5 kg

Chapter 4

1. a = g tan θ, F = (M1 +M2)a = (M1 +M2)g tan θ2.a Ff = mg sin 18 = 60.56N , b. θc = 21.8o, c. a = 4.04ms2 .3.b (ax, ay) = (4.0ms2 ,−8.2ms2 ), c. (ax, 0) = (4.0ms2 , 0)5. a = µsg = 4.9ms , F = (M1 +M2)a = (M1 +M2)µsg = 73.5N

6. M = mµs

tan 60o = 20kg0.6 tan 60 = 57.7kg. c. Ff = T2 = T3 sin θ = mg tan 60 = 339.5N

11. tan θ = 1µs

12. Fmin = Mg√1+µ2

s

13. If the block slips up a =(M2−M1 sin θ−µkM1 cos θ

M1+M2

)g

19. If you don’t push hard enough, the box could slip down the incline, then friction will point up the incline andhelp you hold the box in place; Fmin =M2g

sin θ−mus cos θcos θ+µs sin θ

Chapter 5

1. v =√tan 20 sin 20Rg

2. θ = cos−1 gRω2

5. ωmin =√

gµsR

7. vmax =√Rg

8. vmax =√Rg

9. v0 =√

kg

10. v0 = λ2π

√gh

Chapter 6

1.a. There is a potential since ∂Fx

∂y =∂Fy

∂x

1.b. There is no potential since ∂Fx

∂y = ∂Fy

∂x

2. There is no potential W =∫ 1

0(16 + 16t2)dt = (16t+ 16

3 t3∣∣∣10= 64

3 J

3. There is a potential V = 4(x2 − y2)7. a. −µkMg d, b. 0, c. −mgd tan 60, d. Fd

cos 60 , e.12mv

2 = −µkMg d−mgd tan 60 + Fdcos 60

8. a. W1 = Td, b. W2 = −Td, c. WT = W1 +W2 = 0, d. WG = (M2 −M1)gd, e.12 (M1 +M2)v

2 − 0 =∑W =

(M2 −M1)gd

12. a. It is conservative, V = −x2y + 13y

3 + C ′, W = −(Vf − Vi) = −(C ′ − ( 132

3 + C ′))= 8

3J .

b. Not conservative.13. W = −

(Vf − Vi

)= 0 14.a. W =

∫ 1

0F · v dt = −9.8

∫ 1

0v·v|v| dt = −9.8

∫ 1

0|v| dt = −9.8

∫ 1

0(2π) dt = −19.6πJ

17. hmin = 2.5R

335

18. v =√2gh(1− µk)

19. x = h(1−

√µ2k

1+µ2k

)20. Increasing your speed five percent increases your fuel consumption rate by (over) ten percent.

21. a. W =∫ 1

0

(2 ·4−3 ·16t

)dt, b. W =

∫ 1

0

(2 ·4t ·4−3 ·16t ·8(1− t2)

)dt, c. W =

∫ 1

0

(2 ·8(1− t2) ·4−3 ·16t ·4t

)dt,

d. W = −0.1∫ 1

0v|v| · v dt = −0.1

∫ 1

0

√42 + (16t)2 dt

22. WF =Mgh(1− cos θ)23. ve = 1.12× 104ms24. µk = 0.555525. F = (M1 +M2)

gµs

26. ϵ = 0.949.27. h′ =

(g−ϖm )

(g+ϖm ) h

33. a = −nb2x−2n−1 = F/m34. tan θ = 1, θ = 45o

Chapter 7

1. Inelastic.2. x1 = kx2

2µkgM1(1+M1M2

), x2 = kx2

2µkgM2(1+M2M1

)

3. x =

√µ2kM

2g2

k2 +Mv20k − µkMg

k

4. v0 =√(µ2s + 2µsµk)

Mg2

k

5. v1 =√

kx2

M1(1+M1M2

)

10. Run this through REDUCE; EQ1 := 2∗v1∗sqrt(3)/2+w−v0;EQ2 := w2+2∗v12−v02; solve(EQ1, EQ2, v1, w);11. x = 0.227m

12. v1 = cos θ√

2M2ghM1 cos2 θ+M2

15.d. yCOM = 110

(4R+ 3(R+

√3R) + 2(R+ 2

√3R) + 1(R+ 3

√3R)

)= (1 +

√3)R

18. ∆E = mv2(

14 cos2 ϕ − 1

2

)7.21. δp =

∫ 1

0F dt = 1

2 (2N)(1 s) = 1.0kgms , δp =∫ 2

0F dt = 2 1

2 (2N)(1 s) = 2.0kgms7.22.It is because the force is not applied to the water leaving (or the water that has left) the tank, only the waterin the tank.

Chapter 8

5. a = 5Fri8mr0

6.. cos θ = riro

= 23

13. T = ma+ µsmg = 83µsmg = 23.52N

14. It will slip; a = T−µkmgm = 8.04ms2

15. It will roll a = Tm+I/R2 = 2.5ms2

16. I = 12mR

2 + 6(

12mR

2 +m(2R)2)

17. I = 4(

112mℓ

2 +m( ℓ2 )2)

19. v =√

8hg7 , I = 3

4mR2

21. v =√

2kx2

3m =√2x a, a = kx

32m

23. I = 13 (3m)(3ℓ)2 +Mℓ2 +M(2ℓ)2 +M(3ℓ)2


26. F = mg2 = Fh

41. Fh1 = µMg1+µ2 , Fv1 = Mg

1+µ2 , Fv2 = µ2Mg1+µ2 , Iα = µ2+µ

µ2+1MgR.

Chapter 9

1. ω =√12 = 3.46 rads , d = 0.289m

3. ω2 = 2I1I1+I2

ω0, ω1 = I1−I2I1+I2

ω0

5. mv0z = ( 13mℓ2 +mx2)ω, 1

2 (13mℓ

2 +mx2)ω2 = mg ℓ2 (1− cos θ) +mgx((1− cos θ)

6. ycom = mℓ+M 0m+M , v1 = mv0

m+M

9. ωf =v0+

12Rω0

3R = 8.33rad/s

13. Ω = MgRIω .

Chapter 10

1. h = 0.8m2. R = 3.6m3. Fx = Patmℓh+ 1

2ρwgh2ℓ

4. P ′ = P + 12ρ(1− 16)v2 = P − 15

2 ρv2

6. vf =√v20 + 2ρwgh

7. ρ2 = (1.0 g/cm3) 588. For water h = 10.3m, for mercury h = 0.7614m10. (Pin − Pout)A = F = 1

2ρair · v2A

11. Flift = Fbelow − Fabove ≈ 12ρair(α

2 − 1)v2A17. vterm = mg

6πµa

19. z = zc +12ω2

g r2

Chapter 11

3. v =√20/0.01 = 44.721m/s, ω =

√100/0.01 = 100rad/s, x(t) = 0.44721m, sin

(100 t

)5. ω =

√gR

6. ω2 = mgℓ75MR2+m(R−ℓ)2

7. E = 12m((1− 2

π

)2R2θ2 + 1

2 (12mR

2)θ2 +mg(

12 (1−

2π )Rθ

2)

8. E = 12kA

2 = 12 (m1 +m2)v

21 = 1

2m2

2

m1+m2v20 , x =

√m2

2v20

k(m1+m2)sin(

√k

m1+m2t− π)

10. A =√(0.10)2 + (0.05)(5.0)2/(100)m

11. amax = ω2A = 1000.05

√(0.05)2 + (0.05)(0.4)2/(100) ms

15. ω =√

ρW gρBL

17. ω = 63.24 rads ,E = 2 12 (200)(0.02)

2 + 12 (300)(0.04)

2 J

23. ω =

√k(1+

M1M2

)

M1+M2

1M2

+I1R2 +

I2R2

M21

M22

40.√

mgxIcom+mx2 , x

2 = Icom/m.

41. xeq = −b/c,√b2/mc.

337

Chapter 12

4. =√

16π2R3

29mG

5. ϵ2 = 1− (m−2µ)(m+2µ)2

(m+µ)3 ≈ µm

9. Fm =∫ π0−MGρRdθ

R2

(sin θi+ cos θj)

11. At aphelion/perihelion we have r = 0, E = −µMG2a = 1

2µv2a −

µMGa(1+ϵ) , v

2a = MG

a

(1−ϵ1+ϵ

)14. R =

√3

4πρGv = 65, 536m

15. V = −mM1Gx − mM2G

(R−x) , xe =R

1+√

M2M1

16. a = (1.666 + 1.382)/2 = 1.524AU , ϵ = (1.666− 1.382)/(1.666 + 1.382) = 0.09317. Tm = 1.87398 yr

23. Fr =mGρ 4π

3 R3

r2 − mGρ 4π3 (R/2)3

(r−R2 )2

, M = ρ4π3

(R3 − (R/2)3)

)

Chapter 13

1. 839W3. 9381W , 938.1W4. T == 259K5. t = 4030 s6. You will end up with 137.5 g of water and 12.5 g of ice, all at 0C7. 31.5C8. x = L0

√α∆T/2 = 2.0m

√13.0× 10−6 · 100/2 = 0.05m

9. 819, 000N10. 0.03kgs

11. |v| = 10/13m/s,vrms =√

1413 m/s

13. U = 12kT

(2 + 3 + 2

), C = 7

2k

14. E =∫∞0CE4 e−E/kT dE = 4!C (kT )4 = 4kT

15. n = 4×106

1×106 = 417. R = 1015352m = 1015 km23. E = 3

2NkT

24. 1T = dS

dE = C(Em − 2E) < 0 if E > Em/235. 7826.60ms

Chapter 14

1. T1 = 1000k, T2 = 250K, T3 = 125K2. 313K, 0.13 calK3. 428Js4. ∆H = −483.6kJ

6. T1 = P1V1

8.31 = (1.0×105)(0.1)8.31 = 1203,T2 = T3 = P3V3

8.31 = (2.667×104)(0.1)8.31 = 321, V2 = V1

(T1

T2

) 32

= 0.726m3

8. −29.3 calK9. −9.43 calK10. 37 s11. 0.5656m3

12. −R ln 0.20.3

13.



1 → 232 (8.314)(4000− 2000) 5

2 (8.314)(4000− 2000) (8.314)(4000− 2000) 52 (8.314) ln

40002000

2 → 332 (8.314)(2000− 4000) 0 3

2 (8.314)(4000− 2000) 03 → 1

0 (8.314)(2000) ln 0.21.13 (8.314)(2000) ln 0.2

1.13 (8.314) ln 0.21.13

14.Line ∆U ∆Q ∆W

1 → 2 3(100)(0.3− 0.1) 4(100)(0.3− 0.1) (100)(0.3− 0.1)2 → 3 3(0.3)(25− 100) 3(0.3)(25− 100) 03 → 4 3(25)(0.1− 0.3) 4(25)(0.1− 0.3) (25)(0.1− 0.3)4 → 1 3(0.1)(100− 25) 3(0.1)(100− 25) 0

15. 0.2828 JK17. 0.6918. T1 = P1V1

R = 1202,T2 = 3T1 = 3606K, T4 = T1/2 = 601K,T3 = 3T1/2 = 1804K

∆U ∆Q ∆W1 → 2 (3/2)(8.314)(3606− 1202) 1× 104(3− 1) + (3/2)(8.314)(3606− 1202) 1× 104(3− 1)2 → 3 (3/2)(8.314)(1804− 3606) (3/2)(8.314)(1804− 3606) 03 → 4 (3/2)(8.314)(601− 1804) 5× 103(1− 3) + (3/2)(8.314)(601− 1804) 1× 104(3− 1)4 → 1 (3/2)(8.314)(1202− 601) (3/2)(8.314)(1202− 601) 0

20. [Z++][Cu++] = 1.97× 1037

22. dS = d(

4b3 V T

3)

23.∆U ∆Q ∆W ∆S

1 → 2 32R(578− 1202) 0 −3

2R(578− 1202) 02 → 3 3

2R(144− 578) 32R(144− 578) 0 3

2R ln(144/578)3 → 4 3

2R(300− 144) 0 −32R(300− 144) 0

4 → 1 32R(1202− 300) 3

2R(1202− 300) 0 32R ln(1202/300)

24.∆U ∆Q ∆W ∆S

1 → 2 32R(2404− 1202) 3

2R(2404− 1202) + 1× 105(0.2− 0.1) 1× 105(0.2− 0.1) 52R ln(2404/1202)

2 → 3 32R(1382− 2404) 0 −3

2R(1382− 2404) 03 → 4 3

2R(691− 1382) 32R(691− 1382) + 2.5× 104(0.2297− 0.45947) 2.5× 103(0.2297− 0.45947) 5

2R ln(691/1382)4 → 1 3

2R(1202− 691) 0 32R(1202− 691) 0

26. ∆G = −2 · 96, 368 · 1.1/4.184 = −50, 800 cal27. E = 1.685V28. 0.74V , −34, 200 cal.29. c(z) = c(0) eqEz/kT

31. ∆S = N1k lnV1+V2

V1+N2k ln

V1+V2

V2> 0

32. ∆S = 034. 0.297 JK

39. V = nRTP ,

(∂V∂P

)T= −nRT

P 2 ,KT = nRTP 2V = 1

P

40. Const. P ; dQ = dU + PdV = 32nRdT + P dV = 3

2nRdT + P d(nRTP

)P= 3

2nRdT + P(nRdTT

P

)= 5

2nRdT

48. xA = xB = n−n′

2n−n′ , xAB = n′

2n−n′

51. x = 5/58.4435/58.44+1000/10 = 0.00599.

52. (373)2·8.314·0.005992257 = 3.07K.

54. dS = d(2aV T

)

Chapter 16

Appendix

16.1 Appendix I. Mathematical formulas

sin(a+ b) = sin a cos b+ cos a sin b, cos(a+ b) = cos a cos b− sin a sin b, tan(a+ b) = tan a+tan b1−tan a tan b

sin(a− b) = sin a cos b− cos a sin b, cos(a− b) = cos a cos b+ sin a sin b, tan(a− b) = tan a−tan b1+tan a tan b

sin θ = opphyp , cos θ = adj

hyp , tan θ = oppadj

sin(2x) = 2 sin(x) cos(x) cos(2x) = cos2(x)− sin2(x) tan(2x) = 2 tan(x)1−tan2(x)

sin(x/2) =√

1−cos(x)2 cos(x/2) =

√1+cos(x)

2 tan(x/2) =√

1−cos(x)1+cos(x)

ddx sinx = cosx, d

dx cosx = − sinx, ddx tanx = 1

cos2 x∫sinx dx = − cosx+ c,

∫cosx dx = sinx+ c,

∫tanx dx = − ln(cosx) + c

ddx lnx = 1

x

∫ y1dxx = ln y d

dxeax = a eax∫

eax dx = eax

a + c ddxx

n = nxn−1∫xn dx = xn+1

n+1 + c

ln(ab) = ln(a) + ln(b) ln(a/b) = ln(a)− ln(b) ln(ab) = b ln(a)

ea+b = ea eb ea−b = ea

ebeln(a) = a

eiθ = cos(θ) + i sin θ i =√−1 a+ ib =

√a2 + b2 ei tan

−1( ba )

∫ badf(x) = f(a)− f(b) ∂f(x,y)

∂x = limdx→0f(x+dx,y)−f(x,y)

dxddx

∫ xaf(x′) dx′ = f(x)

16.2 Appendix II. Periodic table

Metals are in blue, transition metals in light blue, non-metals in red, noble gases are green, and semi-metals arepink. Groups run from left to right (H to He), and periods run top to bottom (H to Fr). At STP of 1.01×105Paand 298K Mercury (Hg), and Bromine (Br) are liquids, and H, N, O, F, Cl, He, Ne, Ar, Kr, Xe, Rn are gases.Electronic structures violating Madelung’s rule are in red (and are not the correct structure but are the Madelungrule prediction). Elements without stable nuclides have the atomic mass of the longest-lived isotop in parenthesis.

339

340 CHAPTER 16. APPENDIX

Hydrogen

H1s1

11.0

079

Heliu

m

He

1s2

24.0

026

Lithiu

m

Li

2s1

36.9

41

Beryliu

m

Be

2s2

49.0

122

Boron

B2p1

510.8

11

Carbon

C2p2

612.0

11

Nitrogen

N2p3

714.0

07

Oxygen

O2p4

815.9

99

Flu

orin

e

F2p5

918.9

98

Neon

Ne

2p6

10

20.1

80

Sodiu

m

Na

3s1

11

22.9

90

Magnesiu

m

Mg

3s2

12

24.3

05

Alu

min

um

Al

3p1

13

26.9

82

Silic

on

Si

3p2

14

28.0

86

Phosphorus

P3p3

15

30.9

74

Sulp

hur

S3p4

16

32.0

65

Chlo

rin

e

Cl

3p5

17

35.4

53

Argon

Ar

3p6

18

39.9

48

Potassiu

m

K4s1

19

39.0

98

Calc

ium

Ca

4s2

20

40.0

78

Scandiu

m

Sc

3d1

21

44.9

56

Titaniu

m

Ti

3d2

22

47.8

67

Vanadiu

m

V3d3

23

50.9

42

Chrom

ium

Cr

4d4

24

51.9

96

Manganese

Mn

3d5

25

54.9

38

Iron

Fe

3d6

26

55.8

45

Cobalt

Co

3d7

27

58.9

33

Nickel

Ni

3d8

28

58.6

93

Copper

Cu

3d9

29

63.5

46

Zin

c

Zn

3d10

30

63.3

9

Galliu

m

Ga

4p1

31

69.7

23

Germ

aniu

m

Ge

4p2

32

72.6

4

Arsenic

As

4p3

33

74.9

22

Sele

niu

m

Se

4p4

34

78.9

6

Brom

ine

Br

4p5

35

79.9

04

Krypton

Kr

4p6

36

83.8

Rubid

ium

Rb

5s1

37

85.4

68

Strontiu

m

Sr

5s2

38

87.6

2

Yttriu

m

Y4d1

39

88.9

06

Zirconiu

m

Zr

4d2

40

91.2

24

Nio

biu

m

Nb

4d3

41

92.9

06

Moly

bdenum

Mo

4d4

42

95.9

4

Technetiu

m

Tc

4d5

43

96

Rutheniu

m

Ru

4d6

44

101.0

7

Rhodiu

m

Rh

4d7

45

102.9

1

Palla

diu

m

Pd

4d8

46

106.4

2

Silv

er

Ag

4d9

47

107.8

7

Cadm

ium

Cd

4d10

48

112.4

1

Indiu

m

In5p1

49

114.8

2

Tin

Sn

5p2

50

118.7

1

Antim

ony

Sb

5p3

51

121.7

6

Tellu

riu

m

Te

5p4

52

127.6

Iodin

e

I5p5

53

126.9

Xenon

Xe

5p6

54

131.2

9

Cesiu

m

Cs

6s1

55

132.9

1

Bariu

m

Ba

6s2

56

137.3

3

Lanthanid

e

La-Lu

4f1

−5d1

57–71

Hafn

ium

Hf

5d2

72

178.4

9

Tantalu

m

Ta

5d3

73

180.9

5

Tungsten

W5d4

74

183.8

4

Rheniu

m

Re

5d5

75

186.2

1

Osm

ium

Os

5d6

76

190.2

3

Irid

ium

Ir5d7

77

192.2

2

Pla

tin

um

Pt

5d8

78

195.0

8

Gold

Au

5d9

79

196.9

7

Mercury

Hg

5d10

80

200.5

9

Thalliu

m

Tl

6p1

81

204.3

8

Lead

Pb

6p2

82

207.2

Bism

uth

Bi

6p3

83

208.9

6

Polo

niu

m

Po

6p4

84

(209)

Astatin

e

At

6p5

85

(210)

Radon

Rn

6p6

86

(222)

Franciu

m

Fr

7s1

87

(223)

Radiu

m

Ra

7s2

88

(226)

Actin

ide

Ac-L

r5f1

−6d1

89-1

03

Rutherfo

rd

Rf

6d2

104

(261)

Dubniu

m

Db

6d3

105

(262)

Seaborgiu

m

Sg

6d4

106

(266)

Bohriu

m

Bh

6d5

107

(264)

Hassiu

m

Hs

6d6

108

(277)

Meitneriu

m

Mt

6d7

109

(268)

Uun

6d8

110

(281)

Uuu

6d9

111

(272)

Uub

6d10

112

(285)

Uuq

114

(289)

Lanthanum

La

4f1

57

138.9

1

Ceriu

m

Ce

4f2

58

140.1

2

Praseodym

ium

Pr

4f3

59

140.9

1

Neodym

ium

Nd

4f4

60

144.2

4

Prom

ethiu

m

Pm

4f5

61

145

Sam

ariu

m

Sm

4f6

62

150.3

6

Europiu

m

Eu

4f7

63

151.9

6

Gadolin

ium

Gd

4f8

64

157.2

5

Terbiu

m

Tb

4f9

65

158.9

3

Dysprosiu

m

Dy

4f10

66

162.5

Holm

ium

Ho

4f11

67

164.9

3

Erbiu

m

Er

4f12

68

167.2

6

Thuliu

m

Tm

4f13

69

168.9

3

Ytterbiu

m

Yb

4f14

70

173.0

4

Lutetiu

m

Lu

5d1

71

174.9

7

Actin

ium

Ac

5f1

89

(227)

Thoriu

m

Th

5f2

90

232.0

4

Protactin

ium

Pa

5f3

91

231.0

4

Uraniu

m

U5f4

92

238.0

3

Neptuniu

m

Np

5f5

93

(237)

Plu

toniu

m

Pu

5f6

94

(244)

Am

eric

um

Am

5f7

95

(243)

Curiu

m

Cm

5f8

96

(247)

Berkeliu

m

Bk

5f9

97

(247)

Califo

rniu

m

Cf

5f10

98

(251)

Ein

stein

ium

Es

5f11

99

(252)

Ferm

ium

Fm

5f12

100

(257)

Medele

viu

m

Md

5f13

101

(258)

Nobeliu

m

No

5f14

102

(259)

Lawrenciu

m

Lr

6d1

103

(262)

16.3. REDUCE EXAMPLES 341

16.3 REDUCE examples

Use it as a calculator

x:=(sqrt(2)-4)/(sqrt(3)-2);

1/x;

on rounded;

x;

1/x;

x^2;

Dot/cross products

infix dot;

procedure v1 dot v2;

begin

return(for i:=1:3 sum part(v1,i)*part(v2,i));

end;

infix cross;

procedure v1 cross v2;

begin

v3x:=part(v1,2)*part(v2,3)-part(v1,3)*part(v2,2);

v3y:=part(v1,3)*part(v2,1)-part(v1,1)*part(v2,3);

v3z:=part(v1,1)*part(v2,2)-part(v1,2)*part(v2,1);

retval:=v3x,v3y,v3z;

return(retval);

end;

% try it on the 201 hw

1,-3,0 dot 1,1,1;

1,-3,0 cross 3,1,2;

Differentiate a function

depend f, x;

f:=1/(x^2+a^2);

df(f,x);

df(f,x,2);

Partial derivative of a function

depend f, x, y;

f:=1/(x^2+y^2);

df(f,x);

df(f,y);

Gradient (user defined)


procedure grad(fun);

begin

return(-df(fun,x), -df(fun,y));

end;

depend f, x, y;

f:=1/(x^2+y^2);

Efield:=grad(f);

Find the minimum of a function

procedure f(x);

begin

return(q1*k/(x-a)-q2*k/x);

end;

solve(df(f(x),x),x);

Divergence and curl

procedure divergence(vect);

begin

retval:=df(part(vect,1),x)+df(part(vect,2),y)+df(part(vect,3),z);

return(retval);

end;

procedure curl(vect);

begin

retval1:=df(part(vect,3),y)-df(part(vect,2),z);

retval2:=-df(part(vect,3),x)+df(part(vect,1),z);

retval3:=df(part(vect,2),x)-df(part(vect,1),y);

return(retval1,retval2,retval3);

end;

% try it

Evect:=2*x-3*y,2*y+x,2*x*y;

divergence(Evect);

curl(Evect);

Solve a differential equation

load_package odesolve;

odesolve(df(y,x,2)+4*y,y,x);

Laplace transformation

load_package laplace;

laplace(exp(-x),x,s);

on div;

invlap(1/(s+a)^2,s,t);


Integrate a function

int(1/(x^2+a^2),x);

% definite integrals work best if you

% define the integrand first

f:=(sin(x))^8;

norm:=int(f,x,0,pi);

% and/or load defint

load_package defint;

f:=(sin(x))^8;

norm:=int(f,x,0,pi);

% complex integrals can be done by computing residues

load_package residue;

MyInt:=2*pi*i*residue(exp(-i*s*z)/(z^2+1),z,i);

Compute a limit

load_package limits;

limit(sin(x)/x,x,0);

Expand a function in a series


f:=1/(x^2+a^2);

taylor(f,x,1,4);

g:=e^(-x);

taylor(g,x,0,5);

Taylor series expand Planck distribution


Ebar:=hf/(e^(hf*x)-1);

taylor(Ebar,x,0,5);

Solve polynomial equation numerically

load_package roots;

rootacc(10);

roots(x^4+x^3+x^2+x+1);

Solve nonlinear equation numerically

load_package numeric;

EQUATION:=tan(pi*x)-x=0;

% the 1 is a trial solution

num_solve(EQUATION,x=1, accuracy=10);

Solve linear (circuit law) equations


solve(R1*I1+R3*I3-V0,R1*I1-R2*I2,I1+I2-I3,I1,I2,I3);

Work with vectors/curvilinear coordinates

load_package avector;

vec u,v,w,u_cross_v;

u:=avec(u1,u2,u3);

v:=avec(v1,v2,v3);

w:=avec(w1,w2,w3);

u_cross_v:=u^v;

u_cross_v;

u_dot_v:=u.v;

coordinates r,th,phi;

scalefactors(1,r,r*sin(th));

dependf,v1,v2,v3, r, th,phi;

grad(f);

delsq(f);

div(v);

curl(v);

getcsystem ’cylindrical;

% r will be rho of course

g:=r*z*sin(phi);

v:=grad(g);

% create a locus of points parameterized by t

% z comp has const vel a

% phi comp has const vel 1

u:=avec(r0,a*t,t);

df(u,t);

% create a vector field with only a phi component

v:=avec(0,0,v0);

%integrate it around the curve u

deflineint(v,u,t,0,2*pi);

% create a vector field with only a rho component

v:=avec(r,0,0);

%integrate it around the curve u

deflineint(v,u,t,0,2*pi);

Solve Laplace equation by relaxation

%REDUCE version using matrices V(x,y)

procedure relax(N)$

begin

matrix V(10,10),tmpV(10,10)$

% set all to one volt

for m:=1:10 do for k:=1:10 do V(m,k):=1;

% set boundary voltages


for m:=1:10 do <<

V(1,m):=10;

V(10,m):=0;

V(m,1):=0;

V(m,10):=0 >>;

% now sweep through the interior points

for m:=1:N do <<

for j:=2:9 do for k:=2:9 do

tmpV(j,k):=(V(j-1,k)+V(j+1,k)+V(j,k-1)+V(j,k+1))/4;

for j:=2:9 do for k:=2:9 do

V(j,k):=tmpV(j,k);

% end the "time" loop

>>;

% just print the voltage in the middle

return(V(5,5));

end;

% turn on floating point conversion

on rounded;

relax(10);

relax(100);

relax(200);

Compute work done in moving charge through a field

%define our electric field

Ex:=y/(x^2+y^2);

Ey:=-x/(x^2+y^2);

% define our straigh path from (x0,y0) to (x1,y1)

let x=x1+(1-t)*x0;

let y=y1+(1-t)*y0;

% velocity for path

vx:=df(x,t);

vy:=df(y,t);

% compute the work

dW:=-Ex*vx-Ey*vy;

W=int(dW,t,0,1);

A few REDUCE tips

REDUCE computes numbers exactly and will not divide out common factors unless you tell it to, with switches suchas

% perform floating point (numbers with decimal points) comp.

on rounded;

% expand in powrrs of reverse precision such as

% 1+x+x^2+...

on revpri;

% use complex numbers

on complex;

% divide out all factors

on div;


% factor on variable x

factor x;

% create and use an array

ARRAY F(20); F(1):=F(2):=1;

FOR I:=3:20 DO F(I):=F(I-1)+F(I-2);

F(8);

% clear a variable so it can be reassigned

clear x;

% find largest/smallest power of a variable

Y:= a*x**2+b*x+c;

COEFF(Y,X);

%returns c,b,a

HIPOW!*;

% returns 2

LOWPOW!*;

%returns 0

% declare that y depends on x (so dy/dx can be computed

depend y x;

% extract first idem from a list

l:=1,2,3;

first(l);

second(l);

Index

absolute temperature, 280acceleration, 5acid, 313adhesive force, 80adiabatic curves, 278adiabatic line, 281adiabatic process, 276air resistance, 38, 42air-conditioning, 296ammonia molecule, 118amplitude, 196, 198amplitude-dependent frequency, 205amusement park ride, 79angle, 10, 77angular momentum, 145, 165angular speed, 77anode, 302aphelion, 241Archimedes principle, 181area element, 13area integrals, 116axiom, 26

ballistic pendulum, 122ballistics, 37base, 313battery, 303Benard cell, 269Bernoulli’s law, 181biharmonic equation, 187, 218billiard ball, 168billiard ball collision, 121binary star, 233binomial theorem, 205black hole, 236block and tackle, 60blood flow, 185boiling point elevation, 307Boltzmann entropy, 263bowling ball, 151brake, 170buffering, 313bulk modulus, 229

cannon, 115capillary flow, 185Carnot cycle, 278

Carnot’s theorem, 280CAS syntax, 26cathode, 302center of mass, 114, 116centrifugal potential, 305centrifuge, 306centripetal acceleration, 78, 136chemical equilibrium, 303chemical potential, 299, 317Clausius’ theorem, 283clutch, 170coefficient of friction, 66coefficient of thermal expansion, 288collinearity, 14collisions, 93common ion effect, 312complexions, 263components, 12compound interest, 35compressibility, 330compressional wave, 215computer algebra system (CAS), 26concentration, 304, 308, 309conduction, 273conservation law, 308conservation of energy, 277conservative force, 84constant of integration, 83contact force, 54convection cell, 269critical angle, 67cross product, 13, 137current, 308cycle, 279

damped oscillation, 207Dark Matter, 251diatomic molecule, 231diffusion, 193, 308disk, 140distributive law, 13dot product, 10double yo-yo, 149drag, 184

eccentricity, 241efficiency, 279

347

348 INDEX

elastic collision, 93Electrophoresis, 329ellipse, 241elliptic curve, 35elliptic function, 205empirical temperature, 276energy, 83enthalpy, 283, 287entropy, 281, 285entropy representation, 317equal areas, 242equation of motion, 102equation of state, 275equilibrium, 198, 275equilibrium condition, 224equilibrium constant, 300, 303equilibrium point, 203escape velocity, 236Euler’s theorem, 207excluded volume, 318exhaustion, 285exponential function, 207exponential growth, 35extensive variable, 299external forces, 114extrema, 284

Family mule, the, 26Fick’s law, 308financial math, 35first law, 277flux, 177force constant, 209formality, 312Free-body diagram, 54friction, 65, 91

gas processes, 278Gibbs free energy, 283, 303Gibbs-Duhem relation, 317global variable, 276gradient, 84graduated cylinder, 304Graham’s law, 308gravitational potential, 89, 235, 238gravitational sling-shot, 248gravity, 304gyroscope, 175

half-cell, 303harmonic equation, 187harmonic oscillator equation, 203heat, 253, 277heat bath, 278heat engine, 278Helmholtz free energy, 283

Hess’s law, 287high-altitude projectiles, 246Hookes law, 195hydraulics, 190hydrolysis, 314

Ice berg, 115ideal system, 304impact parameter, 121impulse, 126, 168impulsive force, 168incline, 57integration by parts, 101intensive variable, 299inter-atomic potential, 224inter-atomic vibration, 224inter-molecular potential, 224internal forces, 114inversion temperature, 319irreversible process, 276, 281isotherm, 276, 281

Joule experiment, 277Joule-Thompson effect, 319

Kelvin’s second law, 279Kepler’s laws, 243kinematics, 4kinetic energy, 84, 139kinetic friction, 66

ladder, 153Lame’ constants, 213Laplace equation, 187Laplace transform, 102launch parameters, 40law of exponents, 24Law of gravitation, 233Lechatelier’s principle, 302Legendre transformation, 316lift, 192line-integral, 87local variable, 276

maxima, 26maximal entropy principle, 263maximal frictional force, 67maximum altitude, 38Maxwell relation, 316mechanical energy, 253molarity, 311mole fraction, 299molecular vibration, 201molecular weight, 307moment of inertia, 139, 143momentum, 113, 114Morse potential, 224

INDEX 349

Na Cl crystal, 117Newton’s laws, 53non-linear oscillation, 206normal acceleration, 82normal force, 54normal mode, 200number theory, 35

Onsager’s theory, 308oscillations, 195osmosis, 306osmotic pressure, 306oxidation, 302

parallel axis theorem, 142parameterization, 87, 88partial pressures, 299partition function, 263pendulum, 122, 125penguin, 80, 123, 153, 154penguins, 115perihelion, 241periodic motion, 195periodicity, 198perpendicularity, 11pH, 314phase, 196, 198photon gas, 297photons, 327physical pendulum, 202, 228Pitot tube, 182planetary orbit, 241position graph, 5potential energy, 84potential function, 86, 94power, 132Prandtl flow, 192pressure, 177projectiles, 37pulley, 58Pythagorean theorem, 9

quadratic formula, 25quanta, 327quasi-static, 276

radial vector, 78radius of curvature, 82range, 37Rayleigh number, 268REDUCE, 26reduction, 302refrigeration, 296relative velocity, 6resolving vectors, 12reversible process, 276

Reynolds number, 185right hand rule, 166ring, 140rocking motion, 225rod COM, 117rolling motion, 144, 168rope tension, 54, 149rotational motion, 135

salts, 314scalar, 9second law, 279sedimentation, 304semi-major axis, 241series, 102shear force, 184shear forces, 182shear modulus, 208shear wave, 215simple harmonic motion, 199simple harmonic oscillation, 197simple pendulum, 204small amplitude oscillation, 203solution, 304specific heat, 288sphere (moment of inertia), 141spin rate, 79spontaneity, 285, 303spool, 146spring, 195state, 275Static friction, 66static friction, 67statics, 152stoichiometric coefficients, 327stoichiometry, 303storm windows, 273strain, 208stream function, 186stress, 208stress-momentum tensor, 180strong acid, 313substitution method, 102system, 275systems of particles, 114

tangential acceleration, 82, 136tangential vector, 78Taylor series, 102tension, 54tensor, 177terminal velocity, 44terrible puns, 218thermal expansion, 330thermodynamic potentials, 283Third law, 53

350 INDEX

thrust, 126time of flight, 37torque, 146, 165torsion, 209train wreck, 6trajectory, 39trajectory program, 39triatomic molecule, 272trig function, 207

unit vector, 78, 91

Van der Waals gas, 298, 330variable change, 102variable changes, 102vector, 9vector resolution, 67viscosity, 182, 183, 185vortex strength, 179vorticity, 178

work, 83, 277work reservoir, 286work-energy theorem, 84

yo-yo, 149Young’s modulus, 208

Zeroth law of thermodynamics, 275

university physics

Documents