University Physics

University Physics I

Author: Jeffrey R. Schmidt

Affiliation: University of Wisconsin

c 2001, revision August 27, 2012 ⃝

2

Forward Many of us, faculty and students, are weary of being forced to migrate to a new edition of university physics textbooks every year, differing only in price and weight. There is virtually no difference between edition X and edition X + 1 of any of the standard texts, they are all very much the same, and the same from edition to edition, in all respects except for price and weight (I recently got an evaluation copy of a physics 201/202 text that weighs 15 pounds, and runs over 200.00 retail price). The only way out of this charade is to provide students with a free alternative, that I can update every year at no cost to anyone, and so here it is, University Physics I, a very bare-bones, work in progress, 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 and cut out all of “filler” and pointless anecdotes and superfluous photos, the results will be highly mathematical in appearance, and will be about the length of this book. I have concentrated on problem-solving and expression of the basic concepts, since that is what the student really needs, and that is what testing and assessment focus on. I will remind the student that doing the homework is not optional, but is mandatory. You can’t master something by watching others do it. Times are definitely changing. In 1961 the number of hours per week spent on academic study by full-time college students was 40 hours per week outside of class. In 2004 that number had dropped to 27 hours per week 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), and by 2007 the average over 40% were awarded A-grades, and only 7% were awarded C-grades, with private universities and colleges leading the way in grade inflation, awarding 50% A-grades on average. Polls of students conducted in 2007 reveal that 40% of students expect to get B-grades for simply attending the course, without regard to performance. 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 time because 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 they take their jobs and any other aspects of their lives. The current generation of students is not unique, they don’t face hardships that previous generations did not, and nothing more is expected of them than what was expected of prior generations. And nothing less is expected of them! In fact the current generation has a lot of time-saving technology at its disposal, and good use should be made of this advantage that generations before did not have. I have 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 involving free software such as REDUCE, axiom, maxima, and ode into the course, partly as a way of providing math assistance to students, and to provide the tools for pushing the basic physics forward into other disciplines, in support of those disciplines and the students studying them. Problems and examples labeled with ∗ are very similar to problems 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 in lecture, we only have time for the basics, and much of this optional material requires some mathematical background beyond that required for the standard portion of the course. This optional material is in chapters 10, 11, 13 and 14 and is clearly marked as “not covered in 201”, but is there for you if you need it. I regard the required mathematical background for the standard 201 course to be rather minimal (one semester of calculus, what was called freshman math when your parents were in college), others might say one semester is inadequate, but one semester 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. You must 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 I have combined it with as many tools as I can to make your job easier. But easier is not easy, and you will have to spend 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, Donn Henriksen. You will always treasure the short time that you spend in college, four (or five!) very short but important years. For me they passed quickly, I graduated, and lost track of many good people as we began our careers and families and adult lives. After two decades, I was able to re-established contact with some, Donn included, thanks to 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 through these notes, solved the problems, and uncovered quite a few typos in the winter of 2006/2007 as he refreshed his understanding of the physics that we learned as classmates way back in 1976, and endured his treatments. These notes, which will soon become a real book (but a free one) are dedicated to the memory of my friend Donn, whose desire to keep learning as a life-long pursuit should serve as a model to which all students should aspire.

Contents Forward

i

1 Kinematics 1.1 Motion on a line . . . . . . . . . . . . . 1.1.1 Motion at constant velocity . . . 1.1.2 Motion at constant acceleration . 1.2 Arbitrary motion . . . . . . . . . . . . . 1.3 Vectors . . . . . . . . . . . . . . . . . . 1.3.1 Basic vector operations . . . . . 1.3.2 Examples . . . . . . . . . . . . . 1.4 Appendix. Differential calculus . . . . . 1.4.1 Partial derivatives versus total . 1.4.2 Slopes of curves without calculus 1.4.3 The rules of differentiation . . . 1.4.4 The binomial theorem . . . . . . 1.4.5 Series expansion . . . . . . . . . 1.4.6 Rational functions . . . . . . . . 1.4.7 Radicals . . . . . . . . . . . . . . 1.4.8 Basic integration . . . . . . . . . 1.4.9 The logarithm and exponential . 1.4.10 Quadratic forms . . . . . . . . . 1.4.11 Complex numbers . . . . . . . . 1.4.12 Calculus assistance . . . . . . . . 1.5 Problems . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . .

1 1 4 4 7 9 10 14 15 16 16 18 18 19 21 21 22 24 24 25 26 28

2 Projectile motion 2.1 Basic examples of ballistics 2.2 Using the computer . . . . 2.2.1 Using ode . . . . . . 2.3 Air resistance . . . . . . . . 2.4 Self-propelled projectiles . . 2.5 Obstacles and targets . . . 2.6 Problem solving suggestions 2.7 Appendix. REDUCE code . 2.8 Problems . . . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

37 37 39 44 44 46 47 47 48 49

3 Newton’s Laws. Dynamics 3.1 Free-body diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 The equations of motion (EOMS) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

53 54 55 59

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

4 Frictional Forces 65 4.1 Problem solving strategy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70 4.2 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70 iii

iv

CONTENTS

5 Circular motion at constant speed 77 5.1 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80 6 Work and Energy 6.1 Kinetic energy . . . . . . . . . . . . . . . . . . . . . . . 6.2 Potential energy . . . . . . . . . . . . . . . . . . . . . . 6.3 Total energy . . . . . . . . . . . . . . . . . . . . . . . . . 6.4 The master energy equation . . . . . . . . . . . . . . . . 6.5 Computing work done by forces . . . . . . . . . . . . . . 6.5.1 Gravity is conservative . . . . . . . . . . . . . . . 6.5.2 Friction is non-conservative . . . . . . . . . . . . 6.6 When does a force not have a potential? . . . . . . . . . 6.7 Advantages and techniques of the work/energy theorem 6.7.1 Principle of infinitismal work . . . . . . . . . . . 6.7.2 Solving for accelerations . . . . . . . . . . . . . . 6.8 Appendix. Integral calculus techniques . . . . . . . . . . 6.8.1 Integration by parts . . . . . . . . . . . . . . . . 6.8.2 Variable Changes . . . . . . . . . . . . . . . . . . 6.9 Appendix. Computer assistance . . . . . . . . . . . . . . 6.10 Problems . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . .

83 83 84 86 86 87 89 89 94 95 97 99 101 101 102 104 105

7 Conservation of Momentum 7.1 Elastic and Inelastic Collisions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2 Forces, momenta and impulse . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

113 . 119 . 125 . 127

8 Rotational motion 8.0.1 The acceleration components . . . . . . . . 8.0.2 The case of constant tangential acceleration 8.1 Kinetic energy of a rotating body . . . . . . . . . . 8.1.1 Parallel axis theorem . . . . . . . . . . . . . 8.1.2 Rolling motion . . . . . . . . . . . . . . . . 8.2 Rotational dynamics and torque . . . . . . . . . . 8.3 Statics . . . . . . . . . . . . . . . . . . . . . . . . . 8.4 Work and torque . . . . . . . . . . . . . . . . . . . 8.5 Problems . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

135 136 136 139 142 144 145 152 154 156

9 Angular Momentum 165 9.1 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172 10 Fluids 10.1 Velocity fields of fluids . . . . . . . . . . . . 10.2 Newton’s laws applied to fluids . . . . . . . 10.2.1 Incompressible, irrotational fluids . . 10.3 Advanced topics . . . . . . . . . . . . . . . 10.3.1 Incompressible flow around obstacles 10.3.2 The divergence theorem . . . . . . . 10.4 Appendix . . . . . . . . . . . . . . . . . . . 10.5 Problems . . . . . . . . . . . . . . . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

177 177 179 180 183 187 188 189 190

11 Oscillations 11.1 The simple pendulum . . . . . . . . . . 11.2 Oscillations in general . . . . . . . . . . 11.3 Classical elastica. Stress and strain . . . 11.3.1 Torsional strain . . . . . . . . . . 11.3.2 Complete treatment of elasticity

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

195 204 206 208 209 210

. . . . .

. . . . .

CONTENTS

v

11.3.3 Architectural examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217 11.3.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 224 12 Gravitation and Planetary Motion 12.1 Newton’s universal law . . . . . . . 12.2 Gravitational potential energy . . . 12.3 Planetary orbits . . . . . . . . . . 12.3.1 Orbital corrections . . . . . 12.3.2 High-altitude projectiles . . 12.4 Problems . . . . . . . . . . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

233 233 235 241 244 246 247

13 Heat and temperature 13.1 Early experiments . . . . . . . . . . . . . . . . . . . . . 13.1.1 Joule-Thompson experiment . . . . . . . . . . . . 13.1.2 Calorimetry . . . . . . . . . . . . . . . . . . . . . 13.2 Phase Changes . . . . . . . . . . . . . . . . . . . . . . . 13.3 Heat flow . . . . . . . . . . . . . . . . . . . . . . . . . . 13.4 Radiative cooling and heating . . . . . . . . . . . . . . . 13.5 Statistical mechanics. Kinetic theory . . . . . . . . . . . 13.5.1 Kronig-Clausius kinetic theory . . . . . . . . . . 13.5.2 Boltzmann’s theory . . . . . . . . . . . . . . . . 13.5.3 The meaning of temperature . . . . . . . . . . . 13.5.4 Boltzmann entropy . . . . . . . . . . . . . . . . . 13.6 Convection . . . . . . . . . . . . . . . . . . . . . . . . . 13.7 Appendix. Some numerical values of material properties 13.8 Problems . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . .

. . . . . . . . . . . . . .

. . . . . . . . . . . . . .

. . . . . . . . . . . . . .

. . . . . . . . . . . . . .

. . . . . . . . . . . . . .

. . . . . . . . . . . . . .

. . . . . . . . . . . . . .

. . . . . . . . . . . . . .

. . . . . . . . . . . . . .

. . . . . . . . . . . . . .

. . . . . . . . . . . . . .

. . . . . . . . . . . . . .

. . . . . . . . . . . . . .

. . . . . . . . . . . . . .

. . . . . . . . . . . . . .

. . . . . . . . . . . . . .

. . . . . . . . . . . . . .

. . . . . . . . . . . . . .

. . . . . . . . . . . . . .

. . . . . . . . . . . . . .

. . . . . . . . . . . . . .

. . . . . . . . . . . . . .

. . . . . . . . . . . . . .

. . . . . . . . . . . . . .

. . . . . . . . . . . . . .

253 253 253 254 256 257 258 260 260 261 262 263 264 270 271

14 Equilibrium thermodynamics 14.1 Core definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.2 The laws of thermodynamics . . . . . . . . . . . . . . . . . . . . . 14.2.1 First Law of thermodynamics . . . . . . . . . . . . . . . . . 14.2.2 Kelvin’s Second Law . . . . . . . . . . . . . . . . . . . . . . 14.2.3 Carnot’s Theorem . . . . . . . . . . . . . . . . . . . . . . . 14.2.4 Non-reversible Process . . . . . . . . . . . . . . . . . . . . . 14.2.5 Clausius’ Theorem . . . . . . . . . . . . . . . . . . . . . . . 14.3 Thermodynamic Potentials . . . . . . . . . . . . . . . . . . . . . . 14.3.1 The meaning of F ; the F-Theorem; . . . . . . . . . . . . . . 14.3.2 The meaning of G; the G-Theorem; . . . . . . . . . . . . . 14.3.3 The meaning of entropy . . . . . . . . . . . . . . . . . . . . 14.3.4 The meaning of enthalpy . . . . . . . . . . . . . . . . . . . 14.3.5 Material properties of matter . . . . . . . . . . . . . . . . . 14.4 Ideal gases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.5 P V -diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.5.1 The Carnot cycle . . . . . . . . . . . . . . . . . . . . . . . . 14.5.2 Arbitrary systems . . . . . . . . . . . . . . . . . . . . . . . 14.6 Non-P V work. Chemical and electrical . . . . . . . . . . . . . . . . 14.6.1 Chemical potential for ideal gases . . . . . . . . . . . . . . . 14.6.2 Electrochemical work . . . . . . . . . . . . . . . . . . . . . 14.7 More applications of µ . . . . . . . . . . . . . . . . . . . . . . . . . 14.7.1 The centrifuge . . . . . . . . . . . . . . . . . . . . . . . . . 14.7.2 Osmosis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.7.3 Boiling point elevation . . . . . . . . . . . . . . . . . . . . . 14.8 Non-equilibrium applications. Diffusion, Graham’s and Fick’s laws 14.9 Freshmen chemistry crammed into the chemical potential nutshell 14.9.1 Ionic equilibria . . . . . . . . . . . . . . . . . . . . . . . . . 14.9.2 Acids and bases. pH . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . .

275 275 277 277 279 280 281 283 283 284 284 285 287 288 288 291 293 297 299 301 302 304 305 306 307 308 309 311 312

. . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

vi

CONTENTS 14.9.3 Buffering . . . . . . . . . . . . . . . . . 14.9.4 Hydrolysis . . . . . . . . . . . . . . . . . 14.9.5 Chemistry glossary . . . . . . . . . . . . 14.10Advanced topics . . . . . . . . . . . . . . . . . 14.10.1 Maxwell relations . . . . . . . . . . . . . 14.10.2 The Gibbs-Duhem relation . . . . . . . 14.10.3 The Van der Waals gas . . . . . . . . . 14.10.4 An application of enthalpy; refrigeration 14.11Problems . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

15 Answers to the problems 16 Appendix 16.1 Appendix I. Mathematical formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.2 Appendix II. Periodic table . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.3 REDUCE examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . .

313 314 314 315 315 317 318 319 322 333

339 . 339 . 339 . 341

Chapter 1

Kinematics Kinematics is the description of motion without any consideration as to what causes the motion. It is simply the mathematical 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 position versus time, or as a mathematical function, position as a function of time. We will start with the very simplest type of 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 an object 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 ( ) lim x(t + dt) − x(t) = 0, ∀t (1.1) dt→0

so that if you are at x = 1.0 m (for example) at t = 0 s, you can’t spontaneously jump to some other point in an infinitismal time (no teleportation, no transporter beams, and so forth). We do not necessarily require x(t) to be smooth, which would to say ( x(t + dt) − x(t) ) lim ∀t (1.2) dt→0 dt

10 exists and is bounded. For example the motion  0≤t≤1  1.0 + 2.0 t 3 + 4(t − 1) 1 ≤t≤2 x(t) =  7 + 3(t − 2) 2 ≤ t

8

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.

6

x(t), m

(1.3)

4

This brings us to a set of important ideas. The displacement ∆x of an object between times t1 and t2 is the change in its position ∆x = x(t2 ) − x(t1 )

2

(1.4)

and the average velocity of the object over this interval is the rate with which its displacement changed

0 0.0

1

0.5

1.0

1.5 t, sec

2.0

2.5

3.0

2

CHAPTER 1. KINEMATICS

x(t2 ) − x(t1 ) (1.5) t2 − t1 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 magnitude called the average speed x(t ) − x(t ) 2 1 (1.6) s¯(t1 , t2 ) = t2 − t1 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 ∫ t2 √ ∫ t2 √ s¯(t1 , t2 ) = v 2 dt, s¯(t1 , t2 ) = vx2 + vy2 dt (1.7) v¯(t1 , t2 ) =

t1

t1

√ and you might need to be a bit careful in your calculation of v 2 .

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

dt→0

dx x(t1 + dt) − x(t1 ) = (t1 ) dt dt

(1.8)

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

v(t1 + dt) − v(t1 ) dv = (t1 ) (1.10) dt→0 dt dt A position function must be continuous and smooth if you want to find its instantaneous velocity at any time. a(t1 ) = lim

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   2.0 0 ≤ t ≤ 1 4 1≤t≤2 (1.11) v(t) =  3 2≤t and integrate from ti = 0 to tf = 3 , breaking up the integral (see the Chapter 1 appendix, Eq. 1.131) into parts over which v(t) is constant ∫ 3 ∫ 1 ∫ 2 ∫ 3 ∆ 9 m v(t) dt = 2 dt + 4 dt + 3 dt = 9 m, v¯ = ∆= = =3 (1.12) T 3 s 0 0 1 2 (can you see why s¯ = v¯ for this example?). Example 2. Lets extend this example, suppose that   2.0 0 ≤ t ≤ 1 −4 1 ≤ t ≤ 2 v(t) =  3 2≤t Then

∫ ∆

=

1 ( 3s

∫

1

0

∫

3

4 dt + 1

∫

1

v(t) dt = 0

∫

2

2 dt −

v(t) dt = 0

s¯ =

∫

3

∫

2 dt + 0

3

3 dt = 1 m, 2

2

∫

3

4 dt + 1

(1.13)

2

v¯ =

) m 3 dt = 3 s

1m ∆ = T 3 s (1.14)

1.1. MOTION ON A LINE

3

Example 3. Consider the one-dimensional motion  0≤t≤1  3.0 t2 3 + 6(t − 1) 1 ≤ t ≤ 4 x(t) =  15 3≤t

(1.15)

  6.0 t 0 ≤ t ≤ 1 6 1≤t≤4 v(t) =  0 3≤t

(1.16)

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

Lets use gnuplot as described in problem 1.35 to graph both functions 16 f3(x) 14

12

10

f1(x)=(x1) ? 3+6*(x-1) : f1(x) f3(x)=(x>3) ? 15 : f2(x) plot[0:4][-1:7] f3(x)

8

6

4

2

0 0

0.5

1

1.5

2

2.5

3

3.5

4

7 f3(x) 6

5

4

f1(x)=(x1) ? 6 : f1(x) f3(x)=(x>3) ? 0 : f2(x) plot[0:4] f3(x)

3

2

1

0

-1 0

0.5

1

1.5

2

2.5

3

3.5

4

4

1.1.1

CHAPTER 1. KINEMATICS

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 + v0 t2 ) − (x0 + v0 t1 ) = v0 t2 − t1

(1.18)

The object moves at a constant rate (speed) in the same direction, forever. The instantaneous velocity is exactly the same (x0 + v0 (t1 + dt)) − (x0 + v0 t1 ) v(t1 ) = lim = v0 (1.19) dt→0 (t1 + dt) − t1 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 applied to a body), the position function has the generic form 1 x(t) = x0 + v0 t + at2 2

(1.20)

Compute the average velocity v¯(t1 , t2 ) =

(x0 + v0 t2 + 21 at22 ) − (x0 + v0 t1 + 12 at21 ) t2 + t1 = v0 + a( ) t2 − t1 2

(1.21)

and instantaneous (x0 + v0 (t1 + dt) + 12 a(t1 + dt)2 ) − (x0 + v0 t1 + 12 at21 ) = v0 + a t 1 dt→0 (t1 + dt) − t1

v(t1 ) = lim

(1.22)

The average and instantaneous accelerations are the same a ¯(t1 , t2 ) =

(v0 + at2 ) − (v0 + at1 ) d = a = (v0 + at) t2 − t1 dt t=t1

(1.23)

What is the meaning of x0 ? It is the position when the clock reads zero, x(0) = x0 + v0 · 0 + 12 a · 02 = x0 . What is the 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, where and 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 the same speed. At what height do they collide. The first object, launched at t = 0, will have height versus time of 1 y1 (t) = v0 t − gt2 2

(1.24)

The second ball travels the same path, but reaches each height at a time T after the first ball reaches that same height, and so 1 (1.25) y2 (t) = v0 (t − T ) − g(t − T )2 2

1.1. MOTION ON A LINE

5

60

y1(t), y2(t), (m)

50

Plots of these for v0 = 30 m s and T = 2s are illustrated. From the graph we can immediately obtain a useful bit of data; the collision takes place when the first projectile is on its way back, having already peaked in height. The algebraic solution is gotten by finding the time at which both projectiles reach the same height;

40 30 20

y1 (t∗ ) = y2 (t∗ ) 1 or v0 t∗ − gt2∗ = v0 (t∗ − T ) 2 1 − g(t∗ − T )2 (1.26) 2

10 0

0

2

4 t, (s)

6

8

solving, we discover that the collision occurs at t∗ =

v0 T + , g 2

1 v2 T gT 2 y∗ = v0 t∗ − gt2∗ = 0 − 2 2g 8

at a height

(1.27)

200 Example 5. A car stopped at a traffic light begins at rest. At time t = 0 the light turns green and the car accelerates with acceleration a. At just that instant a car traveling at speed v0 speeds through the light. How much time does it take the accelerating car to catch the speeder, and how far from the light does this occur? A graph of both position versus time curves for the vehicles 1 y1 (t) = at2 (1.28) 2 and y2 (t) = v0 t (1.29)

y1(t), y2(t), (m)

150

100

50

m are below for v0 = 30 m s and a = 10 s2 .

0

0

2

4 t, (s)

6

8

We can see from the slope of the graph that the accelerating car is traveling much faster than the speeder when they finally meet, which takes place when they arrive at the same point at the same time; y1 (t∗ ) = y2 (t∗ ),

or v0 t∗ =

1 2 at 2 ∗

giving t∗ =

2v0 , a

and location

y ∗ = v0 t ∗ =

2v02 a

(1.30)

at which time the accelerating car has velocity v1 (t∗ ) = at∗ = 2v0

(1.31)

6

CHAPTER 1. KINEMATICS

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 B enters the track traveling in the same direction but at a greater speed vB . What is the minimal acceleration that the engineer of train B must apply in order to avoid a collision?

200 The position versus time of the trains is xA (t) = D + vA t,

xA(t), xB(t), (m)

150

1 xB (t) = vb t − at2 2 (1.32)

graphs of these for vB = 2vA = 30 m s and a = 2 sm2 ,D = 50 are illustrated. Our new condition is that when

100

that

50

xA (t∗ ) = vA (t∗ ) =

xB (t∗ ) vB (t∗ )

(1.33)

so that there is no relative velocity between the trains, hence no collision. This means vA

0

and

or

0

2

4 t, (s)

6

t∗

8

= vB − at∗ ) vB − vA = a

(1.34)

200

xA(t), xB(t), (m)

150

inserting this into 1 D + vA t∗ = vb t∗ − at2∗ 2

100

results in

(vB − vA )2 (1.36) 2D The slopes of the two curves must also match up when the trains meet, they will be tangent to one another as in the figure. a=

50

0

(1.35)

0

2

4 t, (s)

6

8

1.2. ARBITRARY MOTION

1.2

7

Arbitrary motion1

We will study motion at constant acceleration (due to constant forces acting on a body) for the first half of the course, but you have to realize that acceleration may not be constant, it can be arbitrary. The problem of describing the position of a body given its acceleration (given the forces acting on it) is one of the fundamental problems of this course. Mathematically, it amounts to solving differential equations, in fact this very problem is what calculus was invented for. Let me illustrate two rather simple and universal approaches based on the idea that the motion will 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 or restoring force a(t) = −ω 2 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 powerful tools of calculus, the series expansion (see the appendix to this chapter), assume that all of the derivatives of x exist, and write 1 1 1 1 x(t) = x0 + x1 t + x2 t2 + x3 t3 + x4 t4 + x5 t5 + · · · (1.38) 2 3! 4! 5! in which xi are all constants. Note first that x(0) = x0 = 1 Then v(t) =

dx 1 1 (t) = x1 + x2 t + x3 t2 + x4 t3 + · · · dt 2! 3!

(1.39)

and again examine things at t = 0 v(0) = x1 = 0 and compute a(t) d2 x 1 1 (t) = x2 + x3 t + x4 t2 + x5 t3 + · · · (1.40) dt2 2! 3! and simply put these expansions into the equation of motion and solve in succession for the constants by matching coefficients of like powers of t a(t) =

−ω 2 x(t) ( ) 1 1 1 1 = −ω 2 1 + 0 t + x2 t2 + x3 t3 + x4 t4 + x5 t5 + · · · 2 3! 4! 5!

a(t) = x2 + x3 t +

1 1 x4 t2 + x5 t3 + · · · 2! 3!

(1.41)

We obtain coefficient of t0 ; coefficient of t1 ; coefficient of t2 ; coefficient of t3 ;

x2 x3 1 x4 2! 1 x5 3!

= = = =

−ω 2 0 1 −ω 2 x2 = ω 4 2 2 1 −ω x3 = 0 3!

···

(1.42)

and if we put things together, and actually paid attention in our calculus courses, we might even recognize that the series adds up to some well-known function x(t) = 1 −

ω2 2 ω4 4 t + t + · · · = cos(ωt) 2! 4!

(1.43)

This technique almost never fails, and is very simple and mechanical to carry out, its only drawback is that the answer is a series, which you might not recognize how to sum up. 1 This

section is optional, but I strongly suggest reading it.

8

CHAPTER 1. KINEMATICS

The second approach (very powerful and universal, see Chapter 6) makes use of the product rule of calculus. Begin with the same equation, −ω 2 x(t) −ω 2 x(t) v(t) d (1 2 ) = −ω 2 x (t) dt 2 ∫ t ( d 1 2 ) = −ω 2 x (t) dt 0 dt 2 ) ω2 ( 2 = − x (t) − x2 (0) 2

a(t) multiply by v(t) a(t) v(t) d (1 2 ) both sides are t derivatives v (t) dt 2 ∫ t ( d 1 2 ) integrate v (t) dt 0 dt 2 ) 1( 2 v (t) − v 2 (0) 2

= =

(1.44)

and solve for v(t) dx v(t) = dt ∫ dt

√ = ∫

( ) v 2 (0) − ω 2 x2 (t) − x2 (0) dx √ ( ) v 2 (0) − ω 2 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 (the energy method) which we will expand upon when the concept of energy is introduced. Suppose that v(0) = 0 and x(0) = 1; ∫ t ∫ x dx 1 √ ( dt = t = cos−1 x, x = cos(ωt) ), ω 0 1 2 2 ω 1−x 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;

v(t) = ∫

∫

dv dt

= a(t),

dx dt

= v(0) +

dx

t

t

= 0

a(t′ ) dt′

0

a(t′ ) dt′ ∫ t (∫

v0 dt +

x(0)

x(t) =

v(0)

0

∫

t

dv = ∫

x(t)

∫

v(t)

0

t′

) a(t′′ ) dt′′ dt′

0

∫ t (∫

t′

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

) a(t′′ ) dt′′ dt′

(1.46)

0

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

t′

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

∫ ) α t′′ dt′′ dt′ = x(0) + v(0) t +

t 0

0

α ′2 ′ α t dt = x(0) + v(0) t + t3 2 6

or suppose a(t) = a0 , a constant ∫ t(∫

t′

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

′′

a0 dt 0

0

)

′

∫

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

t

a0 t′ dt′ = x(0) + v(0) t +

a0 2 t 2

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

1.3. VECTORS

1.3

9

Vectors

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

Typical vectors are forces, positions, velocities and momenta. The common representation of a vector is as a point in space such as in the figure. The point is specified by giving its x and y coordinates, either as an ordered pair (triple) or as multiples of unit vectors. We could specify the vector in the figure by giving Cartesian coordinate of its tip v

= (vx , vy ) = vx i + vy j

v i

= (vx , vy , vz ) = vx i + vy j + vz k = (1, 0, 0)

j = (0, 1, 0) k = (0, 0, 1) (1.47)

from the Pythagorean theorem we obtain its length (magnitude) |v| =

√ vx2 + vy2 ,

|v| =

or

√ vx2 + vy2 + vz2

(1.48)

and its direction relative to the x-axis θ = tan−1

vy vx

(1.49)

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

and similarly in three dimensions a = (ax , ay , az ),

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

b = (bx , by , bz )

(1.51)

10

CHAPTER 1. KINEMATICS

a + b = (ax + bx , ay + by , az + bz ) (1.52) It will prove to be very useful to be able to compute the angle between two vectors, from the figure below we see that by ay − tan−1 ax bx (1.53) take the cosine of both sides of this equation and use θab = θa − θb = tan−1

cos(x − y) cos(θab )

= cos(x) cos(y) + sin(x) sin(y) = cos(θa − θb ) = cos(θa ) cos(θb ) + sin(θa ) sin(θb ) (1.54)

cos(θab ) =

Ay By Ax Bx + |A||B| |A||B|

(1.55)

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

ax bx + ay by ax bx + ay by + az bz (1.56)

and we see that the cosine of the angle between two vectors is cos(θab ) =

a·b |a||b|

(1.57)

which is very useful since it involves only the 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 length of the vector but not its direction. s r = s (rx , ry , rz ) = (srx , sry , srz )

(1.58)

for r 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 ) The length of a vector is computed via the Pythagorean theorem; √ |a| = a2x + a2y + a2z

(1.60)

(1.61)

Find a vector ⊥ to another vector Given a vector in two dimensions such as a = (ax , ay ) = ax i + ay j

(1.62)

any vector b perpendicular to it is a vector such that cos θab = 0 =

a·b = 0, |a||b|

a·b=0

(1.63)

Write this equation out in components ax bx + ay by = 0

(1.64)

This is one equation in two unknowns bx , by , and so cannot be uniquely solved. As long as both components ax and ay are non-zero, we simply pick a value for bx and proceed; Let bx = 1,

ax + ay by = 0,

and therefore b = (1, −

by = −

ax ay

(1.65)

ax ) ay

(1.66)

However if b ⊥ a, then so is s b 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 = ax bx + ay by + az bz = 0,

if b ⊥ a

(1.69)

There will be many such vectors, so for example let bx = 0, by = 1; 0 + ay + az bz = 0, and then any vector sb = (0, s, −

bz = −

ay az

say )⊥a az

(1.70)

(1.71)

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

sax )⊥a az

(1.72)

12

CHAPTER 1. KINEMATICS

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 simply divide by x, since numbers form a field, and division by anything but zero is legitimate in a field. But what about solving v1 + a v2 = 0? You can’t divide vectors because they do not form a field. It might be possible that the 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 are linearly 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 that the 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 ̸= s b

(1.76)

c = (cx , cy )

(1.77)

c = αa + βb

(1.78)

Pick another vector and determine the two scalars α, β such that 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)

(cx , cy ) = (αax , αay ) + (βbx , βby )

(1.80)

(cx , cy ) = (αax + βbx , αay + βby )

(1.81)

use the scalar multiplication rule use the vector addition rule 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

and put it into the second cy = αay +

(1.83)

( c − αa ) x x by bx

and get α by itself; cy bx − cx by = α(ay bx − ax by ), and go back and get β; β=

α=

(1.84) cy bx − cx by ay bx − ax by

cy ax − cx ay ay bx − ax by

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

(1.85)

(1.86)

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 · k = k · i = 0, and the fact that it is distributive;

i·j =j·i=0 k·j =j·j =0

(1.87)

( ) a· b+c =a·b+a·c

(1.88)

Demonstration of these facts from the definition a · b = ax bx + ay by + az bz

(1.89)

I will leave to you. Cross products. The cross product is another vector product invented as a tool for computing areas. We know that the area of a rectangle is base × height, and the same is true for a parallelogram. What if we want to compute the area of a parallelogram whose sides are given by two vectors A = ax i + ay j and B = bx i + by j? Of course we can do it the long way and resolve the vectors into components, find the angle between them and construct the base and 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. Since there will be no area at all if A = B, we can ensure that the product gives us zero when we apply it to two copies of 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 = (ax i + ay j) × (bx i + by j) = ax bx i × i + ax by i × j + ay bx j × i + ay by j × j = (ax by − ay bx )i × j

(1.91)

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

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

(ax by − ay bx ) (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 the status 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, k × i = −i × k = j,

i × j = −j × i = k 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 vectors entirely in terms of its components, but also the sine of the angle between the vectors sin(θa − θb ) =

|A × B| |A||B|

(1.95)

14

CHAPTER 1. KINEMATICS

and provides us with a test of whether or not two vectors are parallel to one another; A×B=0

A || B

if

(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 )

Solve these three equations s=

(1.98)

ax ay az = = bx by bz

(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 the vectors 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 = 0 the 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-products above, 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, ( ) z · w = (1)(1) + (−1)(3) = −2 = αu + βv + γw · w = γ|w|2 = 2γ,

we find β = 2 so that

γ = −1

1.4. APPENDIX. DIFFERENTIAL CALCULUS

1.4

15

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 derivatives 2. expansions of functions, parameterization of curves 3. 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 trigonometry while we are at it. There is no substitute for a full year of calculus instruction, and so math 221 is an absolute prerequisite for physics 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) f (x + dx) − f (x) = lim dx→0 dx dx

exists (is a number) at x.

y=f(x) The limit-taking process is very simple; we expand the numerator in ascending powers of dx, perform the division, and take the limit by setting dx = 0 afterwards (in a nutshell). The geometrical interpretation of the derivative of f (x) at x0 is that f ′ (x0 ) is the slope of the line tangent to f (x) at x0 ; Therefore a useful formula for translating calculus to geometry is tan θ =

( df (x) ) dx

(1.100) x=x0

for the tangent to the curve at x0 .

θ

x0 The limit taking process is most easily handled for polynomial functions such as f (x) =

N ∑

an xn

n=0

by use of the following simple rule;

) df (x) dg(x) d ( f (x) + g(x) = + dx dx 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 the derivatives of the summands. Example 11. Let f (x) = x3 , the steps taken in computing the derivative are; Step 1. Write out the fraction

16

CHAPTER 1. KINEMATICS

f (x + dx) − f (x) dx

(x + dx)3 − x3 dx 3 (x + 3x2 dx + 3x (dx)2 + (dx)3 ) − x3 dx 3x2 dx + 3x (dx)2 + (dx)3 dx 3x2 + 3x dx + (dx)2 ( ) lim 3x2 + 3x dx + (dx)2 = 3x2 (1.102)

= =

Perform the division

(x3 + 3x2 dx + 3x (dx)2 + (dx)3 ) − x3 dx

= =

3

Perform the limit (set dx = 0);

dx dx

=

dx→0

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(x2 + 2x dx + (dx)2 )) − (a + bx2 ) = dx ( ) d a + bx2 = Perform the limit (set dx = 0); dx

Perform the division;

(a + b(x + dx)2 ) − (a + bx2 ) dx (a + b(x2 + 2x dx + (dx)2 )) − (a + bx2 ) dx 2bx dx + b(dx)2 = 2bx + b dx dx ( ) lim 2bx + b dx = 2bx (1.103)

dx→0

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 more than one variable with respect to one variable. If the function depends on one variable, these derivatives are the same. The definition is very simple ( ∂f (x, y) ) ∂x

f (x + dx, y) − f (x, y) , dx→0 dx

= lim y

( ∂f (x, y) ) ∂y

= lim x

dy→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) = 3 x y f (x, y) =

x , y

( ∂f (x, y) )

= 1,

∂x y ( ∂f (x, y) ) = 3y, ∂x y ( ∂f (x, y) ) 1 = , ∂x y y

( ∂f (x, y) )

=2 ∂y x ( ∂f (x, y) ) = 3x ∂y x ( ∂f (x, y) ) x =− 2 ∂y y x

(1.105)

in which the basic rules for differentiating powers, products and quotients (which we are about to list) apply to partial 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.

1.4. APPENDIX. DIFFERENTIAL CALCULUS

17

60

40

20 o 0

-20

-40

-60 -4

-3

-2

-1

0

1

2

3

4

Consider for example the curve y = ax2 + bx + c. Suppose that you want to find the slope of the line tangent to the curve at some point x0 . You can draw three types of straight lines relative to this curve; ones that don’t intersect the curve, ones that intersect it at one point, and lines that intersect the curve at two points. “Generic” lines will intersect the curve at two points, but only vertical lines (infinite slope) and tangent lines intersect at one point. This is 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 other words m = 2au + b which is exactly the derivative of our curve at the point (u, v) d ( 2 ax + bx + c = 2au + b dx x=u Example. Find the slope of the line tangent to a circle x2 + y 2 = R2 at the point (x0 , y0 ) on the circle.

18

CHAPTER 1. KINEMATICS

Once again you see three types of lines, those that intersect the circle at 0, 1, 2 points. The lines through a single point of the circle are the tangent (or the vertical, which we don’t care about) lines. Play the same game; y 2 + x2 − R2 = 0,

(mt + v)2 + (u + t)2 − R2 = 0,

(m2 t2 + 2mvt + t2 + 2ut) + (v 2 + 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 = −

u v

Verify this with calculus; y 2 = R2 − x2 ,

dy dy 2 = 2y = 0 − 2x, dx dx

dy x u =− =− dx x=u,y=v y x=u,y=v v

These methods work (the ideas are from the field of algebraic geometry) because the derivative has a geometrical meaning; it is the slope of the line tangent to the curve, the slope of the line through two points that are taken to be 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 = u1 + 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 + mt) − 1 = 0, u

mt2 + (mu +

a quadratic equation with a double-rrot t t = 0 if m = − u12 , the correct derivative

1.4.3

1 )t = 0 u

dy dx

= − x12 at x = u.

The rules of differentiation

Derivatives of a product f (x) g(x) can be easily computed from the definition, ) d ( f (x + dx)g(x + dx) − f (x)g(x) f (x)g(x) = lim dx→0 dx dx by replacing f (x + dx) with f (x + dx) − f (x) + f (x) and rearranging ( ) ) f (x + dx) − f (x) + f (x) g(x + dx) − f (x)g(x) d ( f (x)g(x) = lim dx→0 dx dx ( ) ( ) f (x + dx) − f (x) g(x + dx) + f (x) g(x + dx) − g(x) = lim dx→0 dx ( f (x + dx) − f (x) ) ( g(x + dx) − g(x) ) = lim g(x + dx) + f (x) lim dx→0 dx→0 dx dx dg(x) df (x) g(x) + f (x) = dx 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. ) df (x) d ( dg(x) f (x)g(x) = g(x) + f (x) dx dx dx

1.4.4

(1.107)

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 ( ) ∑ N m N −m a b m m=0

(1.108)

1.4. APPENDIX. DIFFERENTIAL CALCULUS

19

( ) N! N = m m!(N − m)!

where the number

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

1.4.5

(a + b)2 (a + b)3

= a2 + 2ab + b2 = a3 + 3a2 b + 3ab2 + b3

(a + b)4

= a4 + 4a3 b + 6a2 b2 + 4ab3 + b3

(1.109)

Series expansion

This last calculation was a little on the tricky side, but there exists a powerful tool for performing most of the operations 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 be small, so that x is close to x0 . The idea of the series expansion is that in the neighborhood of x0 , we could replace f (x) with a polynomial N ∑ an Pf (x) = (x − x0 )n (1.110) n! n=0 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 on the following concept: the function f (x) and polynomial Pf (x) agree at x0 , and have the same derivative at x0 , and the same second derivative, and so on up to the N th derivative. We would call Pf (x) an N th order series 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 ) +

a2 aN (x0 − x0 )2 + ... + (x0 − x0 )N = a0 2! N!

requires that a0 = f (x0 ). d Step 2. Both f ′ (x) and dx Pf (x) agree at x0 ; f ′ (x0 ) = 0 + a1 + 2

a2 a3 aN (x0 − x0 ) + 3 (x0 − x0 )2 + ... + N (x0 − x0 )N −1 = a1 2! 3! N!

df requires that a1 = f ′ (x0 ) = dx . x=x0 d ′′ Step 3. Both f (x) and dx Pf (x) agree at x0 ;

f ′′ (x0 ) = 0 + 0 + 2 requires that a2 = f ′′ (x0 ) =

a2 a3 aN + 3 · 2 (x0 − x0 ) + ... + N · (N − 1) (x0 − x0 )N −2 = a2 2! 3! N!

d2 f dx2 x=x0 .

For ninety percent of all of the calculus applications in our physics text, this is enough; the polynomial Pf (x) that agrees with f (x) up to two derivatives in the neighborhood of x0 is 1 Pf (x) = f (x0 ) + f ′ (x0 ) (x − x0 ) + f ′′ (x0 ) (x − x0 )2 + ... 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 .

20

CHAPTER 1. KINEMATICS

What do we use it for? For starters it can be used to get formulas for derivatives of products and quotients. In most applications you only need to keep one or two terms in a Taylor series. For example, let x = t + dt and x0 = t, then 1 Pf (t + dt) = f (t) + f ′ (t) dt + f ′′ (t) (dt)2 + ... (1.112) 2 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( f (t)g(t) = dt

( lim

dt→0

f (t + dt)g(t + dt) − f (t)g(t) dt

(

′

)(

f (t) + f (t) dt + ... =

lim

)

) g(t) + g ′ (t) dt + ... − f (t)g(t)

dt g(t) f ′ (t) dt + g ′ (t) f (t) dt + ... = lim = g(t) f ′ (t) + g ′ (t) f (t) dt→0 dt dt→0

(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 limit become zero. Example 14. l’Hospitals rule is a formula for computing the limit of the ratio of two functions that both vanish at x0 , lim f (x) = 0 = lim g(x) x→x0

x→x0

The limit of the ratio is then lim

x→x0

f (x0 ) + f ′ (x0 ) (x − x0 ) + 12 f ′′ (x0 )(x − x0 )2 + ... f (x) = lim g(x) x→x0 g(x0 ) + g ′ (x0 ) (x − x0 ) + 12 g ′′ (x0 )(x − x0 )2 + ...

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

x→x0

f ′ (x0 ) (x − x0 ) + 12 f ′′ (x0 )(x − x0 )2 + ... f (x) = lim ′ g(x) x→x0 g (x0 ) (x − x0 ) + 12 g ′′ (x0 )(x − x0 )2 + ...

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

x→x0

f ′ (x0 ) + 21 f ′′ (x0 )(x − x0 ) + ... g ′ (x0 ) + 12 g ′′ (x0 )(x − x0 ) + ...

and in the limit x → x0 , (x − x0 ) → 0, lim

x→x0

f ′ (x0 ) + 21 f ′′ (x0 )(x − x0 ) + ... f ′ (x0 ) = g ′ (x0 ) g ′ (x0 ) + 12 g ′′ (x0 )(x − x0 ) + ...

We restate this as l’Hospitals rule; lim

x→x0

f (x) f ′ (x0 ) = ′ g(x) 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 noting that in f ′ (x0 ) + 12 f ′′ (x0 )(x − x0 ) + ... lim ′ x→x0 g (x0 ) + 1 g ′′ (x0 )(x − x0 ) + ... 2 the first term in both numerator and denominator are zero and we can divide out another factor of (x − x0 ); lim

x→x0

0 + 12 f ′′ (x0 ) + ... f ′′ (x0 ) f (x) = ′′ = lim , 1 ′′ g(x) x→x0 0 + 2 g (x0 ) + ... g (x0 )

if

f (x0 ) = g(x0 ) = f ′ (x0 ) = g ′ (x0 )

(1.115)

1.4. APPENDIX. DIFFERENTIAL CALCULUS

1.4.6

21

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 d ( h(x) g(x) = f (x) = f ′ (x), dx dx

h′ (x) g(x) + h(x) g ′ (x) = f ′ (x)

and rearrange h′ (x) =

(x) ′ f ′ (x) − fg(x) g (x) f ′ (x) g(x) − f (x) g ′ (x) f ′ (x) − h(x) g ′ (x) = = g(x) g(x) g 2 (x)

(1.116)

which we will call the quotient rule.

1.4.7

Radicals

Consider the radical f (x) =

√ n x

To compute its derivative, first raise both sides to the nth power f n (x) = f (x)f n−1 (x) = x Now differentiate and apply the product rule repeatedly to the left side d n d f (x) = x = 1, dx dx

n f n−1 (x)f ′ (x) = 1

solve for f ′ (x); f ′ (x) =

1 nf n−1 (x)

=

1 nx

n−1 n

=

1 1 −1 xn n

We have shown that

d √ d 1 1 1 n x= x n = x n −1 dx dx n and therefore for any power a, integral, rational or otherwise d a x = a xa−1 dx 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 √ 1 1 x= √ , dx 2 x

d2 √ d 1 1 1 1 √ =− 2√ x= 2 dx dx 2 x 2 x3

and so inserting this all into Eq. 5 we find that √ √ 1( 1 1 ) dx (dx)2 1 − 2√ (dx)2 + ... = 2 + − + ... 4 + dx = 4 + √ dx + 2 2 4 64 2 4 43

(1.117)

22

CHAPTER 1. KINEMATICS

This should be written using x = 4 + dx, as √ (x − 4) (x − 4)2 x=2+ − + ... 4 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 √ 1 1 + ... = 2.234375... 5≈2+ − 4 64 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 of differentiation, and so we define ∫ b df (x) dx = f (b) − f (a) (1.118) dx a A Riemann sum will do the trick; consider ∫

b

f (x) dx = lim a

N →∞

N ∑ n=0

f (a +

(b − a) (b − a) n) N N

(1.119)

which is the area under the curve f (x) from a to of b, being made up of little strips of width (b−a) N ). 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

f (x) dx = lim

N →∞

a

N ∑

f (a + n∆) ∆

(1.121)

n=0

and insert into this a differentiated function f (x) =

dg(x) g(x + ∆) − g(x) = lim ∆→0 dx ∆ a

∫

b

( ) f (x) dx = lim ∆ f (a) + f (a + ∆) + ... + f (a + (N − 1)∆) + f (a + N ∆) N →∞

a

( g(a + ∆) − g(a) g(a + 2∆) − g(a + ∆) lim ∆ + + ... N →∞ ∆ ∆ g(a + (N − 1)∆) − g(a + (N − 2)∆) + ∆ g(a + N ∆) − g(a + (N − 1)∆) ) + ∆

b

(1.122)

=

(1.123)

1.4. APPENDIX. DIFFERENTIAL CALCULUS

23

You can see that consecutive terms partially cancel, and so does the ∆ in the denominator, leaving ( ) ( ) (b − a) = lim g(a + N ∆) − g(a) = lim g(a + N ) − g(a) = g(b) − g(a) N →∞ N →∞ N

(1.124)

Integration is a harder problem than differentiation, since the only procedure for performing integrals is to either do the Riemann sum directly, or find a function whose derivative is the integrand, or to perform variable changes that put the integrand into a more readily recognized form. Example 16. It is not hard to show that N ∑

n=

n=1

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)

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

(1.128)

and add the two rows

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

N ∑

b

x dx = lim

N →∞

a

with ∆ =

b−a N .

( (a + n∆) ∆ = lim

N →∞

n=0

(N + 1)a∆ + ∆2

N ) ∑ n

(1.129)

n=0

Put this in; ∫

(

b

x dx = a

=

( b − a )2 ) a(b − a) 1 + N (N + 1) N →∞ N 2 N 1 1 1 a(b − a) + (b − a)2 = b2 − a2 2 2 2 lim

(N + 1)

By considering more difficult Riemann sums we can establish ∫ b ) 1 ( n+1 xn dx = b − an+1 , n+1 a

n ̸= −1

(1.130)

(1.131)

This formula is valid for any n ̸= −1, even irrational values. Like differentiation, integration is a linear operation ∫ b ∫ (f (x) + g(x)) dx = a

∫

b

f (x) dx + a

b

g(x) dx

and another very useful property inherited from the Riemann sum definition is ∫ c ∫ b ∫ c f (x) dx = f (x) dx + f (x) dx a

a

(1.132)

a

(1.133)

b

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 ∫ t dx v(t) = , x(t) = x(t0 ) + v(t) dt (1.134) dt t0

24

1.4.9

CHAPTER 1. KINEMATICS

The logarithm and exponential

In Eq. 1.131 the case n = −1 is special. Consider the integral ∫ x ′ dx g(x) = x′ 1 since for such a function

∫

ab

1

dx = x

∫

a

1

dx + x

(1.135)

∫

ab

a

dx x

(1.136)

and the variable change x = ay turns this into

∫

ab

dx = x

1

∫

a 1

dx + x

∫

b 1

dy , y

(1.137)

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 inverse exponentiation, we need only to establish the base, which we will call “e”. In other words f (x) = ex and g(x) are inverses of one another, g(f (x)) = x = f (g(x)) (1.139) From our integral representation

∫

x

ln x = 1

we see that

dy y

(1.140)

1 d ln x = dx x

(1.141)

Apply the chain rule to Eq. 1.139 ) dx d ( g(f (x) = = 1, dx dx

1 df = 1, f (x) dx

df = f (x) dx

(1.142)

which is the law of differentiation of the antilog or exponential function. We can use our power series methods to determine e; f (x)

= =

e = f (1) =

∞ ∑ 1 ( dn f ) xn n x=0 n! dx n=0

∞ ∞ ) ∑ ∑ 1 n 1( f (x) xn = x n! n! x=0 n=0 n=0 ∞ ∑ 1 1 1 n 1 = 1 + + + · · · = 2.71828 · · · n! 2! 3! n=0

To summarize we call g(x) = ln(x), f (x) = ex and we have ∫ x ′ dx , eln(x) = ln(ex ) = x, ln(x) = x′ 1

1.4.10

d x e = ex , dx

d ax e = a eax dx

(1.143)

(1.144)

Quadratic forms

Solving an equation such as ax2 + 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 c b x2 + x + = 0 = (x − r1 )(x − r2 ) a a

1.4. APPENDIX. DIFFERENTIAL CALCULUS

25

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 + r1 r2 b c = x2 − 2ux + (u2 − v 2 ) = x2 + x + a a and we find that −2u =

b b , u=− a 2a

and u2 − v 2 = and the roots are r1 =

−b +

b2 c − v2 = , 2 4a a

√ b2 − 4ac , 2a

v2 =

b2 − 4ac 4a2

−b −

√ b2 − 4ac 2a

r2 =

(1.145)

which is our familiar Quadratic Formula.

1.4.11

Complex numbers

2 It may √ happen that the2 discriminant b − 4ac of the quadratic form is negative. Denote the imaginary number i = −1. Then 4ac − b is real, and √ √ −b − i 4ac − b2 −b + i 4ac − b2 , r2 = r1 = 2a 2a

Let R and I be real numbers (R, I ∈ R). We call C = R + i I a complex number, with real part R = Re C and imaginary part I = Im C. 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 when one is willing to use complex numbers, especially calculus. Complex numbers enable one to perform practically any definite 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) is complex (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 parts is an important first step in working with f . Example f (x, y) = (x + iy)2 = (x2 − y 2 ) + i(2xy),

U = x2 − y 2 , V = 2xy

Example f (x, y) = (x + iy)3 = (x3 − 3xy 2 ) + i(3x2 y − y 3 ),

U = x3 − 3xy 2 V = 3x2 y − y 3

Example f (x, y) =

1 1 x − iy x y = = 2 −i 2 x + iy x + iy x − iy x + y2 x + y2

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

26

1.4.12

CHAPTER 1. KINEMATICS

Calculus assistance

You should be able to deduce that we will be using calculus extensively, physics 201 is a calculus-based course. If you are rusty, one option is to use the computer to re-build your math skills on an as-needed basis. The physics department research support servers are running REDUCE, axiom and maxima as CGI programs, and of course these symbolic math programs are free software and can be downloaded from your Internets. Both programs are excellent, 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 competition with 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/reducealgebra/) or maxima (http://maxima.sourceforge.net/), or open-axiom http://www.open-axiom.org/, and install them on your own computer. If you don’t want to do this, you can run all four from http://azazelo.uwp.edu/cgi-bin/gnuplot.cgi http://azazelo.uwp.edu/cgi-bin/reduceweb.cgi http://azazelo.uwp.edu/cgi-bin/maxima.cgi http://azazelo.uwp.edu/cgi-bin/axiom.cgi Here is how to perform basic calculus operations in both CAS systems;

Define a function REDUCE

maxima

axiom

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

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

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

Differentiate a function REDUCE

maxima

axiom

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

f: 1/(a*a+x*x); diff(f,x); diff(f,x,2);

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

Partial derivatives REDUCE

maxima

axiom

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

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

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

maxima

axiom

load_package taylor; f:=1/(x^2+a^2); taylor(f,x,2,4);

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

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

1.4. APPENDIX. DIFFERENTIAL CALCULUS

27

Solve a system of linear equations REDUCE

maxima

axiom

solve({R1*I1+R3*I3-V0, R1*I1-R2*I2, I1+I2-I3},{I1,I2,I3});

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]);

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

Perform indefinite integral REDUCE

maxima

axiom

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

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

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

Perform definite integral∗ axiom

REDUCE

maxima

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

integrate(1/(a*a+x*x),x,0,inf); f:=integrate(1/(x^2+1),x= -1 .. 1) nonzero; print f positive; f:=integrate(1/(x^2+1), x= %minusInfinity .. %plusInfinity) print f

Extremize a function REDUCE

maxima

axiom

depend f,x; f:=(q1*k/(x-a)-q2*k/x); solve(df(f,x),x);

f: q1/(x-1)-q2/x; solve(diff(f,x),x);

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 mode you 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 will not be able to use any computer aids on exams, and everything in the sciences will be easier for you if your math is strong, a fact well-understood by those people that you will soon be competing with for jobs.

28

1.5

CHAPTER 1. KINEMATICS

Problems

1. Suppose that you are planning a trip (starting at x = 0 at t = 0) in which you will accelerate for time t1 at acceleration a, cruise at the constant speed that you will have achieved at the end of your acceleration, for a time t2 , and finally you will brake to rest with acceleration −a for time t1 . Draw a picture of x(t) versus t, using a = 1.0 sm2 , t1 = 25 s, and t2 = 100 s. 2. Compute the maximal speed achieved during the trip of the previous problem, and compute the total distance ¯ for the entire trip. Make a graph of the instantaneous velocity v(t) versus traveled. Compute the average velocity v ¯? t 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 acceleration for t4 . That makes her trip-time 2t3 + t4 . Find the acceleration time t3 that makes the total trip time minimal. For how long does this driver move at constant speed? −5 gal·s 4. Suppose that a car consumes gasoline at a rate dG dt = 1.4 × 10 m |v|, where |v| is the speed of your car. This is pretty much how things would work if only mechanical friction was responsible for consumption. Compute the amount of fuel used in traveling for a time 1 hr at a speed of 20 m s . Compute the amount of fuel used in accelerating from rest with a = 2.0 sm2 until you reach 20 m . s

5. An object moves at constant acceleration 1 x(t) = x0 + v0 t + at2 2 Find the average velocity v¯ between the times t1 and t2 , with t2 > t1 . Find the time t∗ between t1 and t2 such that the instantaneous velocity at t∗ equals the average velocity between t1 and t2 . This is one reason why such motion is special. 6. An object moves at constant “jerk” 1 1 x(t) = x0 + v0 t + at2 + Jt3 2 6 Find the average velocity v¯ between the times t1 and t2 , with t2 > t1 . Find the time t∗ between t1 and t2 such that the 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), the velocity 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,

dx dt = , v

∫

∫

t

x

dt = t − t0 = t0

x0

dx v

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

x

t − t0 = x0

dx √ =2 2a(x − x0 )

√

√ x − x0 x 2(x − x0 ) = 2a x0 a

Now invert by squaring and solving for x; (t − t0 )2 =

2(x − x0 ) , a

x − x0 =

1 a(t − t0 )2 2

and note that x = x0 when t = t0 . Suppose that at t = 0 we have x = x0 , and v(x) = motion in which the acceleration is not constant.

√ x20 − x2 . Find x(t). This is an extremely important type of

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. ( ) d d Compute dt a + b t at time t = 3. Compute 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, and ∂y ∂x so forth. In other words ∂x ∂y = ∂z = ∂z = 0. For example x z) 1 z ∂ ( 2 x z+y+ + = 2xz + 0 + − 2 ∂x y x y x Compute

∂ ∂x

( y+

1 x2 +y 2

) .

Compute

∂ ∂y

(

√

1

) Compute

x2 +y 2

∂ ∂z

(

e−xy + x e−zy

)

z

10. A rectangular box of dimensions 3.0, 4.0, 5.0 m oriented as illustrated has four corners labeled as vectors a, b, c, d. A. Find the x, y and z components of all four vectors, list them using unit vector notation.

a c

4

B. C. D. E.

3 b

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

5 x

y

d

1 y

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 m ′ decreasing in length at rate dh dt = −3 s and the string of length h ′ m is decreasing in length at rate dh dt = −2 s . At this instant find the velocity vector of the mass.

h

x h’ 1

30

CHAPTER 1. KINEMATICS

h

12. Imagine a boat on the water being pulled towards the shore by a winch mounted a height h above the water. The winch reels in cable (shortening the hypotenuse) at constant rate u in m s . When the boat is a distance d from the shore, find its speed and acceleration. This is an example of motion at non-uniform acceleration.

d 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 and acceleration vectors. Show that dr = ωk × r dt by explicitly calculating both sides. Show that ( ) d2 r = ωk × ωk × r dt2 by explicitly calculating both sides.

15. An object has position vector r = R cos(ωt + 21 αt2 ) i + R sin(ωt + 12 αt2 ) j, in which R, ω are constants. Find its velocity and acceleration vectors. Show that ) dr ( = ω + αt k × r dt by explicitly calculating both sides. ( ) d2 r = ωk × ωk × r + αk × r 2 dt 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) = dicular to a(t).

da(t) dt

is perpen-

mi mi 20. Suppose that you travel along a straight line for 2.25 hr at 30 mi hr , 3.5 hr at 45 hr , and finally 1.25 hr at 60 hr . Find your displacement and average velocity. mi mi 21. Suppose that you travel along a straight line for 2.25 hr at 30 mi hr , 3.5 hr at −45 hr , and finally 1.25 hr at 60 hr . Find your displacement and average velocity.

22. Suppose that you travel along a straight line (West) for 2.0 hr a (directed) distance of 60mi, 3.5 hr a (directed) distance 210mi, and finally 2.5 hr 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) m s i from t = 0 to t = 10 seconds. Find your average velocity and your average speed. 26. Show from Eq. 1.1 that the motion  0≤t≤1  1.0 + 2.0 t 3 + 4(t − 1) 1 ≤ t ≤ 2 x(t) =  7 + 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  0≤t≤1  1.0 + 2.0 t 4 + 4(t − 1) 1 ≤ t ≤ 2 x(t) =  8 + 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. See problem 1.35 for assistance. 28. Compute

d √ d ( 1 ) d √ 2 3 x + 1, x2 + x, , dx dx dx x3 + x and check your results with REDUCE, or axiom or maxima.

d ( 1 − x2 ) dx x3 + x

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

x30 − x−30 x→1 x − x−1 and check your results with REDUCE, or axiom or maxima lim

%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 20 20 ( d ∑ ) ∑ xn = n = 210 dx n=0 x=1 n=0

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

d g(x) = −f (x) dx

32

CHAPTER 1. KINEMATICS

for any point x. Prove that f 2 (x) + g 2 (x) is a constant for any x Can you think of two functions for which this is true? 33. Find a series expansion for 1

f (x) = x 3 valid near x = 1 that agrees with f (x) up through the third derivative at x = 1, and check your results with REDUCE, or axiom or maxima. √ 34. Solve the equation v(t) = dx a2 − x2 using REDUCE, and determine the value of the constant of integration dt = 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, or download your own copy. To make a graph of continuous functions is pretty easy in gnuplot, but gnuplot wants to use x as the independent variable, whereas we want t to be the independent variable, so you must pretend that x is t. Make a graph of a continuous motion such as motion at constant acceleration, plotting position, velocity and acceleration all together 8 y0+v0*x-0.5*a*x*x v0+a*x a

6

y0=1.0 v0=2.0 a=2.0 plot[0:3] y0+v0*x-0.5*a*x*x, v0+a*x,a

4

2

0

-2 0

0.5

1

1.5

2

2.5

That was fun. In order to graph a piece-wise continuous function such as that on page 1, we use gnuplot’s ability to 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;

3

1.5. PROBLEMS

33 4 f1(x)

3

2

f1(x)=(x2) ? 7+(x-2) : f2(x) plot[0:3] f3(x)

5

4

3

2

1 0

0.5

1

1.5

2

2.5

Here is another example; that of problem 1.1

  1.0 t2 625 + 50 ∗ (t − 25) x(t) =  4375 + 50 ∗ (t − 100) − (t − 100)2

0 ≤ t ≤ 25 25 ≤ t ≤ 100 100 ≤ t ≤ 125

(1.148)

3

34

CHAPTER 1. KINEMATICS 5000 f3(x)

4000

3000 f1(x)=(x25) ? 625+50*(x-25) : f1(x) f3(x)=(x>100) ? 4375+50*(x-100)-(x-100)**2 : f2(x) 2000 plot[0:125] f3(x)

1000

0 0

20

40

60

80

100

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 and travels west at constant velocity v. A bird flies from train to train at constant speed 2v, starting when train #1 is at the 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 same time on a trip to city A at a constant velocity as well. They pass each other at 12:00, and traveler #1 arrives at B at 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 not incalculable 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) into http://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

120

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 to your account, your yearly contribution. Suppose that each time period the account gains a certain fraction C of its contents 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 + f1 ξ n 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 much money do you have? What if you double your yearly contribution, how does this affect your 25-year savings? What if 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 y 2 = x3 + ax + b are used very extensively in cryptography, because it is possible to associate a group with the curve, in the sense that a generic line will intersect a cubic in three points, call them p, 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 MATH 367, 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.

36

CHAPTER 1. KINEMATICS

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 solving equations of motion.

y We now study a series of examples that illustrate the variety of projectile problems and the methods for extracting data from the equations that describe the motion of projectiles in two dimensions. Every problem begins the same way; we write a generic vector version of the position versus time for a body undergoing motion at constant acceleration;

ymax r(t)

=

(x(t), y(t)

1 = r0 + v0 t + a t2 2 =

(vx0, vy0)

=

θ (x0, y0)

x

1 (x0 , y0 ) + (v0x , v0y ) t + (ax , ay ) t2 2 1 1 (x0 + v0x t + ax t2 , y0 + v0y t + ay t2 ) 2 2 (2.1)

xmax=R

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 = 9.8

m s2

(2.2)

The next set-up step is to specify the four constants x0 , y0 , v0x , v0y for each particular problem. We finish each problem 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, range and maximum height attained. We first get the vector v0 = (v0x , v0y ) = (|v0 | cos θ, |v0 | sin θ) 37

(2.3)

38

CHAPTER 2. PROJECTILE MOTION

obtaining the two components in terms of the launch angle and speed. This results in 1 y = v0 sin θt − gt2 2 vy = v0 sin θ − gt

x = v0 cos θt, vx = v0 cos θ,

(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 cos2 θ

(2.6)

2v02

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 ground at these times 2 sin θ cos θv02 x=0 and x= (2.7) g the second of which can be rewritten as R=

v02 sin 2θ g

(2.8)

Notice that the range is maximal when θ = π4 rad, and that the two angles range. By launching the projectile at 45o ( π4 rad) a maximum range of Rmax =

π 4

+ θ and

π 4

− θ both result in the same

v02 g

(2.9)

is attained. Actually with air resistance taken into account, the range of a projectile in actuality is much less than this. 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 =

v02 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 tf light =

2v0 sin θ g

Example 2. Find the speed of the projectile, and the inclination of the velocity to the horizontal. The speed is very simple; √ √ s = |v| = vx2 + vy2 = v02 + g 2 t2 − 2gtv0 sin θ

(2.11)

(2.12)

and calling the velocity at any time v = (vx , vy ) = (s cos ϕ, s sin ϕ) we obtain tan ϕ =

g vy = tan θ − t vx v0 cos θ

(2.13)

which tells us which way the projectile is headed; down if this is negative, up if positive, so we can tell from this quantity whether or not the projectile is ascending or descending. Note that from Eqs. 2.6, 2.13 this is the trajectory slope dy = tan ϕ (2.14) dx (just eliminate t using Eq. 2.4).

2.2. USING THE COMPUTER

39 Example 3. Suppose that a projectile is launched horizontally from a height H at speed v0 . What must be the value of v0 for it to clear an obstacle of height h placed a distance d away?

y

1 y = H − gt2 2

x = v0 t,

(2.15)

are the equations describing the flight. The trajectory is y(x) = H −

(vx0, vy0) (x0, y0)

(2.16)

We want the point (x, y) = (d, h) to be on the trajectory and so h=H−

x

gd2 2v02

(2.17)

giving us a speed √

xmax=R

2.2

gx2 2v02

v0 =

gd2 2(H − h)

(2.18)

Using the computer1

We can use the computer to graph the motion of a projectile by either writing a program to integrate the acceleration equations, using df (x) f (x + dx) − f (x) = = g(x) (2.19) dx dx or df f (x + dx) = f (x) + dx ∗ g(x) = f (x) + dx ∗ (2.20) dx For example if dy vy (t) = = v0y − gt (2.21) dt then y(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 you provide together with initial data. I will describe to you can use ode lastly in this section. You can either use my web 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 #include /* Declare any double precision variables */ 1 This

section is optional

40

CHAPTER 2. PROJECTILE MOTION

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; printf("%f\t%f\n", x,y); } while(y >= 0.0); /* end of program */ }

Output of prog. 1

In order to use the program to draw the graph, we need to first compile the code into an executable file. This requires that you have a C-compiler installed in your computer. If you are a Linux user, you already have one. If you are a Windows user, you do not. Windows users can get a free C/C++/FORTRAN compiler from http://www.mingw.org/download.shtml. You will need two files, download both MSYS1.0.10.exe, and MinGW-3.1.0-1.exe from the web-site, or copy them from the CD-ROM put on reserve in the library for your use. Install MinGW first by double-clicking on it, and then install MSYS. Afterwards you will get an icon that starts up a shell (like the DOS window).

25 20

y, (m)

15 10 5 0 −5

0

20

40

60

80

100

You can open the file in Excel and plot the data, or download the free function plotting software gnuplot and open the file in gnuplot, which will graph it for you. You can print the output, which will look like what you see above.

x, (m) 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 θ but the same v0 . We alter the program so that it accepts parameters;

2.2. USING THE COMPUTER /* prog2.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 #include #include #define PI 3.1415926 /* Declare any double precision variables */ double v0x, v0y, x, y, t, g, dt, v0, theta; /* Declare integer variables used in looping */ 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]); /* Set values for variables that do not change */ v0=20.0, g=9.8, dt=0.01; /* set initial velocity */ v0x=v0*cos(theta), v0y=v0*sin(theta); /* 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; printf("%f\t%f\n", x,y); } while(y >= 0.0); /* end of program */ }

41

42

CHAPTER 2. PROJECTILE MOTION

θ=20, 30, 45, 60, 70o 20

15

and we compile the program to executable prog2.exe and run it by typing things like ./prog2.exe 20 > D20, ./prog2.exe 30 > D30, ./prog2.exe 45 > D45 to create the data files D20, D30, D45, ... which we then plot together.

y (m)

10

5

0

−5

0

10

20

30

40

50

x (m) 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 √ dvy = −g − b vx2 + vy2 vy , dt

√ dvx = −b vx2 + vy2 vx dt

(2.23)

and solving these could be very hard, but the computer can do it numerically with no big changes over our old program. /* prog3.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 #include #include #define PI 3.1415926 /* Declare any double precision variables */ double vx, vy, x, y, t, g, dt, v0, theta, b, rad; /* Declare integer variables used in looping */ int n; main(int argc, char*argv[]){ /* get the angle theta */ if(argc != 3){ printf("./prog2.exe theta(deg) b\n"); exit(1);

2.2. USING THE COMPUTER

43

} theta=(PI/180.0)*atof(argv[1]); b=atof(argv[2]); /* Set values for variables that do not change */ v0=28.4, g=9.8, dt=0.01; /* set initial velocity */ vx=v0*cos(theta), vy=v0*sin(theta); /* 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; 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; printf("%f\t%f\n", x,y); } while(y >= 0.0); /* end of program */ }

b=0.0, 0.01, 0.02, 0.03, 0.04

The results are plotted below for various values of the wind resistance constant b. You can see that the trajectory falls off very fast after reaching the top of the flight, and even mild wind resistance has a dramatic effect on the range. These are for initial launch angle of 45 degrees and speed 28.4 m s .

25 20

y (m)

15 10 5 0 −5

0

20

40

60 x (m)

80

100

Use of the computer to solve problems is strictly optional, but understand that many seemingly innocuous problems in physics may be difficult to solve analytically. Practically any problem can be solved or simulated numerically, and although a numerical solution to a problem is technically no solution at all, it may help you to visualize the behavior of the system under study. You are encouraged to use this tool if you are interested.

44

2.2.1

CHAPTER 2. PROJECTILE MOTION

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 the examples, and try it out. You do not need to know any programming, all you do is have list (define) your constants, dvy ′ your equations of motion (in the required syntax y ′ = dy dt = vy , vy = dt ), 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 d2 x = x′′ = F (t), dt2 (if you eliminate v, you are back to x′′ = F (t)).

−→

a=

x′ = v,

v ′ = 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 too complicated to solve by hand. Don’t be surprised if you get unexpectedly straight-line graphs coming out of it if your equations are so simple that you could solve them by hand! Here are some examples

y(x) for v0=40√2, b=0.1, θ=45 12 10 8

y (m)

# 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 0, but has coefficient of friction µk and µs for x ≤ 0. , at speed v0 . At the origin it encounters a spring of force constant k (and friction). How far will the block compress the spring before it is stopped by the spring and friction combination?

x=0 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 overcome the static friction that engages when the block stops. The block will come to a permanent rest. Compute this v0 .

128

CHAPTER 7. CONSERVATION OF MOMENTUM

M1

M2

5. A block of mass M1 = 0.5kg and a block of mass M2 = 0.8kg compress a spring of force N constant k = 100 m by amount x = 0.1 m. When released and disengaged from the spring the blocks slide without friction at v1 and v2 . Find both v1 and v2 .

v1

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 momentum and kinetic energy at time t = 4 s. B. A force F = 30 Ns ti + 10 sN2 t2 j acts on a mass 1.0 kg that is at the origin, at rest at t = 0. Find its momentum and kinetic energy at time t = 4 s. J C. A conservative force whose potential is V = −3x m acts on a mass 1.0 kg that is at (1, 0) (both in meters), at rest at t = 0. Find its position and velocity at time t = 2 s

7. Lets explore how physics influences the behavior of the various types of balls used in athletics. Ignore gravity in this problem. Suppose that the ball is impossible to deform (like a steel ball) and it hits a similarly hard surface. The surface simply reverses its normal component of momentum. The incoming ball has momentum (mv0 sin θ, −mv0 cos θ) and outgoing has momentum (mv0 sin θ, mv0 cos θ), since there is no mechanism for taking away kinetic energy and turning it into some other form. The ball and surface are in contact for an instant. Is momentum conserved? Explain. Suppose the ball can deform, so that ball and surface are in contact for time T . Ignore any spinning of the ball. The ball slips on the surface against friction (coefficient µk ) for time T . Which way does this force of friction on the ball point? Show that the y-component of the balls momentum is reversed by the bounce.

v0

v0

θ θ

v1

v0

θ φ

o Show that friction reduces the balls x-component of momentum. Let v0 = 20 m s , θ = 45 , 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 of energy 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 entire collision time T . If the ball has mass m = 1.0 kg, compute the magnitude of the normal force. Note that dp F = ma = , dt

∫ ∆p =

F dt

implies that

pf − pi ¯ =F tf − ti

7.3. PROBLEMS

129

9. Re-consider the previous two problems, but incorporate the affects of top-spin. We will neglect the rotational sin θ energy of the ball for now. Suppose that the ball is spinning clock-wise (top-spin) at rate ω = v0 R (its radius is m R) throughout the entire collision process. Explain why you would get v1 = v0 = 20 s , and ϕ = θ = 45o . sin θ Suppose that the ball is spinning clock-wise (top-spin) at rate ω = 2 v0 R (its radius is R) throughout the entire collision process. Recalculate both v1 and ϕ, and show that now v1 > v0 and tan ϕ = tan θ + 2µk .

It appears that friction is giving the ball kinetic energy, contrary to what we know friction should do (it robs you of mechanical energy, and turns it into heat). We will see in a few weeks that friction is robbing you of energy in this problem, reducing the rotational kinetic energy and slightly increasing the ordinary kinetic energy.

y

v1

10. Consider a triple collision between a billiard ball and two identical balls in contact. This cue-ball initial speed is v0 = 5.0 m s . Find the speed w of its recoil, the angle θ that the two target balls move off at, and their speeds v1 after collision. Neglect rotation of the balls, they don’t rotate.

m θ θ

w

x

v0

v2

m1

m3

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 = 5 m gives a “come hither” glance towards the amorous (and tasty) m2 = 50 kg male at the other end. They both waddle towards the center of the iceberg, and its all over in 60 seconds. How far x does the dry-iceberg move?

130

CHAPTER 7. CONSERVATION OF MOMENTUM

M1

v1

M2

M2

θ

θ

M1

12. A box of mass M1 begins at the top of a frictionless ramp of height h which rests on a frictionless surface. The box 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 momentum still conserved?

M1

v2 v1

M2 θ

M2 M1

θ

13. A box of mass M1 speeds towards a frictionless ramp which rests on a frictionless surface. The box slides to the top 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 momentum still conserved?

M1

v1

v2

M3

M2

14. The theory of Relativity shows that energy can be transformed into matter (mass), and vice-versa. This is the mechanism by which new forms of matter (elementary particles) are created in nature. The transformation law is E = mc2 , amount of energy E can be transformed into mass m. Consider this center of momentum collision in which the new particle of mass M3 is created at rest by the collision between M1 and M2 . Compute M3 by conserving total energy and momentum.

15. Find the center of mass (COM) position of each of these mass arrangements relative to the given xy-axes.

7.3. PROBLEMS

131

2.00 kg

1.00 kg 1.5 kg, 1.5 m

1.5 kg, 1.5 m

1.0 m

3.00 kg

6.00 kg

4.00 kg

1.0 kg, 1.0 m

5.00 kg

1.0 kg, radius 0.05 m

1.0 m

5.0 kg

2.0 m

m, v

16. Find the center of mass of a solid hemisphere of base-radius R.

2m

17∗ . Find the center of mass of a right-circular cone of base-radius R and height h.

φ φ

18. A particle of mass m and velocity v = v i collides with a stationary mass 2m as in the upper figure. After the collision the mass m is found to be stationary, and the target mass has split into two equal portions m moving off at identical speeds w at identical angles ϕ. Prove that v w> 2 Is the collision elastic? If not, find the energy loss (as a function of m, v and ϕ).

132

CHAPTER 7. CONSERVATION OF MOMENTUM

m, v

2m

19. A particle of mass m and velocity v = v i collides with a stationary mass 2m as in the upper figure. After the collision the mass m is found to be moving at u = u i, and the target mass has split into two equal portions m moving off at identical speeds w at identical angles ϕ. The collision is elastic. Prove that if ϕ > 45o the projectile mass ends up moving leftwards after the collision.

w φ φ

20. In the previous problem prove that (

u

w = 2v

) cos ϕ 1 + 2 cos2 ϕ

w 3 f3(x)

21∗ . A particle of mass m = 1.0 kg at rest at t = 0 is subjected to this force. Find its momentum at t = 1, 2, 3 s.

2.5

2

F(t), N

1.5

Find the acceleration of the particle for any time t in the interval 0 ≤ t ≤ 1 s.

1

The power ℘ delivered by a force is the rate with which it does work ℘ = dE dt = F · v. Compute the power delivered to the particle for any time t in the interval 0 ≤ t ≤ 1 s.

0.5

0

-0.5

-1 0

0.5

1

1.5

2

2.5

3

t (s)

22. A big tank of water is pushed from rest by a constant force F = F0 along a frictionless track. Water escapes from 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 (m(t) v(t)) dt

Show that at time t, the velocity of the tank is v(t) = − 23∗ . What force does a fire-hose spewing water at rate the water?

F0 α ln(1 − t) α m0 dm dt

m = 10 kg s at nozzle-speed 20 s exert on the fireman directing

24. Find the final speed of a rocket that has initial mass m0 = 700, 000 kg (the first stage of a Russian UR-500 proton rocket) whose mass is 80% fuel mass, with exhaust thrust of 8.5 × 106 N (thats 1.9 million pounds) and a total 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 a hanging over the edge. You hold it in place, and let go at t = 0. Show that the velocity of the chain when length x is over the edge is √ g 2 v= (x − a2 ) ℓ 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 + N m. How fast is the railroad car moving after all cannon 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 cold butts, so the frog leaps at speed u relative to the block onto another nearby block of ice of mass m2 . Find the final speeds of both blocks. 30. Find the initial acceleration (at launch) of a rocket of total thrust 2 × 106 N , burn rate 2000 kg s , 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.

134

CHAPTER 7. CONSERVATION OF MOMENTUM

Chapter 8

Rotational motion y v(t)

a T(t) Up until now we have dealt exclusively with rectilinear motion, and all bodies have been structure-less. Of course nearly any but the simplest objects in nature have spatial dimension and extent. We now consider motion of an extended body about it’s center of mass, regarded as a pivot point. Consider a mass m tethered to the origin with a rigid, massless tie-rod. The axis of rotation is the positive z-axis k and the sense is counter clock-wise.

r(t) ac(t)

θ=θ(t) x

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

if the tie-rod has length R. Compute the velocity and acceleration of the object; ( ) dθ ( ) d dθ dθ v(t) = r(t) = − R sin θ i + R cos θ j = k × R cos θ(t) i + R sin θ(t) j dt dt dt dt (by the use of k × i = j, k × j = −i) or

(8.1)

(8.2)

dθ(t) k × r(t) dt d d2 θ(t) dθ d a(t) = v(t) = k × r(t) + k × r(t) dt dt2 dt dt ( dθ )2 ( ) d2 θ(t) = k × r(t) + k × k × r(t) (8.3) dt2 dt 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) v(t) =

so that

( dθ )2

d2 θ k × r(t) (8.5) dt dt2 This is the decomposition of the acceleration vector into it’s polar components. The first term is the centripetal acceleration, towards the center of the circle (normal to the circle), that we have studied previously ( dθ )2 an = ac = − r(t) (8.6) dt a = ac + at = −

135

r+

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 rotation k.

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 with tangential as well as radial (normal or centripetal) acceleration. The simplest such example is the pendulum.

The position of the pendulum “bob” or mass m is typically labeled by the angle θ that the string (of length ℓ) makes with the vertical.

θ ar

If the pivot-point is the origin, the coordinates of the bob are x = ℓ sin θ, y = −ℓ + ℓ cos θ (8.7)

at

Note that the bob has both radial and tangential acceleration = ℓθ˙ cos θ y˙ = −ℓθ˙ sin θ x ¨ = ℓθ¨ cos θ − ℓθ˙2 sin θ y¨ = −ℓθ¨ sin θ − ℓθ˙2 cos θ y¨ ˙ x − x¨ ˙y an = acent = √ = ℓ θ˙2 (8.8) 2 x˙ + y˙ 2 ) x¨ ˙ x + y˙ y¨ d (√ 2 at = √ = x˙ + y˙ 2 = ℓ θ¨ dt x˙ 2 + y˙ 2 x˙

I have drawn the total acceleration vectors a = an n + at t 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 acceleration be purely radial (normal) to your trajectory, and only when the pendulum stops will it be tangential. A falling body such as the pendulum bob is picking up speed as potential energy is converted into kinetic, and so only at 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 d2 θ =α dt2 integrate once; suppose at t = 0 the angular speed is ∫

t 0

dθ dt |t=0

d ( dθ ) dt = dt dt

(8.9)

= ω0 , then ∫

t

∫0 ω

dω dt dt dω = ω − ω0

= ω0 ∫ t

=

α dt = αt 0

so that

ω

=

dθ = ω0 + αt dt

(8.10)

137 now integrate again, using initial condition θ(t = 0) ∫ θ(t) dθ

= θ0 ∫ t = (ω0 + αt)dt

θ0

0

or

θ(t) =

1 θ0 + ω0 t + αt2 2

(8.11)

this is the most common case to encounter for a rotating body. The purely angular part of the acceleration of the mass m tethered to the origin in the first example becomes under these circumstances ( ) aT = α k × r(t) (8.12) or in magnitude aT = R α

(8.13)

analogous to vtangential = Rω, Eliminate t from ω = ω0 + αt

and

to get θ = θ0 + ω 0 (

vtangential = ωk × r(t) 1 θ = θ0 + ω0 t + αt2 2

ω − ω0 1 ω − ω0 2 ) + α( ) α 2 α

(8.14) (8.15)

(8.16)

and rearrange to 2α(θ − θ0 ) = ω 2 − ω02

(8.17)

2a(x − x0 ) = v 2 − v02

(8.18)

which is the rotational analog of We will not dwell much on angular kinematics, but move directly to dynamics and work both types of problems together. Lets refresh ourselves on the cross-product, which is a central ingredient in the description of rotations. Return to the case of a mass m traveling counterclockwise around a circle of radius R in the xy- plane r(t) = then

v(t) =

(R cos ωt, R sin ωt) d r(t) = (−Rω sin ωt, Rω cos ωt) dt

(8.19)

Define an axial angular velocity vector ωk with magnitude ω but pointing in the z-direction, which is along the axis of rotation ω = (0, 0, ω) = k ω (8.20) then the velocity vector of the mass can be written as v = ωk×r

(8.21)

using the cross product. Recall the basic rules pertaining to the cross product; a×b a × (b + c) i × j = k, and finally

|a × b|

= −b × a = a×b+a×c j × k = i, k×i=j =

|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 whose sides are copies of the two vectors.

138

CHAPTER 8. ROTATIONAL MOTION

In the figure

b a b×a

( ) = |b| cos θb i + sin θb j ( ) = |a| cos θa i + sin θa j ( ) = |a||b| cos θb sin θa i × j + cos θa sin θb j × 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 particle which moves in the plane ⊥ to the rotation axis can be gotten from the chain rule

a = = =

d dω dr (ωn × r) = n × r + ωn × dt dt dt dω n × r + ωn × (ωn × r) dt α × r − ω2 r

(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 magnitudes ω and α pointing in the direction of the axis of rotation n.

8.1. KINETIC ENERGY OF A ROTATING BODY

8.1

139

Kinetic energy of a rotating body ω We can compute the kinetic energy of a body undergoing rotation at a fixed rate ω about a fixed point by placing the origin at that point. Let the axis of rotation point in the n direction (a unit vector). Then a small mass dm in the body will have position vector r

r⊥ dm

r = r⊥ + r||

(8.26)

n · r⊥ = 0

(8.27)

in which

r||

this is the part of the position vector perpendicular to the angular velocity vector.

n It will have velocity (for example the little blue arrow) v = ωn × r = ωn × r⊥ ,

ω ≡ ωn

(8.28)

and kinetic energy d KE =

1 1 1 dm |ω × r|2 = dm ω 2 |n × r⊥ |2 = dm |r⊥ |2 ω 2 2 2 2

and integration gives KE =

1 2

∫ 2 dm(r) r⊥ ω2 =

1 I ω2 2

(8.29)

(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 dynamical inertia; the extent to which an object will “resist” changes in its velocity. You know that out in space where there 1 is no gravity, everything is in free-fall, and is “weight-less”. On the moon the surface gravity is 16 times that on the surface of the earth. However if you want to accelerate an object horizontally from rest on the moon, you need to push with the same force to get it going as you would on the earth. The body has the same inertia, or resistance to acceleration, 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, the moment of inertia is more complex than the mass. It is in fact an example of a tensor, which is a vector-like object very 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 ω about 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) ·

R dθ dθ (length of segment) =m =m (length of ring) 2πR 2π

(8.32)

140

CHAPTER 8. ROTATIONAL MOTION

v=ω n×r

First we compute ( ) ωn × r = ωk × R cos θi + R sin θ j (8.33)

R dθ

) = Rω (− sin θi + cos θ j

r

(8.34)

Our segment has energy 1 1 dm |ωk × r|2 = mR2 ω 2 dθ 2 4π (8.35) and we add up all of the fragments composing the ring; d KE =

θ ωn

∫

2π

1 1 mR2 ω 2 dθ = (mR2 ) ω 2 4π 2 0 (8.36) establishing that KE =

Iring = m R2

(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 axis We already know that a ring of mass dm and radius r will have energy

d KE =

1 dm r2 ω 2 2

(8.38)

Cut the disk up into concentric rings, each of mass dm(r) and radius r. Each ring has differential thickness dr and circumference 2πr, and so represents a fraction

dm (r) =

2πr dr m πR2

(8.39)

of the total disk-mass. Add up the kinetic energies of all of the concentric rings; ∫ KE = 0

R

1 dm(r) r2 ω 2 = 2

∫

R

0

1 2πr dr 1 m r 2 ω 2 = m R2 ω 2 2 πR2 4

(8.40)

and so we have established that

Idisk =

1 m R2 2

for an axis through the center perpendicular to the plane of the disk.

(8.41)

8.1. KINETIC ENERGY OF A ROTATING BODY

141

ω √  r2−z2

Example 3. Compute the rotational kinetic energy of a solid ball of mass m, radius r rotating about an axis through it’s center of mass at angular rate ω.

dz

Since we already know what the moment of inertia of a disk of mass dm and radius r′ is, we carve up the sphere into a stack of pancakes of thickness dz and appropriate radii; the radius of the pancake that is distance z from the equatorial plane of the sphere is √ r′ (z) = r2 − z 2 (8.42)

z

and it has mass

dm(z) = m

(volume of pancake) πr′2 (z) dz =m 4π 3 (volume of sphere) 3 r

(8.43)

The moment of inertia about the axis skewering the whole stack is dI(z) =

1 3m dm(z) r′2 (z) = 3 (r2 − z 2 )2 dz 2 8r

(8.44)

from which we can recover the moment of inertia of a sphere rotating about it’s center of mass axis; ∫ r 3m 2 2 Isphere = (r − z 2 )2 dz = m r2 3 5 −r 8r

(8.45)

and its kinetic energy; KE =

y

m

12 m r2 ω2 25

(8.46)

y

v0 x

x

dm(x) = m

(length of segment) dx =m (length of stick) ℓ

Example 4. Compute the rotational kinetic energy of a rod of mass m, length ℓ rotating about an axis through it’s end (and perpendicular to it) at angular rate ω. Locate a sub-mass dm(x) at r = r⊥ = (x, 0, 0). Let this mass have length dx and mass

(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 dx 1 dm(x) v 2 = m (ωx)2 2 2 ℓ

(8.49)

142

CHAPTER 8. ROTATIONAL MOTION

and the kinetic energy of the stick is ∫ KEstick = 0

ℓ

1 dx 1 m (ωx)2 = mℓ2 ω 2 2 ℓ 6

(8.50)

and so for rotation about a perpendicular axis through its end

Iend =

y

m

1 m ℓ2 3

(8.51)

Example 5. As an illustration that a single body will have many moments of inertia, depending on the axis about which it rotates, compute the kinetic energy of a disk of mass m and radius r spun at ω about an axis along a diameter of the disk. Carve the disk up into rods of varying lengths. The mass of a rod ⊥ to the rotation axis of thickness dz and length √ x(z) = r2 − z 2 (8.52)

y

v0

v1 x

x

is √ ( r2 − z 2 dz) dm(z) = m (πr2 )

(8.53)

and we add up the kinetic energy of all of the rods comprising the disk;

∫

r

KE = 2 −r

8.1.1

√ )2 1 ( 1 ( r2 − z 2 dz) )(√ 2 1 1 2 2 2 m r − z ω2 = mr ω , 2 2 3 (πr ) 2 4

and so Idisk,dia =

1 m r2 4

(8.54)

Parallel axis theorem

In many applications of rotation, for example rolling motion, we will need to be able to compute the moment of inertia of a body with respect to an arbitrary axis, not just an axis through the center of mass. For this purpose we will develop the Parallel Axis Theorem. Consider a planar body with a coordinate origin placed at the center of mass. Then under these circumstances ∫ xcom =

∫ dm x = 0,

ycom =

dm y = 0

(8.55)

8.1. KINETIC ENERGY OF A ROTATING BODY

Run a rotation axis parallel to the rotation axis through the center of mass both axes ⊥ to the plane of the body (to simplify the math), and rotate the body about this new axis at angular speed ω Let a small mass element located at r with respect to the origin have position r′ with respect to the point where the rotation axis hits the planar body, with

dm r

143

r’ ω d

ω, C.O.M.

r = d + r′

(8.56)

where d points from the center of mass axis to the new axis. The kinetic energy of this small mass element when the body is rotated about the new axis is

KE

∫ ∫ 1 1 dm |ω n × r′ |2 = dm |ω n × (r − d)|2 2 2 ∫ ∫ ∫ 1 1 2 dm |ω n × r| − dm (ω n × r) · (ω n × d) + dm |ω n × d|2 2 2

= =

(8.57)

but (ω n × r) · (ω n × d) = ω 2 r · d = ω 2 |d||r| cos θ = ω 2 |d|x (a vector arithmetic identity valid for n ⊥ d, r) and so ∫ ∫ dm (ω n × r) · (ω n × d) = ω 2 |d| dm x = 0 and we find KE

= = =

∫ ∫ 1 1 2 dm |ω n × r| + dm |ω n × d|2 2 2 ∫ ∫ 1 1 dm |r|2 ω 2 + dm |d|2 ω 2 2 2 1 (Icom + m|d|2 )ω 2 2

(8.58)

(8.59)

(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 1 Iend = m ℓ2 (8.62) 3 The new axis is a distance

ℓ 2

from the center of mass and so ICOM =

1 ℓ 1 m ℓ2 − m( )2 = m ℓ2 3 2 12

(8.63)

Example 7. Compute the total kinetic energy of a disk of radius R mass m that rolls without slipping on a smooth surface, its center of mass moving at speed v.

144

8.1.2

CHAPTER 8. ROTATIONAL MOTION

Rolling motion

Notice that if the disk rolls without slipping, the contact point between surface and disk cannot be slipping and therefore has zero velocity with respect to the surface. Divide rolling up into two parts, pure translation of all points at speed v, plus rotation at angular speed ω. We take right to be the x-direction, up to be y, and out of the paper z. The center of mass translational velocity of all points on the disk is

vtrans = (v, 0, 0) = vi

(8.64)

n = (0, 0, −1) = −k

(8.65)

The axis of rotation is

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)

The net speed of the bottom point of the disk with respect to the surface when the two separate motions are combined is v

v

v − Rω

(8.67)

×

yet this must be zero. We discover that for the disk to be rolling

Rω

vcom = Rω,

acom = Rα

(8.68)

These are known as the rolling conditions. Contact point at rest

The total kinetic energy of the disk has two contributions, one from the center of mass motion and one for rotation about the center of mass KE =

1 1 1 mv 2 + ( mR2 )ω 2 2 2 2

(8.69)

but when the rolling condition is inserted to eliminate v we find

KE =

1 1 1 1 3 m(Rω)2 + ( mR2 )ω 2 = ( mR2 )ω 2 , 2 2 2 2 2

and

Icontact =

1 3 mR2 + mR2 = mR2 2 2

(8.70)

which is equivalent to saying that rolling is the same as pure rotation about the contact point, since the moment of inertia about the point of contact is

KE =

) 1 1( 1 1 Icontact ω 2 = Icom + md2 ω 2 = Icom ω 2 + mv 2 2 2 2 2

(8.71)

8.2. ROTATIONAL DYNAMICS AND TORQUE

8.2

145

Rotational dynamics and torque F2, ext Consider a two or more body system, with internal forces F12 = −F21 and external forces F1,ext and F2,ext The equations of motion are

F2,1

m2 at r2

m 1 a1

= F12 + F1,ext

m 2 a2

= F21 + F2,ext

(8.72)

We already know that the sum of these equations results in the expression for the motion of the center of mass of the system

r2’ C.O.M. F1,ext r1’

d 2 m 1 r1 + m 2 r2 ( ) dt2 m1 + m2 = F1,ext + F2,ext (8.73) (m1 + m2 )

m1 at r1

Now decompose the position vectors into center of mass position, and positions relative to the center of mass

F1,2

r1 = rcom + r1 ′ ,

r2 = rcom + r2

′

(8.74)

and cross the relative position into each equation of motion in turn r1 ′ × (m1 acom + m1

d2 r1 ′ ) = r1 ′ × F12 + r1 ′ × F1,ext dt2

(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 ′ × m 1

d2 r1 ′ = r1 ′ × F1,ext , dt2

r2 ′ × m2

d2 r2 ′ = r2 ′ × F2,ext dt2

(8.76)

add these two together, and use the fact that v1 ′ = ω n × r1

′

) d( m1 r1 ′ × (ω n × r1 ′ ) + m2 r2 ′ × (ω n × r2 ′ ) = dt = = =

(8.77) r1 ′ × F1,ext + r2 ′ × F2,ext )( )) d (( m1 (r1′ )2 + m2 (r2′ )2 ω n dt ) d( Icom ω n dt 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)

146

CHAPTER 8. ROTATIONAL MOTION

and torque of force F about point p to be τ p = τ p n = Ip α n = 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 (L) = Iα = τi dt i analogous to ma =

∑

(8.81)

Fi

(8.82)

i

F

Example 8. A spool of mass m radius R of moment of inertia I has thread wrapped around its circumference. It floats freely in space. If the thread is pulled with a force F , describe the resulting motion of the spool. The center of mass of the spool will move in a straight line with acceleration a given by

a X

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 where the force is applied to the disk has r = R j, F = Fi (8.84) and so the applied torque about the center of mass is τ n = 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. It floats freely in space. If the thread is pulled with a force F , describe the resulting motion of the spool. The center of mass of the spool will move in a straight line with acceleration a given by

a

F = ma

α

r

F

(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 the force is applied to the disk has r = −r j,

F = Fi

(8.88)

and so the applied torque about the center of mass is τ n = 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

The rotation is in the opposite sense to that in the previous example.

(8.90)

8.2. ROTATIONAL DYNAMICS AND TORQUE

147

Example 10. A spool of mass m radius R of moment of inertia I has thread wrapped around its circumference. It rests on a surface of unknown coefficient of friction. If the thread is pulled with a maximal force F (assume its 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 straight line with acceleration a given by

a α

F

r

R Ff

F + Ff = (F − Ff )i = m a

(8.91)

We saw in the previous example that the disk will move right, and rotate counterclockwise without the intervention of the ground and the friction it applies. When the ground is introduced, friction will cause the contact point to be stationary, forcing Ff to point left. The disk will therefore rotate clockwise (about n = −k) as it rolls without slipping (unless we pull too hard). Using the same coordinates as in the previous examples (−rj) × (F i) + (−Rj) × (−Ff i) = Icom α(−k),

or

RFf − rF = Icom α

(8.92)

but the disk rolls, and so the rolling condition is satisfied RFf − rF = Icom α = Icom Solving for acceleration we find that a=

a R

(8.93)

R(R − r)F Icom + mR2

(8.94)

and we can compute the static coefficient of friction (it rolls, it doesn’t slide) Ff = µs mg = F − ma =

(Icom + mrR) F (Icom + mR2 )

(8.95)

Example 11. A massive spool of moment of inertia I has thread wrapped around it. Suspended at the end of the thread is a mass m. Gravity acts on the mass. Compute all accelerations. The mass m will accelerate downwards m(−aj) = T j + (−mg)j

(8.96)

the string is tangent to an axle of radius r2 at the point r = r2 j as illustrated, and with x to the right, y up and z out; compute torques about the axis through the center of mass of the disk (

) ) r2 j × (−T i = τ n = r2 T k

(8.97)

is the torque applied by the string tension, rotational acceleration has magnitude α and direction n = k. τ n = r2 T k = Icom αn = Icom αk

(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)

148

CHAPTER 8. ROTATIONAL MOTION

This rolling condition is inserted by hand. We solve the three simultaneous equations above by eliminating T and alpha to get mg a a= , α= (8.100) r2 m + rI2 2

F

Example 12. A spool of moment of inertia I = 21 mR2 is wound with thread and set on a smooth table. The coefficient of friction is unknown and therefore cannot be used in the problem. The thread is pulled with enough force to cause the spool to accelerate with no slipping. Find the acceleration. If the spool were not resting on a table, it would accelerate with F i = mai, F = ma (8.101)

a X

and rotate clockwise with acceleration

Ff (Rj) × (F i) = −F R k =

1 mR2 α n 2

(8.102)

n = −k

(8.103)

and so it spins with angular acceleration α=

2F R , mR2

about axis

Then if it is brought near the horizontal surface, the bottom point of the spool would have acceleration with respect to a fixed point on the surface of ai + (−αk) × (−Rj) =

F 2F −F i + (− k) × (−Rj) = i m mR m

(8.104)

and so the contact point would slip to the left on the surface when the two are brought into contact. The force of friction will oppose this slipping and therefore point right! The equations of motion including this force and its torque are 1 (F + Ff )i = ma i, (Rj) × (F i) + (−Rj) × (Ff i) = mR2 α n (8.105) 2 since F and Ff exert torques in the opposite sense. We find that n = −k is the rotation axis, and we have the rolling condition a = αR (8.106) Solving these by eliminating α and Ff we find 4F 3m This problem illustrates some of the subtleties of rotational dynamics problems. a=

(8.107)

Example 13. Find the acceleration of a disk yo-yo. The equations of motion are 1 mR2 αn 2 (8.108) (we compute torques about the disk center of mass)and the rolling condition a = αR (8.109) ma(−j) = mg(−j) + T j,

(Ri) × (T j) = Icom α n =

resulting in n = k and 2 a= g (8.110) 3 We finish our discussion of rotational dynamics with some nontrivial examples, and an illustration of how carefully applied conservation of energy methods can solve even dynamical problems without the use of Newton’s laws. This

8.2. ROTATIONAL DYNAMICS AND TORQUE

149

is possible since the energy function is the constant of integration obtained when Newton’s laws are integrated once.

Example 14. Consider the Atwood machine below with a massive frictionless pulley. Compute the acceleration.

×

Since only an imbalance of torques can cause an object with a nonzero moment of inertia to accelerate angularly, the tension in the two sides of the rope cannot be the same. Labeling them as in the figure and assuming the the disk rolls on the rope without slipping and has radius R, we find that the usual force equations

T1

T1 j + M1 g (−j) = M1 a j, M2 g (−j) + T2 j = M2 a (−j) (8.111) (−Ri) × (−T1 j) + (Ri) × (−T2 j) = Icom α n = Icom α (−k)

T2

and the condition that the string does not slip a = Rα

(8.112)

which can be solved for a

m1 g a=

m2 g

(M2 − M1 )g com M1 + M2 + IR 2

(8.113)

× Example 15. Compute the acceleration of a double yo-yo as shown in the figure. Both disks are identical.

T

This proceeds pretty much the same way, for the hanging disk mg (−j) + T j = ma (−j) (Ri) × (T j) = T R k = Iα1 k (Ri) × (−T j) =

T

T R (−k) = Iα2 (−k)

(8.114)

from which we conclude that the angular accelerations have the same magnitudes α1 = α2 (8.115) but opposite directions.

mg This would not be true if the disks had different radii. The two torque equations were gotten by computing torques about the centers of mass of each disk, which is why mg has no contribution. It has no lever arm about the center of mass of the lower disk, since the force of gravity always acts on the center of mass, as do all forces. 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)

150

CHAPTER 8. ROTATIONAL MOTION

which results in acceleration a=

mg I m + 2R 2

(8.117)

Newton’s laws are simple to apply if one pays attention to the fact that they are vector equations, and so directions of forces must be carefully taken into account. The only inconvenience is that they result in several equations in several unknowns, which are accelerations and constraint forces such as rope tensions, normal forces and frictional forces. Many times conservation of energy can be used to solve dynamical problems for the acceleration without ever having one of these constraint forces appear in the analysis. Consider an object that accelerates from rest for a time t. It will cover distance 1 d = at2 (8.118) 2 and end up with speed v = at, eliminate t, v 2 = 2ad (8.119) This together with conservation of energy can solve dynamics problems.

Example 16. Find the acceleration of the mass m in the picture below. The rope “unrolls” without slipping from the massive disk. 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 spool has angular speed ω. Since the string unwinds without slipping v = Rω. Apply conservation of energy

× T

mgh =

1 1 mv 2 + Iω 2 , 2 2

or mgh =

1 1 I 2 mv 2 + v (8.120) 2 2 R2

put this into the form with v 2 alone on one side of the equation ( mg ) v2 = 2 h (8.121) m + RI2

T

and you can read off the acceleration a=

mg m + RI2

(8.122)

This is a quick method for checking your answers and can replace Newton’s laws methods in most cases if applied carefully.

mg

N v

R

mg θ

Example 17. A small object of mass m is placed at the top of a frictionless hill of radius R. As it begins to slide down, it reaches a point where it loses contact with the hill. Find the angle at which that happens, and the speed of the object at that point. Apply conservation of energy. When at angle θ the object has speed v; 1 mgR = mgR cos θ + mv 2 2

(8.123)

Examine the radial force equation. At any time the mass is in instantaneous circular motion and so its radial acceleration is centripetal; v2 (8.124) mg cos θ − N = mac = m R

8.2. ROTATIONAL DYNAMICS AND TORQUE

151

It loses contact with the hill when N → 0 which happens at θc v2 R

mg cos θc = m

(8.125)

substitute this into the energy balance equation to get 2 , 3

cos θc =

and v 2 (θc ) =

2Rg 3

(8.126)

is the speed of the mass when this happens.

Example 18. Consider the same basic problem, but now with a rolling object.

N ω

v

×

R

The energy equation is 1 1 mg(r + R) = mg(r + R) cos θ + mv 2 + Iω 2 2 2 (8.127) rolling condition is

mg θ

v = rω

(8.128)

and the radial force balance is

mg cos θ − N = mac = m

v2 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.

ω

v0

Example 19. A bowling ball is thrown onto the alley with initial speed v0 .If the coefficient of friction between the ball and alley is µk , compute the distance traveled by the ball before it begins to roll without slipping.

v

×

Ff

The ball will slip to the right and so the force of friction at the contact point with the alley points left. This will cause the ball to accelerate left, slowing it down

d

∫

dv Ff = ma = m , dt

∫

v

t

dv = −

integrate

µk gdt,

v0

or

v = v0 − µk gt

(8.130)

0

This also provides a torque that causes the ball to begin to roll Ff R = Iα = I integrate

∫

∫

ω

t

dω = 0

0

µk mgR dt, I

dω dt

ω=

(8.131)

µk mgRt I

(8.132)

rolling occurs when v = Rω

(8.133)

which we can solve for the time it takes for rolling to begin t=

v0 µk g(1 +

mR2 I )

=

2v0 7µk g

(8.134)

152

CHAPTER 8. ROTATIONAL MOTION

for a sphere. Insert this into the kinematic relation 1 d = v0 t − at2 , 2

with a = µk g

(8.135)

and we see that the ball slips for d=

12v02 , 49µk g

before it begins to roll.

(8.136)

M Example 20. A disk rolls down an incline of height h, starting at rest. Find its speed and rotation rate at the bottom of the hill.

h θ

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; (1

) ( ) )( v )2 1(1 + mg 0 − 0 + 0 + mg h = 0, mv + mR2 2 2 2 R 2

√ v=

2mgh 3 2m

(8.137)

Example 21. Suppose that the hill is so steep that the disk cannot initially roll, but begins slipping. How far will it 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. If it slips, we discover that αR < a, ωR < v at all times, and it slips all the way down; mg sin θ − µk mg cos θ a v µk mg cos θR α ω

= ma = g(sin θ − µk cos θ) = v0 + at = g t (sin θ − µk cos θ) 1 = mR2 α 2 g = 2µk cos θ R g = ω0 + αt = 2µk t cos θ R

(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 forces acting on a body subject to the constraint that it is not accelerating; its center of mass undergoes no rectilinear acceleration, 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

Example 22. A penguin of mass mp stands a distance d (measured along the ladder) from the center of a ladder of length 2ℓ mass mℓ inclined at angle θ. The coefficient of friction with the ground is µs , but with the vertical wall is zero. Find the smallest value of θ for which the ladder does not slip and collapse.

mp g

mL g N1

θ Ff

The sum of external forces equals zero, this leads to Ff = µs N1 = N2 ,

N1 + N2 − mp g = 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) × (−mℓ gj) + (d cos θi + d sin θj) × (−mp gj) + (2ℓ cos θi + 2ℓ sin θj) × (−N2 i) = 0

(8.142)

or ℓ cos θmℓ g + d cos θmp g = 2ℓ sin θN2

(8.143)

which immediately gives N2 . We can now recover the other forces.

Example 23. A penguin of mass mp stands a distance d (measured along the beam) from the 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 such that the beam does not collapse. The hinge with the wall exerts whatever forces Ff , N1 are necessary for stability.

Ff θ N1

mp g

T

mL g

This is a good example of a mechanical engineering statics problem; we would compute the forces supplied by the hinge, and obtain the rope tension. Horizontal and vertical equilibria imply N1 − T cos θ = 0,

Ff + T sin θ − mp g − mℓ g = 0

(8.144)

Torques about the left end of the beam but add up to zero or the beam rotates (di) × (−mp gj) + (ℓi) × (−mℓ gj) + (2ℓi) × (−T cos θi + T sin θj) = 0 or T =

dmp g + ℓmℓ g , 2ℓ sin θ

N1 =

cos θ(dmp g + ℓmℓ g) 2ℓ sin θ

(8.145)

(8.146)

154

CHAPTER 8. ROTATIONAL MOTION

Example 24. Two small penguins support their emperor penguin on a beam of length 2ℓ and mass mℓ . The emperor, whose mass is mp , stands off-center by amount d closer to the right penguin. Find the forces N1 and N2 exerted by the two supporting birds.

N2

N1 mL g

mp g

Vertical force balancing requires N1 + N2 = (mℓ + mp )g

(8.147)

(ℓi) × (−mℓ gj) + ((ℓ + d)i) × (−mp gj) + (2ℓi) × (N2 j) = 0

(8.148)

Torques about the left end

gives us N2 ; N2 =

ℓmℓ g + (ℓ + d)mp g 2ℓ

Example 25. A penguin of mass mp = 20.0 kg heroically climbs an A-frame ladder (hinged at the top) to a distance d = 7.5 m (measured along the ladder from the bottom left corner) , each side of which has length 2ℓ = 10.0 m. The angle θ = 60o . The ladder rests on ice and is stabilized by a wire connecting the centers of the two halves. Compute the values of the two normal forces N1 , N2 and the tension T in the wire.

T N1

mp g θ

(8.149)

N2

The sum of vertical forces is zero; N1 + N2 − mp g = 0

(8.150)

Take torques of external forces on the whole structure about left leg contact point with ground; 0 = (7.5 m cos 60i + 7.5m sin 60j) × (−mp gj) + (10.0 mi) × (N2 j),

3 3mp g N2 = mp g cos 60 = 4 8

(8.151)

This gives N1 =

5mp g 8

(8.152)

Take torques on the right ladder-leg about the top; (5 m cos 60i − 5 m sin 60j) × (−T i) + (10 m cos 60i − 10 m sin 60j) × (N2 j) = 0

(8.153)

gives

√ 3 T = 2 cot 60 N2 = mp g 4 and thus we conclude our most penguiniferous section.

8.4

(8.154)

Work and torque

Energy methods gave us a useful alternative to pure vector-based force determination from Newton’s third law F = 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

˙ from a rotating body (about some pivot point or axis p) we can always recover the angular speed v = ωR = θR conservation of energy ∑ 1 Ip ω 2 = Wi (8.155) 2 i

θ For example, if we begin with the pendulum illustrated in the horizontal position, and let it swing down to angle θ shown, then 1 ( 2) 2 mℓ ω 2

ar

=

1 ( 2 ) ˙2 mℓ θ 2(

= − 0 − mgℓ sin θ

(8.156) )

at Among other things, this gives you the centripetal part of the acceleration ac = Rω 2 = ℓθ˙2 = 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 calculation in two ways; first by simple derivation using the chain rule ( )( ) d ( 1 2 ˙2 ) ℓ θ = ℓθ˙ ℓθ¨ dt 2 = v at ) d( = gℓ sin θ = gℓ cos θ θ˙ (dt )( ) =

g cos θ

v ,

at = g cos θ

(8.158)

or alternatively from the torque ( Ip α =

(

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 that the velocity of the point where the force is applied is v = ω × r; ℘=

dW dt

W is the work don by a torque.

= F·v ( ) = F· ω×r ( ) = ω· r×F =ω·τ ∫ ∫ = τ · ω dt = τ · n dθ

(8.160)

156

8.5

CHAPTER 8. ROTATIONAL MOTION

Problems 1. A. This ball rolls down the ramp without slipping. Find the largest angle θ = θc for which this is possible in terms of µs . Once the ball reaches the bottom of the hill, determine its speed in terms of the hill height h.

M

h B. If the angle θ < θc , the ball still rolls all the way. Under these circumstances compute |Ff ric |.

θ

2. A sign of dimensions illustrated and mass m = 10.0kg hangs from two cords. If the right cord breaks, find the tension in the left cord instantly afterwards. Let h = w = 0.9 m.

T

D.Wagenheim h

Dentist

Double or nothing if you figure out the literary reference.

w Mg

L θ b

mg

N Ff

3. A disk of mass m = 0.005kg and radius a = 0.01m rolls without slipping on a smooth table. It rolls in a circle of radius b = 0.2 m. There are three forces acting on it; gravity, friction, and the normal force. It spins about its center of mass at rate ω = 20.0 rad s . As it rolls around on its “orbit”, its spin angular momentum vector L makes and angle θ with the vertical If we watch the disk, we see that its contact point with the ground travels around the circle of radius b at speed Ωb, and its spin angular momentum vector tip moves around on a circle at angular rate Ω. The angular momentum vector is

( ) L = Lspin + Lorbital = Icom ω cos θk + sin θ sin(Ωt)i + sin θ cos(Ωt)j + |Lorbital |k

The position vector of the center of mass of the disk is ( ) r = (b − a cos θ) sin(Ωt)i + cos(Ωt)j + a sin θk

Ωn

L⊥

Ωt L

From all of this information compute Ω, and the magnitude of the normal force N. From this formula for the angular momentum vector and from∑ its time derivative (and the torque equation L˙ = i ri × Fi,ext ), determine the rate ω as a function of θ such that the disk rolls without slipping. From this information compute the magnitude of the force of friction that keeps the disk traveling in a circle.

8.5. PROBLEMS

157

4. A rod of mass m and length 2ℓ rests on-end on a smooth frictionless table. It begins to fall, its contact point with the table sliding out from beneath it. At what angle of inclination θ does it lose contact with the table? Note that with gravity and the normal force being the only forces acting on the rod, its center of mass can only move vertically.

θ

mg N

5. A spool with an inner hub of radius ri = 0.2m and an outer hub of radius ro = 0.3 m and mass m = 10 kg and moment of inertia Icom = 35 m ro2 rests on a smooth horizontal surface of unknown coefficients of friction. A string wound around its inner hub is pulled straight upwards with force F = 25 N j. The disk rolls without slipping. Compute the magnitude and direction of the acceleration of the disk and the force of friction exerted by the horizontal surface.

F

α

ri

ro

6. A spool with an inner hub of radius ri = 0.2m and an outer hub of radius ro = 0.3 m and mass m = 10 kg and moment of inertia Icom = 35 m ro2 rests on a smooth horizontal surface of unknown coefficients of friction. A string wound around its inner hub is pulled gently at angle θ yet the disk does not move! Find the angle θ.

a α

F ri

θ

ro

R

M2

F

M1

7. A block of mass m1 = 4.0 kg slides on a frictionless surface. A ball (sphere) of mass m2 = 1.0 kg and radius R = 0.08 m rests on it. A force F = F i = 4.0 N i is applied to the block as shown. The coefficients of friction between ball and block are unknown. Find the acceleration a1 of the block and a2 , α2 n of the sphere if the ball rolls without slipping on the block.

158

CHAPTER 8. ROTATIONAL MOTION

r1

8. A spool with an inner hub of radius r1 = 0.2m and an outer hub of radius r2 = 0.3 m and mass m = 10 kg and moment of inertia Icom = 3 2 5 m r2 rests on a smooth horizontal surface of unknown coefficients of friction. A string wound around its inner hub is attached to a hanging mass M = 1.0 kg. The spool rolls without slipping. Find a for spool and mass, and the force of friction.

r2

M

r1

M2

r2 F

M1

F

r1

r2

M2

9. A block of mass M1 slides on a frictionless surface. A spool of mass M2 and radii r1 , r2 rests on it. A string wound around the spool’s inner radius is gently pulled with force F as shown. The coefficients of friction between spool and block are unknown. Find the acceleration a1 of the block and a2 , α2 n of the spool if the spool rolls without slipping on the block.

10. A block of mass M1 slides on a frictionless surface. A spool of mass M2 and radii r1 , r2 rests on it. A string wound around the spools inner radius is gently pulled with force F as shown. The coefficients of friction between ball and block are unknown. Find the acceleration a1 of the block and a2 , α2 n of the sphere if the sphere rolls without slipping on the block.

M1

F

M2 M1 Θ

11. In this problem the coefficients of friction between wedge M1 and the horizontal table it rests on are zero, but the friction between moving wedge M1 and the spool M2 has coefficients µs = 0.3 and µk = 0.2. Let M1 = 10.0 kg and M2 = 5.0 kg, spool (disk) radius R = 0.1 m, 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

r1

12. A disk with an inner hub of radius ri = 0.2m and an outer hub of radius ro = 0.3 m and mass m = 10 kg and moment of inertia Icom = 35 m ro2 rests on a smooth horizontal surface of unknown coefficients of friction. A string wound around its inner hub is pulled straight downwards by a hanging mass M = 1.0 kg. The disk rolls without slipping. Compute the magnitude and direction of the acceleration of the disk. Compute the force of friction exerted by the horizontal surface.

r2

M

13. This spool of mass M = 3.0 kg and radius R = 0.1 m and moment of inertia 35 M R2 is being pulled by a rope of tension T connected to its axle. The coefficients of friction with the ground are µk = 0.2, µs = 0.3. Find the largest T for which the spool rolls without slipping.

T M, R µs, µk

14. Suppose that the same spool is pulled by a rope of tension T = 30N connected to its axle. Find Ff , a and α for the 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 disks of mass m (each) and radius R about the axis through the center of the middle disk.

17∗ . Find the moment of inertia of this frame made of four rods of mass m length ℓ about an axis through the geometric center of the square interior.

160

CHAPTER 8. ROTATIONAL MOTION

18∗ . A rod of mass m and length ℓ connected to the ground with a frictionless hinge is released from rest and allowed to fall. Find the speed with which the upper end hits the ground.

19∗ . This spool of mass M and radius R rolls without slipping to the bottom of the hill, upon reaching it its √ center of mass speed is

M

measured to be v = of inertia.

h θ

8Hg 7 .

Find its moment

20. A penguin of mass mp = 30.0 kg heroically climbs an A-frame ladder (hinged at the top) each side of which has length ℓ = 5.0 m. The angle θ = 45o . The ladder rests on ice and is stabilized by a wire connecting the centers of the two halves. Find the tension T in the wire, and the forces N1 , N2 . T N1

N2

θ

v

21. This disk of mass m = 2.0 kg and radius R = 0.1 m compresses a spring of force constant k = 20 N m by amount x = 0.1 m. When released and disengaged from the spring the disk rolls without slipping at speed v. Find v. Find its acceleration at the time it is released and begins to move.

22. A sphere of radius R = 0.1 m, mass m2 = 2.0 kg and a block of mass m1 = 3.0 kg compress a spring N of force constant k = 200 m by amount x = 0.1 m. When released and disengaged from the spring the sphere rolls without slipping at v2 and the block slides without friction at v1 . Find both v1 and v2 .

v1

ω

v2

8.5. PROBLEMS

M

161

M m

M

M m

M m

M m

M m

M m

23. Compute the moment of inertia about the origin of this gadget (four masses M at intervals of ℓ on a rod of length 3ℓ mass 3m), and find its kinetic energy when rotated at rate ω. Find the speed of each mass M .

24. Compute the moment of inertia about the origin of this gadget (four masses M at intervals of ℓ on a rod of length 3ℓ mass 3m), and find its kinetic energy when rotated at rate ω. Find the speed of each mass M .

F 25. A stick of length ℓ and mass m ordinarily hangs vertically from its hinge point. What force F must be applied vertically at its center in order to hold it horizontally? What force (magnitude and direction) must the hinge exert?

F 26. A stick of length ℓ and mass m ordinarily hangs vertically from its hinge point. What force F must be applied vertically at its end in order to hold it horizontally? What force (magnitude and direction) must the hinge exert?

162

CHAPTER 8. ROTATIONAL MOTION

27. A stick of length ℓ and mass m ordinarily hangs vertically from its hinge point. What force F must be applied horizontally at its center in order to hold it at an angle θ with respect to vertical? What force (magnitude and direction) must the hinge exert?

28. A stick of length ℓ and mass m ordinarily hangs vertically from its hinge point. What force F must be applied normally at its center in order to hold it at an angle θ with respect to vertical? What force (magnitude and direction) 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 it is 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 it is 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 energy is 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 edge on a table with which it has frictional coefficient µk . Once it begins to roll, find its spin rate and COM speed. How far 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 that an = a · n = ℓθ˙2 ,

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 speed given by 1 ( ˙ )2 m ℓθ = mgℓ sin θ 2 and tangential acceleration given by ( ) m ℓθ¨ = mg cos θ Hint; draw a free-body diagram to get at .

8.5. PROBLEMS

163

34∗ . A pendulum of mass m length ℓ being at rest in the horizontal position, and is allowed to fall to the point θ illustrated. Show that the magnitude of its acceleration at that point is √ √ a2n + a2t = g 1 + 3 sin2 θ

θ

ar

Hint, use at = ℓθ¨ = ℓτ /I,

at

an = ℓθ˙2

and energy conservation.

35∗ . A body of mass m moves on a circle R = 10 m. At a certain time its instantaneous speed is 10 m/s, and the instantaneous rate of change of its speed is 10 m/s2 . At that instant find the angle between its velocity vector and acceleration vector. 36. A body of mass m moves on a circle R. At a certain time its instantaneous speed is ωR, and the instantaneous rate of change of ( its) speed is αR. At that instant show that the angle between its velocity vector and acceleration −1 ω 2 vector is tan α . 37. When θ = 45o in problem 8.34, find the angle between the total acceleration vector and the instantaneous velocity vector of the falling pendulum. 38. Consider a heavy rotating flywheel of mass m = 100 kg and radius R = 0.25 m spinning at ω0 = 100 rad/s. A brake applies a force of friction of 100 N to its rim tangentially. Through what total angle θ does it rotate before stopping?

y

39. Wind a string of length 5 m around the rim of a wheel of mass m = 1.0 kg and radius R = 0.25 m with its axle held fixed. Pull the string off tangentially with a constant force of 10 N . By time the string has rolled off completely, how fast is the wheel rotating? 40∗ . A equilateral triangular frame is made with three sticks each of mass m = 1.0 kg and length L = 1.0 m. One corner is at the origin. Find the cent er of mass position (xcom , ycom ) and compute the moment of inertial for rotation about the origin.

m, L

x

ω

(Fh1, Fv1)

(Fh2, Fv2)

41. A disk of mass m and radius R rotates at speed ω while in contact with both walls in a corner, with coefficients of friction µk . Find the torque on the disk.

164

CHAPTER 8. ROTATIONAL MOTION

42. A massless level table rests on three equally spaced legs placed on its perimeter, forming an equilateral triangle of side a = 0.5 m. A ball rests on the table. The forces exerted upwards by the legs are 10, 20 and 30 N . What is the mass of the ball, and where is it on the table relative to the leftmost leg (the 10 N leg)?

Chapter 9

Angular Momentum

Conservation of angular momentum is one of the most important of all conservation laws in nature. In any system with no torques exerted by outside agents we find that

y

L=mr × v=(0,0, −mvy) v=(v,0,0) × m r=(−x,y,0)

∑ d τi,ext ni = 0 L= dt i

(9.1)

L = L(t) = L(0)

(9.2)

and so

x

and so both magnitude and direction of the angular momentum are conserved. In the absence of externally applied torques about the point, angular momentum is always conserved with respect to the point. Angular momentum is computed with respect to a reference point regarded as a point about which the system is in instantaneous 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 the parallelepiped whose sides are the velocity and position vector. This turns out to be v times the component of r that is perpendicular to v, in other words y. 165

166

CHAPTER 9. ANGULAR MOMENTUM y

This is called the lever arm of the momentum about the origin. The angular momentum vector becomes r’

ω n × r’

L = mr × v = m(−x i + y j) × (v i) = −myv k (9.3) and points into the paper by the right hand rule. Compute the angular momentum of a rolling disk about the origin; the fragment of mass dm at body-relative coordinate r′ has position vector r + r′ and total velocity

v0

rcom x

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 + Icom ωn

(9.8)

which is the angular momentum of the center of mass motion plus the angular momentum about the center of mass.

P

r

v × ω

Example 1. A tin can of moment of inertia Icom about its center of mass axis and mass m, height ℓ sits on a post. It is dealt a blow near its top, delivering momentum P to it. Compute the speed of its center of mass and rotation rate after being struck.

The blow carried momentum P = P i and angular momentum ℓ Pℓ ℓ L = r × P = (− i + j) × (P i) = − k 2 2 2

(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 = m v = mv i

(9.10)

and spin about the center of mass at rate and direction L=−

Pℓ k = Icom ωn = Icom ω (−k) 2

(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 a single nail through its center, allowing it to pivot and freely rotate about its center. A ball of mass m, speed v0 collides with it and sticks to the end. Compute the rotation rate of the whole mess after the collision. The ball has angular momentum ℓ ℓ L = Ln = mr × v = m(−xi + j) × (v0 i) = mv0 (−k) 2 2

y

m

(9.12)

y

v0 x

x

with respect to the pivot point (where we put the origin) before the collision. Since the collision is inelastic, only momenta, angular momenta, can be conserved, and so the angular momentum before the collision equals that after the collision. Linear momentum is not conserved; the nail holding the stick in place exerts forces on the system constraining the center of the stick to remain fixed. The presence of external forces foils the conservation of momentum. Angular momentum contains contributions from the stick and the ball stuck to its end ( ( ℓ ℓ ) ℓ ) mv0 (−k) = Icom + m( )2 ω n = Icom + m( )2 ω (−k) (9.13) 2 2 2 which we can solve for ω. We find that the axis of rotation is n = −k, so rotation is clockwise, and ω=

mv0 2ℓ

(9.14)

Icom + m( 2ℓ )2

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.

y

m

y

v0

v1 x

x

168

CHAPTER 9. ANGULAR MOMENTUM

Conservation of energy says 1 mv 2 = 2 0 let angular momenta into the paper be positive, then

1 1 m v12 + Icom ω 2 (9.15) 2 2 conservation of angular momentum about the pivot point says

ℓ ℓ m(−xi + j) × (v0 i) = m(−x′ i + j) × (−v1 i) + Icom ω(−k) 2 2 or mv0

ℓ ℓ = Icom ω − mv1 2 2

(9.16)

(9.17)

which gives us 2

4v0 , ω= com ℓ + 4Imℓ

v1 =

Icom − m ℓ4

2

Icom + m ℓ4

v0

(9.18)

Example 4. A billiard ball is to be struck with a cue stick such that it rolls without slipping instantly after the blow is struck. How high above the table must the cue strike the ball?

P R

Let the P = Pi slipping collision force,

h

stick impart all of its momentum to the ball. If the ball rolls without then no forces act on it after the (except for gravity and the normal

the presence of static friction is not required, and so the solution is the same whether it is present or not provided the 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 = M v = M vi,

v=

P M

(9.19)

The impulse delivered had angular momentum-magnitude P (h − R) about the center of mass of the ball before the collision. Afterwards the ball rolls with angular speed ω and so has angular momentum Iω about its center. The rolling condition states v = Rω and so using Icom = 52 M R2 for the ball we find L = r × P = (−xi + (h − R)j) × (P i) = = M v(h − R) (−k) =

2 M R2 ω n 5

2 M R2 ω n 5

(9.20)

we see that n = −k;

2 v 7 M R2 (−k), eliminating R we find that H = R (9.21) 5 R 5 is the required height above the table of the blow to be delivered by the cue, and this does not depend on how hard the 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 in contact with the ball is infinitismal, nevertheless in that amount of time the cue delivers a finite impulse (momentum change) into the ball. ∫ δt F (t) dt = P (9.22) L=

0

Write the equations of motion for the ball including friction if there is any F (t)i + Ff (t)(−i) = M a(t) i,

(h − R)j × F (t)i + (−Rj) × (−Ff (t)i) = Iα(t)(−k)

(9.23)

169 solve; F (t) = M a(t) + Ff (t),

F (t) =

2M R2 R α(t) + Ff (t) 5(h − R) h−R

(9.24)

integrate over time ∫

∫

δt

F (t) dt = P ∫

0

∫

δt

= M 0

δt

F (t) dt = P 0

=

δt

a(t) dt +

2M R2 5(h − R)

Ff (t) dt 0

∫

δt 0

R α(t) dt + h−R

∫

δt

Ff (t) dt

(9.25)

0

We obtain ∫

δt

P = M (vf − 0) +

2M R2 R P = (ωf − 0) + 5(h − R) h−R

Ff (t) dt, 0

∫

δt

Ff (t) dt,

set ωf =

0

vf R

(9.26)

but friction has a maximal magnitude µs N , therefore ∫

∫

δt

Ff (t) dt ≤ 0

δt

µs N dt = µs N δt → 0

(9.27)

0

and so provided the impulse is delivered instantaneously and the ball does not slip momentum is conserved 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, in which 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 a ball of mass m, moving perpendicular to the stick at speed v0 . Compute the resulting spin rate and velocity of the stick after the collision.

y

m

y

v0

v1 x

x v2

Labeling velocities before and after as in the figure, with momenta to the right positive, and angular momenta into the paper positive, the momentum, energy and angular momentum about the center of mass if the stick are all conserved ℓ ℓ (9.28) mv0 i = mv1 (−i) + M v2 (i), mv0 (−k) = mv1 (k) + Icom ω (−k) 2 2 and 1 1 1 1 mv 2 = mv12 + M v22 + Icom ω 2 2 0 2 2 2

(9.29)

since the stick has two types of kinetic energy after the collision, center of mass rectilinear and angular. We solve these three equations for v1 , v2 , and ω.

170

CHAPTER 9. ANGULAR MOMENTUM

ycom

y

v0 R m2

×

x

a

v1

b

xcom

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 the center 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,

m 1 a = m2 b

(9.30)

or a=

m2 R, m1 + m2

b=

m1 R m1 + m2

(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 mass is fixed. This requires that the two masses move “equal angles in equal times” on their respective circles, both completing an “orbit” in the same amount of time

T =

2πa 2πb = , v1 v0

v1 =

a m2 v0 = v0 b m1

(9.32)

Computation of angular momentum with respect to the fixed origin of the C.O.M. system leads to m2 bv0 k + m1 av1 k + Icom ωn = 0

(9.33)

requiring that the disk rotate about its center clockwise at rate

ω=

m2 bv0 + m1 av1 m2 (a + b)v0 m2 Rv0 = = Icom Icom Icom

(9.34)

Example 7. Consider two disks on parallel axes. One has an initial spin rate ω0 , the other is stationary. They are brought into contact and friction between the two causes them to slip until eventually they spin at such a rate that no mutual slipping occurs, they roll on one another. Compute this final spin rate.

171

R1 ω0

ω1 ×

× Ff

Ff R2

ω2

This problem is very deceptive. If we try to use angular momentum conservation, we must pick a single point about which 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

τ n = (−R1 j) × (Ff i) = R1 Ff k = I1,com α1 n = I1,com

(9.35) (d ) ω1 n dt

(9.36)

integrating, we find − ω1 (t)k = −ω0 k +

Ff R1 t k, I1,com

ω1 (t) = ω0 −

Ff R 1 t I1,com

(9.37)

and ω2 (t)n = ω2 (t) k,

ω2 (t = 0)n = 0

τ n = (R2 j) × (−Ff i) = Ff R2 k = I2 α2 k = I2,com

(d ) ω2,com (t) k dt

(9.38)

(9.39)

integrate ω2 (t) k = 0 +

Ff R2 t k, I2,com

ω2 (t) =

Ff R 2 t I2,com

(9.40)

The rolling condition is ω1,f R1 = ω2,f R2

(9.41)

for the final non-slipping motions, eliminating Ff and t we obtain ω1 =

ω0 1+

R12 I2 R22 I1

(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 see that it is not necessary to know N or Ff if we resort to basic principles.

172

CHAPTER 9. ANGULAR MOMENTUM

9.1

Problems y

m

y

v0

v1 x

x

2. A stick of mass m1 and length ℓ hangs from a perfect hinge at one end. It is struck elastically by a ball of mass m moving horizontally at v0 = v0 i. Determine how far x below the hinge that the ball must strike in order to be stopped dead by the collision. Find the rotation rate ω of the stick instantly after the collision.

ωn

x m v0

3. Two sticks of mass m1 , m2 , length ℓ hang from a common perfect hinge at one end. One (m1 ) is displaced to horizontal, and released. It collides elastically with the second stick. Compute both rotation rates ω1 , ω2 of the two sticks instantly after the collision.

ω1 n ω2 n

ω1 v1

ω2

1∗ . A ball of mass m = 0.5 kg traveling at v0 = 2.0 m s collides elastically with a stick of mass M = 1.0 kg of length ℓ = 1.0 m, which is free to rotate and move horizontally and vertically. How far d above the stick center must the ball hit the stick in order to be stopped dead by the collision? Find the final rotation rate ω of the stick and the speed v of its center of mass. A classic exam problem.

4. Two sticks of mass m1 , m2 , length ℓ float freely in empty space (no gravity). A powerful spring of force constant k is compressed by a very small∗ amount x, placed between the top ends, and released. The sticks are flung apart. Find the velocities v1 , v2 of the centers of mass and both rotation rates ω1 , ω2 of the two sticks. ∗ This lets you assume F is ⊥ to the stick while the spring is acting on it.

× v2

ωn θ x m v0

5. A rod of mass m and length 2ℓ hangs below a fixed end. It is struck inelastically at a point x below the pivot point by a ball of mass m traveling at v0 . Compute the final angle θ to which the rod swings (under the influence of gravity).

9.1. PROBLEMS

173

y

m

y

v0

ω

x

x v1

6. A rod of mass M and length 2ℓ floats freely in space. It is struck inelastically at its end by a ball of mass m traveling at v0 . After the collision the combined object rotates about its center of mass at ω, which moves with velocity v1 . Find ω and v1 .

7. A disk of radius R = 0.25 m, mass m = 10.0 kg rolls towards a step of height h < R at speed v0 . Determine the maximal height h step that the disk can “hop” by snagging on the corner and pivoting (without slipping) onto the top of the step. Hint; angular momentum with respect to the corner of the step is conserved, compute it instantly before and instantly after the collision.

× v0

h

v0

8∗ . Two discs (I = 12 mR2 ) of radii R = 0.1m and mass m = 1.0kg, with one initially moving at v0 = 2.0 m s and the other spinning at ω0 = 20 rad/s are lined with Velcro and are on a collision course. They stick together at a point on the rims. Find the final spin rate ωf about the new center of mass, and the speed vf of the center of mass after the collision.

ω0

This occurs in space, there are no forces present.

ωf

vf

174

CHAPTER 9. ANGULAR MOMENTUM

v0

9∗ . Two discs (I = 12 mR2 ) of radii R = 0.1m and mass m = 1.0kg, with one initially moving at v0 = 2.0 m s and the other spinning at ω0 = 10 rad/s are lined with Velcro and are on a collision course. They stick together at a point on the rims. Find the final spin rate ωf about the new center of mass, and the speed vf of the center of mass after the collision.

ω0

This occurs in space, there are no forces present. Full credit for numerical answer.

ωf

vf

y 10∗ . A person of mass m = 50 kg stands at the rim of a freely rotating disk of mass M = 200 kg, radius R = 5 m, spinning at ω0 = 1 rad She walks s j. inwards to the center of the disk. Find the final spin rate when she arrives at the center.

x

11. In the preceding problem, there were no torques applied to the system (why?) yet there was angular acceleration of the disk. From d d( ) L= Iω dt dt show that if our person riding the disk walks towards the center at rate v, then at the instant that she leaves the rim the angular acceleration of the disk is ( α=

) 2mRv ω0 1 2 2 2 M R + mR

y

m

y

v0 x

ω

x v1

12. Two balls of mass m each connected by a massless string of length 2ℓ floats freely in space. It is struck inelastically at its end by a ball of mass 3m traveling at v0 . After the collision the combined object rotates about its center of mass at ω, which moves with velocity v1 . Find ω and v1 .

9.1. PROBLEMS

175

z

R y x

13. A disk of moment of inertia I forms a gyroscope with a ball-joint (red) a distance R from the disk. The disk rotates at angular speed ω (fast) so that its angular momentum vector at t = 0 is L = Iω j. The torque of gravity acting on the disk causes the angular momentum vector to rotate at rate Ω, so its 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.

Mg 14. Suppose that the gyroscope is nearly vertical, with its axis tipped at angle θ with respect to perfect verticality. Let ( ) (1 ) 1 L = M a2 ω sin θ sin(Ωt) j + sin θ cos(Ωt) j + cos θ k + M a2 + M R2 ωk 2 4 and find the precession rate Ω caused by the torque of gravity. What angle maximizes the precession rate?

15. A mass M is made to travel in a circle of radius r0 at speed v0 on a frictionless horizontal plane with a string under tension T through a hole as illustrated. If the string is pulled down until the radius of the path is reduced to r0 /2, compute the work done on the system.

176

CHAPTER 9. ANGULAR MOMENTUM

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 rates in different places, so this means that v = v(t, r). The velocity will then have two parts to its rate of change, since the position r of a block of fluid also changes r = r(t). Compute the acceleration of a block of fluid d v(t, x(t), y(t), z(t)) dt

∂v ∂v dx ∂v dy ∂v dz + + + ∂t ∂x dt ∂y dt ∂z dt ( ) ( ) ∂v dx ∂v (10.2) = + · ∇v = + v · ∇v ∂t dt ∂t This is the convective derivative. The last idea that we need is pressure, force exerted normally on a surface per 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 that only 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” such as ( ) v = x i + y j + zz /s (10.3) Flow lines (lines tangent to v) are straight lines through the origin dy dx dx dy = y x y = C ′ x,

=

vy y = vx x ln y = ln x + c c ′ = ec

(10.4)

This type of field corresponds to fluid entering space at a regular rate, measure the flux or flow-rate through a sphere of radius R around the origin, which is the point where the fluid is entering space 1 This chapter is skipped in 201, it is included for MCAT and GRE preparation, which requires only section 10.2.1. Problems 1-10 are 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/s n I v · n dS = 4πR3 /s

(10.5)

S

Now multiply by the density ρ, and you get the amount of mass per second of fluid entering space at the origin; I v · n dS = 4πρR3 /s =

ρ S

dm dt

(10.6)

Lets calculate the “source strength” of this field ( ) ∂ρv ∂ρvy ∂ρvz x ∇ · ρv = + + = ρ(1 + 1 + 1)/s = 3ρ/s ∂x ∂y ∂z

(10.7)

Note that this is exactly the correct mass per second per volume entering our sphere; multiply it by the sphere-volume and check for yourself, you get the result of Eq. 10.6. In other words ( ) dρ ± ∇ · ρv = 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 the vector 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 of components   ∂v ∂v ∂v  ∇v = 

which for our flow v = (αx, αy, αz) with α =

1 s

x

x

x

∂x ∂vy ∂x ∂vz ∂x

∂y ∂vy ∂y ∂vz ∂y

∂z ∂vy ∂z ∂vz ∂z

0 α 0

 0 0  α

 

(10.9)

looks like 

α ∇v =  0 0

(10.10)

What would the acceleration of a block of fluid be? Eq. 10.2 says 

  dx α dt ax  ay  =  α dy dt az α dz dt  dx α dt =  α dy dt α dz dt





α + 0 0  + α2 x + α2 y  + α2 z

0 α 0

using the normal rules of matrix multiplication (from linear algebra).

   0 αx 0  ·  αy  α αz (10.11)

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 y dy = −x dy

=

x vy =− vx y 2 2 x + y = 2c (10.13)

Lets measure its “swirliness” (the vortex strength) ω = ∇ × v = −2/s k

(10.14)

The −k says it swirls clockwise (the −) around the z-axis.

10.2

Newton’s laws applied to fluids Fluid elements accelerate because there are different pressures on the different walls of the element (pressure gradients), consider a volume of fluid in a pipe of cross-section A between x and x + dx. Its momentum changes because forces (pressures times areas) are exerted on it

P(x)

) ( ) d( ρA dx v = A P (x)−P (x+dx) i (10.15) dt

P(x+dx)

so the fluid equation of motion is ρ

x x+dx Re-write this using our convective derivative; ρ

( ∂v ∂t

( )) + v · ∇v = −∇P,

d dP v=− i = −∇P dt dx

Euler’s equation

(10.16)

(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 = −∇ · (ρ v), ∂t

mass conservation

(10.18)

Momentum flow and stress. The equation for the ith component of the fluid equation of motion is ρ

( ∂v

i

∂t

+

∑(

vj

j

∂ ) ) ∂P vi = − , ∂xj ∂xi

x1 = x, x2 = y, x3 = z

(10.19)

take the mass conservation equation ∑ ∂ ∑ ∂ρ ∑ ∂vj ∂ρ =− (ρ vj ) = − vj − ρ ∂t ∂xj ∂xj ∂xj j j j multiply by vi vi

∑ ∂ρ ∑ ∂ρ ∂vj =− vi vj − ρ vi ∂t ∂x ∂x j j j j

(10.20)

(10.21)

180

CHAPTER 10. FLUIDS

and add

) ) ) ∑ ∂ ( ∑ ∂ ( ∑ ∂ ∂( ∂P ρ vi = − − ρ v i vj = − P δij + ρ vi vj = − Tij ∂t ∂xi ∂xj ∂xj ∂xj j j j

(10.22)

We have invented the stress-momentum tensor T Tij = P δij + ρvi vj ,

∂( ) ρ v = −∇ · T, ∂t

which in explicit component form looks like    P + ρvx2 Txx Tyx Tzx    ρvx vy Txy Tyy Tzy T= = Txz Tyz Tzz ρvx vz

{ with δij =

ρvx vy P + ρvy2 ρvy vz

1 i=j 0 i= ̸ j

 ρvx vz ρvy vz  P + ρvz2

(10.23)

(10.24)

The left side of this equation is the rate of change of momentum density (momentum per unit volume), the right side is the divergence of the stress momentum, so integrate over volume and apply the divergence theorem2 ∫ I ∂ ρ v d3 x = − T · n dA (10.25) ∂t V0 S the object dp = −T · n dA, dt

dpi = Tij nj dA dt

(10.26)

is the flow of momentum through the area dA (force exerted on the area) whose normal is n (in other words T is the 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

Txy

Tyz

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

i, Txx Txz

The second of Eq. 10.23 applies to all sorts of continuous media, not just fluids, but the explicit form of Tij = P δij + ρvi vj is for fluids only. Elastic media will typically have the components of the tensor as variables to be solved for given specified deformations of the material, or the components of T will be known applied forces and one will solve 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 the gradient of some “velocity potential” Φ v = ∇Φ (10.27) 2 see

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 up through a constriction in the pipe so that the flux stays constant A v · n = A′ v′ · n

(10.28)

Put this into the fluid equation of motion ) ∂v ( ∂v 1 1 + ∇v · v = + ∇(v · v) − v × (∇ × v) = − ∇P ∂t ∂t 2 ρ

(10.29)

∂v 1 ∂ 1 1 + ∇(v · v) − v × (∇ × v) = (∇Φ) + ∇(v · v) − v × (∇ × v) = − ∇P ∂t 2 ∂t 2 ρ both sides are gradients

( ∂Φ 1 P) ∇ + v2 + = 0, ∂t 2 ρ

( ∂Φ

1 P) + v2 + = f (t) ∂t 2 ρ

(10.30)

(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 2 ρ v + P = constant 2 to which we could add external potential energy densities

dϕ dV

(10.32)

(such as gravitational

dϕ dV

=

dm gh dV

= ρgh)

1 2 dϕ ρv + + P = constant 2 dV

(10.33)

Example 2. Find the speed with which water is ejected from a tiny hole a distance h below the waterline of a very large tub of water open to the atmosphere. Bernoulli’s law says that the left side of Eq. 10.33 is constant throughout 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 2 ρ 0 + ρ gh + Patm 2 v

1 2 = ρ v + 0 + Patm 2 √ 2gh =

A

C

B

(10.34)

Example 3. What is the pressure a distance h below the surface of a body of water? This is the hydrostatic pressure. Use the previous example; plug the hole and apply Bernoulli to points “A” and “C”; 1 2 ρ 0 + ρ gh + Patm 2 P

1 2 ρ0 + 0 + P 2 = Patm + ρ gh

=

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, pressurized fluid, and the pressure exerted by the fluid on the four sides of a submerged still cube of side area A. From Eq. 10.22   P 0 0 Tij = P δij , T =  0 P 0  0 0 P   −P A Fright = −T · i A =  0  (10.36) 0

(10.35)

182

CHAPTER 10. FLUIDS



Ftop

 0 = −T · j A =  −P A  0



Flef t

 PA = −T · (−i) A =  0  0

(10.37)

and so you see that all of these forces are directed inwards on the cube walls. The cubic “bubble” would collapse if these 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 molecular 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 ,

ρv 2 T= 0 0

 0 0 0 0 , 0 0



Flef t

 ρv 2 A = −T · (−i) A =  0  , 0

Ftop = 0

(10.38)

Notice that there are no shear forces exerted by the fluid (again).

Example 6. This is a common GRE problem; A block of fluid of width dx, area A carries momentum dp = dm v = (ρ A dx) v. Of it is stopped by the wall placed in its way (and made to turn as illustrated), within time dt the momentum imparted to the wall per unit time (force on the wall) is dp dx 2 dt = (ρ A dt ) v = ρ v A.

A

dx

Example 7. A Pitot tube is a simple analog device for measuring air-speed of an aircraft. Air (fluid #1) at point “A” at Patm rushes through the pipe at vair , but is still at point “B”, so for the air 1 2 ρair vair + Patm = PB 2

D

For the mercury column (fluid #2), the pressure on top at “D” is Pcabin and comparing two formulas for the hydrostatic pressure at “C”

A B

PC

h1

h2 C

and eliminate PC and PB to get

PB + ρm g h1 Pcabin + ρm g h2

we simplify things Pcabin = Patm √

vair =

= =

ρm g(h2 − h1 ) ρa

by

letting

(10.39)

10.3. ADVANCED TOPICS

10.3

183

Advanced topics3

Lets take Euler’s equation and separate off the pressure term ρ

dv = −∇P + ∇ · T′ dt

(10.40)

from T, calling what is left over T′ . What follows is a very typical model-building approach for describing fluid flow 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 in these (freshman physics) notes. Lets assume that T′ depends on the fluid velocity and other scalar properties possessed by the matter that the fluid is 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 single index label) that involve the components of v. We can come up with three ( ) Tij′ = ρ bi vj + cj vi Tij′

=

Tij′

=

ρα vi vj ∂vi ρµ ∂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 to describe flow that has different velocities for different layers, so there should be velocity gradients, spatial derivatives 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 ( ∂v ∂vy ∂vz ∂vx ∂vy ∂vx ) z − , − , − = (0, 0, 0) ∂y ∂z ∂x ∂z ∂x ∂y Lets write T′ as

(10.42)



T′

 x x x ρµ ∂v ρµ ∂v ρµ ∂v ∂x ∂y ∂z   ∂v ∂v ∂v =  ρµ ∂xy ρµ ∂yy ρµ ∂zy  z z z ρµ ∂v ρµ ∂v ρµ ∂v ∂x ∂y ∂z ) ) )  ( ( ( ∂vy ∂vx ∂vx ∂vz x ρµ ∂v ρµ + ρµ + ∂x ∂y ∂x ∂x ) ( ) ( ∂z 1  ∂vy ∂vy ∂vy ∂vx ∂vz ρµ = ρµ + ρµ +  ∂x ) ∂y ∂z ∂y 2 ( ∂y ( ) ∂vy ∂vz ∂vx ∂vz ∂vz ρµ ∂z + ∂x ρµ ∂z + ∂y ρµ ∂z  ) ( ( ∂v x x ρµ ∂v 0 ρµ ∂v − ∂xy − ∂y ( ) ( ∂z 1  ∂vy ∂v x + 0 ρµ ∂zy −  ρµ − ∂v ∂y + ∂x ) 2 ( ) ( ∂v ∂vz z x ρµ − ∂zy + ∂v 0 ρµ − ∂v ∂z + ∂x ∂y

     ) 

∂vz ∂x ) ∂vz ∂y

   

(10.43)

Notice that for an irrotational fluid the last part is zero. We have refined our constitutive description of an irrotational, viscous fluid to 1 ( ∂vi ∂vj ) dv Tij = 2ρµ Dij , Dij = + , ρ = −∇P + ∇ · T′ (10.44) 2 ∂xj ∂xi dt 3 This

is for the benefit of physics and hydrology students. It is not part of the 201 curriculum, it was part of our 401 course.

184

CHAPTER 10. FLUIDS

which is called the linearized Navier-Stokes equation. Do you need some spare cash? The Clay Institute is offering a $1, 000, 000.00 for a general solution to the fully-developed equations containing the convective term v ·∇v.

h Example 8. Couette flow. Consider such a viscous, irrotational 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 The fluid layer touching the bottom plate must be stationary, the fluid touching the top plate moves with it because of 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; dv dt = 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 automatic transmission in a car works. Then 

T′

∇ · T′

dv ρµ dy 0 0  ∂ ∂ ∂ = ( , , )· ∂x ∂y ∂z

0 dv =  ρµ dy 0



 0 2 d v  =  ρµ dy 2 0

 0 0  0 0 dv ρµ dy 0

dv ρµ dy 0 0

 0 0  0 (10.46)

and the Navier-Stokes equation gives us (∇P0 = 0)

0 = ρµ

d2 v , dy 2

v = a + by,

v=

u y h

(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 the plate. This example tells us very clearly what µ represents; viscosity is basically friction 

0 T′ = −  ρµ uh 0

ρµ uh 0 0

 0 0 , 0



0 F = −T′ · n A =  ρµ uh 0

ρµ uh 0 0

     0 −ρµA uh 0  0 · 1  A= 0 0 0 0

(10.48)

10.3. ADVANCED TOPICS

185

h Example 10. Poisseville or capillary flow. In this case we will have a fluid trapped between two surfaces at y = ±h, but let it be forced with a pressure gradient ∇P = −Ci,

x

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)

−h −

1 ∂P d2 v + µ 2 = 0, ρ ∂x dy

which has solution v=−

d2 v C =− dy 2 ρµ

(10.51)

C 2 y + Ay + B 2ρµ

(10.52)

) C ( 2 y − h2 2ρµ

(10.53)

which reduces to v=−

after enforcing our boundary conditions. There must be quite a high pressure at the left end of the “pipe” to push the flow through at this rate. Lets say the “pipe” has length ℓ in the x-direction and square cross section; the flow rate is ∫ h ∫ h dm CA2 4Ch4 =− = (10.54) dz dy ρ v(y) = dt 3µ 12µ −h −h in which A = (2h)2 is the cross-sectional area of the “pipe”. Lets suppose that this flow rate is constant, since P −P C = lef t ℓ right , we find that there can be a huge pressure drop in the pipe Plef t = Pright +

12µℓ dm A2 dt

(10.55)

This has obvious biomedical ramifications; the pressure drop in an artery is proportional to its length, the blood viscosity, 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, warts and all) ∂v ρ + ρv · ∇v = ρv · ∇v = −∇P + µρ∇2 v (10.56) ∂t Let u, ℓ, p be typical order of magnitude velocity, length and pressure scales for the fluid, so that we can create dimensionless 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 (µ) ( p ) ∇π + ∇2 w w · ∇w = − ρu2 uℓ The number R=

uℓ µ

(10.58)

(10.59)

is called the Reynolds number, it is a ratio of inertial to viscous forces. There can be layers in a fluid near stationary surfaces where viscous forces dominate the fluid motion, even though elsewhere in the fluid viscosity my

186

CHAPTER 10. FLUIDS

play little or no role at all in the fluid motion.

For example if an otherwise constant, uninterrupted fluid hits a stationary planar edge, like a wing, within a layer of thickness δ ≈ √ℓR viscous forces are too large to neglect, whereas beyond this distance we can completely neglect them. This means that you can solve the F.D.N.-S. equations in the two regions separately, and blend the solutions along the boundary of the two regions.

δ(L)

x=L

Outside of this envelope, vx = u(x) and vy = 0, within the envelope ∂vx ∂vx + vy ∂x ∂y ∂vy ∂vy + vy vx ∂x ∂y

vx

( ∂2v 1 ∂P ∂ 2 vx ) x +µ + ρ ∂x ∂x2 ∂y 2 ( 1 ∂P ∂ 2 vy ∂ 2 vy ) = − +µ + ρ ∂y ∂x2 ∂y 2 = −

(10.60)

These can’t be solved exactly (welcome to my world), we can try a sequence of successive approximations, starting with Bernoulli to get rid of the pressure, assuming that it is the same in both regions 1 P + ρu2x = constant, 2

∂P =0 ∂x

(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 ∂ 2 vx ∂x2 ∂ 2 vx ∂y 2

scales like scales like

u ℓ2 u δ2

(10.62) (10.63)

which means that the second term is the largest, so we discard the first term, and to lowest approximation, for large y we have a uniform flow, so ux = u0 , we should have little or no pressure variations in the fluid as well, reducing things down to vx

∂vx ∂vx + vy ∂x ∂y ∂vy ∂vx + ∂x ∂y

≈ µ = 0

∂ 2 vx ∂y 2 (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; nonlinear but somewhat simpler. Incompressibility can be enforced by assuming that v can be gotten from a stream function ψ, which is much like a potential function ∂ψ ∂ψ vx = , vy = − (10.65) ∂y ∂x which gives us a third-order non-linear equation to solve ( ∂ψ )( ∂ 2 ψ ) ( ∂ψ )( ∂ 2 ψ ) ( ∂3ψ ) − = µ ∂y ∂x∂y ∂x ∂y 2 ∂y 3

(10.66)

You sure do have to break a lot of eggs to do science (this is aeronautical and hydrodynamical engineering) at any level 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 √ u0 √ ξ=y , ψ = µxu0 Ψ(ξ) µx

(10.67)

10.3. ADVANCED TOPICS

187

to get the vastly simpler but still most heinous equation Ψ

d2 Ψ d3 Ψ + 2 = 0, dξ 2 dξ 3

Ψ(0) = 0,

dΨ = 0, dξ 0

dΨ =1 dξ ∞

(10.68)

which can be solved numerically. This is a real physics problem, its solution is necessary for the design of supersonic wing 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∇2 v

(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∇2 v ( ) 0 − µ∇2 ∇ × v = For two-dimensional flow ωz =

∂vy ∂x

vx =

−

∂vx ∂y ,

∂ψ , ∂y

µ∇2 ω,

ω =∇×v

(10.70)

and if we do our familiar trick of using a stream function ψ vy = −

∂ψ , ∂x

ω=−

∂2ψ ∂2ψ − = −∇2 ψ ∂x2 ∂y 2

(10.71)

and our fluid velocity obeys the biharmonic equation ∇2 ∇2 ψ = 0

(10.72)

which is fourth order, but a sufficient condition that ψ obeys this is the much simpler harmonic or Laplace equation ∇2 ψ = 0

(10.73)

(this is irrotational, if we want constant rotation ωz = c we look for solutions ∇2 ψ = c). As an example, lets consider flow around a cylinder. Far from the cylinder the flow is undisturbed lim v = −v0 i

r→∞

and so lim ψ

r→∞

= −v0 y = −v0 r sin θ

188

CHAPTER 10. FLUIDS

Because our obstacle is a cylinder, it might be a good idea to switch to cylindrical coordinates, in which our equation becomes ( 1 ∂ ( ∂ψ ) 1 ∂2ψ ) ∇2 ψ = (10.74) r + 2 2 =0 r ∂r ∂r r ∂θ It is easy to verify that the following stream function is a solution, and it obeys Eq. 10.74 ( α) ψ = −v0 r + sin θ (10.75) r We can’t have fluid flow through the obstacle, it must go around it; this is our second boundary condition that we use to get the unknown constant α vr (r = a) = 0, From r= and

∂r x = = cos θ, ∂x r

vr = vx cos θ + vy sin θ

√ x2 + y 2 ,

∂r y = = sin θ, ∂y r

θ = tan−1

(10.76)

y x

∂θ x cos θ = 2 = , ∂y x + y2 r

(10.77) ∂θ sin θ =− ∂x r

(10.78)

together with the chain rule vr

= vx cos θ + vy sin θ ∂ψ ∂ψ = cos θ − sin θ ∂y ∂x ( ∂ψ ∂r ( ∂ψ ∂r ∂ψ ∂θ ) ∂ψ ∂θ ) = + cos θ − + sin θ ∂r ∂y ∂θ ∂y ∂r ∂x ∂θ ∂x

(10.79)

you can show that vr =

1 ∂ψ r ∂θ

(10.80)

and deduce that the stream function is ψ

( a2 ) = −v0 r − sin θ r ( a2 y ) = −v0 y − 2 x + y2

(10.81)

and from it find the velocity components that I have illustrated on the previous page, and compute the drag on the cylinder or the pressure in the fluid around it. Don’t be surprised if the drag is zero, we threw out the baby when we 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 your interest. If it does, come and see me, we have something to talk about. If it does not, well, we just don’t have anything 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 v through a closed surface S can be expressed as an integral of the source strength of v over the volume enclosed by S. I ∫ FS = v · n dA = ∇ · v d3 x (10.82) S

V

It makes some sense when stated in English; the flux counts the fluid flow outward through S, and this is related to how many sources of fluid there are within S.

10.4. APPENDIX

10.4

189

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);

190

10.5

CHAPTER 10. FLUIDS

Problems

g g 1. The density of ice is 0.92 cm 3 and of liquid water 1.0 cm3 . What height of a floating cubic block of ice of side a = 10.0 m will be above the surface of the water?

2. A balloon payload of total mass M = 200 kg is attached to a spherical balloon to be filled with helium. The density of helium is kg kg 0.1786 m 3 , and of air 1.2 m3 . 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 area hℓ holding back water of density ρ and height h. Hint, divide the wall into strips of height dy, depth y below the surface, 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 constriction in the pipe where the cross-sectional area drops to A/4. Suppose that both regions of the pipe are at the same height, 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 fluid mechanics. A vessel of fluid of density ρ connects two pistons of areas A1 and A2 sealed at the top with movable piston heads. Show that if forces F1 and F2 , presumably one of these being the weight of a heavy object that you wish to elevate, are applied to the piston-heads, the condition for static equilibrium is F2 F1 + ρg h1 = + ρg h2 A1 A2 so that a small F1 can be used to support or raise a large F2 . This is how the lifts in an auto repair shop work.

F1 F2

A1 A2 h1

h2

10.5. PROBLEMS

191

v0, R0 6. Water leaks from a faucet of radius R0 = 1.0 cm at the rate of 1.0 gs . Find the speed v0 with which it exits the faucet. As water falls, the stream narrows as illustrated. By the time it has descended h = 5.0 cm, find the radius of the stream Rf and the velocity vf of the fluid. At some point the fluid stream will break up into droplets.

h

vf, Rf

h1

P1

h2

7∗ . Here is a crude “density-meter” for immiscible liquids. It is initially filled with a fluid of density ρ1 (light) presumably known. A darker fluid of density ρ2 is poured into one arm, and the fluids are allowed to settle down. the heights h1 , h2 are measured. Suppose g ρ1 = 1.0 cm 3 , h1 = 5 cm and h2 = 8 cm. Find ρ2 .

P0 h

8. The manometer can be used to measure pressures. Typically one end (say the right) is sealed at the top after a fluid of density ρ is put into the tube. This makes P0 = 0. The pressure P1 . is gotten by measuring 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 the fluid 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 2 ρair vwind A 2

192

CHAPTER 10. FLUIDS

10. Show that on a windy day you must exert a force

P, v

|F| =

1 2 ρair vair A 4

on a doorknob at the edge of the door of area A in order to keep it closed. Hint; compute the torque on the door.

Patm

11. Bernoulli’s law is responsible for aerodynamic lift. A wing cross section is designed to cause air to flow faster over the top surface than it does over the bottom surface. When the airplane has airspeed v, suppose that the speed of air-flow over the top surface is v ′ = α v where α is slightly larger than one. If the surface area of the wing is A for both top and bottom surface (approximately), show that the upward-directed force on the wing (the lift) is |F | = 12. Show that a stream function vx =

1 ρair (α2 − 1) v 2 A 2

∂ψ , ∂y

vy = −

∂ψ ∂x

will ensure that the fluid is incompressible ∇ · v = 0. 13. Show that a for Prandtl flow vx =

∂ψ ∂Ψ = u0 ∂y ∂ξ

Use ode to generate a solution to the Prandtl flow equation Eq. 10.66

1.2 1.0

# 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

vx

0.8 0.6

# our initial data psi=0 p=0 v=0.47 # this makes v_x/u_0-->1 for large y

0.4 0.2 0.0

print x,p step 0,6

0

1

2

3 ξ

4

5

6 2

You must fool around with the initial value of v = ddξΨ 2 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 √ xµ vx ≈ 0.99 u0 is at δ ≈ 5 u0 , 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 diffusion equation ∂f ∂2f =J 2 ∂t ∂x First define τ = J t to get ∂f ∂2f = ∂τ ∂x2 Now note that

1 x 1 f (x, τ ) = √ g( √ ) = √ g(ξ), τ τ τ

x ξ=√ τ

Show that d2 g ξ dg 1 + + g=0 2 dξ 2 dξ 2 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 transversely to the direction it travels in? For a circular-cross section obstacle of radius a, fluid flowing (far away) at velocity v0 i has velocity components vx vy

( ) a2 = −v0 1 − 2 cos(2θ) r a2 = v0 2 sin(2θ) r

In the fluid and bubble the pressure is P0 at rest and at equilibrium. Use Bernoulli’s law to compute the pressure on its surface, explaining why it will deform as I have illustrated.

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.

194

CHAPTER 10. FLUIDS

18. Consider Poisseville flow in a long cylinder of radius a with a pressure gradient ∇P = c k. Look for a solution c v = v(r)k that vanishes on the wall. Show that v(r) = 4µ (r2 − a2 ), and find the mass-flow rate dm = dt

∫

∫

2π

dθ 0

a

(

) ρ v(r) r dr

0

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 =

∆P ℓ

.

19∗ . Consider a rotating cylindrical vessel of fluid, with rotation vector ω = ω k. If you co-rotate with it, a bit of matter in the fluid a distance r from the center would move radially outward were it not for the forces exerted by surrounding matter forcing it to travel in a circle of radius r. This means that in the rotating frame there is an outward directed force of magnitude Fr = mω 2 r on it, which has a centrifugal potential 1 Vc (r) = − mω 2 r2 2 Use Bernoulli’s law to find the shape of the fluid surface in the rotating frame under the combined influences of the gravitational and centrifugal potentials.

Chapter 11

Oscillations Oscillatory motion is periodic motion; the pattern of displacements of a body under the influence of forces is periodic of 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 force depending only on position d2 x m 2 = Fext (x) (11.2) dt to a first order equation by a simple transformation equivalent to constructing a conserved energy; multiply both sides by dx dt ( d2 x )( dx ) dx d ( 1 ( dx )2 ) m = F (x) = m (11.3) 2 dt dt dt dt 2 dt and if there exists a function V (x) such that dV dx then the chain rule allows us to write the differential equation as F (x) = −

(11.4)

d ( 1 ( dx )2 ) d m + V (x) = 0 dt 2 dt dt

(11.5)

dx d since dt V (x) = dV dx dt , 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 ( dx )2 m + V (x) = E 2 dt and invert this to get

∫

x(t)

t−0=

√

x(0)

(11.6)

dx′ 2 m (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 2 kx 2

(11.8)

we can perform this integral, and the result is a periodic function. Starting with ∫

x(t)

t= x(0)

√

dx′ 2 m (E

195

− 12 kx′2 )

(11.9)

196

CHAPTER 11. OSCILLATIONS √

make the variable substitution ′

x =

√

2E sin θ, k

′

dx =

2E cos θ dθ k

(11.10)

and the integral becomes ∫

√ θ(t)

t=

√

θ0

2E k

2 m (E

√

cos θ dθ − E sin θ)

=

2

∫

m k

We solve this equation;

√

θ(t)

dθ = θ0

√ θ(t) = θ0 +

m (θ(t) − θ0 ) k

k t m

(11.11)

(11.12)

and insert it into the equation for x; √ x(t) =

2E sin θ = k

√

( 2E sin θ0 + k

√

k ) t m

(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) implies that

√

( 2E sin θ0 + k

√

(11.14)

√ √ ( ) k ) 2E k t = sin θ0 + (t + T ) m k m

and we can use the trig identity √ √ √ ) ( (√ k ) ( (√ k ) ( k k ) k ) (t + T ) = sin θ0 + t cos T + cos θ0 + t sin T sin θ0 + m m m m m Periodicity will be met if cos and this is true if

(√ k m √

) T

= 1,

k T = 2π, m

sin

(√ k m

(11.15)

(11.16)

) T

√ T = 2π

=0 m k

(11.17)

(11.18)

We often say that the motion repeats with frequency 1 ω f= = , T 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 displacement of the mass is zero but its velocity is v0 . From our work above we start with √ k t) (11.21) x(t) = C sin(θ0 + m and first enforce x(0) = 0 = C sin(θ0 ) (11.22)

197 which gives us θ0 = 0. Next we compute √ √ dx k k v(t) = =C cos( t), dt m m

√ v(0) = v0 = C √

and we arrive at the solution v0 x(t) = √

sin(

k m

k cos(0) = C m

√

k m

k t) m

(11.23)

(11.24)

Example 2.Find the amplitude and phase for a mass m connected to a spring k such that at t = 0 the displacement of the mass is x0 but its velocity is 0. Again we start with √ k x(t) = C sin(θ0 + t) (11.25) m and first enforce x0 x(0) = x0 = C sin(θ0 ), θ0 = sin−1 (11.26) C then √ √ √ k k k x0 dx =C cos(θ0 + t), v(0) = 0 = C cos(sin−1 ) (11.27) v(t) = dt m m m C this indicates that x0 π π sin−1 = , x0 = C sin = C (11.28) C 2 2 and we are finished √ √ π k k x(t) = x0 sin( + t) = x0 cos( t) (11.29) 2 m m 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)

x0 = C sin θ0 ,

v0 = C ω cos θ0

(11.31)

x0 ω = tan θ0 , v0

θ0 = tan−1

x0 ω v0

(11.32)

then x(t) = C sin(θ0 + ω t) reduces to

dividing

instead, square and add x20

v2 + 02 = C 2 sin2 θ0 + C 2 cos2 θ0 = C 2 , ω √

and the master formula is x(t) =

x20 +

√ C=

x20 +

v02 ω2

v02 x0 ω sin(ωt + tan−1 ) ω2 v0

(11.33)

(11.34)

Example 4. How are these factors (amplitude and phase) related to the energy of the oscillating mass? The total mechanical energy is the sum of kinetic and potential, and is conserved; Emech =

1 1 1 1 mv 2 (t) + kx2 (t) = mv 2 (0) + kx2 (0) 2 2 2 2

(11.35)

Multiply by two, divide by k v 2 (t) 2E v 2 (t) v 2 (0) = k + x2 (t) = + x2 (t) = + x2 (0) = C 2 2 2 k ω ω m and we discover that the oscillator energy depends only on its amplitude √ √ x0 ω 2E v02 −1 x0 ω 2 )= sin(ωt + tan−1 ) x(t) = x0 + 2 sin(ωt + tan ω v0 k v0

(11.36)

(11.37)

198

CHAPTER 11. OSCILLATIONS

Example 5. A mass suspended from a spring is pictured below in an equilibrium configuration, with the hanging mass m at rest. The un-stretched spring has a length L. Under these conditions all vertical forces (x−direction) on the mass are in balance

Fx = 0 = −k x0 + m g,

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 equilibrium position will satisfy Fx = m ax = m

d2 x = −k (x0 + x) + m g = −k x dt2

(11.39)

when the mass is displaced by x. This equation has general solution (√ k ) (√ k (√ k ) √ ) x(t) = C1 cos t + C2 sin t = C12 + C22 sin t+ϕ m m m

(11.40)

with C1 = tan ϕ C2

(11.41)

For our particular problem, with displacement x equal to A at t = 0 when the mass is released from rest vx = dx dt = 0 as in the figure below. We find that √ ( dx(t) ) k A = x(0) = C1 , 0= = C2 (11.42) dt t=0 m and so

(√ k ) x(t) = A sin t m

(11.43)

describes the motion of the mass. This is called simply periodic motion of frequency ω 1 f= = 2π 2π

√

k m

and of period T = f1 , which is the time needed to complete one full cycle of the motion.

(11.44)

199

Any object that executes such periodic motion is called a simple harmonic oscillator. A more detailed analysis shows that the mass of the spring ms has a significant affect on the period of the motion, and if properly taken into account, the entire system executes periodic motion with √ (m + m3s ) T = 2π (11.45) k T2 =

4π 2 4π 2 ms m+ k 3k

(11.46)

and so a plot of T 2 versus the hanging mass 2 m should be a straight line of slope 4πk .

This together with an independent measurement of k with which to make a comparison provides us with a means of testing that the motion of a real mass and spring is actually simple harmonic motion.

y

Example 6. The frequency of oscillations can be very simply found by reducing equations of motion to the standard form for oscillation

x

x ¨ = −ω 2 x

Consider a pair of springs k1 and k2 joined at a massless junction (the red ball m1 = 0) connected to a mass m. The procedure to compute ω is to displace the mass by x and find its Newtonian equation of motion. Displace m by x, then spring k1 stretches by x1 and k2 by x2 , with x1 + x2 = x (11.48)

y

−k2 x2

−k1 x1

(11.47)

x

k2 x2

Each spring pulls in opposite directions on m1 = 0 m1 a1 = 0 = −k1 x1 + k2 x2 ,

k1 x1 = k2 x2

(11.49)

Solving we find that x1 =

k2 x, k1 + k2

x2 =

k1 x k1 + k2

(11.50)

200

CHAPTER 11. OSCILLATIONS

Only k2 is attached to m; ma = m

d2 x k1 k2 = m¨ x = −k2 x2 = − x dt2 k1 + k2

(11.51)

Compare this to the oscillator equation and extract the frequency of oscillation x ¨ = −ω 2 x,

ω2 =

k1 k2 m(k1 + k2 )

(11.52)

y

M1

Example 7. These two masses slide on a frictionless surface, and are attached to nothing but one another, and so there are no applied external forces. Therefore momentum is conserved. Displace m1 to the right by x1 and m2 to the left by x2 , stretching the spring by x = x1 +x2 . Apply conservation of momentum;

M2 x

0 = m1

dx1 dx2 − m2 , dt dt

(11.53)

Solving

y

x1 =

m2 x, m1 + m2

x kx

m1 a1,x = −m1

x2

We obtain − m1

x2 =

m1 x (11.54) m1 + m2

Pick one (say m1 ) and write its equation of motion

−k x

x1

m1 x1 = m2 x2

m2 x ¨ = kx, m1 + m2

ω2 =

d2 x1 = F1,x = kx dt2

(m1 + m2 )k m1 m2

(11.55)

(11.56)

y

M

M

Example 8. Momentum is not conserved here; the walls exert external forces through the springs. This system (many masses) can oscillate in two modes each with a different frequency. Displace each mass symmetrically by x, stretching the middle spring k1 by 2x and compressing each outer spring k2 by x. The equation of motion for the right mass is

x

y

m k2 x

−k1 2x

x

−k2 x

d2 x = −2k1 x − k2 x, dt2

ω2 =

k2 + 2k1 m (11.57)

k1 2x

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 (two atoms of mass m, one atom of mass 2m can have two linear oscillatory modes. No external forces are applied, so all motions preserve the center of mass position. Displace each light mass to the left by x and the heavy one to the right by x; this conserves momentum. The right spring is compressed by 2x, the left stretched by 2x Pick any mass, say the far right one, and write its equation of motion; max = −m

d2 x = k(2x), dt2

ω2 =

2k m (11.58)

Example 10. The other oscillatory mode of this linear molecule must also involve displacements that conserve momentum (leave the center of mass fixed). For the m, 2m, m molecule the only other such mode has the middle atom stationary, the outer atoms displace away from it by x; for the rightmost atom the equation of motion is m

d2 x = −kx, dt2

ω2 =

k m

(11.59)

202

CHAPTER 11. OSCILLATIONS

Example 11. This disk of mass m and radius R rolls without slipping. It is drawn with spring stretched by x, and so m is displaced from equilibrium by x to the right. We write the force, torque and rolling conditions

v −k x

− kxi − Ff i = ma = m

d2 i dt2

Ff R(−k) = Icom α(−k)

Ff and a=

d2 x = Rα dt2

(11.60) (11.61)

(11.62)

and eliminate the force of friction, and install the rolling condition. Compare the result to x ¨ = −ω 2 x; − kx −

d2 x Icom d2 x = m , R2 dt2 dt2

ω2 =

k com m + IR 2

(11.63)

Fy Fx

Example 12. This is an example of a physical pendulum, a rigid by pivoted about a fixed point, allowed to execute periodic oscillations. It is illustrated with the rod (mass m length 2ℓ) displaced by some small amount x from equilibrium (hanging vertically with the spring unstretched). The force equations are complex, momentum is not conserved, and the hinge exerts unknown forces that we could in principle compute. To find the frequency of oscillations we hammer the torque equation about the pivot point into the oscillator generic form x ¨ = −ω 2 x. Assuming that x 0 since the bar stretches but e11 , e22 < 0 so the cross-section shrinks, and we have recovered (better yet, obtained from first principles) the summarized version of the first page of section 11.3;

ν

≡

E

≡

(11.154)

e11 λ = , Poisson’s ratio e33 2(λ + µ) t33 (3λ + 2µ) =µ Young’s modulus (11.155) e33 λ+µ

Poisson’s ratio tells us how much transverse dimensions contract when we stretch or squeeze an object (the shrink or bulging of the sides and cross-section), and Young’s modulus is like a spring-constant for Hooke’s law, the relation between applied forces/stresses and the resulting deformations/strains. Example 18. Elastic longitudinal waves (a 202 example). Consider a semi-infinite elastic medium bounded by the plane x = 0 that suffers a normal displacement { f (t) i = 1 ui = 0 i = 2, 3

x2

then u1 = u(X 1 , t), uj = 0 for j = 2, 3 in ρ dv dt = ∇ · t; ρ

x1

∂ 2 u1 ∂2t

∂ 2 uj ∂ 2 u1 + µ ∂X 1 ∂X j ∂X j ∂X j ( 1 ∂ ∂u ∂u2 ∂u3 ) = (λ + µ) + + 1 1 2 ∂X ∂X ∂X ∂X 3 ( ∂ 2 u1 2 1 2 1 ) ∂ u ∂ u + µ + + 1 2 2 2 ∂(X ) ∂(X ) ∂(X 3 )2 2 1 ∂ u = (λ + 2µ) ∂(X 1 )2 = (λ + µ)

(11.156)

11.3. CLASSICAL ELASTICA. STRESS AND STRAIN

215

which has wave-like solutions u1 = f (t − X 1 /c) + g(t + X 1 /c),

c2 =

λ + 2µ ρ

Lets consider instead a sudden shear displacement of the surface x = 0 { f (t) i = 2 ui = 0 i = 1, 3

x2

ρ

∂ 2 u2 ∂2t

x1

∂ 2 uj ∂ 2 u2 + µ ∂X 2 ∂X j ∂X j ∂X j ( 1 ∂u ∂u2 ∂u3 ) ∂ + + = (λ + µ) ∂X 2 ∂X 1 ∂X 2 ∂X 3 ( ∂ 2 u2 2 2 2 2 ) ∂ u ∂ u + µ + + ∂(X 1 )2 ∂(X 2 )2 ∂(X 3 )2 ∂ 2 u2 = µ ∂(X 1 )2 = (λ + µ)

(11.157)

which has wave-like solutions u2 = f (t − X 1 /c′ ) + g(t + X 1 /c′ ),

c′2 =

µ < c2 ρ

which is very interesting, these waves are slower than the previous group of waves (in an earthquake there are both types 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

=

e22

=

e33

=

e12

=

e23

=

e13

=

( ) 1 (2λ + 2µ)t11 − λ(t22 + t33 ) 2µ(3λ + 2µ) ( ) 1 (2λ + 2µ)t22 − λ(t11 + t33 ) 2µ(3λ + 2µ) ( ) 1 (2λ + 2µ)t33 − λ(t11 + t22 ) 2µ(3λ + 2µ) 3λ + 2µ t12 2µ(3λ + 2µ) 3λ + 2µ t23 2µ(3λ + 2µ) 3λ + 2µ t13 2µ(3λ + 2µ)

(11.158)

Example 19. The twisting of a bar. Let X 1 = x,

X 2 = y,

X 3 = 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 angle that depends on how far the cross-section is from the end, and the cross-sections themselves warp out of plane; ( ) ( )( 1 ) ( )( ) u cos θ − 1 sin θ X 0 αz x = ≈ v − sin θ cos θ − 1 X2 −αz 0 y using small angle approximations; let the beam twist at a fixed rate along its length, so at the end of the bar it is twisted by (see the figure of section 11.3) Rθ(ℓ) = tan γ ≈ γ, ℓ

θ(ℓ) =

ℓ γ R

(11.159)

216

CHAPTER 11. OSCILLATIONS

and at some point z from the top

z γ ≈ αz, R

θ(z) =

α=

γ R

(11.160)

Lets look for cross sections with u = −αzy,

v = αzx,

w = αψ(x, y)

For static warping ∂ 2 ui =0 = ∂t2 ∂ 2 u3 ρ 2 =0 = ∂t

∂ 2 uj ∂ 2 ui + µ ∂X i ∂X j ∂X i ∂X j (∂ ( ∂ ( ) ) ∂ ( ∂ ( )) (λ + µ) − αzy + αzx + αψ ∂z ∂x ∂y ∂z ( ∂2ψ ∂2ψ ∂2ψ ) + + + µα ∂x2 ∂y 2 ∂z 2 ∂2ψ ∂2ψ = + , (∇2 ψ = 0) ∂x2 ∂y 2

ρ

0

(λ + µ)

(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?  ( )  ∂ψ 1 0 0 − αy + α 2 ( ∂x  )    ∂ψ 1 e= 0 0 αx + α (11.162)  2 ∂y  ( ) ( )  ∂ψ ∂ψ 1 1 0 2 − αy + α ∂x 2 αx + α ∂y and for this problem t = 2µe gives    t= 

(

0

0

0

0

µ − αy + α ∂ψ ∂x

)

(

µ αx + α ∂ψ ∂y

)

( ) µ − αy + α ∂ψ ∂x ( ) µ αx + α ∂ψ ∂y

    

(11.163)

0

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 = ty = 0 =

nx txx + ny txy nx tyx + ny tyy

tz = 0

nx tzx + ny tzy =

=

( ∂ψ

) ( ∂ψ ) − y nx + + x ny ∂x ∂y

(11.164)

For a circular cross-section rod the entire set txz tyz Fx Fy ∇ ψ 2

= µα

( ∂ψ ∂x ( ∂ψ

−y

)

) = µα +x ∂y ∫ ) ( ∂ψ − y dA = 0 = µα ∂x A = 0 = 0

(11.165)

can be satisfied with ψ = 0! (no warping), since y = r sin θ and ∫

∫

R

r dr 0

2π

dθ(−µα r sin θ) = 0 0

11.3. CLASSICAL ELASTICA. STRESS AND STRAIN

217

and the same for x. The cross-sections remain planar circles, with end-torque ∫ M = µα (x2 + y 2 ) dA A R

∫

∫ 2

= µα

r r dr 0

2π

dθ = 0

πR4 µα 2

= µα J

(11.166)

The stresses are then simply M M y, tyz = µαa = x J J which maxes out on the surface. These are the equations gotten in section 11.3.1; txz = −yµα = −

Eq. 11.135

πR4 θ(ℓ) 2ℓ πR4 ℓγ = G 2ℓ R πR4 πR4 πR4 ℓ αR = Gα = µα = G 2ℓ R 2 2

τ =M

(11.167)

= G

(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 to be extremely rigid in one or more directions. This will be sufficient to cover the only really important elementary architectural 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 ( exx + eyy λ + 2µ

Our set of strains Eq. 11.158 collapses to exx eyy ezz exy

1 ν txx − tyy E E ν 1 tyy − txx = E ( E ) ν = − txx + tyy E 1+ν = txy E =

(11.169)

The lack of all body forces results in ∇ · t = 0 or ∂txx ∂txy + = 0, ∂x ∂y

∂txy ∂tyy + =0 ∂x ∂y

(11.170)

Introduce the Airy stress function txx =

∂2ϕ , ∂y 2

tyy =

∂2ϕ , ∂x2

txy = −

∂2ϕ ∂x∂y

(11.171)

218

CHAPTER 11. OSCILLATIONS

which automatically satisfies ∇ · t = 0, however it makes a big assumption about the stress components, and we should question whether or not they are compatible with boundary conditions. In other words Eq. 11.171 is sufficient to satisfy Eq. 11.170 but may not be compatible with typical boundary conditions. Let X 1 = x and X 2 = y, and recall that ∂ui ∂Xj ∂ 2 ui ∂Xj ∂Xk

= eij + wij ∂eij ∂wij + ∂Xk ∂Xk ∂wik ∂eik + ∂Xj ∂Xj

= =

(11.172)

since the order of the derivatives on the left side is irrelevant ∂ ∂ui ∂ ∂ui = ∂Xk ∂Xj ∂Xj ∂Xk therefore ∂eik ∂eij − ∂Xk ∂Xj

= = =

∂wik ∂wij − ∂Xj ∂Xk ( ( ∂ 1 ∂ui ∂uk )) ∂ ( 1 ( ∂ui ∂uj ) − − − ∂Xj 2 ∂Xk ∂Xi ∂Xk 2 ∂Xj ∂Xi ∂wjk ∂Xi

(11.173)

Then ∂ ( ∂eij ∂eik ) − = ∂Xℓ ∂Xk ∂Xj = =

∂ ( ∂wjk ) ∂Xℓ ∂Xi ∂ ( ∂wjk ) ∂Xi ∂Xℓ ∂eℓk ) ∂ ( ∂eℓj − ∂Xi ∂Xk ∂Xj

(11.174)

which is the compatibility relation that we are after; if we specify stresses, the solution of the equation Eq. 11.171 gives rise to a consistent strain tensor if these are true. In the present case this means that ∂ 2 exy ∂x∂y (1 + ν) ∂ 2 txy 2 E ∂x∂y ∂4ϕ −2 2 2 ∂x ∂y Commonly written as ∇2 ∇2 ϕ 2

∂ 2 exx ∂ 2 eyy + ∂2y ∂2x 2 ( ∂ ∂ 2 ( tyy − νtxx ) txx − νtyy ) = + 2 ∂y E ∂x2 E 4 4 ∂ ϕ ∂ ϕ = + 4 ∂x4 ∂y = 0 =

(11.175)

the biharmonic equation, probably the highest order equation that you will see in undergraduate physics. This equation 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 ∇2 ϕ = 0 subject to your problem’s boundary conditions, which should be the applied loads on surfaces. The resulting ϕ gives t everywhere within the block of elastic matter, and from Eq. 11.169 you obtain the stresses.

11.3. CLASSICAL ELASTICA. STRESS AND STRAIN

219

Typical boundary conditions. Construct torques and load forces; ∫ Nxx = txx dA ∫A Nxy = txy dA A ∫ Mxx = y txx dA

(11.176)

A

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 Nx x = 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 ∫ h ( ∂txy ) ∂txx + = 0 (11.177) dz dy ∂x ∂y −b −h ∫ b ∫ h ∫ b ( ) ∂ dz dy txx = − dz txy (h) − txy (−h) = 0 ∂x −b −h −b ∂Nxx = 0 ∂x ∫ b ∫ h ( ∂tyy ) ∂txy + = 0 (11.178) dz dy ∂x ∂y −b −h ∫ b ∫ h ∫ b ( ) ∂ dz dy txy = − dz tyy (h) − tyy (−h) = −q(x) ∂y −b −h −b ∂Nxy (11.179) = −q(x) ∂y and

∫

∫

b

h

dz −b

dy y −h

∂ ∂x

∫

( ∂t

b

xx

∂x ∫ dz

−b

+

∂txy ) = 0 ∂y

h

−h

dy ytxx

= −

(11.180) ∫

b −b

∫ y=h dz ytxy − (

y=−h

)

h

−h

dy txy

∂Mxx = Nxy (11.181) ∂x Example 20. Pressure loading of a beam. We load the top surface of the beam with a normal load force, the bottom of the beam is not loaded;

y

tx = txy = 0,

ty = tyy = −

tx = txx = ty = tyy = 0

x

P 2b

for

for y = h y = −h

attempt a polynomial solution ∑∑ ϕ= am,n xn y m = ϕ2 (y) + xϕ3 (y) + x2 ϕ1 (y) m=0 n=0

because we should have a symmetric solution ϕ(x, y) = ϕ(−x, y) for a uniform beam supported at both ends as I have illustrated, we reduce this to ϕ = ϕ2 (y) + x2 ϕ1 (y) (11.182)

220

CHAPTER 11. OSCILLATIONS

This is the usual strategy taken in solving a complicated equation, we guess a solution containing coefficients and parameters, substitute it into the equation, solve for the coefficients and parameters. It is low-tech, and requires little or no advanced training in mathematical techniques. Put it into ∇2 ∇2 ϕ = 0 and we obtain 4

∂ 2 ϕ1 ∂ 4 ϕ1 ∂ 4 ϕ2 + x2 + = 0, 2 4 ∂y ∂y ∂y 4

∂ 4 ϕ1 =0 ∂y 4

(11.183)

so that ϕ1 = a + c y + b y 2 + e y 3 which in turn specifies 8b + 24 ey = − Now employ the surface conditions

txy

=0=

y=±h

so

∂ϕ ∂y

∂ 4 ϕ2 ∂y 4

∂ϕ (±h) ∂y

(11.184)

must be even in y, forcing b = 0, then 24 ey = −

∂ 4 ϕ2 , ∂y 4

e ϕ2 = − y 5 + B y 4 + C y 3 + D y 2 + G y + H 5

(11.185)

2 but ∂ϕ ∂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

e f g ϕ2 = − y 5 + y 3 + y, 5 6 2

ϕ1 = a + c y + e y 3

(11.186)

after renaming two constants. The stresses are txx

= 2e y(3x2 − y 2 ) + f y + g

tyy txy

= 2(a + c y + e y 3 ) = −2x(c + 3e y 2 )

(11.187)

Boundary conditions on y = ±h make a=−

P , 8b

c=−

3P , 16bh

e=

P 16bh3

(11.188)

with stresses (3P x2 − 2P y 2 + 48bh3 f ) 8bh3 P (y 3 − 3h2 y − 2h3 ) 8bh3 3P x(h2 − y 2 ) 8bh3

txx

= y

tyy

=

txy

=

(11.189)

but both f and g are unspecified (and g is irrelevant) unless end-conditions at x = ± 2ℓ are specified. For example if it d2 v is simply supported (as I have illustrated it in the figure), we have displacement conditions v(±ℓ/2) = dx 2 (±ℓ/2) = 0, and u(±ℓ/2) = 0. Example 21. A cantilevered beam. Consider an asymmetrical (left/right) solution;

which is automatically biharmonic if

4

d ϕ2 dy 4

ϕ = ϕ1 (y) + xϕ2 (y)

(11.190)

ϕ2 = A + By + Cy 2 + Dy 3

(11.191)

= 0,

11.3. CLASSICAL ELASTICA. STRESS AND STRAIN

221

and we can try to discard ϕ1 altogether and see if we can solve some meaningful problem with just ϕ = x(A + By + Cy 2 + Dy 3 )

(11.192)

alone. −(B + 2Cy + 3Dy 2 )

txy

=

txx tyy

= x(2C + 6Dy) = 0 ∫ h = (2b) txx dy = 8bhC x

Nxx

∫ Nxy

=

Mxx

=

−h h

(2b) −h ∫ h

(2b) −h

txy dy = −4bh(B + Dh2 ) y 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 the beam. We can place no shear loads on these surfaces if txy (±h) = 0. There is no x-directed force on the left end at x = 0, with the choice C = 0 we have no net x-directed forces at the right end either. Then no shear loads on the top and bottom surface will set B + 3Dh2 = 0. This leaves us with txy

= −B(1 −

txx

= −2

tyy

=

0

Nxx

=

(2b)

Nxy

=

B xy h2 ∫

h

−h ∫ h

(2b) ∫

Mxx

=

y2 ) h2

−h h

(2b) −h

txx dy = 0 txy dy = −

8bh B 3

y txx dy =

8bh2 Bx 3

(11.194)

Lets choose the coefficients remaining to create an applied downward force F on the left end, ( )( ) ∫ b ∫ h ∫ h 2 0 −B(1 − hy 2 ) −1 dz dy t · (−i) = 2b dy 2 0 −B(1 − hy 2 ) 0 −b −h −h ) ( ) ( ∫ h 0 0 = 2b dy = 8bhB y2 B(1 − ) −h 3 h2 ( ) 0 = (11.195) −F 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 F xy + xy 3 8bh 8bh3

(11.196)

What is going on at the right end? Compute strains and displacements txx − νtyy E tyy − νtxx eyy = E 1+ν txy exy = E

exx =

3F xy 4bh3 E 3F ν = − xy 4bh3 E 3F y2 = (1 + ν) (1 − 2 ) 8bhE h =

(11.197)

222

CHAPTER 11. OSCILLATIONS

Integrate to get displacements u and v; ∫ u= ∫ v=

3F x2 y + F (y) 8bh3 E 3F = − xy 2 + G(x) 8bh3 E

exx dx = eyy dy

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

(11.198)

11.3. CLASSICAL ELASTICA. STRESS AND STRAIN

223

We have (let b = 1/2, I did this to simplify things) u

=

v

=

) F ( 2 3 3 x y − (2 + ν)y + C y + C 1 2 4Eh3 ) F ( 2 3 − 3ν xy − x + D x + D 1 2 4Eh3

(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 both ∂v and v should vanish there, but you can see this can’t be done for all y, The best we can do is to demand u, ∂x that 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

=

0

=

0

=

) F ( 2 3 3 ℓ y − (2 + ν)y + C y + C 1 2 4Eh3 ) F ( 2 2 − 3ν y − 3ℓ + D 1 4Eh3 ) F ( 2 3 − 3ν ℓy − ℓ + D ℓ + D 1 2 4Eh3

(11.200)

all for y = 0 give D1 = 3ℓ2 , C2 = 0, D2 = −2ℓ3 . You have to conclude that our original Airy function is only an approximate solution to this problem, but it is still rather good because it is simple (a polynomial). Our final displacements are u

=

v

=

) F ( 2 3 3 x y − (2 + ν)y + C y 1 4Eh3 ) F ( 2 3 2 3 − 3ν xy − x + 3ℓ x − 2ℓ 4Eh3

(11.201)

and the moral of the story is that simple polynomial Airy functions are good approximate solutions to a variety of beam and support problems, in general most but not all boundary conditions can be satisfied by them.

224

11.3.4

CHAPTER 11. OSCILLATIONS

Problems

1. Show that the speed of a harmonic oscillator of energy E and amplitude A and frequency ω when its displacement is x is √ v = ω A2 − x2 ( 2. The displacement of a mass m = 0.01 kg is x(t) = 0.25 m sin energy, and its speed at t = 0.

62.83 s t

) − 0.785398 . Find its amplitude, its total

N 3. A mass m = 0.01 kg attached to a spring k = 100 m initially at rest at zero displacement is given an impulse causing 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

4. Very small amplitude deviations from equilibrium (zero-force condition) for any potential will result in simple harmonic motion. For example a commonly used inter-molecular potential (Morse) is

Solid; V(x) = 2a e−2ax−a e −ax

V (x) = 2ae−2ax − ae−ax

Dotted; V(x) = V(x0)+1/2 k (x−x0)2 + ...

Show that a mass subjected to this potential has no force acting on it if it is located at x0 = a1 ln 4, and that if it is allowed to deviate slightly from this position by amount x, it undergoes oscillations of frequency

√ ω=

8a3 e−2ax0 − a3 e−ax0 = m

√

k , m

k=

d2 V (x) dx2 x=x0

y

x

R

y x

5. A very small mass m, to be thought of as a point mass, slides without friction under the influence of gravity in a bowl of radius R as shown. The equilibrium point is at the bottom of the bowl. For very small amplitude displacements x from equilibrium, find the frequency of the periodic motion that it executes. Solve the problem if the mass is a sphere of radius r that rolls without slipping.

11.3. CLASSICAL ELASTICA. STRESS AND STRAIN

225

6. Rocking motion. A solid disk of mass M and radius R has a mass m embedded in it at a distance ℓ from its center. It is placed on-edge on a smooth surface. The dotted outline shows its initial position and orientation. It is rolled slightly (by amount x) to the left and released. It rolls back and forth about the equilibrium initial position. Find the frequency of this rocking motion.

x

θ

y

7. Find the center of mass of a half-ring of radius R and mass m. If this is placed on a flat surface and allowed to rock (roll) back and forth, find the frequency of the motion. x

8. A mass m1 connected to a spring of force constant k is at rest at equilibrium at the origin. It is struck inelastically by a mass m2 moving at speed v0 at t = 0. Find the frequency, amplitude and phase of the resulting simple harmonic motion

v0 m1

See the previous problem.

m2

x(t) = A sin(ωt − ϕ) Solve the problem again if the collision is elastic.

226

CHAPTER 11. OSCILLATIONS

k m2 m1

9. A mass m1 rests on a frictionless surface, and a mass m2 rests upon m1 . There is friction with coefficients µk , µs between the blocks. The top block m2 , shown at rest at equilibrium, is connected to a wall by a spring of force constant k. Determine the maximal amplitude of simple harmonic motion such that both blocks co-move with 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 a m spring of force constant k = 100 N m with initial displacement x(0) = 0.1 m and initial velocity v(0) = 5.0 s .

11. Find the maximal speed and acceleration of the simple harmonic motion executed by a mass m = 0.05 kg conN nected to a spring of force constant k = 100 m with initial displacement x(0) = 0.1 m 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 4r 6 r 12 m in potential V (x) = x012 − x60 . Hints;

V:=(r0/x)^(12)-4(r0/x)^4; solve(df(V,x),x); % will give xeq load_package taylor; taylor(V,x,xeq,4); % will expand around xeq

120 π 13. A mass m connected to a spring executes simple harmonic motion with velocity v(t) = 6.0 m s cos( s t − 12 ) kg when connected to a spring of force-constant 100 s2 . Compute m and the total mechanical energy possessed by the system.

N has initial velocity 14. A mass m = 0.05 kg connected to a fixed support with a spring of force constant k = 100 m m v0 = 0.4 s and initial displacement x0 = 0.05 m. Determine the amplitude of the simple harmonic motion that it executes.

11.3. CLASSICAL ELASTICA. STRESS AND STRAIN

227

FB d

ρB

L−d

ρW

15. Compute the length d of an iceberg of volume L3 that is out of the water when in static equilibrium g g (ρB = 0.92 cm 3 , ρw = 1.0 cm3 ). If the iceberg is pushed down by an amount x and released, find the frequency with which it bobs up and down in the water.

FG

x

16. A Si O2 molecule has a linear oscillatory mode that is illustrated in the figure. The atomic masses are mO = 15.999 Amu, mSi = 28.086 Amu and the chemical bond between Si and O has force constant −9 k = 120Amu s. Find the fre(ps2 ) , where 1 ps = 1.0 × 10 quency of the oscillatory mode. 1 Amu = 1.661×10−27 kg.

y

M

M x

y

x

17. Two blocks of mass M = 0.2 kg are connected as illustrated with springs. The far left and far right springs N N have k = 200 m , the middle spring has k = 300 m . The masses are each displaced 2.0 cm away from equilibrium (top figure) in opposite directions and released. Find the frequency of the resulting periodic motion. Find the amplitude A and total energy E of the oscillations.

228

CHAPTER 11. OSCILLATIONS

ωn θ

18. This is a physical pendulum; an object allowed to pivot about a fixed point under the influence of gravity. It consists of a stick of mass M of length ℓ with a sliding mass m attached a distance d below its center. Find the frequency of its oscillations.

E 19∗ . A disk of mass M and radius R is fixed to the wall with a nail through it at the rim. If displaced (center moved away from a vertical line through the nail) and released, it acts like a physical pendulum. Find its frequency of oscillation.

C 20∗ . A hoop of mass M and radius R is fixed to the wall with a nail through it at the rim. If displaced (center moved away from a vertical line through the nail) and released, it acts like a physical pendulum. Find its frequency of oscillation.

D

21. To your left is a torsional pendulum, a steel disk (A) of radius R = 15 cm and mass M = 4.0 kg suspended from a steel circular cross section wire (B) of radius r = 0.1 cm and length ℓ = 1.0 m. The disk is rotated slightly about its axis and released. Find the frequency of the torsional oscillations.

B

By how much is the steel wire stretched by the disk?

22. Compact bone has approximately the same Young’s modulus as aluminum. By how much does the lower leg-bone (radius 2.0cm, length 1.0 m) of a 1200 pound thoroughbred racehorse compress supporting its weight at rest?

A F

M1

M2 k

23. The spring is stretched by some small amount and released. The disks roll back and forth at what frequency?

11.3. CLASSICAL ELASTICA. STRESS AND STRAIN

k

229

24. A mass m1 rests on a frictionless surface, and a mass m2 rests upon m1 . There is friction with coefficients µk , µs between the blocks. The bottom block m1 , shown at rest at equilibrium, is connected to a wall by a spring of force constant k. Determine the maximal amplitude of simple harmonic motion such that both blocks co-move with no slipping of m2 on m1 .

m2 m1

25. Determine the shearing force Ts needed to cut through a 1.0 cm radius circular 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 under its own weight? Think of any cross-section as supporting the weight of all of the matter above it, and sub-divide it into thin slices. Add up the 2 deformations of all of the slices. Show that you get ρgh 2E .

27. By how much does the end of a cantilevered beam of material of length ℓ, cross-sectional area A and shear modulus G and density ρ deflect under its own weigh? Think of any cross-section as supporting the shear of all of the matter to the right of it, with the left end attached firmly to a wall, and sub-divide it into thin slices. Add up 2 the deformations of all of the slices. Show that the right end is ρgh 2G lower than the left end. This assumes that the beam 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, G compute 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 = 0 is automatically satisfied. Does this remind you of the potential function from which all force components are gotten by taking derivatives? 30. Show that the free end of a cantilevered beam with end-load F is vertically deflected by solution. 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

2 κ=λ+ µ 3

F ℓ2 8bh3 E

from the Airy

230

CHAPTER 11. OSCILLATIONS

32. A traveling wave of wavelength λ moving at speed c on a string is a displacement of a point x on the string at time t ) 2π ∆x = A sin( (x − ct) λ 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 (a free-standing square cross-section column). Show that e12 = e13 = e23 = 0,

e11 = e22 =

ν ρg z, E

e33 = −

ρg z E

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 + 5Cy 3 ) x2 + Dy 3 − Cy 5 , which we have shown is biharmonic, and apply it 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 locked into 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 1 2γ dX 1 = dx2 0 1 dX 2 36. Find the strains and stresses on an h × h block of matter with moduli E, ν subjected to a pure planar isotropic dilation ( ) ( )( ) dx1 1+ϵ 0 dX 1 = dx2 1 1+ϵ dX 2

11.3. CLASSICAL ELASTICA. STRESS AND STRAIN

231

37. Most types of wood cannot suffer through a maximal compression of more than 1.5% of the length of the piece without rupturing on the compressed side, or maximal elongation of more than 2.0% of the length of the piece without rupturing on the stretched side. Consider a piece of wood h × h × ℓ as illustrated. Bend it into an arc of a circle subtending total angle θ, so that its median line (dotted) remains ℓ in length. Find the 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 the plank on the median line and follow the median, and y be ⊥ to the median, with origin at the plank center. Find the stresses in the plank. 39. The energy method. Suppose that you can write the energy of a system as ( dx )2 E=D + Bx2 + C dt then you can solve for x ∫ √

√

dx (E−C) D

inverting this to get

−

= B 2 Dx

( D sin−1 B

√

) ∫ B x = dt = t − t0 E−C

√

(√ B ) E−C x(t) = sin (t − t0 ) B D √ √ B from which you can read off both amplitude A = E−C and frequency ω = B D . 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 ω for rolling spin-rate and for frequency of oscillation, which are two different things) √ 1 1 ( v )2 1 2 k 2 E = mv + I + kx , ω= 2 2 R 2 m + RI2

Example. A diatomic molecule in space moves freely with a stationary center of mass. This requires that when the atoms displace, −m1 x + m2 y = 0 x

y

The energy is E=

or E=

( m )2 ) 1( 1 ( m 1 )2 2 1 m1 + m2 x˙ 2 + k 1 + x , 2 m2 2 m2

1 1 1 m1 x˙ 2 + m2 y˙ 2 + k(x + y)2 2 2 2 ω2 =

k(m1 + m2 ) m1 m2

40. Re-examine the physical pendulum (problem 11.18). Suppose that a planar body of mass m has moment of inertial Icom about an axis through its center of mass. You pass a nail through it a distance x from the COM about which it will swing as a physical pendulum under gravity. Find its frequency ω as a function of x. Find the value of x that maximizes the frequency of oscillation. 41. A mass m is subjected to potential energy function V = 2b x3 + 4c x4 with b, c > 0. Find the equilibrium position xeq where the body experiences no force, and the frequency of small-amplitude oscillations about this equilibrium point.

232

CHAPTER 11. OSCILLATIONS

42. In analogy with a spring, which stores potential energy U=

1 1 k (ℓ − ℓ0 )2 = k (∆ℓ)2 2 2

when stretched or compressed, show that an elastic bar of length ℓ0 and cross-sectional area A0 stores energy U=

1 EA0 (∆ℓ)2 2 ℓ0

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 of planets around the sun, and the force of gravity. By physical and purely mental experimentation Newton determined that the force of gravity exerted on mass M1 by mass M2 satisfies

y F12

M1 F1,2

G =

r1

(12.1)

This was an amazing deduction that can easily be taken for granted today, but at the time at which it was made, it ranked as the most important physical discovery of the century. The magnitude of the force between two masses varies with the inverse square of the distance between them, and is always attractive, pulling one mass towards the other. This force acts vectorially just like any other vector forces.

F2,1 r2

(r1 − r2 ) |r1 − r2 |3 N m2 6.67 × 10−11 kg 2

= −M1 M2 G

M2 x

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. ( (0, 0) − (0, d) (0, 0) − (d, 0) (0, 0) − (d, d) ) √ F = −m2 G + + d3 d3 ( 2d)3 which can be simplified a bit to F=

1 m2 G ( 1 ) (1 + √ ), (1 + √ ) 2 d 8 8

(12.2)

(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 center of mass is at the origin then the planets travel in circles of radii r1 and r2 that satisfy 0 = M1 r1 − M2 r2

(12.4)

r = r1 + r2

(12.5)

and are separated by a distance

233

234

CHAPTER 12. GRAVITATION AND PLANETARY MOTION y

Examine planet number one. It is in a circular orbit and therefore has centripetal acceleration ac v2 ac = 1 (12.6) r1

v1 M1

r1

and the centripetal force must be supplied by the gravitational attraction to mass M2 x

M1

r2

M2

M1 M2 G v12 = r1 (r1 + r2 )2

(12.7)

The period is the time needed to orbit the center of mass once v2

T =

√

we find that M2 r r1 = , M1 + M2

v1 =

which leads to

2πr1 v1

M22 G r(M1 + M2 )

(12.8)

(12.9)

3

2πr 2 T =√ G(M1 + M2 )

(12.10)

which we have written in terms of the interstellar distance r. y v

M

F2

F1 R

x

Example 3. Compute the period of a triple star system of masses M and equal radii R. Such a star system again consists of the stars in circular orbits about their center of mass. If the center of mass is at the origin then the planets travel in circles of √ radii R. The interstellar distance is R12 = R 3. The acceleration vector of the top star is ac = −

v2 j R

(12.11)

and this is produced by the forces exerted by the two lower stars;

F1,2 + F1,3

) ) M 2G ( M 2G ( √ sin 30o i − cos 30o j + √ − sin 30o i − cos 30o j ( 3R)2 ( 3R)2 2 M G = −√ j 3 R2

=

(12.12)

12.2. GRAVITATIONAL POTENTIAL ENERGY and so

235

( v2 ) ( 22 π 2 R2 ) M 2G √ j M − j =M − j = − R T2 R 3 R2

The period T of the motion is therefore T2 =

12.2

√ 4π 2 3R3 MG

(12.13)

(12.14)

Gravitational potential energy

We can construct a potential energy function for the gravitational field by equating the difference in potential energy between points at ri and rf with the work done against the gravitational field in moving from the first to the second point. ( ) ∫ rf mM Gr · dr (12.15) WFg = − V (rf ) − V (ri ) = − r3 ri where the mass M at the origin exerts gravitational forces on the test mass m that we are moving through the force field of stationary mass M . We can rewrite this by introducing the vector derivative or gradient ( ∂ ∂ ∂ ) ∂ ∂ ∂ ∇= , , =i +j +k (12.16) ∂x ∂y ∂z ∂x ∂y ∂z and by noting that

x dx + y dy + z dz r · dr 1 1 √ = ) · dr = −(∇ ) · dr = −(∇ √ 3 r r ( x2 + y 2 + z 2 )3 x2 + y 2 + z 2

and the integral becomes an integral of an exact differential ∫ ( ) − V (rf ) − V (ri ) =

( M mG ) ∇ − · dr r ri M mG rf = − r ri mM G M mG + = − |rf | |ri |

(12.17)

rf

(12.18)

we discover that the point mass M placed at the origin produces a gravitational potential energy field whose value depends only on how far one is from the origin V (r) = V (r) = −

M mG |r|

(12.19)

This formula completely replaces Vg = mgh in any calculation in which the altitude h is a significant distance compared 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 1 mM G M mG = mv 2 − (12.20) − 2Re 2 Re so the impact speed will be √ MG v= (12.21) Re 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 at rest. 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; M1 v1 = M2 v2 ,

1 M1 M2 G M1 M2 G 1 M 1 v1 + M 2 v2 − =0+0− 2 2 (R1 + R2 ) R

(12.22)

236

CHAPTER 12. GRAVITATION AND PLANETARY MOTION

Solving we get v12 =

2M22 G ( 1 1) − (M1 + M2 ) R1 + R2 R

(12.23)

Example 6. At what speed must a projectile be launched from the earth’s surface at in order to barely escape the earth’s gravitational field? This is the escape velocity, and is found from the condition that v = 0 when R → ∞, since the gravitational field reaches to infinity. Conserve energy; 1 mMe G 1 2 mMe G v − = m02 − =0 m e Re 2 “∞′′ and we obtain

√ ve =

(12.24)

2Me G Re

(12.25)

Example 7. A black hole is a star that has collapsed under its own gravitational field. This happens when energy production in the stellar core is too slow to provide enough heat and pressure to support the star from within. The event-horizon is a roughly spherical shell around the hole upon which the escape velocity is the speed of light. For a one-solar mass black hole, find its radius. We have 2 √ (2)(2 × 1030 kg)(6.67 × 10−11 Nkgm2 ) 2M⊙ G 2M⊙ G c= , Reh = = = 3.0 km (12.26) 2 Reh c2 (3 × 108 m s ) This is the physical size of a black hole of the same mass as the sun. Black holes do not form from stars whose initial mass is below three solar masses, although in the process some mass is blown off in the explosion and collapse.

M1

a

M3

b

M2

Example 8. Find the gravitational force exerted on M3 , and find the work needed to move it from the top position to the lower. For the top figure F=−

M1 M3 G M2 M3 G i+ i a2 b2

(12.27)

and −(Vf − Vi )) ( (M M G M M G) 1 3 2 3 b M3 a M2 M1 = − − + b a ( M M G M M G )) 1 3 2 3 + + (12.28) a b Example 9. Two stationary masses (held in place) M are located at points (x, y, z) = (0, b, 0), (0, −b, 0). A third mass 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; Wf

=

−mM G −mM G W = −(Vf − Vi ) = −(0 + 0 − √ −√ ) 2 2 a +b a2 + b2

(12.29)

(you add up the potentials between m and each other mass). The force is ) ) mM G ( a b mM G ( a −b √ −√ i+ √ j + √ −√ i+ √ j ( a2 + b2 )2 a2 + b2 a2 + b2 ( a2 + b2 )2 a2 + b2 a2 + b2 mM G −a √ = 2 2 i 2 a +b a2 + b2

F =

(12.30)

12.2. GRAVITATIONAL POTENTIAL ENERGY

237

y

Example 10. Find the gravitational force exerted by a stick of mass m length 2ℓ on a point mass M held a distance x from the stick center. All we have to go on is the form of the force between two point-masses, so divide the stick into point masses of length dy and mass

dm=ρ dy

dm = ρ dy =

y x x

M

m dy 2ℓ

(12.31)

The fragment illustrated exerts force ) dm M G ( dF = √ − xi + yj ( x2 + y 2 )3

(12.32)

on the point-mass M , and integration gives a total force ∫ ℓ ( ) m dy MG √ F= − xi + yj (12.33) 2 2 3 −ℓ 2ℓ ( x + y )

y

Example 11. Find the gravitational force exerted by a stick of mass m length 2ℓ on a point mass M held a distance x from the stick end. divide the stick into point masses of length dy and mass m dm = ρ dy = dy (12.34) 2ℓ The fragment illustrated exerts force x dm=ρ dy

y

x

dF =

M

) dm M G ( − (x + y)i ((x + y)3

(12.35)

and integration gives ∫ F = =

) m dy M G ( − (x + y)i 2ℓ((x + y)3 0 mM G ( 1 1) − i (12.36) 2ℓ x + 2ℓ x 2ℓ

238

CHAPTER 12. GRAVITATION AND PLANETARY MOTION

Example 12. Find the gravitational force exerted by a ring of mass m radius R on a point mass M held a distance z from the center. Divide the ring into point masses of length R dθ and m mass ρ R dθ, ρ = 2πR . The fragment at location ( ) r1 = R cos θi + R sin θj (12.37)

M

z

R dm=ρ R dθ θ

is a distance

√ z 2 + R2 from our mass, and so exerts

) dmM G ( dF = √ R cos θi+R sin θj−zk (12.38) ( z 2 + R 2 )3 on it. Two of three integrals are zero; ∫ F= 0

2π

) ρ2πRM G z mM G z ρRdθM G ( √ R cos θi + R sin θj − zk = − √ k=− √ k 2 2 3 2 2 3 ( z +R ) ( z +R ) ( z 2 + R2 )3

(12.39)

Example 13. Compute the gravitational potential energy function for a ring of mass m radius R, at a distance z from 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 dm M G dV (z) = − √ z 2 + 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 the potential energies of M with all of the ring fragments; ∫ V (z) = −

dm M G mM G √ = −√ 2 2 z +R z 2 + R2

(12.41)

for the total. Furthermore we can compute Fz from the fact that the gravitational force is conservative; Fz = −

∂V (z) mM G z =− √ ∂z ( z 2 + R2 )3

(12.42)

in agreement with the previous example. Example 14. Compute the gravitational force on a mass M exerted by a flat disk of matter of mass m a distance z away. Divide the disk into concentric rings of thickness dr, m radius r and mass dm = πR 2 2πr dr. Use the expression for the force exerted on M by a ring;

M

z

∫

dm=ρ 2π r dr

F r

dr

R

MG z m − √ 2πr dr k 2 2 2 3 πR ( z +r ) 0 ) 2mM G ( z z √ = − − k (12.43) R2 |z| z 2 + R2 =

12.2. GRAVITATIONAL POTENTIAL ENERGY

239

Example 15. Compute the gravitational potential energy of a mass M inside of a thin-skinned hollow spherical shell of radius R and mass m. Subdivide the shell into a stack of rings, one seen edge-on in the figure. Each ring has radius r = R sin θ, is a distance z + R cos θ from M , and if cut and laid flat would be a strip of area dA = (R dθ)(2πR sin θ) and mass

M

z

θ

m m sin θ dθ 2π R2 sin θ dθ = 4πR2 2 (12.44) The potential energy of M in the field of the stack of rings is ∫ π dm M G (12.45) V (z) = −√ 0 (z + R cos θ)2 + R2 sin2 θ dm = ρ dA =

R R sinθ dm

∫

π

= 0

) MG m sin θ dθ mM G ( −√ = |R − z| − |R + z| 2 2Rz (z + R cos θ)2 + R2 sin2 θ

(12.46)

) mM G ( mM G (R − z) − (R + z) = − 2Rz R

(12.47)

If z < R, this results in

V (z) =

a result completely independent of z, therefore inside of this hollow sphere there is no gravitational field and M experiences no force ∂V =0 ∂z

(12.48)

) mM G mM G ( (z − R) − (R + z) = − 2Rz z

(12.49)

mM G ∂V =− ∂z z2

(12.50)

Fz = −

If z > R we obtain

V (z) =

and m experiences a force

Fz = −

which is exactly the same as if all of the mass M comprising the shell were concentrated at is center.

240

CHAPTER 12. GRAVITATION AND PLANETARY MOTION

M Example 16. Compute the force exerted by a ball of radius r, mass m on a point mass M a distance z > R away (measured center to center). Subdivide the sphere into a stack of disks, the one shown is z − ζ away from M , has radius √ r2 − ζ 2 and has mass

z−ζ √  r2−ζ2

dζ

ζ

dm =

m

√

4π 3 2π( 3 r

r2 − ζ 2 )2 dζ

(12.51)

It exerts a force ) 2 dm M G ( z−ζ √ 1 − k r2 (z − ζ)2 + r2 (12.52) on M , and integration results in dF = −

∫

r

F= −r

−

) m √ 2 MG( mM G z−ζ √ 2π( r2 − ζ 2 )2 dζ k = − 1 − k 4π 2 3 r z2 (z − ζ)2 + r2 3 r

(12.53)

This is the same result that we would get by replacing the entire ball with a point mass m and placing it 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 mass M radius R on a point mass m located inside of 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) = ρ2 dρ m (4πρ2 )(dρ) 4πR 3 = 3m R3 , each acting like a point mass concentrated at the center of the sphere and exerting 3

dFr =

−m dm(ρ)G r2

if

ρ m of radius R. Determine the speed v of its orbit. It is struck inelastically by a fast-moving comet of mass µ T1 . Place them in thermal contact, and let dQ flow from the hot to the cold. Then for the whole irreversible process dQ −dQ + ≥0 T1 T2

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 appropriate form of the energy useful for computation when a specific set of conditions are met. 1. Helmholtz Free Energy F = U − TS

(14.42)

284

CHAPTER 14. EQUILIBRIUM THERMODYNAMICS

2. Gibbs Free Energy G = U − TS + PV

(14.43)

H = G + TS

(14.44)

3. Enthalpy

14.3.1

The meaning of F ; the F-Theorem;

In a mechanically isolated system at constant temperature, F never increases and thermal equilibrium is a state of extremal F . Proof; dQ ≤ ∆S T

(14.45)

∆W ≤ −∆U + T ∆S

(14.46)

− ∆F = −∆U + T ∆S

(14.47)

at constant temperature, and so but 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 contact with a heat reservoir at constant volume is the decrease in F . Chemists tend to use the Gibbs free energy since all of their experiments are conducted in the open atmosphere, at constant P . Physicists use the Helmholtz free 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 ideal monatomic gas at volume 10 ℓ, on the other side is one mole of the same gas at volume 2 ℓ. The entire apparatus is immersed in a heat bath that maintains both gases at 273 K. Compute the Helmholtz free energy change as the partition 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 ( T 3 V ) F F0 = − ln ( ) 2 nRT nRT0 T0 V0

(14.50)

∆F = −nRT (ln 6 − ln 10 + ln 6 − ln 2) = −∆W

(14.51)

and so The work delivered in a reversible process at constant temperature for a system in contact with a heat bath is the decrease 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)

∆G = P ∆V + ∆U − T ∆S

(14.53)

at constant pressure, and

14.3. THERMODYNAMIC POTENTIALS

285

at constant T . From the first law, for a constant T process ∆Q ≤ ∆S T

(14.54)

P ∆V = ∆W ≤ −∆U + T ∆S

(14.55)

0 ≤ −P ∆V − ∆U + T ∆S = −∆G

(14.56)

substituting this we find that and so 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 differentials dF = dU − T dS − S dT = dQ − P dV − T dS − S dT

(14.59)

dQ = dS T

(14.60)

dF = −P dV − S dT

(14.61)

but for a reversible process

which causes two terms to cancel, leaving only

This tells us how the derivatives of F are related to the properties of the system ( ∂F ) ( ∂F ) P =− , S=− ∂V T ∂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-P V work done by a system in a reversible, constant P constant T process, but any exploration of this must wait until we discuss non-P V work, such as electrical, magnetic or chemical work.

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 this statement 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 spontaneous change. We will discover that in a spontaneous or irreversible process, the entropy always increases. Some chemistry people say that entropy is a measure of exhaustion, in that it increases as a systems ability to undergo further spontaneous change decreases. The entropy function and second law gives us a powerful tool for the computation of equilibrium temperatures in the case of two or more systems brought into thermal contact. Consider two identical systems of completely unknown characters, with constant specific heats and respective initial temperatures T1 and T2 . If they are brought into thermal contact and allowed to equilibrate by reversible heat exchange, they will reach a temperature Tf determined by 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; ∫ Tf ∫ Tf ∆U1 + ∆U2 = 0 = C dT + C dT (14.63) T1

T2

286

CHAPTER 14. EQUILIBRIUM THERMODYNAMICS

and the final temperature is found to be T1 + T2 2

Tf = and entropy change

∫

Tf

∆S = T1

dT C + T

∫

(14.64)

Tf

C T2

dT >0 T

(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 the volumes fixed will increase or decrease. The final temperature is determined by the second law ∫

Tf

∆S = 0 = T1

dT C + T

giving, in the case of identical specific heats Tf =

∫

Tf

C T2

dT T

√ T1 T2

(14.66)

(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 P V 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 ∫

Tf

∆U1 = ∆Q1 + P1 ∆V1 = ∆Q1 =

C dT

(14.70)

C dT

(14.71)

T1

∫

Tf

∆U2 = ∆Q2 + P2 ∆V2 = ∆Q2 = T2

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 with a 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 be extracted from a change of state of a system. Overall energy must be conserved, taking all forms into account, and entropy can at best be constant, and in any but a reversible process, must increase. The most work that a state change could deliver then will be delivered when the state change is carried out reversibly, so dS = 0 and heat transfer is minimized. This is a principle of maximal work, a consequence of the increase of entropy. Depending on exactly how an irreversible process is carried out, the entropy changes and heat transfers can vary, and therefore so can the work delivered to a work source. When the heat transfer is zero though, so is ∆S and the delivered work is 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.

14.3. THERMODYNAMIC POTENTIALS

287

V0 = 5.0 liters

p1

p2

p3

p4

p5

p6

There are six partitions, between each is an additional 1 ℓ of empty space. We can compute the entropy change as each partition is removed, and we know that once removed, the molecules will not ignore the new volume opened up to them, they will spontaneously expand into it. Entropy calculations require equations of state if you wish to carry them out. For an ideal gas P V = nRT,

U = nCV T

(14.73)

You can’t increase the gas energy by removing a partition, so dU

= 0 = n CV dT,

∆S

Remove p2 ;

∆S

Remove p3 ;

∆S

dT = 0 P dV dS = dV = nR T V

T dS − P dV, ⇒ ∫ 6ℓ dV = = R ln(6/5) = 0.182 R R V 5ℓ ∫ 7ℓ dV = R = R ln(7/6) = 0.154 R V 6ℓ ∫ 8ℓ dV = R = R ln(8/7) = 0.134 R V 7ℓ

dU = 0 = Remove p1 ;

⇒

(14.74)

In each case ∆S > 0, indicating that the change is spontaneous, but in each case ∆S is smaller than in the previous step, 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 measuring heats of reaction; the amount of heat needed to initiate a reaction or the amount liberated by a reaction. If the amount of heat absorbed in a chemical reaction depended upon the thermodynamic path, rather than just initial and final states, experiments would have to be conducted accordingly. In 1840 Hess demonstrated that the heat absorbed in a chemical reaction was independent of the number of steps in the reaction; different reaction paths gave 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 quantity H = H(S, P ),

H = U + P V,

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)

288

CHAPTER 14. EQUILIBRIUM THERMODYNAMICS

from CaO + H2 O ⇌ Ca(OH)2 ,

cal mole cal = −68315 mole cal = −151800 mole

∆H18o C = −15260

1 H2 + O2 ⇌ H2 O, 2 1 Ca + O2 ⇌ CaO, 2

∆H18o C ∆H18o C

∆H18o C = −235375

cal mole

(14.79)

cal adding all of it together we obtain ∆H18o C = −235375 mole 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 = isobaric heat capacity is CP =

( ∂Q )

( ∂Q )

∂T

=

( ∂U )

V

∂T

( ∂V )

+

( ∂U )

(14.81) ∂T P ∂T P ∂T P These quantities tell us how much the energy must change by with an appropriate variable held fixed, in order for the temperature to change by a specific amount. If CP is high, like it is for water, a small temperature change could require quite a bit of energy. The coefficient of thermal expansion is 1 ( ∂V ) (14.82) 3α = V ∂T P which measures an objects size changes at constant pressure as its temperature is changed.

14.4

=P

(14.80) V

Ideal gases

The ideal gas is one of the few systems for which one can very easily construct a detailed and exact picture of thermodynamical processes. This is because the two equations of state P V = N kT = n NA kT = nRT,

U=

3 3 N kT = nNA kT = nCV T 2 2

(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 P V = nRT, P dV + V dP = nR dT

U = nCV T dU = nCv dT

(14.84)

2. The first law dU = dQ − dW specialized to reversible (equilibrium) processes without any form of work other than P V , 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 applications of 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 subjected to. 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 = nCV

dT , T

∫

S2

∫

T2

dS = S2 − S1 =

nCV

S1

T1

T2 dT = nCV ln T T1

(14.85)

Example 2. Compute the increase in internal energy of an ideal gas that is allowed to expand from V1 to V2 at constant 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 constant temperature. Use the first law with dT = 0, note that the second equation of state implies dU = 0; dS = nR

dV , V

∫

S2

∫

V2

dS = S2 − S1 =

S1

nR V1

V2 dV = nR ln V V1

(14.87)

Example 4. Compute the increase in internal energy of an ideal gas that is heated from T1 to T2 at constant volume. Use ∫ U2 ∫ T2 ndU = nCV dT, dU = U2 − U1 = nCV dT = CV (T2 − T1 ) (14.88) U1

T1

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 and adiabatically 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 gas law nRT dT nR dV 2 dV dU = nCV dT = − dV, =− =− (14.90) V T nCV V 3 V and integrate ∫ T2 ∫ dT T2 2 V2 dV 2 V2 = ln =− = − ln (14.91) T T 3 V 3 V1 1 T1 V1 rearrange (remember the second of these, it is very handy) ln

T23 V12 = ln , T13 V22

2

2

T2 V23 = T1 V13

for adiabatic processes

(14.92)

290

CHAPTER 14. EQUILIBRIUM THERMODYNAMICS

and we get ∆U = nCV (T2 − T1 ) = nCV T1

(( V ) 23 1

) −1

(14.93) V2 If you use the ideal gas law to eliminate the temperature in the so-called adiabatic relation, you get a second adiabatic relation 2 2 5 5 T2 V23 = T1 V13 , P2 V23 = P1 V13 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 from V1 to V2 . Start with dU = δQ − P dV = T dS − P dV (14.95) and use F = U − T S,

dF = dU − T dS − S dT = −S dT − P dV

(14.96)

For isothermal expansion dT = 0, and insert the ideal gas law dV , dF = −P dV = nRT V

∫ F2 − F1 = −nRT

V2

V1

V2 dV = −nRT ln V V1

(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 use G = U − T S + P V,

dG = dU − T dS − S dT + P dV + V dP = −S dT + V dP

For isothermal expansion dT = 0, and insert the ideal gas law ∫ P2 P2 nRT G2 − G1 = dP = nRT ln P P1 P1

(14.99)

(14.100)

Example 8. Compute the heat that flows into a gas during a constant pressure heating process in which the temperature 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 ( nRT ) nR dT nRT dV dP = 0 = d = − , V V V2

dT dV = T V

(14.102)

therefore inserting this and the ideal gas law; δQ = dU + P dV = dU + P ∫

This leads us to ∆Q =

δQ =

3 V dT = dU + nR dT = nR dT + nR dT T 2

(14.103)

5 nR(T2 − T1 ) = CP (T2 − T1 ) 2

(14.104)

Example 9. There is a master entropy formula for the ideal gas, from which you can get ∆S for any state-to-state process. Find it! Rearrange the first law dV dT + nR T V can be integrated to a final entropy formula for the ideal gas. We first note that it can be written as ( ∂S ) ( ∂S ) dT dV dS = nCV + nR = dT + dV T V ∂T V ∂V T dS = nCV

(14.105)

(14.106)

14.5. P V -DIAGRAMS

291

Divide by dσ

( ∂S ) dT ( ∂S ) dV dS = + dσ ∂T V dσ ∂V T dσ

and integrate

∫

dS dσ = dσ

∫ ( ∫ ( ∂S ) dT ∂S ) dV dσ + dσ ∂T V dσ ∂V T dσ

(14.107)

(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 and V = V1 . We can change variables in each of the three integrals and perform each one over a separate, appropriate variable; ∫ σ1 ∫ S1 ∫ σ1 ( ∫ T1 ( dS ∂S ) dT ∂S ) dσ = dS = S1 − S0 , dσ = dT (14.109) ∂T V dσ ∂T V σ0 dσ S0 σ0 T0 and

∫

σ1

σ0

∫ V1 ( ( ∂S ) dV ∂S ) dσ = dV ∂V T dσ ∂V T V0

(14.110)

Performing these last two explicitly for the ideal gas; ∫

T1

∫

( ∂S ) ∂T

T0

T1

dT = V

nCV T0

T1 dT = nCV ln T T0

(14.111)

dV V1 = nR ln V V0

(14.112)

Performing these last two explicitly for the ideal gas; ∫

V1

V0

∫

( ∂S ) ∂V

and we arrive at S1 − S0 = nCV ln

V1

dV = T

nR V0

(( T ) 32 ( V )) V1 T1 1 1 + nR ln = nR ln T0 V0 T0 V0

(14.113)

Example 10. There is a master free energy formula for the ideal gas, from which you can get ∆F for any state-tostate process. Find it! A master Helmholtz free energy equation can be obtained by exactly the same means. Starting with the definition F U = −S T T

(14.114)

F (T1 , V1 ) U (T1 , V1 ) = − S(T1 , V1 ) T1 T1

(14.115)

(( T ) 32 ( V )) F (T1 , V1 ) F (T0 , V0 ) 1 1 − = S(T1 , V1 ) − S(T0 , V0 ) = nR ln T1 T0 T0 V0

(14.116)

F = U − T S, we obtain F (T0 , V0 ) U (T0 , V0 ) = − S(T0 , V0 ), T0 T0 subtract these and use

14.5

U T

= CV = 32 nR for an ideal gas

P V -diagrams

Because the first equation of state is a constraint f (V, T, P ) = 0, the state of the ideal gas is specified by two quantities, 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 a process on a P V -diagram. The most technically important processes are cycles, in which the gas is subjected to a sequence of sub-processes that ultimately return it to its original state. These sub-processes could be arbitrary, but we will study only constant V , constant P , isothermal and adiabatic

292

CHAPTER 14. EQUILIBRIUM THERMODYNAMICS

Process

∆U

∆Q

∆W

Constant V Constant T Adiabatic Constant P

nCV (Tf − Ti ) 0 nCV (Tf − Ti ) nCV (Tf − Ti )

nCV (Tf − Ti ) V nRT ln Vfi 0 n(CV + R)(Tf − Ti )

0 V nRT ln Vfi −nCv (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 diagram and 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 all changes over the cycle sub-processes must be zero, because the cycle returns the gas to its original state

∆X =

∑

∆Xi = 0,

X = U, S, F, G, H,

for any cycle

(14.118)

i

Example 11. Consider the thermodynamic cycle carried out on one mole of monatomic ideal gas illustrated to the left, a → b is constant P , b → c is constant T , c → d is constant P , d → a is constant T . Fill in the table of N thermodynamic quantities below. Pa = 1.0 × 104 m 2, 3 3 4 N Va = 1.0 m , Vb = 2.0 m , Pc = 0.5 × 10 m2 . Note that Vd ̸= Vb . The first thing you do is get the temperatures; Pa Va R Pb Vb Tb = R Tc Td

Ta =

(1.0 × 104 )(1.0) = 1203 K 8.314 (2.0 × 104 )(1.0) = = 2406 K 8.314 = Tb = 2406 K = Ta = 1203 K (14.119) =

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 = 4 m3

Now its all pretty easy; for example ∆Ua−>b

=

∆Wa−>b

= =

∆Wb−>c

= =

3 1 · (8.314)(2406 − 1203) = 15, 000 J 2 Pa (Vb − Va ) = (1 × 104 )(2.0 − 1.0) 10, 000J Vc 1 · RTb ln = (8.314)(2406) ln(2) Vb 13863 J (14.120)

14.5. P V -DIAGRAMS

293 Example 12. Consider the thermodynamic cycle carried out on one mole of monatomic ideal gas illustrated to the left, a → b is adiabatic, b → c is constant T , c → a is constant V . Fill in the table of thermodyN namic quantities below. Pa = 1.0 × 104 m 2, 3 4 N Va = 1.0 m , Pb = 0.25 × 10 m2 . Get the temperatures Ta Vb Tb Tc

Pa Va = 1203 K R ( P ) 35 a = Va · = 2.3 m3 Pb ( V ) 23 a = Ta = 690 K Vb = Tb = 690 K (14.121) =

Once you have the temperatures, its very easy to wrap it all up.

3 (8.314)(690 − 1203) 2 = −6397J

∆Ua−>b

=

Process

∆U

∆Q

∆W

∆S

∆Sa−>b

= 0

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

∆Wb−>c ∆Sb−>c

(adiabatic) (V ) c = RTb ln = −4778J Vb (V ) c = R ln = −6.92J/K Vb (14.122)

Always use the first law ∆U = ∆Q − ∆W ; apply it to a row in a table, you need only two out of three entries per row, 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 =

Qh Qc − Th Tc

(14.124)

The figure below details a four step cycle for a heat engine in which two isothermal processes perfectly turn heat into work, and cool the gas by requiring work input to expel unwanted residual heat in the gas. The cycle is completed with adiabatic steps. Heat in/out is with thermal (conductive) contact with heat sources/sinks

294

CHAPTER 14. EQUILIBRIUM THERMODYNAMICS

It is a trivial matter to compute the work, heat and 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. Notice first Pa Va = nRTh = Pb Vb ,

Pc Vc = nRTc = Pd Vd

∆Qa→b = ∆Wa→b ∆Ub→c = −∆Wb→c

and so forth, using the first law of thermodynamics. process a→b b→c c→d d→a

∆U 0 3 2 nR(Tc − Th ) 0 3 2 nR(Th − Tc )

∆Q nRTh ln( VVab ) 0 nRTc ln( VVdc ) 0

∆W nRTh ln( VVab ) − 32 nR(Tc − Th ) nRTc ln( VVdc ) − 32 nR(Th − Tc )

∆S nR ln( VVab ) 0 nR ln( VVdc ) 0

Notice that the sum of all of the energy changes for the entire cycle is zero. This is because the gas returns to its original state, and therefore original temperature, at the end of the cycle. Since the energy of an ideal gas depends only on the temperature, the energy change per cycle is zero. So what does the cycle do? It extracts heat from the high temperature heat bath, which could be burning fuel or any heat source, converts some of it into net out-flowing work (notice that the net area under the curve=area within the cycle is positive) and discards unused heat as waste

14.5. P V -DIAGRAMS

295

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. The entropy calculation is not immediately obvious, but Ta Td

2

2

Tb Vb3 2 3

divide

Td Vd ( Tc Vc ) 23 Td Vd (V ) c or Vd

Therefore

1

∆S

a → b isothermal c → d isothermal

= Tb , = Tc , = Tc Vc3 ,

b → c adiabatic

2 3

= Ta Va , d → a adiabatic 2 ( ) Tb Vb 3 = Ta Va (V ) b = Va (V ) (V ) b d = nR ln + nR ln =0 Va Vc

(14.125)

2

Example 13. The Carnot cycle is the simplest in the S − T plane, since the four sub-processes will leave either T or S unchanged.

S

One mole of a monatomic ideal gas is made to undergo the reversible Carnot cycle in the figure. J J T1 = 200K, T2 = 500 K, S1 = 5.0 K and S4 = 3.0 K .

4

All of the calculations are particularly simple because every row has a zero in it.

3

T

Line

∆U

∆Q

∆W

∆S

1→2

(3/2)(1)(8.31)(500-200) =3740 J

0

-3740 J

0

2→3

0

(500)(3-5)=-1000J

(500)(3-5)=-1000J

(3-5)J/K

3→4

(3/2)(1)(8.31)(200-500) =-3740 J

0

3740 J

0

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, causing the gas trapped in the cylinder to expand, doing work that can be exported or harnessed to perform a mechanical task. This is the power-stroke part of the engine cycle. Afterwards the gas must be brought back to its original

296

CHAPTER 14. EQUILIBRIUM THERMODYNAMICS

state, so that the process can be repeated. This is done by thermally isolating the gas from b → c and allowing it to expand, which causes a temperature drop. Now heat is extracted and discarded into a heat sink and the gas contracts isothermally to a much lower volume, after which it is again thermally isolated from d → a while it is compressed 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 turn it 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 = 0 means dQ = dW ; input heat becomes output work. But this can’t go on forever; the spontaneous expansion will lead 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. This results 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

but therefore

∆S ϵ

Wout Qin Qin − Qout Qout = =1− Qin Qin Qout Qin − = 0= Th Tc Tc = 1− Th =

(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 a high-T reservoir) as waste than you input as fuel (from a low-T reservoir); this is a refrigerator. A measure of how well a reversed Carnot cycle is at extracting heat is K but

Qin + Win K

Qin Win = Qout 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 Win =

(T − T ) h c Qin , Tc

℘=

( T − T ) dQ h c in Tc dt

dQc dt

(14.128)

Example 14. Suppose that you own a poorly insulated small house, with outside walls ℓ = 0.25 m thick, total outer W wall plus roof surface area 210 m2 (about 1992 f t2 ) and your walls have a thermal conductivity of 1.3 m·K (limestone). Calculate how much power your (Carnot) air conditioner will require to keep your house at 293 K when the outside temperature is 308 K. You must remove the heat that gets in through the walls (1.3)(210)(308 − 293) dQc = = 16, 380J, dt 0.25

( 308 ) dW =℘= − 1 16, 380J = 838 W dt 293

14.5. P V -DIAGRAMS

14.5.2

297

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.

1

Light (electromagnetic radiation) trapped in a cavity of volume V in thermal equilibrium with hot conducting walls at temperature T has equations of state

2

U = a V T 4,

P

a = 7.56 × 10−16

J m3 K 4

U = 3P V

4

Suppose that it is made to undergo the reversible process N 3 3 illustrated, with P1 = 100 m 2 , V1 = 0.1m , V2 = 0.3m N and P4 = 25 m2 . Compute ∆U, ∆Q, ∆W for each process line on the cycle.

3

Everything can be calculated with just dU = 3P dV or dU = 3V dP since all processes are constant P or V , along with ∆W = P ∆V or ∆W = 0. V

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 = 1 → 2 and 3 → 4 are isothermal.

a 4 3T ,

so constant P processes like

Entropy problems just go back to basics as well; suppose the “photon gas” undergoes isothermal expansion from N (V1 , P1 ) = (0.1m3 , 100 m 2 ) to (3 V1 , P1 ). Find ∆S. Since P = a3 T 4 , constant T means constant P ; so dU = 3P dV dS

∆S

dU P + dV T T P P = 3 dV + dV T T P = 4 dV T P (100) = 4 ∆V = 4 √ 3(100) T 4 =

5.67×10−8

(0.3 − 0.1) = 0.297

J K

(14.129)

298

CHAPTER 14. EQUILIBRIUM THERMODYNAMICS

What about ∆S2→3 ? Once again we go back to basics. Note that dV = 0 for 2 → 3, and dU = 3d(P V ) = 3V dP dU P + dV T T V = 3 dP T ∫ P4 dP √ = 3V 4 3P/a P3 ) 3 4V ( 34 4 = √ P − P 3 2 4 3/a

dS

=

(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 the correct equations of state are substituted. The Van der Waals equations of state ( ) n2 a P + 2 (V − nb) = nRT, V

U=

3 n2 a nRT − 2 V

(14.131)

is a better description of the behavior of a gas at low temperatures than the ideal equations of state (which describe gasses 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 (3 n2 a ) ( 3 n2 a ) n2 a n2 a ∆U = nRT2 − − nRT1 − = − ̸= 0 2 V2 2 V1 V1 V2 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 2

3 n a nR dT + 2 dV 2 V 3 dT 2 T ( T ) 32 f Ti

= dQ − P dV = −P dV ( nRT n2 a ) = − − 2 dV V − nb V dV = − V − nb ( V − nb ) i = Vf − nb

(14.132)

14.6. NON-P V WORK. CHEMICAL AND ELECTRICAL

14.6

299

Non-P V work. Chemical and electrical1

Work in general has the form dW = (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 extensive quantities (like V ) do. For very simple systems such as an ideal gas, work is easy to compute (its just P dV ). Keep in 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 charge dc 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 our abilities 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 recombination of molecules occur. The chemical potential µi is the work done in adding another particle of species i to the system. In a system with multiple chemical species, the first law reads ∑ dU = dQ − P dV + µi dNi + E dc (14.135) i

The total number of particles of any element must be conserved in a chemical process, let N = over the species, and if we write Ni in moles, we can work with mole-fractions xi =

∑ i

Ni where we sum

Ni N

(14.136)

For example in an ideal gas (in these formulas k is Boltzmann’s constant, related to the ideal gas constant by Avagadro’s number R = k NA ) ( ∑ ) kT ∑ N kT ∑ = xi = Pi (14.137) P = Ni V V i i i which is the law of partial pressures; the pressure exerted by species i alone is Pi = xi NVkT . Example 15. For example in a gas of H-atoms we could allow for bonding to H2 to occur and include the work done by adding a chemical work term such as dWc = µH dNH + 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 + µi dNi + E dc (14.140) i

which we invert and write as dS =

(1) T

dU +

(P ) V

dV −

∑ ( µi ) i

T

dNi −

E dc T

(14.141)

1 It 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.

300

CHAPTER 14. EQUILIBRIUM THERMODYNAMICS

In order to study chemical thermodynamics, we will need to figure out what µ depends on. To do this, we limit ourselves of course to reversible, equilubrium processes, and employ the maximal entropy principle, which is one of the most powerful and insightful concepts in physics; the equilibrium state is the most probable state.

Consider a universe U of volume V divided into a system S to be studied, of volume v, and an environment E of volume ∪ V − v. If there are 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 particle will be found in S is ℘S = Vv , and in E ℘E = 1 − Vv , so the probability that m of the n particles will be found is S is

E, V

℘=

S, v

( v )m ( n! v )n−m 1− m!(n − m)! V V

(14.142)

There is a value m∗ of m that maximizes this, lets us the fact that for large N ; ln N ! ≈ N ln N is a very good approximation

d℘ d ( n ln n−n−m ln m+m−(n−m) ln(n−m)+(n−m)+m ln(v/V )+(n−m) ln(1−v/V ) (14.143) =0= dm m∗ dm m∗ and we get − ln m∗ + ln(n − m∗ ) + ln v − ln(V − v) = 0 There are two useful and insightful way to write this. Firstly we write it as m∗ n − m∗ ln = ln v V −v

or

m∗ n − m∗ = v V −v

(14.144)

(14.145)

which says that the equilibrium state is one in which E and S have equal particle densities. This is something that we all feel is correct at an intuitive level. Secondly we leave it as m∗ n − m∗ ln = ln , ln cs = ln cE (14.146) v V −v introducing particle concentrations or densities, and return to the First Law of Thermodynamics, eq. 14.141, dS =

(1) T

dU +

(P ) V

dV −

(µ ) S

T

dm −

(µ ) E

T

d(n − m)

(14.147)

which says that if we maximize S with respect to particle distribution, entropy is maximal when dS µS µE =0= = dm T T

(14.148)

S = k ln ℘

(14.149)

µS = µ0 + kT ln cS

(14.150)

recalling the Boltzmann definition of entropy we obtain our formula for the chemical potential

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 − T S + P V for the following reason. Differentiate this thing dG = dU − T dS − SdT + P dV + V dP

(14.151)

14.6. NON-P V WORK. CHEMICAL AND ELECTRICAL

301

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 lose a B and gain an AB dNA = dNB = −dNAB (14.155) This makes all of the dN ’s cancel 0 = µA + µB − µAB

(14.156)

or by putting in Eq.14.150 − µ0A − µ0B + µ0AB = kT ln(

0 0 −(−µ0 cAB A −µB +µAB ) kT =e = Keq cA cB

cA cB ), cAB

(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 and pressure. 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 ideal gas is Eq. 14.113 (( T ) CRv ( V )) T1 V1 1 1 S1 − S0 = nCV ln + nR ln = nR ln (14.158) T0 V0 T0 V0 Lets carefully seperate out the dependence on N, V and T ; let V0 = N ν0 and V = N ν where ν, ν0 are volume/reference volume per particle, and pick ν0 and T0 so that S is zero when T = T0 and ν = ν0 S = N k ln

(( T ) CRv ( ν )) T0 ν0

and into this, for gaseous chemical species i, we insert P = S = N k ln

N kT V

=

kT ν

, and P0 =

(( T ) CRv +1 ( P )) 0

T0

P

(14.159) N kT0 V0

=

kT0 ν0 ;

(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 entropy out front as the coefficient, which is very nice. Our next step is to start with the first law; dU dS ( ∂S ) ∂N

U,V

= T dS − P dV + µ dN P µ dU + dV − dN = T T T 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 ) CRv +1 ( P )) 0

T0

P

(14.162)

302

CHAPTER 14. EQUILIBRIUM THERMODYNAMICS

Since both Cv and P for a gas of species i in a mixture could depend on species (P is the pressure exerted by the species, a partial pressure), for a mixture of gases we have µi = −kT ln

(( T ) CRv,i +1 ( P )) 0

T0

Pi

(( T ) CRv,i +1 ( P )) 0 = −kT ln = µ0i + kT ln xi T0 xi P

(14.163)

from which we get an actual formula displaying the temperature and pressure dependence of µ0i that you see in Eq. 14.157 This formula is extremely useful for understanding principles layed out (without proof) in freshman chemistry, such as Lechatelier’s principle. For example, in the reaction A + B ⇌ AB, our equilibrium constant would be ( ) µ0A + µ0B − µ0AB ln Keq = − kT (( T ) Cv,A ( P )) (( T ) Cv,B ( P )) (( T ) Cv,AB +1 ( P )) R +1 R +1 R 0 0 0 = ln + ln − ln T0 P T0 P T0 P −Cv,AB (( T ) Cv,A +Cv,B ( )) +1 P0 R = ln (14.164) T0 P so that

−Cv,AB ( T ) Cv,A +Cv,B +1 ( P ) xA xB R 0 = xAB T0 P

(14.165)

If you increase P , the right-side gets smaller, meaning that xAB gets bigger, in other words increasing the pressure pushes A + B ⇌ AB to the right. Thermodynamics gives you formulas for Keq !

14.6.2

Electrochemical work

H2 gas

When a known amount of charge dc passes through a known 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 an electrical charge dc some potential energy ( ) − Vf − Vi = dW = E dc (14.166) The cell pictured has a zinc strip undergoing an oxidation reaction (loss of electrons) on its anode Zn (s) → Zn++ (aq) + 2e− 0 Eanode = +0.76 V

Pt (cat.) Zn++, 1 M

H3O+, 1 M

The other electrode (cathode, where chemical reduction occurs) is a solution of 1 M H + created by bubbling H2 gas over a platinum (inert) strip. This hydrogen half-cell reaction is the zero-point against which all other half-cell reactions are measured, being assigned a cell-voltage of zero H2 (g) → 2H + (aq) + 2e− ,

Zn (anode)

What do these voltages

0 Ecat = 0.0 V

The entire cell reaction is the sum of the half-cell reactions 0 Ecathode ,

0 Eanode

2H + (aq) + Zn (s) → Zn++ (aq) + H2 (g) mean? How can this device actually supply voltage?

14.6. NON-P V WORK. CHEMICAL AND ELECTRICAL

303

Put this device together with chemical concentrations nowhere near the equilibrium values. Then the chemical reaction will run spontaneously and electrons will be driven one way or the other through the connecting wire. It takes voltage to drive the electrons along, the voltage of the battery. So how much voltage is this? Put your own external battery into the circuit where the meter is, and tune it up to a sufficient value to stop the flow of charge. 0 0 At that value it is exactly negating the battery voltage. These are the half-cell voltages Eanode and Ecathode , each is separately measured by setting up a half-cell with an external variable battery and tuning the external voltage until no current flows. This must be done by momentarily completing the circuit, so no major chemical changes occur in the cells while under measurement. In a typical chemical process such as (note the sign convention) ν1 X1 + ν2 X2 ⇌ ν3 X3 + ν4 X4 ,

∑

more generally written

νi Xi = 0,

νprod > 0,

νreact < 0

(14.167)

i

the stoichiometric coefficients play a big role in the thermodynamics because the total number of atoms must be conserved (they simply rearrange in different molecular combinations), so that there exists a number ξ such that the number of particles of species i changes as Ni = Ni,0 + νi ξ (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 whatever direction minimizes the Gibbs free energy (so ∆G = Gf − Gi < 0). This is the natural tendency. So instead we set it up so that the chemical concentrations are at or near equilibrium values. What are those values? If we work at constant T and P 2H + (aq) + Zn (s) ⇌ Zn++ (aq) + H2 (g) (14.170) dG = −S dT + V dP +

) ) ∑ ( ∑ ( νi µ0i + kT ln ci dξ → 0 = νi µ0i + kT ln ci i

or −

(14.171)

i

(∏ ) 1 ∑ νi µ0i = ln cνi i , kT i i

∏

cνi i = Keq = e− kT 1

∑ i

νi µ0i

(14.172)

i

Solids are generally taken to have zero total chemical potential in such reactions, they are treated simply as particle reservoirs, not as chemical reactants. Using the stoichiometric numbers for our reaction eq. 14.170 we can find these concentrations ci . Under equilibrium operation 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 battery until equilibrium is restored, and no current flows. The battery is doing reversible work on the system, and since we are at equilibrium, if an infinitismal shift dξ occurs in the participating chemical species numbers, 2dξ electrons will be transferred; let F = NA e = 96500 C be the charge on a mole of electrons ) ∑ ( dG = −S dT + V dP + Eext 2e dξ + νi µ0i + kT ln ci dξ = 0 ( 0 = or

− 2eEext

=

i

)) ∑ ( Eext 2e + νi µ0i + kT ln ci dξ

∑

i

νi µi0

= −kT ln Keq

i

Keq

= e

2eEext kT

=e

nF Eext RT

(14.173)

(if n electrons are transferred) so by reading the external battery voltage needed to restore reversibility, we obtain the equilibrium constant.

304

CHAPTER 14. EQUILIBRIUM THERMODYNAMICS

We generally write eq. 14.173 as dGuniv = 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 Li+ ; Li K +; K Al+++ ; Al Cd++ ; Cd − Cl ; AgCl(s) , Ag Cu+ ; Cu Cu++ ; Cu

E0 -3.045 V -2.925 V -1.66 V -0.403 V 0.2224 V 0.521 V 0.337 V

Reaction Li+ + e− → Li K + + e− → K 1 +++ + e− → 13 Al 3 Al 1 ++ + e− → 12 Cd 2 Cd − AgCl + e → Ag + Cl−1 Cu+ + e− → Cu 1 ++ + e− → 12 Cu 2 Cu

Chemists use some simple conventions to take the guessing out of computing some of these quantities. First { > 0 Spontaneous (∆G < 0) 0 0 0 0 Ecell = Eright − Elef t = Ered (cat) − Ered (an) = (14.176) < 0 non-spontaneous and second; the reaction on the left electrode is written as an oxidation, X → X + + e− , and on the right electrode we have a reduction 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 chemical transformation. The process of sedimentation (this requires some force/potential) will make the concentration of a chemical species vary with position. The simplest sedimentation is caused by gravity.

The chemical potential of the ideal solute will be of the form µ(z) = µ0 + kT ln c(z)

z+dz z

z

(14.177)

where c(z) is the concentration. Chemical potential is Gibbs free energy per particle, and we have seen several times that Gibbs free energy is a measure of work, in particular −µ dN is the work done in moving another dN particles into the 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 of mass 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 −

c(z) + dc(z) c(z + dz) 1 dc(z) mg dz dz dz + · · · = ln = ln( )≈ dz kT c(z) c(z) c(z) dz

(14.179)

14.7. MORE APPLICATIONS OF µ

305

and so dc(z) mg dz =− , c(z) kT or

∫

c(z)

c(0)

dc(z) = c(z)

c(z) = c(0) e−

∫

z

− 0

mg dz kT

mg z kT

(14.180) (14.181)

and the concentration is greatest at the bottom of the cylinder. The work done against the gravitational potential is non-P V work.

14.7.1

The centrifuge

Example 18. The centrifuge works by the same principle. Macro and bio-molecules can be seperated from solvents by application of a much greater force than gravity.

dr

r

Consider two thin “cells” of solution in a testtube in the machine, one at radius r, the other at r+dr, and compute the work done in moving a solute molecule of mass m from the inner cell into 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 test tube, relative to the stationary xy-coordinate system outside of the tube,

for which x = r cos ωt, y = r sin ωt where ω is the spin-rate of the centrifuge. The energy of such a molecule is (see problem 10.19) 1 1 Ustat. = m(x˙ 2 + y˙ 2 ) = m(r˙ 2 + ω 2 r2 ) (14.184) 2 2 Now think about what the molecule experiences with respect to the coordinate system inside of the tube. The molecule only moves radially within the tube, relative to a co-rotating set of coordinate axes, and so has energy Urot. =

1 1 2 mr˙ = Ustat. − mω 2 r2 2 2

(14.185)

so even if it is not moving radially, it has “potential energy” 1 Vcent. (r) = − mω 2 r2 2

(14.186)

which acts like a simulated force of gravity within the tube. Equilibrium (with respect to particle exchange) between the adjacent cells illustrated is then ) ( ) ( 1 1 (14.187) dG = 0 = 1 · kT ln c(r + dr) − m(r + dr)2 ω 2 − 1 · kT ln c(r) − mr2 ω 2 2 2 or ( c(r) + dc(r) dr + . . . ) dr = mrω 2 dr (14.188) kT ln c(r)

306

CHAPTER 14. EQUILIBRIUM THERMODYNAMICS

and finishing the problem by the methods of the previous example results in c(r) = c(0) e

mω 2 r 2 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 side of a two-part cell, with pure solvent in the right side and a solvent-permeable membrane seperating the two sides. Solvent molecules will flow through the membrane from pure-solvent to solution side, because the chemical potential of the solvent is lower on the left side (“matter flows from 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 has chemical potential µ = µ0solv + kT ln(1 − X) (14.191) Let the solute mole fraction change by dX, and the pressure change by dP , then ∂µsolv ∂µsolv dX + dP ∂X ∂P

dµsolv =

(14.192)

but the first part is easy for an ideal solvent ∂µsolv kT =− ∂X 1−X

(14.193)

The chemical potential is the Gibbs free energy per particle, so ∂G ∂µ = N =v ∂P ∂P

(14.194)

and for a mixture, this will be the molal volume per particle, so that if the particle exchange and pressure change are performed in such a way that equilibrium is maintained, µsolvent, lef t = µsolvent, right , we find that 0 = v dP − kT and we integrate

∫

dX 1−X ∫

Patm +π

Xf

v dP = kT Patm

dµsolvent = 0

0

(14.195)

(14.196)

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 This can be written as π ≈ kT

Xf Xf RT = RT = RT [c] = [C] v Nv Msolute

(14.198)

(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 molecular weight 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 a gas 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 X of 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 = R ln(1 − X) T

(14.202)

Differentiate with respect to temperature Keeping pressure fixed µ0 µ ∂ Tg ∂ Tℓ ∂ ln(1 − X) − =R ∂T P ∂T P ∂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 − T S + P V = H − T S ∂G ∂G dG = −S dT + V dP, = −S, =V ∂T P ∂P T and so

(14.204) (14.205)

G H H ∂G = −S = + T T T ∂T P

(14.206)

∂G ∂1 1 ∂G 1 ∂G G T = +G T = − ∂T P T ∂T P ∂T T ∂T P T 2

(14.207)

∂G G H S T =− − 2 =− 2 ∂T P T T T

(14.208)

and this means that

and we obtain

Hv H0 ∂ ln(1 − X) + 2ℓ = R (14.209) 2 T T ∂T P but the difference in the heat of formation of the pure liquid and the heat of formation of the gas is the heat of vaporization ∆Hvap ∂ ln(1 − X) − = R (14.210) T2 ∂T P This can be integrated if ∆Hvap is a known function of T . Lets pretend that it is constant ∫ T ∫ X dT dX − ∆Hvap = −R (14.211) 2 T 1 −X T0 X0 −

308

CHAPTER 14. EQUILIBRIUM THERMODYNAMICS − ∆Hvap

(1 T

−

1) (1 − X) = −R ln T0 (1 − X0 )

(14.212)

Let X0 = 0, and let X be small (dilute solution), this leads to ∆Hvap

(1 1) − ≈ RX T0 T

(14.213)

giving us the change in the temperature at which gas/liquid equilibrium occurs as a function of changes in the concentration of a solute present in the liquid.

14.8

Non-equilibrium applications. laws

Diffusion, Graham’s and Fick’s

Consider a tube of cross-sectional area A, filled with matter. Let JN (x) be the flux or rate per unit area with which matter crosses cross-sections of the tube, moving to the right.

x

Consider how the number of particles dN in the middle cell could change, by a current of matter into the cell from the left, our out into the cell to the right;

x+dx

d( ) dN = A JN (x) − A JN (x + dx), dt This is a matter conservation law, call n(x) = volume. We have a matter conservation law

d ( 1 dN ) JN (x) − JN (x + dx) d = = − JN (x) dt A dx dx dx 1 dN A dx ,

(14.214)

which is the concentration or number of particles per unit

dn d = − JN dt dx

(14.215)

Now lets figure out what potential is responsible for our matter flow. Examine the energy formula for candidates dF = −S dT + µ dN + E dc

(14.216)

This exhibits the couplings (Onsager’s theory) between quantities that flow (T, N, c) and the potential responsible for its flow. It is clearly the chemical potential that is responsible for matter flow, and so in the simplest linear relationship between current and potential gradient we would suppose that JN = −ν

d µ dx

(14.217)

where ν is some constant (that can actually be determined from kinetic theory). Lets suppose that the matter in the cells, liquids or gases, are ideal, so that we can use our formula for the chemical potential of a normal system µ(x) = kT ln n(x) + f (T ),

1 dn dµ = kT dx n dx

(14.218)

then we obtain Fick’s law JN = − and

νkT dn n dx

( 1 d2 n dn 1 dn 2 ) = νkT − ( ) dt n dx2 n2 dx

(14.219)

(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 of chemistry and biology depends dc(x) νkT d2 c(x) ≈ + ··· (14.221) dt n0 dx2

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 of diffusion is obtained by illustrating that the solution to the diffusion equation, subject to all of the particles being at the origin at t = 0, is x2 − νkT 1 4 t c(x) = √ e n0 (14.222) νkT 4π n0 t and so the distance d that molecules diffuse through in time t is √ νkT d= 4 t n0

(14.223)

Within equal times t, changing the temperature from T0 to T results in a ratio of average diffusion distances of √ d T = (14.224) d0 T0

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 provide a very short review of the most basic concepts of freshman chemistry. A typical chemical reaction such as (14.225) A + 12 B2 ⇌ AB ∑ can be written in terms of stoichiometric coefficients νi as i νi Xi = 0, for example in this case νA = 1, νB2 = 12 and νAB = −1. The reaction formula looks like a mathematical equation, and can be thought of one. It is a constraint on the changes in the numbers of each reactant or product as the reaction proceeds. Suppose that you put 0 NA0 moles of A, NB0 2 moles of B2 and NAB moles of AB into a reaction vessel, and allow the reaction to occur. As time and the reaction progresses, we see the numbers of molecules of each species change in accordance with the chemical reaction formula.

Time step 0 1 2 3

NA NA0 0 N A − ξ1 NA0 − ξ2 NA0 − ξ3

NB2 NB0 2 0 NB2 − 12 ξ1 NB0 2 − 21 ξ2 NB0 2 − 21 ξ3

NAB 0 NAB 0 NAB + ξ1 0 NAB + ξ2 0 NAB + ξ3

0 Consider another example, 3A + 2B2 + C ⇌ CA2 B2 + AB2 . Suppose that we begin with NA0 , NB0 2 , NC0 , NCA 2 B2 0 and NAB2 moles of each species, and from there the reaction proceeds to create an additional ξ moles of AB2 . The chemical reaction formula tells us that for every mole of AB2 created, a mole of CA2 B2 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 of moles of AB2 created as a reaction progress indicator, we find that the number of molecules of each species present in the reaction chamber have changed in proportion to the stoichiometric coefficients;

NA0 − NA = 3ξ,

NB0 2 − NB2 = 2ξ,

NC0 − NC = ξ,

0 NCA − NCA2 B2 = −ξ, 2 B2

0 NAB − NAB2 = −ξ 2

(14.226)

Therefore when we are given a chemical reaction’s stoichiometric equation we can write down the changes in the mole numbers very easily in terms of some reaction progress variable ξ of our choosing ∑ νi Xi = 0, Ni0 − Ni = νi ξ, −dNi = νi dξ (14.227) i

At constant temperature (dT = 0) and pressure (dP = 0), the condition for chemical equilibrium is ∑ ∑ dG = −S dT + V dP + dNi µi = dNi µi = 0 i

i

(14.228)

310

CHAPTER 14. EQUILIBRIUM THERMODYNAMICS

which becomes 0 = −dξ

(∑

) νi µi ,

0=

(∑

i

) νi µi

(14.229)

i

and for ideal reactants and products, µi = µ0i + kT ln xi , which leads to the equilibrium formula ∏( ) ∑ ∑ − νi µ0i = kT ln xνi i = kT ln xνi i i

which chemists write as

e− kT 1

∑ i

(14.230)

i

i

µ0i (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 (( P )( T ) CVk+k )) 0 µi = −kT ln = ϕ0i (T ) + kT ln P + kT ln xi xi P T0 to get P−

∑ i

νi

e− kT 1

∑ i

ϕ0i (T )

= xν11 xν22 · · · xνNN

(14.232)

(14.233)

Example 20. Study the gas reaction N2 + 3H2 ⇌ 2N H3 . The equilibrium equation is P −1−3+2 e− kT (ϕN2 +3ϕH2 −2ϕN H3 ) = 1

0

0

0

xN2 x3H2 x2N H3

(14.234)

or in terms of partial pressures pi = xi P (remember that K is product/reactant); e− kT (ϕN2 (T )+3ϕH2 (T )−2ϕN H3 (T )) = K −1 (T ) = 1

0

0

0

pN2 p3H2 p2N H3

(14.235)

A chemistry student conducting the experiment in a sealed vessel of volume V = 5 × 10−3 m3 at T = 723 K finds that at equilibrium, with an internal pressure of 3.35 × 107 P a, that the partial pressures of the constituents are pH2 = pN H3 = 1.44 × 107 P a, pN2 = 0.48 × 107 P a. Determine the equilibrium constant K(T ). In terms of P a this is simple K −1 (723) =

(0.48 × 107 P a)(1.44 × 107 P a)3 = 6.912 × 1013 P a2 (1.44 × 107 P a)2

(14.236)

Example 21. The same wacky chemist puts a mole of N2 and three moles of H2 in a bomb and maintains its temperature at 723 K. Find the mole fractions of all reactants at a pressure of 100 atm = 1.01 × 107 P a. Suppose ξ moles of ammonia form. That means that there are 1 − 12 ξ moles of nitrogen, and 3 − 3 12 ξ moles of hydrogen in the mix. Therefore the mole fractions are xN2 =

1 − 12 ξ 1 − 12 ξ = , 1 1 4−ξ (ξ) + (1 − 2 ξ) + (3 − 3 2 ξ)

xN H3 =

ξ , 4−ξ

6.912 × 1013 P a2 · (1.01 × 107 P a)−2 = 27

xH 2 =

3(1 − 12 ξ) 4−ξ

(1 − 12 ξ)4 ξ 2 (4 − ξ)2

(14.237)

(14.238)

which should be solved numerically. You can see that since none of the molecules can exist in a negative mole fraction, 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 constants in terms of molar concentrations, and so re-embarks on his quest to understand the reaction N2 + 3H2 ⇌ 2N H3 . Returning to the original equilibrium mixture in a V = 5 × 10−3 m3 vessel at T = 723 K with an internal pressure of 3.35 × 107 P a, and partial pressures of the constituents are pH2 = pN H3 = 1.44 × 107 P a, pN2 = 0.48 × 107 P a, he computes actual numbers of moles using nN2 =

(0.48 × 107 P a)(5 × 10−3 m3 ) pN2 V = = 4.0 moles, RT (8.314J/mole · K)(723K)

nH2 = nN H3 = 12 moles

(14.239)

14.9. FRESHMEN CHEMISTRY CRAMMED INTO THE CHEMICAL POTENTIAL NUTSHELL

311

and then computes the concentrations in moles per liter [N2 ] =

(4.0 moles) moles = 0.8 , (5.0ℓ) ℓ [N H3 ] =

and therefore K −1 =

[H2 ] =

(12.0 moles) moles = 2.4 (5.0ℓ) ℓ

(14.240)

(12.0 moles) moles = 2.4 (5.0ℓ) ℓ

(14.241)

[H2 ]3 [N2 ] moles2 = 1.92 [N H3 ]2 ℓ2

(14.242)

Example 23. In molarity units the equilibrium constant for CO + H2 O ⇌ CO2 + H2 ,

is 0.25 =

[CO][H2 O] [H2 ][CO2 ]

(14.243)

mole At equilibrium one finds [CO] = 0.6 mole and [CO2 ] = 0.5 mole ℓ , [H2 O] = 0.2 ℓ ℓ . Our intrepid experimenter deduces (0.6)(0.2) that the molarity of H2 in the equilibrium mixture is [H2 ] = 4 (0.5) = 0.96 moles ℓ .

Example 24. One wonders what the precise relationship is between the equilibrium constants expressed in molarities to those written in terms of partial pressures. Since [Ni ] = we obtain

Ni pi xi P = = V RT RT

(14.244)

( P )∑i νi ∏ ∏ ∏ ∑ νi [Ni ] = xνi i = (RT )− i νi pνi i RT i i i

(14.245)

In our first example with N2 +3H2 ⇌ 2N H3 we found the equilibrium constant in pressure units to be 6.912×1013 P a2 (6.912×1013 P a2 ) mole2 8 mole2 at 723 K. Therefore it should be (8.314·723 J/mole)2 = 1.913 × 10 m3 = 1.912 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 of CH4 and 0.1 mole of H2 S into the calorimeter, and maintains it at 1000 K while the reaction CH4 + 2H2 S ⇌ 4H2 + CS2

(14.246)

equilibrates at a total pressure of 9.2 × 105 P a. The partial pressure of the hydrogen is found to be 2.0 × 105 P a. Assuming that ξ moles of hydrogen and 14 ξ moles of CS2 have formed, leaving 0.2 − 14 ξ moles of methane and 0.1 − 12 ξ moles of H2 S, Ilya computes the mole fraction of of hydrogen xH2 =

2.0 ξ pH2 = = , 1 P 9.2 ξ + 4 ξ + 0.1 − 12 ξ + 0.2 − 41 ξ

ξ = 0.0732

(14.247)

and therefore the vessel contains 0.3 + 21 ξ = 0.3366 moles 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 decompose as µi = µ0i + kT ln[i]. It is conventional for chemists to refer to the formality of a solute as the number of moles of 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

312

CHAPTER 14. EQUILIBRIUM THERMODYNAMICS

ionization process XY ⇌ X − + Y + is a chemical process whose equilibrium at constant T and P is again determined by 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 0.01 F solution of XY is 5% ionized at equilibrium. What is K(T )? ( ) This means that the actual molarity of XY is [XY ] = 0.01 − (0.01)(0.05) = 9.5 × 10−3 M , and [X − ] = [Y + ] = 5.0 × 10−4 , so K = 2.63 × 10−5 M . 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 = K = 2.63 × 10−5 M ξ − 0.001ξ

(14.249)

and we solve for the formality ξ = 26.27 F 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 0.1 F solution of XY and 0.2F XZ? Letting ξ = [Y + ] at equilibrium, the two reactions XY ⇌ X − + Y + ,

K = 2.63 × 10−5 M,

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) = 2.63 × 10−5 M, 0.1 − ξ

ξ ≈ 1.32 × 10−5 M

(14.251)

Example 28. Complex equilibrium calculations involve multiple reactions, for example a formality f solution of B2 C undergoes B2 C ⇌ B + + BC − ,

[B + ][BC − ] = K1 , [B2 C]

BC − ⇌ B + + C −− ,

[B + ][C −− ] = K2 [BC − ]

(14.252)

It would be imprudent to contemplate a net reaction, since the equilibrium constants are radically different. Let x moles of B2 C ionize, leaving [B2 C] = f − x. Then in the second stage y moles of BC − ionize, so the solution contains [B2 C] = f − x, [B + ] = x + y, [BC − ] = x − y, [C −− ] = y (14.253) then

(x + y)(x − y) = K1 , f −x

y(x + y) = K2 x−y

(14.254)

Eliminate y between the two equations to get ( √(x + K )2 + 4xK − (x + K ) )2 2 2 2 x + K1 x − f K1 = 2 2

(14.255)

This we solve graphically by plotting each side versus x, over √ the range 0 ≤ x ≤ f . What if K2 2) ? 7+3*(x-2) : f2(x) plot[0:3] f3(x)

Chapter 2 √ 2v 2 sin 20o 1. R = g0 cos 1 + 4 tan2 20o 2 20o , v = v0 m 2. v0 = 29.9 s . ϕ = 0.313 rad, v0 = 46.7 m s , descending. √ 9.8 m 3. v0 = 20 20 s 4. v = 35.9 m s , tan ( ϕ = 0.66. ) 2v 2

θ−tan α 5. R = g0 cos2 θ tan cos α √ x 2 2 7. s = v0 + (g v0 ) 12. ymax = 95.7 m, R = 191.32 m 20. v0 = 50.34 m s 26. It looks like θ = 0.7 rad, about 40.1o gives the longest range.

Chapter 3 1. 2. 3. 4.

(0.3 sm2 , 0.4 sm2 ) (27 N, 26 N ) m (6 m s , 9 s ) at t = 3 s N = mg − F sin θ = 93 N,

a = 0.866 sm2 333

334

CHAPTER 15. ANSWERS TO THE PROBLEMS

5. Ff = −3.0 N i, Ff = −4.0 N i M1 M2 g 2g 7. a = MM , T =M 1 +M2 1 +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 ′′ = 392 N , Ff = T = T ′′ sin 60 = 339.5N 13. 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 = M21 = 5 kg

Chapter 4 1. a = g tan θ, F = (M1 + M2 )a = (M1 + M2 )g tan θ 2.a Ff = mg sin 18 = 60.56N , b. θc = 21.8o , c. a = 4.04 sm2 . 3.b (ax , ay ) = (4.0 sm2 , −8.2 sm2 ), c. (ax , 0) = (4.0 sm2 , 0) 5. a = µs g = 4.9 m F = (M1 + M2 )a = (M1 + M2 )µs g = 73.5N s , 6. M = µms tan 60o = 20kg 0.6 tan 60 = 57.7kg. c. Ff = T2 = T3 sin θ = mg tan 60 = 339.5N 1 11. tan θ = µs 12. Fmin = √M g 2 1+µs ( ) sin θ−µk M1 cos θ 13. If the block slips up a = M2 −M1 M g 1 +M2 19. If you don’t push hard enough, the box could slip down the incline, then friction will point up the incline and θ−mus cos θ help you hold the box in place; Fmin = M2 g sin cos θ+µs sin θ

Chapter 5 √ 1. v = tan 20 sin 20 Rg g 2. θ = cos−1 Rω √ 2 5. ωmin = µsgR √ 7. vmax = √Rg 8. vmax √ = Rg 9. v0 = kg √g λ 10. v0 = 2π h

Chapter 6 ∂Fy ∂Fx ∂y = ∂x ∂Fy x is no potential since ∂F ∂y ̸= ∂x ∫1 no potential W = 0 (16 + 16t2 )dt 2 2

1.a. There is a potential since 1.b. There 2. There is

= (16t +

1 16 3 t 3 0

=

64 3 J

3. There is a potential V = 4(x − y ) 1 Fd Fd 2 7. a. −µk M g d, b. 0, c. −mgd tan 60, d. cos 60 , e. 2 mv = −µk M g d − mgd tan 60 + cos 60 ∑ 1 8. a. W1 = T d, b. W2 = −T d, c. WT = W1 + W2 = 0, d. WG = (M2 − M1 )gd, e. 2 (M1 + M2 )v 2 − 0 = W = (M2 − M1 )gd ( ) 12. a. It is conservative, V = −x2 y + 13 y 3 + C ′ , W = −(Vf − Vi ) = − C ′ − ( 13 23 + C ′ ) = 83 J. b. Not conservative. ( ) ∫1 ∫1 ∫1 ∫1 13. W = − Vf − Vi = 0 14.a. W = 0 F · v dt = −9.8 0 v·v |v| dt = −9.8 0 |v| dt = −9.8 0 (2π) dt = −19.6πJ 17. hmin = 2.5 R

335 √ 2gh(1 √− µk ) ) ( µ2k 19. x = h 1 − 1+µ 2

18. v =

k

20. Increasing your speed five your fuel consumption) rate by (over) ten percent. ) percent increases ) ∫1( ∫1( ∫1( 2 21. a. W = 0 2 · 4 − 3 · 16t dt, b. W = 0 2 · 4t · 4 − 3 · 16t · 8(1 − t ) dt, c. W = 0 2 · 8(1 − t2 ) · 4 − 3 · 16t · 4t dt, ∫1 v ∫1√ d. W = −0.1 0 |v| · v dt = −0.1 0 42 + (16t)2 dt 22. WF = M gh(1 − cos θ) 23. ve = 1.12 × 104 m s 24. µk = 0.5555 25. F = (M1 + M2 ) µgs 26. ϵ = 0.949. (g− ϖ ) h 27. h′ = (g+ m ϖ m) 33. a = −nb2 x−2n−1 = F/m 34. tan θ = 1, θ = 45o

Chapter 7 1. Inelastic. kx2 kx2 2. x1 = , x2 = M M 2µk gM1 (1+ M1 ) 2µk gM2 (1+ M2 ) 2 1 √ 2 µk M 2 g 2 M v02 µk M g 3. x = + k − k 2 √ k 2 4. v0 = (µ2s + 2µs µk ) Mkg √ kx2 5. v1 = M1 M1 (1+ M ) 2

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 √ 2M2 gh 12. v1 = cos θ M1 cos 2 θ+M 2 ( ) √ √ √ √ 1 15.d. yCOM = 10 4R + 3(R + 3R) + 2(R + 2 3R) + 1(R + 3 3R) = (1 + 3)R ( ) 18. ∆E = mv 2 4 cos1 2 ϕ − 12 ∫1 ∫2 kgm 1 7.21. δp = 0 F dt = 12 (2 N )(1 s) = 1.0 kgm s , δp = 0 F dt = 2 2 (2 N )(1 s) = 2.0 s 7.22.It is because the force is not applied to the water leaving (or the water that has left) the tank, only the water in the tank.

Chapter 8 5F ri 5. a = 8mr 0 6.. cos θ = rroi = 23 13. T = ma + µs mg = 83 µs mg = 23.52N 14. It will slip; a = T −µmk mg = 8.04 sm2 T m 15. It will roll a = m+I/R 2 = 2.5 s2 ) ( 16. I = 12 mR2 + 6 12 mR2 + m(2R)2 ( ) 1 17. I = 4 12 mℓ2 + m( 2ℓ )2 √ , I = 43 mR2 19. v = 8hg √ 7 √ 2 21. v = 2kx 2x a, a = 3kx 3m = m 2

23. I = 31 (3m)(3ℓ)2 + M ℓ2 + M (2ℓ)2 + M (3ℓ)2

336

CHAPTER 15. ANSWERS TO THE PROBLEMS mg 2 = Fh µM g Fh1 = 1+µ 2 , Fv1

26. F = 41.

=

Mg 1+µ2 , Fv2

=

µ2 M g 1+µ2 ,

Iα =

µ2 +µ µ2 +1 M gR.

Chapter 9 1. 3. 5. 6.

√ ω = 12 = 3.46 rad s , d = 0.289m 2 ω2 = I12I+I1 2 ω0 , ω1 = II11 −I +I2 ω0 1 1 1 2 2 mv0 z = ( 3 mℓ + mx )ω, 2 ( 3 mℓ2 + mx2 )ω 2 = mg 2ℓ (1 − cos θ) + mgx((1 − cos θ) mv0 0 ycom = mℓ+M m+M , v1 = m+M

9. ωf = 13. Ω =

v0 + 12 Rω0 3R M gR Iω .

= 8.33rad/s

Chapter 10 1. h = 0.8 m 2. R = 3.6 m 3. Fx = Patm ℓh + 21 ρw gh2 ℓ 2 4. P ′ = P + 12 ρ(1 − 16)v 2 = P − 15 2 ρv √ 6. vf = v02 + 2ρw gh 7. ρ2 = (1.0 g/cm3 ) 58 8. For water h = 10.3 m, for mercury h = 0.7614 m 10. (Pin − Pout )A = F = 12 ρair · v 2 A 11. Flif t = Fbelow − Fabove ≈ 12 ρair (α2 − 1)v 2 A mg 17. vterm = 6πµa 19. z = zc +

1 ω2 2 2 g r

Chapter 11 3. v =

√ 20/0.01 = 44.721m/s, √g

5. ω = 6. ω 2 =

R

ω=

( ) √ 100/0.01 = 100rad/s, x(t) = 0.44721 m, sin 100 t

mgℓ

7 2 2 5 M R +m(R−ℓ)

(

7. E = 12 m (1 −

2 π

)2

R2 θ˙2 + 12 ( 12 mR2 )θ˙2 + mg m2

(

) − π2 )R θ2 √ √ m2 v02 k x = k(m12+m sin( m1 +m2 t − π) 2)

1 2 (1

2 v2 , 8. E = 12 kA2 = 12 (m1 + m2 )v12 = 12 m1 +m 2 0 √ 2 /(100) m 10. A = (0.10)2 + (0.05)(5.0) √ 100 2 11. amax = ω A = 0.05 (0.05)2 + (0.05)(0.4)2 /(100) m s √ g 15. ω = ρρW BL

1 1 2 2 17. ω = 63.24 rad s ,E = 2 2 (200)(0.02) + 2 (300)(0.04) J √ M

23. ω = √ 40.

k(1+ M1 )

M2 M1 + M1 2

mgx Icom +mx2 ,

41. xeq

2

I

I

+ R12 + R22

2 M1 2 M2

x2 = Icom /m. √ = −b/c, b2 /mc.

337

Chapter 12 √

16π 2 R3 29mG 2 µ 2 ϵ = 1 − (m−2µ)(m+2µ) ≈m (m+µ)3 ( ∫π dθ Fm = 0 − M GρR sin θi + R2

4. = 5. 9.

cos θj)

G 1 2 11. At aphelion/perihelion we have r˙ = 0, E = − µM 2a = 2 µva − √ 3 14. R = 4πρG v = 65, 536m

15. V = − mMx1 G −

mM2 G (R−x) ,

xe =

R √ 1+

µM G a(1+ϵ) ,

va2 =

MG a

(

1−ϵ 1+ϵ

)

M2 M1

16. a = (1.666 + 1.382)/2 = 1.524 AU , ϵ = (1.666 − 1.382)/(1.666 + 1.382) = 0.093 17. Tm = 1.87398 yr ( ) 3 3 mGρ 4π mGρ 4π 4π 3 3 3 R 3 (R/2) 23. Fr = − R − (R/2) ) , M = ρ 2 R r 3 (r− )2 2

Chapter 13 1. 839 W 3. 9381 W , 938.1 W 4. T == 259 K 5. t = 4030 s 6. You will end up with 137.5 g of water and 12.5 g of ice, all at 0 C 7. 31.5 C √ √ 8. x = L0 α∆T /2 = 2.0m 13.0 × 10−6 · 100/2 = 0.05 m 9. 819, 000 N 10. 0.03 kg s √ 11. |v| = 10/13 m/s,vrms = 14 13 m/s ( ) C = 72 k 13. U = 21 kT 2 + 3 + 2 , ∫∞ ¯= 14. E CE 4 e−E/kT dE = 4!C (kT )4 = 4kT 0

15. 17. 23. 24. 35.

6

n ¯ = 4×10 1×106 = 4 R = 1015352 m = 1015 km ¯ = 3 N kT E 2 1 dS T = dE = C(Em − 2E) < 0 if 7826.60 m s

E > Em /2

Chapter 14 1. 2. 3. 4.

T1 = 1000k, T2 = 250 K, T3 = 125 K 313K, 0.13 cal K 428 Js ∆H = −483.6kJ P1 V1 8.31 −29.3 cal K −9.43 cal K

6. T1 =

8. 9. 10. 37 s 11. 0.5656m3 12. −R ln 0.2 0.3 13.

=

(1.0×105 )(0.1) 8.31

= 1203,T2 = T3 =

P3 V3 8.31

=

(2.667×104 )(0.1) 8.31

( = 321, V2 = V1

T1 T2

) 32

= 0.726m3

338

CHAPTER 15. ANSWERS TO THE PROBLEMS

Line 1→2 2→3 3→1

∆U

∆Q

14. Line

∆S

(8.314)(4000 − 2000)

5 4000 2 (8.314) ln 2000

3 2 (8.314)(4000

− 2000)

5 2 (8.314)(4000

3 2 (8.314)(2000

− 4000)

0

3 2 (8.314)(4000

0.2 (8.314)(2000) ln 1.13

0.2 (8.314)(2000) ln 1.13

0 ∆U

− 2000)

∆W

∆Q

− 2000)

0 0.2 (8.314) ln 1.13

∆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 J 15. 0.2828 K 17. 0.69 18. T1 = P1RV1 = 1202,T2 = 3T1 = 3606K, T4 = T1 /2 = 601 K, T3 = 3T1 /2 = 1804 K ∆U ∆Q ∆W 1 → 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) 0 3 → 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 [Z ++ ] [Cu++ ]

= 1.97 × 1037 ( ) 3 22. dS = d 4b 3 VT 23. ∆U ∆Q ∆W ∆S 1 → 2 32 R(578 − 1202) 0 − 32 R(578 − 1202) 0 3 3 2 → 3 32 R(144 − 578) 0 2 R(144 − 578) 2 R ln(144/578) 3 3 3 → 4 2 R(300 − 144) 0 − 2 R(300 − 144) 0 3 4 → 1 32 R(1202 − 300) 32 R(1202 − 300) 0 2 R ln(1202/300) 24. ∆U ∆Q ∆W ∆S 5 1 → 2 32 R(2404 − 1202) 32 R(2404 − 1202) + 1 × 105 (0.2 − 0.1) 1 × 105 (0.2 − 0.1) 2 R ln(2404/1202) 3 3 2 → 3 2 R(1382 − 2404) 0 − 2 R(1382 − 2404) 0 3 5 4 3 3 → 4 32 R(691 − 1382) R(691 − 1382) + 2.5 × 10 (0.2297 − 0.45947) 2.5 × 10 (0.2297 − 0.45947) 2 2 R ln(691/1382) 3 3 4 → 1 2 R(1202 − 691) 0 0 2 R(1202 − 691) 26. ∆G = −2 · 96, 368 · 1.1/4.184 = −50, 800 cal 27. E = 1.685 V 28. 0.74V , −34, 200 cal. 29. c(z) = c(0) eqEz/kT 2 2 31. ∆S = N1 k ln V1V+V + N2 k ln V1V+V >0 1 2 32. ∆S = 0 J 34. 0.297 K ( ) 20.

39. V =

nRT P

,

∂V ∂P

= − nRT P 2 ,KT =

nRT P 2V

=

1 P

( ) ( ) 40. Const. P ; dQ = dU + P dV = 32 nRdT + P dV = 32 nRdT + P d nRT = 32 nRdT + P nRPdT T = 52 nR dT P n−n′ n′ 2n−n′ , xAB = 2n−n′ 5/58.44 x = 35/58.44+1000/10 = 0.00599. (373)2 ·8.314·0.00599 2257 ( ) = 3.07 K.

48. xA = xB = 51. 52.

T

54. dS = d 2aV T

P

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 b 1−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 b 1+tan a tan b

sin θ =

opp hyp ,

cos θ =

sin(2x) = 2 sin(x) cos(x) √ sin(x/2) = d dx

∫

ln x =

√ cos(x/2) = d dx

∫

+c

1+cos(x) 2

cos x = − sin x,

dx x

d n dx x

ln(ab) = ln(a) + ln(b)

ln(a/b) = ln(a) − ln(b)

ea+b = ea eb

ea−b =

eiθ = cos(θ) + i sin θ

i=

a

df (x) = f (a) − f (b)

√

ea eb

∫

tan x =

1−cos(x) 1+cos(x)

1 cos2 x

tan x dx = − ln(cos x) + c = a eax

xn dx =

xn+1 n+1

+c

ln(ab ) = b ln(a) eln(a) = a

−1

∂f (x,y) ∂x

√

∫

= n xn−1

2 tan(x) 1−tan2 (x)

tan(x/2) =

d ax dx e

= ln y

opp adj

tan(2x) =

d dx

cos x dx = sin x + c,

1 eax a

tan θ =

cos(2x) = cos2 (x) − sin2 (x)

∫y

1 x

eax dx =

∫b

16.2

sin x = cos x,

sin x dx = − cos x + c,

d dx

∫

1−cos(x) 2

adj hyp ,

a + ib =

= limdx→0

f (x+dx,y)−f (x,y) dx

d dx

∫x a

√ −1 b a2 + b2 ei tan ( a )

f (x′ ) dx′ = f (x)

Appendix II. Periodic table

Metals are in blue, transition metals in light blue, non-metals in red, noble gases are green, and semi-metals are pink. Groups run from left to right (H to He), and periods run top to bottom (H to F r). At STP of 1.01 × 105 P a and 298 K Mercury (Hg), and Bromine (Br) are liquids, and H, N, O, F, Cl, He, N e, Ar, Kr, Xe, Rn are gases. Electronic structures violating Madelung’s rule are in red (and are not the correct structure but are the Madelung rule prediction). Elements without stable nuclides have the atomic mass of the longest-lived isotop in parenthesis.

339

CHAPTER 16. APPENDIX 340

8 15.999

F

9 18.998

Ne

10 20.180

2 4.0026

7 14.007

O

1 1.0079

6 12.011

N

He

5 10.811

C

H

4 9.0122

B

1s2 Helium

3 6.941

Be

1s1 Hydrogen

Li

Fluorine

3p6 Argon

Neon

2p4 Oxygen

2p6

2p3 Nitrogen

Chlorine

36 83.8

2p5 Carbon

18 39.948

3p4 Sulphur

35 79.904

2p2

17 35.453

Ar

3p3 Phosphorus

34 78.96

Boron

16 32.065

Cl

2p1

15 30.974

S

2s2 Berylium

14 28.086

P

2s1 Lithium

Si

13 26.982

Al

12 24.305

Mg

11 22.990

Na

Silicon

33 74.922

3p5

32 72.64

3p2

31 69.723

3p1

30 63.39

Aluminum 29 63.546

3s2 Magnesium

28 58.693

3s1 Sodium

27 58.933

26 55.845

Kr

25 54.938

Br

24 51.996

Se

23 50.942

As

22 47.867

Ge

21 44.956

Ga

20 40.078

Zn

19 39.098

Cu

Fe

Ni

Mn

Co

Cr

4p6 Krypton

V

4p5

Ti

Bromine

Sc

4p4

Ca

Selenium

K

Arsenic

6d8

Uun

6d9

Uuu

65 158.93

6d10

Uub

Indium

6p1 Thallium

66 162.5

6p3

Polonium

6p4

70 173.04

Astatine

6p5

71 174.97

Radon

6p6

Xenon

Bismuth

69 168.93

Iodine

68 167.26

Lu

Uuq

114 (289)

Lead

67 164.93

Yb

5d1 Lutetium

Tm

4f 14

Er

4f 13

103 (262)

Ho

5p2

4p2

4p3 3d7 Cobalt

Germanium

3d6 Iron

Gallium

3d3 4d4 3d5 Vanadium Chromium Manganese

4p1

3d2 Titanium

3d10 Zinc

3d1 Scandium

3d9 Copper

4s2 Calcium

46 106.42

3d8 Nickel

4s1 Potassium

45 102.91

54 131.29

44 101.07

53 126.9

43 96

52 127.6

42 95.94

51 121.76

41 92.906

50 118.71

40 91.224

49 114.82

39 88.906

48 112.41

38 87.62

Pd

47 107.87

37 85.468

Rh

Xe

Ru

I

Tc

Te

Mo

Mt

64 157.25

Dy

4f 12

102 (259)

Erbium

Ytterbium

100 (257)

Thulium 99 (252)

Md

101 (258)

98 (251)

Fm

4f 10 4f 11 Dysprosium Holmium 97 (247)

Es

Lr

Cf

No

Bk

5f 9 5f 10 5f 11 5f 12 5f 13 5f 14 6d1 Berkelium CaliforniumEinsteinium Fermium Medelevium Nobelium Lawrencium

Terbium

5p6

Sb

Nb

Hs

6d7 Meitnerium

63 151.96

Tb

96 (247)

5p5

Sn

Zr

Bh

6d6 Hassium

62 150.36

Gd

95 (243)

5p4

In

Y

4d9 Silver

Cd

Sr

4d8 Palladium

6d5 Bohrium

61 145

Eu

Ag

Rb

4d7 Rhodium

Tellurium

4d3 4d4 4d5 4d6 Niobium MolybdenumTechnetium Ruthenium

5p3 Antimony

4d2 Zirconium

79 196.97

Tin

4d1 Yttrium

78 195.08

5p1

5s2 Strontium

77 192.22

6d3 6d4 Dubnium Seaborgium

60 144.24

Sm

Hg

59 140.91

Pm

94 (244)

6p2

4d10 Cadmium

5s1 Rubidium

76 190.23

Au

5d10 Mercury

86 (222) 75 186.21

Pt

5d9 Gold

85 (210) 74 183.84

Ir

5d8 Platinum

112 (285)

84 (209)

73 180.95

Os

5d7 Iridium

111 (272)

83 208.96

72 178.49

Re

5d6 Osmium

110 (281)

82 207.2

57–71

W

5d5 Rhenium

109 (268)

81 204.38

56 137.33

Ta

5d4 Tungsten

108 (277)

80 200.59

55 132.91

Hf

5d3 Tantalum

107 (264)

Rn

La-Lu 5d2 Hafnium

106 (266)

At

4f 1 − 5d1

105 (262)

Po

Ba Lanthanide

104 (261)

Db

Bi

Cs 6s2 Barium 89-103

Rf

Pb

6s1 Cesium 88 (226)

Ac-Lr

6d2 Rutherford

Tl

87 (223)

Ra

5f 1 − 6d1

58 140.12

Nd

Sg

Fr 7s2 Radium

Actinide

7s1 Francium

57 138.91

Pr

93 (237)

Cm

4f 9

Ce

92 238.03

Am

Curium

5f 8

Pu

5f 5 5f 6 5f 7 Neptunium Plutonium Americum

Np

4f 7 4f 8 Europium Gadolinium

La 4f 1

91 231.04

U

4f 3 4f 4 4f 5 4f 6 4f 2 NeodymiumPromethium Samarium Cerium Praseodymium 90 232.04

Pa

Lanthanum 89 (227)

Th 5f 2

Ac

Thorium Protactinium Uranium

5f 4

5f 1

5f 3 Actinium

16.3. REDUCE EXAMPLES

16.3

341

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)

342

CHAPTER 16. APPENDIX

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);

16.3. REDUCE EXAMPLES

343

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 load_package taylor; f:=1/(x^2+a^2); taylor(f,x,1,4); g:=e^(-x); taylor(g,x,0,5);

Taylor series expand Planck distribution load_package taylor; 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

344

CHAPTER 16. APPENDIX

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)); depend{f,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

16.3. REDUCE EXAMPLES

345

for m:=1:10 do >; % now sweep through the interior points for m:=1:N do >; % 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 such as % 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;

346 % 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);

CHAPTER 16. APPENDIX

Index absolute temperature, 280 acceleration, 5 acid, 313 adhesive force, 80 adiabatic curves, 278 adiabatic line, 281 adiabatic process, 276 air resistance, 38, 42 air-conditioning, 296 ammonia molecule, 118 amplitude, 196, 198 amplitude-dependent frequency, 205 amusement park ride, 79 angle, 10, 77 angular momentum, 145, 165 angular speed, 77 anode, 302 aphelion, 241 Archimedes principle, 181 area element, 13 area integrals, 116 axiom, 26 ballistic pendulum, 122 ballistics, 37 base, 313 battery, 303 Benard cell, 269 Bernoulli’s law, 181 biharmonic equation, 187, 218 billiard ball, 168 billiard ball collision, 121 binary star, 233 binomial theorem, 205 black hole, 236 block and tackle, 60 blood flow, 185 boiling point elevation, 307 Boltzmann entropy, 263 bowling ball, 151 brake, 170 buffering, 313 bulk modulus, 229 cannon, 115 capillary flow, 185 Carnot cycle, 278

Carnot’s theorem, 280 CAS syntax, 26 cathode, 302 center of mass, 114, 116 centrifugal potential, 305 centrifuge, 306 centripetal acceleration, 78, 136 chemical equilibrium, 303 chemical potential, 299, 317 Clausius’ theorem, 283 clutch, 170 coefficient of friction, 66 coefficient of thermal expansion, 288 collinearity, 14 collisions, 93 common ion effect, 312 complexions, 263 components, 12 compound interest, 35 compressibility, 330 compressional wave, 215 computer algebra system (CAS), 26 concentration, 304, 308, 309 conduction, 273 conservation law, 308 conservation of energy, 277 conservative force, 84 constant of integration, 83 contact force, 54 convection cell, 269 critical angle, 67 cross product, 13, 137 current, 308 cycle, 279 damped oscillation, 207 Dark Matter, 251 diatomic molecule, 231 diffusion, 193, 308 disk, 140 distributive law, 13 dot product, 10 double yo-yo, 149 drag, 184 eccentricity, 241 efficiency, 279 347

348 elastic collision, 93 Electrophoresis, 329 ellipse, 241 elliptic curve, 35 elliptic function, 205 empirical temperature, 276 energy, 83 enthalpy, 283, 287 entropy, 281, 285 entropy representation, 317 equal areas, 242 equation of motion, 102 equation of state, 275 equilibrium, 198, 275 equilibrium condition, 224 equilibrium constant, 300, 303 equilibrium point, 203 escape velocity, 236 Euler’s theorem, 207 excluded volume, 318 exhaustion, 285 exponential function, 207 exponential growth, 35 extensive variable, 299 external forces, 114 extrema, 284 Family mule, the, 26 Fick’s law, 308 financial math, 35 first law, 277 flux, 177 force constant, 209 formality, 312 Free-body diagram, 54 friction, 65, 91 gas processes, 278 Gibbs free energy, 283, 303 Gibbs-Duhem relation, 317 global variable, 276 gradient, 84 graduated cylinder, 304 Graham’s law, 308 gravitational potential, 89, 235, 238 gravitational sling-shot, 248 gravity, 304 gyroscope, 175 half-cell, 303 harmonic equation, 187 harmonic oscillator equation, 203 heat, 253, 277 heat bath, 278 heat engine, 278 Helmholtz free energy, 283

INDEX Hess’s law, 287 high-altitude projectiles, 246 Hookes law, 195 hydraulics, 190 hydrolysis, 314 Ice berg, 115 ideal system, 304 impact parameter, 121 impulse, 126, 168 impulsive force, 168 incline, 57 integration by parts, 101 intensive variable, 299 inter-atomic potential, 224 inter-atomic vibration, 224 inter-molecular potential, 224 internal forces, 114 inversion temperature, 319 irreversible process, 276, 281 isotherm, 276, 281 Joule experiment, 277 Joule-Thompson effect, 319 Kelvin’s second law, 279 Kepler’s laws, 243 kinematics, 4 kinetic energy, 84, 139 kinetic friction, 66 ladder, 153 Lame’ constants, 213 Laplace equation, 187 Laplace transform, 102 launch parameters, 40 law of exponents, 24 Law of gravitation, 233 Lechatelier’s principle, 302 Legendre transformation, 316 lift, 192 line-integral, 87 local variable, 276 maxima, 26 maximal entropy principle, 263 maximal frictional force, 67 maximum altitude, 38 Maxwell relation, 316 mechanical energy, 253 molarity, 311 mole fraction, 299 molecular vibration, 201 molecular weight, 307 moment of inertia, 139, 143 momentum, 113, 114 Morse potential, 224

INDEX Na Cl crystal, 117 Newton’s laws, 53 non-linear oscillation, 206 normal acceleration, 82 normal force, 54 normal mode, 200 number theory, 35 Onsager’s theory, 308 oscillations, 195 osmosis, 306 osmotic pressure, 306 oxidation, 302

349 Reynolds number, 185 right hand rule, 166 ring, 140 rocking motion, 225 rod COM, 117 rolling motion, 144, 168 rope tension, 54, 149 rotational motion, 135

quadratic formula, 25 quanta, 327 quasi-static, 276

salts, 314 scalar, 9 second law, 279 sedimentation, 304 semi-major axis, 241 series, 102 shear force, 184 shear forces, 182 shear modulus, 208 shear wave, 215 simple harmonic motion, 199 simple harmonic oscillation, 197 simple pendulum, 204 small amplitude oscillation, 203 solution, 304 specific heat, 288 sphere (moment of inertia), 141 spin rate, 79 spontaneity, 285, 303 spool, 146 spring, 195 state, 275 Static friction, 66 static friction, 67 statics, 152 stoichiometric coefficients, 327 stoichiometry, 303 storm windows, 273 strain, 208 stream function, 186 stress, 208 stress-momentum tensor, 180 strong acid, 313 substitution method, 102 system, 275 systems of particles, 114

radial vector, 78 radius of curvature, 82 range, 37 Rayleigh number, 268 REDUCE, 26 reduction, 302 refrigeration, 296 relative velocity, 6 resolving vectors, 12 reversible process, 276

tangential acceleration, 82, 136 tangential vector, 78 Taylor series, 102 tension, 54 tensor, 177 terminal velocity, 44 terrible puns, 218 thermal expansion, 330 thermodynamic potentials, 283 Third law, 53

parallel axis theorem, 142 parameterization, 87, 88 partial pressures, 299 partition function, 263 pendulum, 122, 125 penguin, 80, 123, 153, 154 penguins, 115 perihelion, 241 periodic motion, 195 periodicity, 198 perpendicularity, 11 pH, 314 phase, 196, 198 photon gas, 297 photons, 327 physical pendulum, 202, 228 Pitot tube, 182 planetary orbit, 241 position graph, 5 potential energy, 84 potential function, 86, 94 power, 132 Prandtl flow, 192 pressure, 177 projectiles, 37 pulley, 58 Pythagorean theorem, 9

350 thrust, 126 time of flight, 37 torque, 146, 165 torsion, 209 train wreck, 6 trajectory, 39 trajectory program, 39 triatomic molecule, 272 trig function, 207 unit vector, 78, 91 Van der Waals gas, 298, 330 variable change, 102 variable changes, 102 vector, 9 vector resolution, 67 viscosity, 182, 183, 185 vortex strength, 179 vorticity, 178 work, 83, 277 work reservoir, 286 work-energy theorem, 84 yo-yo, 149 Young’s modulus, 208 Zeroth law of thermodynamics, 275

INDEX