Cover
Half Title
Title Page
Dedication
Preface
Acknowledgments
Illustrations
Julia R. D’Rozario
Chapter 1: Newton’s Laws
1.1 Theory
1.1.1 Vectors
1.1.1.1 Space and Time
1.1.1.2 Position, Velocity, and Acceleration
1.1.1.3 Scalar (Dot) and Vector (Cross) Product
1.1.1.5 Line Integral
1.1.1.6 Surface Integral
1.1.1.7 Volume Integral
1.1.1.8 Kronecker Delta
1.1.2 Coordinate Systems
1.1.2.1 Cartesian Coordinates
1.1.2.2 Cylindrical Polar Coordinates
1.1.2.3 Spherical Polar Coordinates
1.1.3 Newton’s Laws
1.1.3.1 Newton’s First Law – The Law of Inertia
1.1.3.2 Newton’s Second Law
1.1.3.3 Newton’s Third Law (Action–Reaction)
1.2 Problems and Solutions
Chapter 2: Motion with Air Resistance
2.1 Theory
2.1.1 Drag Force of Air Resistance
2.2 Problems and Solutions
Chapter 3: Momentum and Angular Momentum
3.1 Theory
3.1.1 Linear Momentum
3.1.2 Rockets
3.1.3 Center of Mass
3.1.4 Moment of Inertia
3.1.5 Principle of Conservation of Angular Momentum
3.1.6 Principle of Conservation of the Angular Momentum for a System of N Particles
3.2 Problems and Solutions
Chapter 4: Energy
4.1 Theory
4.1.1 Work Kinetic Energy Theorem
4.1.2 Conservative Forces
4.1.3 Obtaining the Equation of the Motion from the Conservation of the Energy
4.2 Problems and Solutions
Chapter 5: Oscillations
5.1 Theory
5.1.1 Hooke’s Law
5.1.2 Simple Harmonic Motion
5.1.3 Energy
5.1.4 Particular Types of Oscillations and the Differential Equations Associated with Them
5.1.4.1 Damped Oscillations
5.1.4.2 Weak Damping β < ω0
5.1.4.3 Critical Damping β = ω0
5.1.4.4 Strong Damping β > ω0
5.1.4.5 Driven Damped Oscillations
5.2 Problems and Solutions
Chapter 6: Lagrangian Formalism
6.1 Theory
6.1.1 The Lagrangian
6.1.2 Hamilton’s Principle
6.2 Problems and Solutions
Chapter 7: Hamiltonian Formalism
7.1 Theory
7.1.1 The Hamiltonian
7.1.2 Example – One-Dimensional Systems
7.2 Problems and Solutions
Chapter 8: Coupled Oscillators and Normal Modes
8.1 Theory
8.2 Problems and Solutions
Chapter 9: Nonlinear Dynamics and Chaos
9.1 Theory
9.2 Problems and Solutions
Chapter 10: Special Relativity
10.1 Theory
10.1.1 Galileo’s Transformations
10.1.2 Postulates of the Theory of Relativity
10.1.3 Lorentz Transformations
10.1.4 Length Contraction, Time Dilation
10.1.5 Composing Velocities
10.1.6 Relativistic Dynamics
10.1.7 Doppler Shift
10.1.7.1 Redshift
10.1.7.2 Blueshift
10.2 Problems and Solutions
Appendix: Differential Equations
Separable Equations
First-Order Equations with an Integrating Factor
Second-Order Homogeneous Equations
Bibliography
Index

##### Citation preview

Classical Mechanics This book of problems and solutions in classical mechanics is dedicated to junior or senior undergraduate students in physics, engineering, applied mathematics, astronomy, or chemistry who may want to improve their problem-solving skills, or to freshman graduate students who may be seeking a refresh of the material. The book is structured in 10 chapters, starting with Newton’s Laws, motion with air resistance, conservation laws, oscillations, and the Lagrangian and Hamiltonian Formalisms. The last two chapters introduce some ideas in nonlinear dynamics, chaos, and special relativity. Each chapter starts with a brief theoretical outline and continues with problems and detailed solutions. A concise presentation of differential equations can be found in the appendix. A variety of problems are presented, from the standard classical mechanics problems, to context-rich problems and more challenging problems. Key Features: • Presents a theoretical outline for each chapter • Motivates the students with standard mechanics problems with step-by-step explanations • Challenges the students with more complex problems with detailed solutions

Classical Mechanics Problems and Solutions

Carolina C. Ilie Zachariah S. Schrecengost Elina M. van Kempen

To my students: be extraordinary!! To my family, mentors, and friends: thank you for the journey!! -CCI To my family, friends, and mentors. -ZSS To my family and SUNY Oswego friends. -Elina van Kempen

Contents Preface........................................................................................................................xi Acknowledgments....................................................................................................xiii About the Authors..................................................................................................... xv Chapter 1

Newton’s Laws...................................................................................... 1 1.1

Theory.......................................................................................... 1 1.1.1 Vectors............................................................................. 1 1.1.1.1 Space and Time............................................... 1 1.1.1.2 Position, Velocity, and Acceleration................ 1 1.1.1.3 Scalar (Dot) and Vector (Cross) Product......... 1 1.1.1.4 Gradient........................................................... 2 1.1.1.5 Line Integral.................................................... 2 1.1.1.6 Surface Integral............................................... 2 1.1.1.7 Volume Integral............................................... 3 1.1.1.8 Kronecker Delta.............................................. 3 1.1.2 Coordinate Systems........................................................ 3 1.1.2.1 Cartesian Coordinates..................................... 3 1.1.2.2 Cylindrical Polar Coordinates......................... 4 1.1.2.3 Spherical Polar Coordinates............................ 4 1.1.3 Newton’s Laws................................................................ 5 1.1.3.1 Newton’s First Law – The Law of Inertia....... 5 1.1.3.2 Newton’s Second Law.................................... 5 1.1.3.3 Newton’s Third Law (Action–Reaction)......... 6 1.2 Problems and Solutions............................................................... 7 Chapter 2

Motion with Air Resistance................................................................. 43 2.1

Theory........................................................................................ 43 2.1.1 Drag Force of Air Resistance........................................ 43 2.2 Problems and Solutions............................................................. 44 Chapter 3

Momentum and Angular Momentum.................................................. 69 3.1

Theory��������������������������������������������������������������������������������������� 69 3.1.1 Linear Momentum........................................................ 69 3.1.2 Rockets.......................................................................... 69 3.1.3 Center of Mass.............................................................. 69 3.1.4 Moment of Inertia......................................................... 70 3.1.5 Principle of Conservation of Angular Momentum........ 70 3.1.6 Principle of Conservation of the Angular Momentum for a System of N Particles........................ 71 3.2 Problems and Solutions............................................................. 71 vii

viiiContents

Chapter 4

Energy................................................................................................. 91 4.1

Theory........................................................................................ 91 4 .1.1 Work Kinetic Energy Theorem..................................... 91 4.1.2 Conservative Forces...................................................... 91 4.1.3 Obtaining the Equation of the Motion from the Conservation of the Energy........................................... 92 4.2 Problems and Solutions............................................................. 93 Chapter 5

Oscillations........................................................................................ 115 5.1

Theory...................................................................................... 115 5.1.1 Hooke’s Law............................................................... 115 5.1.2 Simple Harmonic Motion........................................... 115 5.1.3 Energy......................................................................... 116 5.1.4 Particular Types of Oscillations and the Differential Equations Associated with Them............ 116 5.1.4.1 Damped Oscillations................................... 116 5.1.4.2 Weak Damping β < ω0................................ 117 5.1.4.3 Critical Damping β = ω0............................. 117 5.1.4.4 Strong Damping β > ω0............................... 117 5.1.4.5 Driven Damped Oscillations....................... 117 5.2 Problems and Solutions........................................................... 118 Chapter 6

Lagrangian Formalism...................................................................... 143 6.1

Theory...................................................................................... 143 6.1.1 The Lagrangian........................................................... 143 6.1.2 Hamilton’s Principle................................................... 143 6.2 Problems and Solutions........................................................... 144 Chapter 7

Hamiltonian Formalism..................................................................... 167 7.1

Theory...................................................................................... 167 7.1.1 The Hamiltonian......................................................... 167 7.1.2 Example – One-Dimensional Systems........................ 167 7.2 Problems and Solutions........................................................... 168 Chapter 8

Coupled Oscillators and Normal Modes........................................... 197 8.1 Theory...................................................................................... 197 8.2 Problems and Solutions........................................................... 198

Chapter 9

Nonlinear Dynamics and Chaos........................................................ 221 9.1 Theory...................................................................................... 221 9.2 Problems and Solutions........................................................... 221

Contents

ix

Chapter 10 Special Relativity.............................................................................. 241 10.1 Theory...................................................................................... 241 10.1.1 Galileo’s Transformations........................................... 241 10.1.2 Postulates of the Theory of Relativity......................... 242 10.1.3 Lorentz Transformations............................................. 242 10.1.4 Length Contraction, Time Dilation............................. 243 10.1.5 Composing Velocities.................................................244 10.1.6 Relativistic Dynamics................................................. 246 10.1.7 Doppler Shift............................................................... 247 10.1.7.1 Redshift....................................................... 247 10.1.7.2 Blueshift...................................................... 247 10.2 Problems and Solutions........................................................... 247 Appendix: Differential Equations.......................................................................... 255 Bibliography.......................................................................................................... 259 Index....................................................................................................................... 261

Preface This book of problems and solutions is written for undergraduate students in physics, mechanical engineering, applied mathematics, and chemistry, who may want to improve their skills in solving classical mechanics problems, or for first-year graduate students who may need a refresher. For a comprehensive textbook, the authors recommend “Classical Mechanics” by John R. Taylor, and for graduate level, “Classical Mechanics” by Herbert Goldstein. The book is structured into ten chapters. The first eight chapters span the traditional spectrum of classical mechanics: starting with Newton’s Laws, conservation laws, and oscillations, and ending with Lagrangian and Hamiltonian Formalisms, along with coupled oscillations. The last two chapters introduce some ideas in nonlinear dynamics, chaos, and special relativity. Each chapter starts with a brief theoretical outline. The problems are organized from simple, standard pedagogical problems (e.g., Atwood machine, free fall, etc.) useful for beginners, and solved in great detail, to more involved problems, and each chapter ends with more challenging, interesting problems. The authors tried to keep a balance between simpler problems, contextrich problems, and more complex problems. Good problem-solving skills can be acquired by following a few steps – becoming familiar with the theory and the notations, making sure the mathematical background is solid, then solving the problems without checking the solution, and finally, solution check. For solidifying the material, it may be useful to work on the problems in a few days and gradually add more difficult problems.

xi

Acknowledgments We thank Dr. Ilie’s classical mechanics students for pointing out some of the more challenging concepts and motivating us to write this book. We are especially grateful to our illustrator, Dr. Julia R. D’Rozario, for creating the figures during a very busy time just before and after her Ph.D. defense. We thank Victor Sabirianov, who volunteered to do the final figure editing. Elina van Kempen thanks Mathieu van Kempen for his efficient and valuable help with proofreading and typing. Dr. Ilie would like to thank the wonderful co-authors Zachariah Schrecengost and Elina van Kempen for becoming a great, efficient, and fun writing team; working together was exciting and inspiring! Dr. Ilie is grateful to Dr. Peter Dowben, from the University of Nebraska at Lincoln for brilliant mentoring, summer research opportunities, coffee, and for inspiring her to write books. Many thanks to the administration at SUNY Oswego for overall support and for recognizing students and faculty scholarly and creative endeavors. Many thanks to our kind and professional editors, Dr. Danny Kielty, Rebecca Hodges-Davies, project manager Thivya Vasudevan (Straive), production editor Kari Budyk, and the full editing team from CRC Press and Taylor & Francis Group. The authors thank our readers for their curiosity and joy in learning and encourage them to take time to enjoy challenge and discovery. Last, but not least, the authors would like to thank their families and friends for their love, joy, and funny jokes, which made this journey exciting.

xiii

About the Authors Carolina C. Ilie is a Sigma Xi Fellow, Full Professor with tenure at the State University of New York at Oswego. She taught Classical Mechanics for more than ten years and she designed various problems for her students. Dr. Ilie obtained her Ph.D. in Physics and Astronomy at the University of Nebraska at Lincoln, an M.Sc. in Physics at the Ohio State University, and another M.Sc. in Physics at the University of Bucharest, Romania. She received the President’s Award for Teaching Excellence in 2016 and the Provost Award for Mentoring in Scholarly and Creative Activity in 2013. Her research is focused on condensed matter physics: dynamics at surfaces, capillary condensation, perovskites photovoltaics. She lives in Central New York with her spouse, also a physicist, and their two sons. Her hobbies are classical music, learning languages, and writing. Zachariah S. Schrecengost is a State University of New York alumnus. He graduated summa cum laude with a B.S. degree having completed majors in Physics, Software Engineering, and Applied Mathematics. He took the Advanced Mechanics course with Dr. Ilie and loved to be involved in this project. He brings to the project the experience of writing other two books of problems, but also the fresh perspective of the student taking classical mechanics and the enthusiasm and talent of an alumnus who is a physics and upper-level mathematics aficionado. Mr. Schrecengost works as software engineer in Syracuse while working toward his Ph.D. in Physics at Syracuse University. Elina M. van Kempen graduated summa cum laude from the State University of New York at Oswego, with a double major in Physics and Applied Mathematics, and a minor in Computer Science. She is now working on her Ph.D. in Computer Science at the University of California, Irvine, with a focus on Security and Privacy. At SUNY Oswego, she had the great opportunity to take several courses taught by Dr. Carolina Ilie, including her course on Advanced Mechanics. Elina M. van Kempen also tutored for three years at SUNY Oswego, in Physics, Mathematics, and Computer Science and enjoyed helping students to understand and succeed in their classes. She loves traveling, cooking, and swimming. xv

xvi

ILLUSTRATIONS Julia R. D’Rozario Julia R. D’Rozario (illustrator) graduated with a Ph.D. in Microsystems Engineering at Rochester Institute of Technology in Rochester, NY, in May 2022. Her doctoral research includes the study of light manipulation in optoelectronic devices, including III–V compound semiconductor space photovoltaics and micrometer-scale LEDs to improve device performance. Dr. D’Rozario aims to utilize the skillset she developed throughout grad school to help innovate technological advancements in her future career. Also by Ilie and Schrecengost, with illustrations by Julia D’Rozario:

1. Carolina C. Ilie, Zachariah S. Schrecengost, Electromagnetism: Problems and Solutions, Institute of Physics IOP Science, UK, Morgan & Claypool Publishers, CA, USA; November 2016; Online ISBN: 978-1-6817-4429-2 • Print ISBN: 978-1-6817-4428-5 2. Carolina C. Ilie, Zachariah S. Schrecengost, Electrodynamics: Problems and Solutions, Institute of Physics IOP Science, UK and Morgan and Claypool Publishers, CA, USA; May 2018; Online ISBN: 978-1-6817-4931-0 • Print ISBN: 978-1-6817-4928-0

1

Newton’s Laws

1.1  THEORY This chapter introduces the three Newton’s Laws and problems solved in a system of coordinates appropriate for each case. The authors hope that this chapter will bridge the knowledge and skills acquired in lower-­level classical mechanics courses and bring the refinement of differential equations in different systems of coordinates and a broader vision on problem solving.

1.1.1 Vectors 1.1.1.1  Space and Time The motion of an object is analyzed by placing the object in a system of coordinates with a clock measuring the time t. The system of coordinates is chosen depending on the type of motion: a cartesian system of coordinates ( x, y, z) , a two-­dimensional polar one (r, φ ) , a cylindrical one (s, φ , z) , or a spherical system of coordinates (r,  ,  ) . 1.1.1.2  Position, Velocity, and Acceleration   For an object moving from the initial position r0 at time t0 to the final position r at time t, the velocity (instantaneous) is defined as   dr  = v = r dt

and the acceleration (instantaneous) as

dv d  dr  d 2r     2 r a dt dt  dt  dt

1.1.1.3  Scalar (Dot) and Vector (Cross) Product   Given vectors A  Ax xˆ  Ay yˆ  Az zˆ and B  Bx xˆ  By yˆ  Bz zˆ The scalar product is a scalar equal to

  A  B  Ax Bx  Ay By  Az Bz  AB cos

DOI: 10.1201/9781003365709-1

1

2

Classical Mechanics

The vector product is a vector equal to

xˆ   A  B  Ax Bx

yˆ Ay By

zˆ Az Bz

with

  A  B  AB sin

  where A  ｜ A ｜ Ax2  Ay2  Az2 , B  ｜ B ｜ Bx2  By2  Bz2 , and θ is the angle   between A and B. 1.1.1.4  Gradient Given a scalar function T, the gradients for various coordinate systems are given below. Gradient in cartesian coordinates ( x, y, z) :

T

T T T xˆ  yˆ  zˆ x y z

Gradient in cylindrical coordinates (s, φ , z) :

T

T 1 T ˆ T sˆ   zˆ s s  z

Gradient in spherical coordinates (r ,  ,  ) :

T

T 1 T ˆ 1 T ˆ rˆ    r r  r sin  

1.1.1.5  Line Integral  Given vector function v and path  , a line integral is given by  b

v  d

a

   where a and b are the end points and d is the infinitesimal displacement vector along path  . In Cartesian coordinates d   dx xˆ  dy yˆ  dz zˆ . 1.1.1.6  Surface Integral  Given vector function v and surface  , a surface integral is given by 

v  da 

3

Newton’s Laws

 where da is the infinitesimal area vector with direction normal to the surface. Note  that da always depends on the surface involved. 1.1.1.7  Volume Integral Given scalar function T and volume  , a volume integral is given by

∫T dV

where dV is the infinitesimal volume element. In Cartesian coordinates dV = dx dy dz. 1.1.1.8  Kronecker Delta 0 if i  j  ij   1 if i  j

For example,

  A B

n

n

Ai ij B j

i , j 1

A B i

i

i 1

Levi-­Civita symbol in three dimensions,

ijk

1 if (i, j, k ) is an even permutation of (1, 2, 3), i.e., (1, 2, 3), (2, 3, 1), or (3, 1, 2)   1 if (i, j, k ) is an odd permutation of (1, 2, 3), i.e., (1, 3, 2), (2, 1, 3), or (3, 2, 1) 0 if i  j, or j  k, or k  i 

In three dimensions, Levi-­Civita is related to the Kronecker delta in the “contracted epsilon identity” 3

ijk imn

 jm kn   jn km

i 1

1.1.2 Coordinate Systems 1.1.2.1  Cartesian Coordinates Here (Figure 1.1), our infinitesimal quantities are

d   dx xˆ  dy yˆ  dz zˆ

4

Classical Mechanics z

z y

y

0

x x

FIGURE 1.1  Cartesian system of coordinates.

The element of volume is expressed in the following way in Cartesian coordinates: dV = dx dy dz

1.1.2.2  Cylindrical Polar Coordinates Here (Figure 1.2), our infinitesimal quantities are

d   ds sˆ  s d ˆ  dz zˆ

and element of volume dV  s ds d dz

1.1.2.3  Spherical Polar Coordinates Here (Figure 1.3), our infinitesimal quantities are

d   dr rˆ  r d ˆ  r sin  d ˆ z

S

y x

FIGURE 1.2  Cylindrical system of coordinates.

5

Newton’s Laws z

r

0

y

x

FIGURE 1.3  Spherical system of coordinates.

and the element of volume is

dV  r 2 sin  dr d d

1.1.3 Newton’s Laws Isaac Newton was a student at Cambridge University when in 1665 plague swept through England. The university was closed and Newton decided to move back to his family’s farm in Lincolnshire, where he lived for 18 months before moving back to Cambridge. That was for Newton his annus mirabilis, his miraculous year, when he discovered the nature of light, made a breakthrough in calculus more than ten years before Leibniz, and discovered gravity. Part of his results, what is known as Newton’s Laws, was published in 1687 under the name Principia Mathematica. 1.1.3.1  Newton’s First Law – The Law of Inertia A particle maintains the state of rest or the state of motion with constant velocity as long as no net force is acting on it. Inertia is the property of objects with mass to maintain their state of motion with constant velocity (the speed is constant and the direction is straight, so the velocity vector is constant), or their state of rest until an external force is acting on them to change this state. 1.1.3.2  Newton’s Second Law The acceleration of a particle is proportional to the net force acting on the particle and inversely proportional to the mass of the particle.

  F = ma

  F a= m

6

Classical Mechanics

  In terms of linear momentum, p = mv.

= p mv = ma (Here, it is assumed the mass is constant.)

So, F = p . In classical mechanics, these two forms of Newton’s Second Law are equivalent, with the second form more general than the first. However, in relativity, the derivative of momentum is different, because the mass is not constant at speeds close to the speed of light. In different systems of coordinates, the Second Newton’s Law is written in various forms. In vector form,   F = mr

In cartesian coordinates ( x, y, z) ,

Fx = mx

Fy = my

Fz = mz In two-­dimensional polar coordinates (r , φ ) ,

Fr  m( r  r2 )

F  m(r  2r) In cylindrical coordinates (s, φ , z ),

Fs  m( s  s2 )

F  m(s  2s)

Fz = mz

1.1.3.3  Newton’s Third Law (Action–Reaction)  If an object 1 acts on object 2 with a force F12 (Figure 1.4), then the object 2 acts on object 1 with a force equal in magnitude, but in opposite direction: F21   F12 . Note that in relativity, Newton’s Third Law does not hold due to problems related to simultaneity.

7

Newton’s Laws

F21

F12

1

2

FIGURE 1.4  Two objects acting upon each other with forces equal in magnitude, but in opposite directions.

1.2  PROBLEMS AND SOLUTIONS PROBLEM 1.1    Given vectors A , B, C prove the following vector relations using summation notation:

      A  ( B  C )  C  ( A  B)

         A  ( B  C )  B( A  C )  C ( A  B)

SOLUTION 1.1 a. Looking at this in terms of components

   A  ( B  C )  Ai ij ( BlCm lmj )  Ai BlCm lmi  Cm Ai Bl ilm     Ck kl Ai Bl lmi  C  ( A  B) as desired. Note there is a third relation which can be shown in a similar way, yielding

         A  ( B  C )  C  ( A  B)  B  (C  A) b. Looking at a particular component    [ A  ( B  C )]k  Ai ( BlCm lmj ) ijk  Ai BlCm lmj kij  Ai BlCm ( lk mi   li mk )  Bl lk ( AiCm mi )  Cm mk ( Ai Bl li )  Bk ( AiCm mi )  Ck ( Ai Bl li )            Bk ( A  C )  Ck ( A  B)  [ B( A  C )  C ( A  B)]k          Since this holds for all k, we have shown A  ( B  C )  B( A  C )  C ( A  B).

8

Classical Mechanics

PROBLEM 1.2 A mass m is hanging from a massless string of length R (Figure 1.5). The mass is moving in a circular motion with constant angular velocity ω . Using Newton’s Second Law in polar coordinates, find the tension in the string. SOLUTION 1.2 Newton’s Second Law is applied (Figure 1.6):   F = ma

As specified, the polar components of acceleration, ar   r  r2 and a  2r  r, are used. Thus,

Fr  m( r  r2 )

R

m

FIGURE 1.5  Mass hanging from the string.

R

T

m mg cos

mg

FIGURE 1.6  Mass hanging from the string, with gravitational force and tension in the string shown.

9

Newton’s Laws

and F  m(2r  r)

As shown in Figure 1.6, the sum of the forces on the radial axis is Fr  T  mg cos

From this equation and Newton’s Second Law, an expression for the tension in the string is obtained. T  mg cos   m(r  r2 )

Solving for T :

T  mg cos   m( r  r2 )

From the given information, this expression can be simplified. The mass is attached to a massless string of length R. Since the mass is always at distance R, it follows that and r = R and r= r= 0. It is also given that the mass moves with constant angular speed ω. Thus,    . The tension in the string is

T  m( g cos   R 2 )

PROBLEM 1.3 A rock of mass m falls down a cliff of height H (Figure 1.7). The rock has initial velocity v0 , directed horizontally. Assume no air resistance. a. Find the position of the rock as a function of time. b. What is the velocity of the rock when it touches the ground? V0

H

FIGURE 1.7  Rock on top of a cliff of height H.

10

Classical Mechanics

SOLUTION 1.3 a. Newton’s Second Law:   F = ma

In the x direction (Figure 1.8), no forces are acting on the rock Fx= 0= mx

In the y direction, the weight of the rock is considered Fy  mg  my

First, by integrating is obtained:

x dx   0 dt , the x component of the rock’s velocity, x , x = C1 (1.1)

x (0) v= v0 . So C1 = v0 . Now, solving for x is possible. At t = 0 , = x (0 )

x dx   v

0

dt

x  v0t  C2

At t = 0 , x(0) = 0 . So C2 = 0 . Solving for y as a function of time is done similarly by integrating.

my dy   mg dt

y

V0

H

0

FIGURE 1.8  Rock on top of a cliff of height H.

X

11

Newton’s Laws

y   gt  C3 (1.2)

y (0) v= At t = 0 , = 0 . So C3 = 0 . y (0 )

y dy   gt dt

y

gt 2  C4 2

At t = 0 , y(0) = H . So C4 = H . The position of the rock as a function of time is described by  x(t )  v0t   gt 2 H  y( t )   2 

b. The rock touches the ground at y = 0. First, the time at which the rock touches the ground is determined.

y(tground )  0

tground =

2 gtground H 2

2H g

We substitute tground in Equations (1.1) and (1.2) and obtain the velocity of the rock when it reaches the ground:

vx (tground )  v0  2H   vy (tground )   g g   2 gH 

PROBLEM 1.4 Consider a block of mass m sliding down a frictionless ramp at an incline θ (Figure 1.9). Find the velocity of the block at time t if the block is stationary at t = 0 . SOLUTION 1.4 To find the velocity, consider

Fext  p

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Classical Mechanics

FIGURE 1.9  Block placed at the top of a ramp. FN y

X

Fg

FIGURE 1.10  Forces on the block with respect to the coordinate system.

The coordinate system and forces can be taken as shown in Figure 1.10. The forces are then given by  Fg  mgyˆ

and

FN  mg(sin  x  cos  y ) cos

with

2      p  F N  Fg  mg(sin  x  cos  y ) cos   mg y  mg(sin  cos  x  (cos  1) y ) Using cos2   sin 2   1

p  mg sin  (cos x  sin  y ) From this, the momentum is given by

p  mgt sin  (cos x  sin  y )  C

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Newton’s Laws

  Since the velocity at t = 0 is zero, p(0) = 0 , which means C = 0 . Thus,  p  mgt sin  (cos x  sin  y )

From this, the velocity is given by   p v   gt sin  (cos x  sin  y ) m

the magnitude of which is v  gt sin

as desired.

PROBLEM 1.5 A crow kicks a frictionless plastic plate (which does not rotate) with initial speed v0 , so that it slides straight up a snowy slippery roof that is inclined at an angle θ above the horizontal, then the crow flies to the plate and uses it as a sled down the roof. a. Write down Newton’s Second Law for the lid and solve it to give its velocity as a function of time. b. Find the time needed for the lid to reach the highest point on the roof. c. Find how long the lid will take to return to its starting point, assuming the crow arrives to the lid exactly when it stops on the roof on its way up. Neglect friction and air drag. d. Find the position as a function of time. SOLUTION 1.5 a. The x-­axis is chosen to be up the incline, as in the figure, and y-­axis perpendicular to the plane (Figure 1.11). N

mg sin  mg cos

y

x

mg

FIGURE 1.11  Inclined plane with the gravitational force decomposed on two perpendicular directions.

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Newton’s Second Law is applied in vectorial form and then written separately on each axis, noting that the acceleration on y-­axis is zero.   F = ma

On y-­axis, the resultant is zero (no acceleration). Fy  N  mg cos   ma y  0

On x-­axis, there is acceleration: Fx  mg sin

Also, = Fx ma = mv . From both equations, and after dividing with the x mass, it follows that: v   g sin

Considering v =

dv , dt dv   g sin  dt

By integration it follows that

dv   g sin  dt From here it is easy to obtain the velocity as

v(t )   gt sin   C where C is the constant of integration to be obtained from the initial conditions when the clock starts: = t 0= s, v(t 0) = v0 . Therefore,

v(0)  v0   g  0  sin   C  C and C = v0 . From here the velocity is

v(t )  v0  gt sin  It is important to note that, since the friction is neglected, the velocity is independent of mass.

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Newton’s Laws

b. In order to find the time t up it is just needed to write that the velocity at that time will become zero, since the object stops. 0  v0  gt up sin

So indeed,

t up

v0 g sin

Also, the time t up is mass independent. c. Since there is no friction force and no air drag, the problem is symmetric (time upward is the same as time downward), and it is very easy to find the total time: t total = 2 t up

Of course, there is another solution as well, by starting from the general formula for the velocity as a function of time: v(t )  v0  gt sin

After the plate goes up and down (crow or no crow), the velocity on the way up will be of the same magnitude as the initial velocity, but in opposite direction −v0 (here, symmetry is assumed as well), and the equation becomes v0  v0  gt total sin

t total

t total = 2t up

2v0  2t up g sin

d. The position versus time is obtained by starting from the velocity formula dx(t ) v( t ) = and by integrating one more time. dt

v(t )  v0  gt sin

dx(t )  v0  gt sin  dt so

dx   (v  gt sin  ) dt 0

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Classical Mechanics

And the position as a function of time follows, with another constant of integration C′ x(t )  v0t

gt 2 sin   C  2

From the initial conditions, C′ can be determined. At = t 0= ,x 0 x(t  0)  C  so C  0 The position is

x(t )  v0t

gt 2 sin  2

PROBLEM 1.6 A ball of mass m is sliding up a frictionless ramp of length l and at an angle θ . Determine the maximum initial velocity of the ball such that it doesn’t slide off the ramp. Consider the following two cases and compare the answers: a. The ball is initially released parallel to the ramp (Figure 1.12). b. The ball is initially released parallel to the floor (Figure 1.13).

L

V0

FIGURE 1.12  Ball traveling up frictionless ramp.

L

FIGURE 1.13  Ball traveling toward frictionless ramp.

V0

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Newton’s Laws

x

y

FIGURE 1.14  Coordinate system with respect to the ramp.

SOLUTION 1.6 a. The coordinate system is set up as shown in Figure 1.14. So the initial velocity is given by  v0 = vc xˆ

The force of gravity on the ball is given by

Fg  mg sin  x  mg cos  y  mg(sin  x  cos  y ) and the normal force is given by  FN  mg cos

Using these forces, in Newton’s Second Law, the ball’s acceleration can be determined

     F  Fg  FN  ma

mg(sin  x  cos  y )  mg cos  y  ma

a   g sin  x   Since a = v ,  v   g sin  x

By integration,

v   gt sin  x  C1

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Classical Mechanics

   where C1 is some constant vector. Given that at t = 0 , = v v= v0 xˆ , one has 0  v  (v0  gt sin  ) x

 Since it is required that v = 0 at the top of the ramp, this takes t

v0 g sin

And v = r , so  r  (v0  gt sin  ) x

By integration,

1  r  C2   v0t  gt 2 sin   x 2    where C2 is another constant vector. The initial position can be taken to be zero so   1  r   v0t  gt 2 sin   x 2  

 At the top of the ramp, r = Lxˆ . Since the time for the ball to reach the top of the ramp is known, everything can be plugged in and solved for v0 . Everything is in xˆ so the unit vector can be dropped: L  v0t

1 2 gt sin  2

By substituting the time, 2

L  v0

L

1  v0  v0  g  sin  g sin  2  g sin  

1 v02 v02  g sin  2 g sin

v0  2 gL sin

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Newton’s Laws y x

V0

FIGURE 1.15  The initial velocity of the ball with respect to the coordinate system.

b. Now, the initial velocity is not parallel to the ramp. Keeping the same coordinate system as part a, the initial velocity is given in Figure 1.15. So  v0  v0 cos  x  v0 sin  y  v0 (sin  x  cos  y ) Since the ball does not move in the yˆ direction once it is on the ramp, only the  xˆ part of the initial velocity must be considered. From part a, the constant C1 was just the initial velocity. Therefore, the velocity is then given by  v  (v0 cos   gt sin  ) x

Again, it takes

t

v0 cos  g sin

to reach the top. Also, suppose the ball hits the ramp at t = 0 , then the initial position is 0 and

r  v0t cos

1 2 gt sin  2

 At the top r = Lxˆ and again, vo can be solved for

L

v02 cos2  1 v02 cos2   g sin  2 g sin

v0

1 cos

2 gL sin

Note that as θ increases, the ball can be thrown faster and faster approach ing   where v0   and the ball will never go past the ramp. 2

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PROBLEM 1.7 A particle with initial mass m0 and initial velocity v0 begins losing mass according to the equation m(t )  m0e  t where α is constant. If there are no external forces, find an expression for the velocity. SOLUTION 1.7 Since there are no external forces:   Fext= p= 0

Now that there is a changing mass, care must be taken with p .  d   mv p  (mv)  mv dt

Therefore,

  mv  0 mv

Using m(t) yields

m0 e  t v  m0e  t v  0

After dividing both sides by m0e  t , v   v  0

dv  v dt

Separation of variables,

dv   dt v

ln v   t  C

v(t )  Ae t

where A and C are constants. Given v(0) = v0 , A = v0 . Therefore, the velocity is given by

v(t )  v0e t

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Newton’s Laws

This makes sense: the particle exponentially loses mass so it must exponentially gain velocity. Note as m → 0 , v   , which clearly cannot happen considering the velocity must be less than the speed of light. However, it is always assumed to be in a regime where v  c. PROBLEM 1.8 A particle of mass m moves with a velocity v(t )  v( sin( t ) x  cos( t ) y ) . Is there a net external force acting on the particle? If so, find it and comment. SOLUTION 1.8  If there is a net external force acting on the particle, p ≠ 0. The momentum is given by  p(t )  mv( sin( t ) x  cos( t ) y )

so

p (t )  m v(cos( t ) x  sin( t ) y )

Note p ≠ 0 for all t so there is a net external force acting on the particle given by  Fext  m v(cos( t ) x  sin( t ) y )

  Consider v ⋅ Fext ,

  v  Fext  m v 2 ( sin( t ) cos( t )  cos( t ) sin( t ))  0

  Therefore, v and Fext are orthogonal and the external force drives the particle in a circular motion. PROBLEM 1.9 A soccer ball is hit from the ground level at an angle θ and an initial velocity v0 . Discuss the motion, considering that the air resistance is negligible. Obtain: (a) the position versus time, (b) the time necessary for the ball to hit the ground, (c) the maximum height. Also: (d) discuss for which initial angle is the range maximized, (e) determine the equation of the trajectory in xy plane, that is, find y( x ) . SOLUTION 1.9 a. The x direction is chosen on the horizontal and y direction on the vertical. The z direction is not interesting, as no forces or velocities act in that direc  tion. Newton’s Second Law is applied, F = mr .

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The forces acting on the soccer ball are identified in each direction. No air resistance, therefore, Fx = 0 . Gravitational force acts in y direction, and considering the y -axis upward, Fy  mg

  so F  (0,  mg, 0) , therefore, the acceleration is  r  (0,  g, 0) . The initial conditions are as follows: the initial position is chosen at the  origin of the Cartesian system of coordinates, so x0 = (0, 0, 0) , while the  initial velocity is v0  (v0 cos  , v0 sin  , 0). The next step is to integrate on x and y directions separately and, by considering the initial conditions, the velocities are obtained.

x   x dt  0  v0 x  v0 cos

y   y dt   g dt   gt  v0 y   gt  v0 sin 

By another integration, x (t )

x dt   v cos dt  v t cos  x 0

0

0

v0t cos

Similarly,

y( t )

y dt  ( gt  v0 sin  ) dt

gt 2 gt 2  v0t sin   y0    v0t sin  2 2

b. The time necessary for the ball to hit the ground can be easily found by imposing the condition y(t ) = 0. It follows that

gt 2  v0t sin   0 2

so

gt  t  v0 sin    0  2  and the two solutions are t = 0 at the origin (when the ball is hit) and 2v sin  t 0 , which is the total falling time. g

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Newton’s Laws

c. It is important to recall that the vertical velocity is zero at the maximum height, otherwise the ball will go even higher. This is used to obtain the maximum height.

y (t )   gt  v0 sin

y (t ) = 0

gt  v0 sin   0 so t

v0 sin  g

which is half of the total falling time, which is expected due to the symmetry of the problem. Back to the y component y( t )

gt 2  v0t sin  2

The time necessary for the ball to reach the highest point is substituted and the highest position is obtained. y

v02 sin 2  2g

Another way to find the maximum height is using Galileo’s equation vy2  v02y  2 gy, and by replacing vy = 0 , the maximum height it is obtained as y

v02y v02 sin 2  , as before.  2g 2g

d. The range is calculated first. x(t )  v0t cos

and using the total time t

R( )  v0

2v0 sin  , the range is given by g

2v0 sin  v 2 2 sin  cos  v02 cos   0  sin 2 g g g

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Classical Mechanics

The range is maximum for a given initial speed when sin 2θ is maximum, which happens when sin 2  1

so

45

e. In order to find the equation of motion, the time is eliminated between the two equations for x and y.

x(t )  v0t cos  y( t )

gt 2  v0t sin  2

x and after subv0 cos  stitution in the second equation the position y as a function of x is obtained: From the first equation, the time is, as before, t

2

g x  x  v0 sin  y( x )     2  v0 cos   v0 cos  g g  x 2  x tan    x 2  x tan  2(v0 cos  )2 2(v0 x )2 The position y is quadratic in x , which is expected since the trajectory is a parabola. Here, the angle θ is the initial angle.

PROBLEM 1.10 A mass m is suspended vertically to a spring with constant k . It is released from rest at y(0) = yr at t = 0 . Describe the motion of the mass. SOLUTION 1.10   my . The forces acting on the mass m are F mr Using Newton’s Second Law, = =  F  mg  ky . Thus, my  mg  ky

 y

k yg m

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Newton’s Laws

The solution of this differential equation is of the form y  yn  y p , where yn is a k k y  y  0, and y p is a function satisfying  y y  g. solution to  m m mg k k y  y  g . Solving for  y  y  0, yp = is a function that satisfies  k m m  k   k  yn  C1 cos  t  C2 sin  t . So,  m   m     

k   k  mg y(t )  C1 cos  t  C2 sin  t   m   m  k     Since y(0) = yr , C1  yr

mg . Since y (0) = 0 , C2 = 0 . The equation of motion is k

k  mg mg   y(t )   yr  cos  t    m  k k    

mg ky0 Note: = = y0 is the shifted equilibrium. If the origin is chosen to be y0 , k k  k  y(t )  yr cos  t .  m    PROBLEM 1.11 A particle of mass m is moving at a velocity v0 . At t = 0 , a force F   v 2 , where α is a constant, acts on the particle. Find the particle’s momentum as a function of time. SOLUTION 1.11 Using Newton’s Second Law

  F = ma

v 2  mv This can be solved for v :

2 dv v  m dt

Separation of variables yields

dt  v 2 dv m

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Classical Mechanics

and integration yields v 1

t C m

At t = 0, v = v0, so 1 =C v0

and

1  1  t v m v0

Therefore,

v

mvo  v0t  m

and the momentum is given by p(t )

m 2 vo  v0t  m

PROBLEM 1.12 A child on a sled, of combined mass m, is going up a snowy incline at an angle θ above the horizontal (Figure 1.16). The coefficient of friction between the sled and the snow is µ . The child has an initial speed v0 , parallel to the incline.  a. Find the position as a function of time r. b. What is the maximum distance traveled upward by the child?

V0

FIGURE 1.16  Child on a sled going up an incline.

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Newton’s Laws N V0 f

mg

FIGURE 1.17  Child on a sled going up an incline.

SOLUTION 1.12 a. We choose the x-­axis in the direction of the incline pointed upward (Figure 1.17). The y-­axis is normal to the incline, pointed up. The z-­axis is perpendicular to the x-­axis andin the plane of the incline. From Newton’s Second Law, F = mr. Component-­wise, we have  Fx  mx   Fy  my  0  F  mz  0  z

The forces acting on the system are  Fx    N  mg sin    Fy  0  F  N  mg cos   z

Thus, N  mg cos   0 , so the value of N is mg cos θ . From Fx , the following relationship is obtained:

mx    mg cos   mg sin  Integration is performed to get x .

x  (  g cos   g sin  ) dt x   (  g cos   g sin  )t  C1 x (0) v= v0 . So C1 = v0 . Integration is now performed to get x . At t = 0 , = x (0 )

x  ((  g cos   g sin  )t  v0 ) dt

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Classical Mechanics

x

(  g cos   g sin  )t 2  v0t  C2 2

At t = 0 , x(0) = 0 . So C2 = 0 . The position as a function of time is    (  g cos   g sin  )t 2  v0t  x r   2  

b. The child is at a maximum distance upward when v= (t ) x= (t ) 0. The time tmax at which this maximum distance is reached is (  g cos   g sin  )t  v0  0

tmax

v0  g cos   g sin

Plugging this value into x(t ), the maximum distance is obtained: 2

 (  g cos   g sin  )  v0 v0 x(tmax )      v0  g cos   g sin  2   g cos   g sin   x(tmax )

v02 2(  g cos   g sin  )

PROBLEM 1.13 Describe the motion of a ladybug sitting on a vinyl disk of radius R rotating with a constant angular velocity ω in xy plane. SOLUTION 1.13 The ladybug is at rest with respect to the disk, but in circular motion with respect to an observer. The trajectory is a circle of radius R and the ladybug has a constant angular velocity of ω . The linear velocity is perpendicular on the trajectory at every point (Figure 1.18) and has the magnitude v   R . The centripetal acceleration is oriented toward the center of the circle (from the Greek centripetos – toward the center) and has the magnitude of

acp

v2  2R R

The position vector in xy plane is

r (t )  R[ x cos( t )  y sin( t )]

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Newton’s Laws

V

x

acp

FIGURE 1.18  The velocity of the ladybug is tangent to the trajectory, while the centripetal acceleration is radial and perpendicular to the velocity.

The velocity can be found by differentiating the position vector with respect to time.

dr r (t )    R[  x sin( t )  y cos( t )] dt The acceleration is obtained by differentiating again:

dr  r (t )    2 R[ x cos( t )  y sin( t )]   2r (t ) dt

Therefore, the acceleration is antiparallel (note the minus sign) to the radius and has indeed the magnitude

acp

v2  2R R

PROBLEM 1.14 An unidentified object/particle of mass m has the following position as a function of time

1 r (t )  e at x  gt 2 y 2

where a is a constant and g is the gravitational acceleration. a) Find the velocity and the acceleration. b) Describe the force acting on this particle and describe how the motion depends on the constant a . c) Find the equation of the trajectory y( x ) .

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Classical Mechanics

SOLUTION 1.14 a. The motion is two dimensional and the gravitational influence can be seen in y direction.

1 r (t )  e at x  gt 2 y 2 The velocity is easily obtained as

dr d 1  d (e at )  1 d v (t )    e at x  gt 2 y   x ( gt 2 ) y  ae at x  gt y dt dt  2 dt 2 dt  Similarly, the acceleration is obtained by one more differentiation as

dv d (ae at )  d ( gt )  a (t )   x y  a 2e at x  g y dt dt dt b. The force acting on the object is   F (t )  ma  m(a 2e at x  g y ) assuming that the mass is constant. The particle is moving in the negative y direction with gravitational acceleration and in the x direction with an acceleration which is exponentially decreasing in time if a is negative, and exponentially increasing in time if a is positive. c. For the equation of the trajectory, let us write the position on each of the two directions:

x(t ) = e at

1 y(t )   gt 2 2 From the first equation, by applying the logarithm, it yields ln x = at ; there1 fore, t = ln x . a By substituting in the equation for y, it is obtained the equation of the trajectory as

y( x )

g ln 2 x 2 2a

The interesting thing about this fictious object/particle is that the motion is under sea level, so it may be a sea animal or a type of a submarine.

31

Newton’s Laws

PROBLEM 1.15 A ball is placed at the top of a frictionless ramp with a height h and incline θ . The ball is also attached to a point 2h above the ground (h above the top of the ramp, as in Figure 1.19) with a string of length 2h and negligible mass. Find the initial velocity, down (parallel to) the ramp, the ball must have so that the highest position it reaches is halfway between the fixed point and the top of the ramp. SOLUTION 1.15 The ball will slide down the ramp until it is a distance of 2h away from the fixed point (Figure 1.20). At this point it will travel upward, fixed in a circular arc due to the string. Consider the motion down the ramp and the pendulum motion separately. fixed point

h

h

FIGURE 1.19  The ball positioned at the top of a ramp, connected to a fixed point.

h

2h y

h Fg

x

(a) 2h h y h Fg

x

(b)

FIGURE 1.20  The ball moving as far down the ramp as possible before it must leave the surface of the ramp (a) and the ball in the final position in the air (b).

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Classical Mechanics FN

Fg sin   Fg sin

Fg

FIGURE 1.21  Forces acting on the ball.

First, it slides down the ramp with initial velocity v0 x . The only forces on the ball are gravity and the normal force (Figure 1.21). During this part of the motion, there is no movement in the yˆ direction, and the initial velocity is only in the xˆ direction. Therefore, one can ignore the yˆ part of the gravitational force and the normal force. Thus,

Fx  ma x

mg sin   ma x

and the x component of acceleration is a x  g sin

Considering v = a ,

v x  g sin

vx  v0 x  gt sin

where v0 x is what is ultimately solved for. In order to move to the next part of the calculation (the swinging portion of the movement), the ball’s velocity the instant it leaves the ramp must be known. This requires the time the ball slides down the ramp. Considering vx = x , then x  v0 x  gt sin

By integration,

x  x0  v0 x t

1 2 gt sin  2

33

Newton’s Laws

Taking the initial position to be zero, the position is given by x  v0 x t

1 2 gt sin  2

and solving for t yields

2x 2v0 x  t  t2 g sin  g sin

2x v0 x  v2  2 0x 2 t    g sin  g sin   g sin  

2 xg sin   v02x v0 x  t    g 2 sin 2   g sin  

2

2

2 xg sin   v02x  v0 x g sin

t

From the triangle in Figure 1.22:

  (2h)2  h2  x 2  2hx cos     2 

4h2  h2  x 2  2hx sin

3h2  x 2  2hx sin

( x  h sin  )2  3h2  h2 sin 2

h

2h π + 2 x

FIGURE 1.22  Geometry of the ball with respect to the ramp angle.

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Classical Mechanics

h

0

y

2h

x Vx

h

FIGURE 1.23  The velocity of the ball as it begins it swing motion.

xh

sin 2   3  sin 



The velocity of the ball right as it loses contact with the ramp is

vx  v0 x  gt sin   2 xg sin   v02x  2 gh sin 



sin 2   3  sin   v02x

For the swing part of the motion, take coordinates shown in Figure 1.23 with the fixed point as the origin. The initial velocity is given by

vi  vx (cos  x  sin  y ) The force of gravity is now given by  Fg  mgyˆ

plus the tension in the string. Note, in switching to polar coordinates, everything in the rˆ direction can be ignored since the motion is only in φˆ . Therefore,

 Ftension = ignorable

 Fg  Fg,  mg sin

 vi  vi,  vx (cos  cos 0  sin  sin 0 )  vx cos(  0 ) Newton’s Second Law can now be used to find vφ

Fg,  ma  mg sin  a   g sin

35

Newton’s Laws

Consider the following v

dv dv d  dv   dt d dt d

It is also known that v  2h . Thus, v dv   g sin  2h d

v

Using separation of variables,

v dv  2hg sin  d

1 2 v  2hg cos   C 2

v2  4hg cos   C

where C is a constant. Since the initial velocity is known, the initial angle is given as follows (Figure 1.24): x 2  h2  (2h)2  2(2h) cos 0

h2



sin 2   3  sin



2

 5h2  4h2 cos 0

cos 0

5



sin 2   3  sin



2

4

0 2h

h

x

FIGURE 1.24  Geometry of the ball with respect to the fixed-point angle.

36

Classical Mechanics

2h

h 2

FIGURE 1.25  Geometry of the ball at the top of the swing motion.

Solving for C yields

vx2 cos2 (  0 )  4hg cos 0  C

C  vx2 cos2 (  0 )  4hg cos 0 The velocity is now given by v2  4hg(cos   cos 0 )  vx2 cos2 (  0 )

When the ball reaches a maximum height halfway between the fixed point and the top of the ramp, v  0 and φ is found as follows (Figure 1.25):

cos1

h 1  cos1 4h 4

Now, solving for vx 0  4hg(cos   cos 0 )  vx2 cos2 (  0 )

vx

2 hg(cos 0  cos  ) cos(  0 )

Also,

cos 0  cos

1 5 4



sin 2   3  sin

  1  14 cos(2 )  2 2

Therefore,

vx

hg cos(2 )  2 sin 2   3  5  cos    cos1  

sin 2   3  sin  4

 2

 

sin 2   3



37

Newton’s Laws

This can now be combined with the expression found for vx before vx  2 gh sin 



sin 2   3  sin   v02x

Finally,

v0 x

hg cos(2 )  2 sin 2   3  5  2 cos    cos1  

sin   3  sin  2

4

2

    

 2 gh sin 

sin 2   3  sin 

PROBLEM 1.16  A plane is flying in the xˆ direction at a velocity vi = v0 xˆ . It is delivering a package which is mysteriously attracted to the plane by the force F   rrˆ, where β is the strength of the attraction, r is the distance from the plane, in rˆ points from the package to the plane. As soon as the package is dropped, the plane immediately begins traveling at v f , 45° above xˆ . Find an expression for the height h the package should be dropped from such that its yˆ velocity is zero when it hits the ground. SOLUTION 1.16  The forces acting on the package are gravity, Fg  mgyˆ , and the attractive force,  two  F   r , which can be written as

   r  rplane  rpackage

Take the origin of the coordinate system to be the spot where the package was  v released. Given the planes new velocity is vplane  f ( x  y ) , the position of the 2 plane is

vt rplane  f ( x  y ) 2 The position of the package can be expressed as

rpackage  x p (t ) x  y p (t ) y

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Classical Mechanics

Using Newton’s Second Law yields   v t   v t    F   r  mgyˆ     f  x p  xˆ   f  y p  yˆ   mgyˆ   2    2  v t  v t      f  x p  xˆ     f  y p   mg  yˆ  m(a x xˆ  a y yˆ )  2     2 

Since the yˆ velocity is the only one in question, the xˆ velocity will be ignored. Therefore, myp

 yp

vf t   y p  mg 2

vf  yp  t  mg m 2

Solving the homogeneous differential equation  yp

yp  0 m

yields       y p,h  A cos  t   B sin  t  m   m     

A particular solution can be obtained considering

y p, p  Ct  D

y = 0. Therefore, where 

vf (Ct  D)  tg m m 2

with

C=

vf 2

39

Newton’s Laws

and D

mg

Combining these solutions yields       vf mg y p  A cos  t  B sin  t  t  m   m   2    

It is known y p (0) = 0 , so 0  A

mg

which yields A : A

mg

and y p (0) = 0 , so 0B

v   f m 2

and B is calculated to be B  v f

m 2

Finally, y p (t )

   vf     mg  m t   1  sin  t  cos  t    m         2  m    

Since it is required the velocity to be zero at the bottom of the motion, consider y p y p   g

  vf    v m t t sin   f cos     m   2 2  m   

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Classical Mechanics

    t as s and cos  t as c, this can be solved for c (and Expressing sin   m   m      eventually t )

v m s  f (1  c)  2

mg 2 2 v 2f s  (1  c)2  2

g

After using s 2  1  c 2 v2 mg 2 (1  c 2 )  f (1  2c  c 2 )  2

v 2f g 2 m  2 2  v 2f g 2 m    c  vf c    0      2  2 Therefore,  v 2 g 2 m   v 2f g 2 m  v 2f  v 4f  4  f       2       2 t  c  cos   m   v2 g2m    2 f      2

v4 g 4m2  v 2f  v 4f  4  f  2     4  2 2 v g m 2 f   2  

 t is bounded between negative one and one, the minus sign is Since cos   m    taken. Therefore,

t

v4 g 4m2  v 2f  v 4f  4  f  2    m  4  2 n cos1 2 2  vf g m   2      2

41

Newton’s Laws

for n  . Since h must be chosen such that y p (t )  h where t is the time above, the package should be dropped from a height  4 2    4  v 2f  v 4f  4  v f  g m2     mg   4 h 1   2 2  vf g m     2      2      4  v g 4m2   v 2f  v 4f  4  f  2   v  m  4 sin  cos1  f  2 2 v 2  g m  2 f     2    

v4 g 4m2 v 2f  v 4f  4  f  2  m  4  cos1 2 2 v  g m 2 f      2

so that y velocity is zero when it lands.

      

      

2

Motion with Air Resistance

2.1  THEORY This chapter introduces the motion of objects and projectiles with air resistance. Air resistance can be linear or quadratic, and in some circumstances, we need to consider both contributions.

2.1.1 Drag Force of Air Resistance

 The drag force f has a magnitude dependent on speed, with its direction opposite to  the velocity v , as in Figure 2.1.  f   f (v) v

For lower speeds (that is, much lower than the speed of light), the drag force has a linear term and a quadratic term, as follows

f (v)  flinear  fquadratic  bv  cv 2

with b and c constants with appropriate units. The constants depend on the shape of the object, for example, for a spherical object, b   D and c   D 2 , with D the diameter of the sphere and β and γ coefficients which depend on the medium. The linear drag is due to the viscous drag of the medium (air, water, oil, etc.) and it depends on the viscosity of the medium and the linear size of the projectile. The quadratic term is proportional to the density of the medium and the cross-­sectional area of the projectile. For some systems, the linear air drag can be neglected, for others, the quadratic air drag can be neglected, but in some circumstances, both linear and quadratic drag should be considered. v f = -f(v)v

FIGURE 2.1  Cart moving to the right with velocity v under the influence of drag force f . DOI: 10.1201/9781003365709-2

43

44

Classical Mechanics

2.2  PROBLEMS AND SOLUTIONS PROBLEM 2.1 A bicycle is cruising in a horizontal motion with initial velocity v0 under the influence of linear air drag. Analyze the motion, that is, find a) the velocity as a function of time and b) the position as a function of time. Neglect the friction forces. SOLUTION 2.1 a. The net force is only the linear air drag, acting horizontally in a direction opposite to the velocity of the bicycle. The vertical forces – the gravitational force and the normal force of the surface – yield a zero vertical force, as in Figure 2.2.

      Fnet  mg  N  flinear  flinear  bv    From Newton’s Second Law, F= ma = mv , and considering only the x net direction, mv  bv

m

dv  bv dt

Separation of variables, dv b   dt v m

The velocity is obtained by integrating on both sides the previous equation over velocity, from v0 to v, and over time, from 0 to t . v

v0

t

dv b  dt v m

0

N

f

mg

FIGURE 2.2  Forces acting on the bicycle.

45

Motion with Air Resistance

And from here it yields v

ln v v0

bt m

Using the properties of the logarithm, ln A  ln B  ln ln

A , yields B

v bt  v0 m

And from here, the speed as a function of time is calculated: v(t )  v0e

bt m

Note that the velocity is exponentially decreasing to zero, reaching zero asymptotically as time approaches infinity. b. The equation of motion can be obtained by integrating the speed obtained in part (a). v=

dx dt

bt m

bt m

v(t )  v0e

dx  v0e x

t

dx  v0 e

dt bt  m

dt

0

x0

bt

m  x  x0   v0e m b

t

0

And the position as a function of time is x (t )  x0

bt   m  v0  1  e m   b  

As time approaches infinity, the object approaches a position equal to

x  x0

m v0 b

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Classical Mechanics

PROBLEM 2.2 A projectile is falling into the atmosphere (gravitational acceleration g ) with initial speed v0 oriented vertically downward and it is subject to linear drag. a) Find the terminal velocity, vter . b) Obtain the speed at time t , v(t ). SOLUTION 2.2 a. The two forces acting on the projectile are the gravitational force, downward, and the drag force upward (the direction opposite to the direction of the velocity, as in Figure 2.3). Newton’s Second Law is written as mv  mg  bv

The terminal speed is the speed for which the drag force is balancing out the gravitational force, and therefore the acceleration becomes zero: 0  mg  bvter

And the terminal speed for linear air drag is vter =

mg b

b. mv  mg  bv

By using vter =

mg , the previous equation becomes b mv  b

mg  bv b

mv  bvter  bv

f = -bv

v0

mg

FIGURE 2.3  Projectile in vertical motion under the influence of gravitation and air drag.

47

Motion with Air Resistance

The speed is obtained from

dv b   (v  vter ) dt m

dv b   dt (v  vter ) m By integrating over speed from initial speed v0 to v and over time from 0 to t v

v0

t

dv b  dt  (v  vter ) m

0

ln(v  vter )  ln(v0  vter )

ln

b t m

(v  vter ) b  t (v0  vter ) m

By using properties of logarithms and exponents, the speed as a function of time is obtained as v(t )  vter  (v0  vter ) e

Introducing

bt m

m , the speed becomes b v(t )  vter  (v0  vter ) e

t

Or rearranging the terms

t   v(t )  vter  1  e  

t     v0 e  

It is easy to see that, when the time goes to infinity, the speed goes asymptotically to terminal speed. If the projectile starts with zero initial velocity, then the speed becomes

t   v(t )  vter  1  e  

  

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Classical Mechanics

PROBLEM 2.3 A ball of mass m is thrown with initial velocity v0 at an angle of 45° from the horizontal. a. Assuming no air resistance, find the trajectory of the ball y(x). b. Assuming linear resistance bv, find the trajectory of the ball y(x). SOLUTION 2.3

 0 and Fy  my  mg. a. According to Newton’s Second Law, F = mx = x First, an expression for x is computed by integrating twice.

x  0 dt

x = C1

Since x (0) = v0 x , C1 = v0 x .

x  v0 x dt

x  v0 x t  C2

x . v0 x Now, an expression for y is computed by integrating twice.

Since x(0) = 0 , C2 = 0 . So x(t ) = v0 x t . Thus, t =

y   g dt

y   gt  C3

Since y (0) = v0 y , C3 = v0 y .

y   gt  v0 y dt 1 y   gt 2  v0 yt  C4 2 1 Since y(0) = 0 , C4 = 0 . So y(t )   gt 2  v0 yt . 2 By plugging t as a function of x in y(t ) , a result for y( x ) is

y( x )

gx 2 v0 y  x 2v02x v0 x

49

Motion with Air Resistance

b. The   process is similar  to part (a). According to Newton’s Second Law, Fx  mx  bvx and Fy  my  mg  bvy . First, an expression for x is computed by integrating twice. vx

v0 x

dvx  vx

v ln  x  v0 x

t

b

m dt 0

b mt 

vx  v0 x e

b t m

After getting an expression for vx , integration is performed a second time to solve for x . t

x  v0 x e

b t m

dt

0

t v0 x m   1  e m  b   b

x Thus,

b   t bx m 1  e    v 0 xm  

and

t

m  bx  ln  1  b  v0 x m 

Now, an expression for y is computed by integrating twice. vy

v0 y

t

dv y  dt  b  g  v y 0 m

m m ln(bvy  mg)  ln(bv0 y  mg)  t b b

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Classical Mechanics

vy  v0 ye

b t m

t mg   1  e m  b   b

After getting an expression for vy , integration is performed a second time to solve for y . t

b   b t  t   gm  y   e m v0 y   1  e m   dt    b   0



b   t   mv gm 2  gm y  1  e m   0 y  2   t   b b  b  

By plugging t as a function of x in y(t ) , a result for y( x ) is y( x )

xb  mv0 y gm 2  2  v0 x m  b b

y( x )

x v0 x

 gm  m  xb     ln  1    b b v 0 xm    

gm  gm 2  bx    v0 y  b   b2 ln  1  v m    0x  

PROBLEM 2.4 A bicycle is cruising in a horizontal motion with initial velocity v0 under the influence of the quadratic air drag. Analyze the motion, that is, find a) the velocity as a function of time and b) the position as a function of time. Neglect the friction forces. SOLUTION 2.4 a. In this case, the net force is the quadratic air drag. As expected, the gravitational force and the normal force of the surface yield a zero vertical force.

Fnet  mg  N  fquadratic  fquadratic  cv 2 v mv  cv 2

m

dv  cv 2 dt dv c   dt v2 m

51

Motion with Air Resistance

The velocity is obtained by integrating on both sides the previous equation over velocity, from v0 to v, and over time, from 0 to t . v

v0

t

dv c  dt  v2 m

0

And from here it yields v

ct 1  v v0 m

1 1 ct   v v0 m

And from here we obtain the speed as a function of time: v(t )  where q =

1 v  0 1 ct 1  qt  v0 m

cv0 . Note that the velocity is decreasing to zero as time goes m

to infinity. b. The equation of motion can be obtained by integrating the speed obtained in part (a). v=

v(t )

dx

dx dt v0 1  qt v0 dt 1  qt

x

t

dx  v0

x0

x  x0

1

1  qt dt 0

t v0 ln (1  qt ) 0 q

And the position as a function of time is

x (t )  x0

v0 v (ln(1  qt )  ln 1)  x0  0 ln(1  qt ) q q

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Classical Mechanics

PROBLEM 2.5 A bobsled of mass m is launched on a straight horizontal plane with initial speed v0 . It is under quadratic air resistance cv 2 . What is the equation of position of the bobsled? Assume no friction. SOLUTION 2.5    mv . Air resistance being the only force act= = Using Newton’s Second Law, F mx x ing on the bobsled, it follows that mv  cv 2 . By integrating, an expression for v is obtained. v

v0

dv  v2

t

c

m dt 0

1 1 c    t v v0 m v

1 1 c  t v0 m

Since v = dx /dt , the position x(t ) of the bobsled is t

x (t )

0

x (t )

1

1  c t dt v0

m

m (ln(cv0t  m)  ln(m)) c

PROBLEM 2.6 A projectile is falling in the atmosphere (gravitational acceleration g) with initial speed v0 oriented vertically downward and is subject to quadratic drag. (a) Find the speed v(t ). (b) Obtain the equation of motion, that is, obtain y(t ). SOLUTION 2.6 a. In this case, the motion is constrained in the y direction, with the gravitational force oriented vertically downward and the quadratic drag upward. Newton’s Second Law is written as

mv  mg  cv 2 As before, the projectile will increase in speed, but the drag force will increase as well (quadratic) until gravitational force will balance out the

53

Motion with Air Resistance

drag force, and the acceleration becomes zero. The maximum speed is the terminal speed vter , which, as expected, will have a different form than in the linear drag case. 2 0  mg  cvter

vter =

mg c

Now, rewriting Newton’s Second Law in terms of the speed and of the terminal speed, it follows m

dv 2  c(vter  v2 ) dt

Generally, for equations of this type, it is a good option to express them in dimensionless quantities as follows: m

dv v2   cvter 2  1  2  dt  vter 

dv c  vter 2 dt  v2  m 1  2   vter 

By substituting terminal velocity on the right side dv  g dt v2 1 2 vter

By integration, v

t

dv  g dt  v2 v0 1  0 vter 2

It yields that

v v  vter  tanh 1  tanh 1 0   gt v v ter ter 

54

Classical Mechanics

If the initial velocity is zero, v0 = 0 . vter tanh 1

v  gt vter

From here, tanh 1

v gt  vter vter

And the speed is obtained as v(t ) = vter tanh

b. By integrating one more time, the equation of motion is obtained, y

t

dy  tanh

gt vter

y0

0

y(t )  y0

gt  dt  vter

(vter )2  gt  ln cosh  g v ter 

If the projectile starts with zero velocity and at zero vertical coordinate y0 = 0 , then

y( t )

(vter )2  gt  ln cosh g vter 

PROBLEM 2.7 Imagine a pendulum consisting of a spherical mass m which is placed in front of a large fan as in Figure 2.4. At t = 0 , the fan begins blowing air at a velocity v0 . Find the equation of motion for the mass considering only quadratic air resistance. SOLUTION 2.7 The mass experiences the following forces:

Ffan  cvo2 x

55

Motion with Air Resistance

v0 l

m

FIGURE 2.4  Pendulum in the presence of a fan.

 [ Ffan ]  cv02 cos

Fg  mgy

 [ Fg ]  mg sin

Fair  cl 22sign()

Considering Newton’s Second Law  F  ml

Using the forces yields

cv02 cos   mg sin   c2l 2sign()  ml

Therefore

2

g cv cl   sign()2   sin   0 cos  m l ml

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Classical Mechanics

PROBLEM 2.8 Consider a pendulum of length l displaced by an angle φ as shown in Figure 2.5. Find the equations of motion if the mass m experiences a. Linear drag. b. Quadratic drag. SOLUTION 2.8 a. In the case of linear drag

F  m  ml

bl  mg sin   ml Therefore b g      sin  m l

b. In the case of quadratic drag

F  m  ml

cl 22 sign()  mg sin   ml Therefore

cl g   2 sign()   sin  m l

m

FIGURE 2.5  Simple pendulum.

57

Motion with Air Resistance

PROBLEM 2.9 A rock of mass m is thrown downward from a height h at an initial velocity v0 . Find the expression for velocity: a. with no air friction b. with quadratic air friction cv 2 on the rock SOLUTION 2.9    mv . With no air friction, the F mx = a. According to Newton’s Second Law, = only force acting on the object is weight. So = F mg = mv . Now solving for v is possible by integrating. v

t

v0

0

dv  gdt

v  v0  gt

v  gt  v0

b. With air friction, there are two forces acting on the rock: F  mg  cv 2  mv. m mg 2 By integrating the equation v   v , an expression for the velocity is c c obtained. v

c

v  mg  m dt

v0

t

dv

c mg



2

c

0

c  c  c  tanh 1  v0  tanh 1  v    t    mg   m  mg 

PROBLEM 2.10 A block of mass m is launched with initial velocity v0 down an incline at an angle  θ with the horizontal. The block starts from rest and is subject to linear drag f = bv. The friction coefficient between the incline and the block is µ . Find an expression for the velocity of the block. SOLUTION 2.10      mv x . The forces acting on  0 and F = = From Newton’s Second Law, my = = y x   F   mx the block (Figure 2.6) are Fy  N  mg cos  and Fx  bvx  mg sin    N . First, a value for N is found.

58

Classical Mechanics

N Ff

f mg sin y

mg cos

mg

x

FIGURE 2.6  Block sliding down an incline, with drag and friction forces.

N  mg cos   0

N  mg cos

An expression for the velocity of the block is obtained by integrating b v x   vx  g sin    g cos  . m vx

dvx

t

b v  g sin   g cos

v0

m

x

 dt  0

 m b b  ln  vx  g sin    g cos   ln  v0  g sin    g cos    t b m m  b vx  g sin    g cos  b m ln  t b m  v0  g sin    g cos  m

b vx  g sin    g cos  b  t m e m b  v0  g sin    g cos  m

b

If

t b b vx  g sin    g cos    v0  g sin    g cos  e m m m

b vx  g sin    g cos   0 m

59

Motion with Air Resistance b

t b b vx   g sin    g cos    v0  g sin    g cos  e m m m  t m b  g sin    g cos    v0  g sin    g cos  e m  b m  b

v x (t )  If

b vx  g sin    g cos   0 m b

t b b vx  g sin    g cos    v0  g sin    g cos  e m m m  t m b  g sin    g cos    v0  g sin    g cos  e m  b m  b

v x (t )

PROBLEM 2.11 A cart is moving on a horizontal surface with initial velocity v0 subject to a drag 3

force of the type f (v)  cv 2 . Find the speed of the cart as a function of time. SOLUTION 2.11 The net force is the drag force.

3      Fnet  mg  N  fdrag  fdrag  cv 2 v 3

mv  cv 2

3

m

dv  cv 2 dt dv

v

3 2

c dt m

The velocity is obtained as before, by integrating on both sides the previous equation over velocity, from v0 to v, and over time, from 0 to t . v

dv

(v)

v0

3 2

t

c dt  m

0

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Classical Mechanics

2 v

v

v0

c t m

1 1 ct   v v0 2m

v( t )  v( t )

1  1 ct      v0 2m 

2

v0  ct v0  1  2m

  

2

As expected, when time approaches infinity, the speed goes to zero. PROBLEM 2.12 An object of mass m is coasting horizontally with initial horizontal speed v0 under the influence of both linear and quadratic air drag. Obtain the speed as a function of time for this system. SOLUTION 2.12 The object is cruising under the influence of both linear and quadratic air drag, which means that both drag forces have the same order or magnitude and neither of them can be neglected. Here, Newton’s Second Law is mv  bv  cv 2

m

m

m

dv  bv  cv 2 dt

dv  dt bv  cv 2

dv  dt b  cv   v  c 

Using the integral (with a ≠ b), 1

1

( x  a)( x  b) dx  b  a [ln(a  x)  ln(b  x)]

61

Motion with Air Resistance

m c

v

t

dv

v  v  b   dt

v0

 

c 

0

v

m 1  b  ln v  ln   v    t  b c   c   v0 c  

m v ln  t  b b v   c    v0

    v0 m v   t  ln ln b   b  b   v v  c 0     c    

v

By using the properties of the logarithms, ln A  ln B  ln

v b    c  v    b t ln v0 m b   c  v0   

b 1 b cv  t ln b 1 m cv0

b   t b 1    1  e m  cv  cv0 

b 

With some algebra rearrangement,

b   b   t  b v(t )   1    1  e m   c   cv0  

1

A , it follows that B

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Classical Mechanics

v( t )

1

b c

b 

b   t 1    1  e m   cv0 

With a little algebra rearrangement, the speed becomes v( t )

b c

b v( t )  c b v( t )  c

1   b 1    1  cv0 

e e

b   t  m 

e

1 e

b   t  m 

b   t  m 

  b   1 cv  0 

b   t  m  b 

1

 t  b e m  cv0

By checking the result for the validity at initial time zero, we obtain the initial velocity (= t 0= ) v0 , as expected. When the time approaches infinity, the velocity is closer to zero, therefore the quadratic air drag becomes negligible compared to the linear drag, which will be the main contribution for velocity close to zero. The speed is exponentially decreasing, similarly to problems in which only the linear air drag is present. PROBLEM 2.13 Suppose there exists a nonstandard sphere which experiences quadratic drag in one orientation and linear drag in an orthogonal one. Given mass m and initial velocity v0 , it is launched at an angle θ . Determine its velocity as a function of time and angle for the orientation of linear drag in the x direction, quadratic drag in the y. Take flin (v) = bv and fquad = cv 2 for constants b , c with the appropriate units. SOLUTION 2.13 Consider the x and y directions separately. In the x direction

Fx  mv x bvx  m

dvx dt

63

Motion with Air Resistance

dvx b   vx dt m

so for a constant A

v x (t )  A e

b t m

Since vx (0)  v0 cos( ) , A  v0 cos( )

Therefore

vx (t )  v0 cos( )e

b t m

Now consider the y direction. This must be split into two parts, the ascent motion and the descent motion. During the ascent, gravity and the drag will be in the same direction, where during the descent, they are in opposite direction. Considering the ascent first  Fy  mv y

mg  cv 2  m

dvy c  gm c       g  v2      v2  dt m  m c   dvy c   dt gm m  v2 c

dvy dt

c  c c vy    t  A tan 1    gm m  gm 

where A is a constant. Considering vy (0)  v0 sin( ) A

c  c tan 1  v0 sin( )   gm  gm  

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Classical Mechanics

Now solving for vy

c   c  c c c vy    t  v0 sin( )  tan 1  tan 1      gm m gm  gm   gm 

c   c  gc v  t  tan 1  v0 sin( )  tan 1   gm y    m    gm    c  c gc vy  tan   t  tan 1  v0 sin( )      gm m  gm  

vy

  gc   c v0 sin    tan  t   m   gm  gm    gc   c  c  1 v0 sin   tan  t   m    gm   

On the descent, the drag force is in the opposite direction. Utilizing the setup of the ascent, dvy c  c  gm 2      g  v2     v  dt m  m c  

dvy c   dt gm 2 m v c

gm  vy c c  tA m gm  vy c

gm  vy gc c  2 tA m gm  vy c

gm  vy 2 c  Ae gm  vy c

1 2

c ln gm

ln

gc t m

65

Motion with Air Resistance

Since the ball is descending, the initial velocity is zero, so at time t = 0

gm c =A gm c

A =1 Also vy ≤ 0 so gm  vy 2 c e gm  vy c

Since terminal velocity is

gm , ｜vy ｜≤ c

gm , c

gm  vy 2 c e gm  vy c

gm  vy  c  2 vy  1  e

gc t m

   

gc t m

gc t m

gm 2 e c

gc t m

gm  2 e c

vye

2

gc t m

gc t m

  1  

gc t m

  gm gc t tanh  c m  

Therefore

v y (t )

2 e gm  c  2 e 

  1   gc  t m  1   gc t m

gm  e  c   e

gc t m gc t m

e e

gc t m

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Classical Mechanics

PROBLEM 2.14 Consider a sailboat with a sail which experiences quadratic drag on one side and linear drag on the other. On its maiden voyage, there is a curious wind blowing at a velocity v(t )  v0 sin( t ) . Determine the velocity of the boat as a function of time if the wind hits the side with the quadratic drag. Assume no friction due to the water. SOLUTION 2.14 The wind is pushing with a force  Fapplied  cv02 sin 2 ( t ) x

And the air resistance is given by  Fair  bv x

Therefore,

Fx  mv

cv02 sin 2 ( t )  bv  mv Since the mass is constant, introduce c  v

c . m

b v  cv02 sin 2 ( t ) m

This can be solved using the method of integrating factors. Consider the integrating factor

e

b

m dt

b

em

t

Now,

d [  v]   cv02 sin 2 ( t ) dt v

cv02

sin (t ) dt

v  cv02e

2

b t m

b

t

e m sin 2 ( t )dt

67

Motion with Air Resistance

This can be solved using integration by parts. It will be helpful to express sin 2 (ω t ) differently using the following equations: cos2 x  sin 2 x  cos(2 x )

cos2 x  1  sin 2 x

so

sin 2 ( t )

1 (1  cos(2 t )) 2

Therefore, the integral to solve is b

b

t 1 2 mt mt cv0 e e  e m cos(2 t ) dt 2

v

1 2 mt m mt 1 2 mt mt cv0 e e  cv0 e e cos(2 t ) dt 2 b 2

v

cv02 m 1 2  m t m t  cv0 e e cos(2 t ) dt 2b 2

b

b

v

b

b

b

Now to solve for the integral tion by parts

b

b

b

t

e m cos(2 t ) dt consider the following for integra-

u  cos(2 t ) , du  2 sin(2 t )dt , v =

b

b

t m mt e , dv = e m dt b

so

b

t

e m cos(2 t ) dt

b

b

m mt 2 m m t e sin(2 t )dt e cos(2 t )  b b

Again, integration by parts will be used to solve

u  sin(2 t ) , du  2 cos(2 t )dt , v =

b

t

e m sin(2 t ) dt . b

b

t m mt e , dv = e m dt b

so

b

t

e m sin(2 t )dt

b

b

m mt 2 m m t e cos(2 t )dt e sin(2 t )  b b

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Classical Mechanics

Combining the two integrations by parts yields b

e m cos(2 t ) dt

e m cos(2 t ) dt

b

t

t

b b b  m mt 2 m  m m t 2 m m t e cos(2 t )  e cos(2 t )dt   e sin(2 t )   b b b b 

b

b

m mt 2 m 2 m t 4 2 m 2 e cos(2 t )  e sin(2 t )  2 b b b2

b

t

e m cos(2 t )dt

b b  4 2 m 2  mb t m mt 2 m 2 m t e  t dt e  t e cos( 2 ) cos( 2 ) sinn(2 t )   1   b2  b b2



b t  b2 e m cos(2 t ) dt   2 2 2  b  4 m

b

b    m mb t 2 m 2 m t e  t e cos( 2 )  sin(2 t )    2  b  b  b

t mb 2 m   mt   e m cos(2 t )dt   2 e  cos(2 t )  sin(2 t )  2 2  b  b  4 m   

An expression for v is now given by b

b

v

2 m cv02 m 1 2  m t  mb  mt    cv0 e  2 e  cos(2 t )  sinn(2 t )  2 2  2b 2 b  b  4 m   

v

2 m cv02 m  1 b   cos(2 t )  sin(2 t )     2 2 2  2  b b  4 m  b 

3

Momentum and Angular Momentum

3.1  THEORY This chapter introduces linear momentum, moment of inertia, center of mass, angular momentum, and the principles of conservation of linear momentum and of angular momentum.

3.1.1 Linear Momentum

dP  The variation of the total linear momentum = P is equal to the external force Fext dt acting on the system:   P = Fext

with P = p1 + p2 +…+ pN =

N

∑p

i

i =0

Conservation of the linear momentum: if the net external force applied on the system is zero, the total linear momentum of the system is conserved.

3.1.2 Rockets A rocket of initial mass m0 and initial velocity v0 (on positive x direction), with a fuel with exhaust velocity vex (on negative x direction) with respect to the rocket and without external forces, acquires a velocity v at the moment the mass becomes m v = v0 + vex ln

m0 m

The thrust of the rocket is defined as the product of the rate at which the mass is  ex , where it is important ejected, −m with the speed of the exhaust vex , thrust = −mv to note that the mass becomes smaller with the ejection of the fuel, and m < 0 .

3.1.3 Center of Mass

For a system of N discrete particles of mass mi and position vector ri , the center of the mass is

1 R= M

m r + m2r2 +…+ mN rN miri = 1 1 M i =0 N

DOI: 10.1201/9781003365709-3

69

70

Classical Mechanics

where M is the total mass of the system, N

M=

∑m

i

i =0

For a continuous system of mass M and density ρ ,

1  1 R= r dm = M M

∫ ρ r dV

3.1.4 Moment of Inertia The moment of inertia for a discrete system is obtained as N

I=

∑m s

2 α α

α =1

with sα the distances of the mass mα from the axis of rotation. For a continuous system, the sum is transformed into an integral, with ρ the density of the object, s the distance from the axis of rotation, and dV the element of volume. I=

∫ ρ s dV 2

Angular momentum is given by

l =r×p  dl  =l The rate of the angular momentum dt

l = r ×F = Γ

where Γ is the net applied torque about the origin.

3.1.5 Principle of Conservation of Angular Momentum

If the net external torque Γ acting  on a particle is zero, then the total angular momentum of the particle is constant, l = constant .

71

Momentum and Angular Momentum

Angular momentum for a system of N particles L is  L=

N

N

li =

∑ ∑r × p i =0

i

i

i =0

with  L=

N

li =

N

ext

∑ ∑r × F = Γ i =0

i

i

i =0

3.1.6 Principle of Conservation of the Angular Momentum for a System of N Particles

If the net external torque Γ ext acting on a system  of N particle is zero, then the total angular momentum of the particle is constant, L = constant .

3.2  PROBLEMS AND SOLUTIONS PROBLEM 3.1 A beginner bodybuilder accidentally put a mass of 20 kg at one end of his bar and a mass of 30 kg at the other end. The bar is 1.5 m long. Find the center of mass of the system. SOLUTION 3.1 This problem consists of finding the center of mass of a two-­ particle system (Figure 3.1). X=

X=

1 M

2

∑m x

α α

α =1

1 (20 kg ⋅ 0 m + 30 kg ⋅ 1.5 m ) 20 kg + 30 kg

X = 0.9 m

72

Classical Mechanics y

x 20 kg

30 kg

FIGURE 3.1  Two masses of 20 kg and 30 kg, respectively, at the ends of a bar.

PROBLEM 3.2 Two blocks of mass m1 and m2 , with initial velocity v1 and v2 , respectively, meet in a perfectly inelastic collision. Then, a third block of mass m3 and velocity v3 hits the two blocks in a perfectly inelastic collision. Using conservation of momentum, find the velocity of the three blocks, assuming no external forces (one dimensional case). SOLUTION 3.2 The collision of m1 and m2 is first considered. Because of the conservation of   momentum, pinitial = pfinal . Thus,

m1v1 + m2 v2 = (m1 + m2 )v(12 ) The velocity of the two blocks is

v(12 ) =

m1v1 + m2 v2 m1 + m2

Now, the collision with the third block is studied. Again, conservation of momen  tum gives pinitial = pfinal

(m1 + m2 )v(12 ) + m3v3 = (m1 + m2 + m3 )v The velocity of the three blocks is

v=

m1v1 + m2 v2 + m3v3 m1 + m2 + m3

73

Momentum and Angular Momentum

PROBLEM 3.3 Consider two particles of mass m1 and m2 . Find the amount of kinetic energy loss   during an inelastic collision if the particles are initially traveling at v1 and v2 . Considering the case when m = m= m , can this energy loss be ignored in any 1 2 situation? SOLUTION 3.3 The initial kinetic energy is given by KEi =

1 (m1v12 + m2 v22 ) 2

and the final kinetic energy is given by KE f =

1 (m1 + m2 )v 2f 2

where v f is the final energy of the system. An expression for v f can be found by considering conservation of momentum:   pi = p f

m1v1 + m2 v2 = (m1 + m2 )v f

with

m v + m2 v2 vf = 1 1 (m1 + m2 )

Therefore,  v 2f =

1 ((m1v1 + m2 v2 ) ⋅ (m1v1 + m2 v2 )) (m1 + m2 )2   1 = (m12 v12 + m22 v22 + 2m1m2 v1 ⋅ v2 ) (m1 + m2 )2

The final kinetic energy is now given by

KE f =

1 1 (m12 v12 + m22 v22 + 2m1m2 v1 ⋅ v2 ) 2 (m1 + m2 )

74

Classical Mechanics

and the energy lost is therefore   1 m12 v12 + m22 v22 + 2m1m2 v1 ⋅ v2  2 2   m1v1 + m2 v2 − 2 m1 + m2    2 2 1  m1m2 v1 + v2 − 2v1 ⋅ v2  =  2 m1 + m2   

KEi − KE f =

(

)

Considering the case where m = m= m , the energy lost becomes 1 2

δ KE = KEi − KE f =

1   1 m(v12 + v22 − 2v1 ⋅ v2 ) = m(v1 − v2 )2 2 2

Considering v1 = v2 + δ v ,

δ KE =

1 mδ v2 2

which is negligible provided ｜δ v2 ｜ 1. This amounts to the boring scenario of two masses traveling in approximately the same direction at approximately the same speed and simply sitting together. This shows that treating inelastic collisions via energy consideration may not be the best idea as the energy lost is not negligible.

PROBLEM 3.4 Analyze the motion of a rocket starting with initial mass m0 , which accelerated from rest. Obtain the momentum versus mass, p(m) and find the mass for which you obtain the maximum momentum. What was the mass of the fuel consumed? Find the maximum momentum. SOLUTION 3.4 The speed of the single stage rocket is dependent on the speed of the exhaust with respect to the rocket vex , initial mass of the rocket and fuel m0 , and the final mass of the rocket m v = vex ln

m0 m

And the linear momentum

p(m = ) mv = mvex ln

m0 m

75

Momentum and Angular Momentum

The momentum is maximum when the derivative with respect to the mass is zero. dp(m) m m  m0   m  = vex ln 0 + mvex − = vex  ln 0 − 1  dm m m0  m 2  m  

The derivative of momentum with respect to mass is zero when  m  vex  ln 0 − 1  = 0  m 

which implies

ln

m0 =1 m

therefore, the mass of the rocket is m=

m0 e

where e is the base of the natural logarithm. The mass of the fuel consumed is

mfuel = m0 − m = m0 −

m0  1 = m0  1 −  ≈ 0.63 m0 e  e

Therefore, for a rocket to reach the highest momentum, 63% of the rocket’s initial mass should be consumed as fuel. In general, the rocket’s velocities are improved by a multiple stage system, when the fuel tank is discarded, as discussed in another rocket problem. The momentum corresponding to the maximum point is m  m m pmax = p  0  = 0 v  0  e  e  e

m0 m0 vex  m0  = e vex ln m = e 0  e

PROBLEM 3.5 Find the center of mass of a. A cone of radius R and height h , with its base on the xy plane. b. A hemisphere of radius R , with its base on the xy plane.

76

Classical Mechanics z

y

0

x

FIGURE 3.2  Cone with its base on the xy plane, centered at origin.

SOLUTION 3.5 a. Because of the cone’s symmetry (Figure 3.2), X = 0 and Y = 0. Z=

1 M

∫ ρ z dV

The integration is performed using cylindrical coordinates, from 0 to h−z r= R over r , from 0 to 2≠ over θ , and from 0 to h over z . h

Z=

ρ M

h−z R h 2π h

∫∫ ∫ zr drdθ dz 0 0

2πρ Z= M

0

h

2

R2  h − z  dz 2  h 

∫ 0

h

Z=

πρ R 2 ( zh2 − 2hz 2 + z 3 )dz Mh2

∫ 0

Z=

πρ R 2h2 12 M

Since ρ = M /V ,

Z=

π R 2h 2 12V

h The volume of a cone is V = π R 2 . 3

77

Momentum and Angular Momentum z

r

R

z 0

FIGURE 3.3  Half sphere with its base on the xy plane.

Z=

π R 2h 2 12π R 2h / 3

Z=

h 4

b. Because of the hemisphere’s symmetry (Figure 3.3), X = 0 and Y = 0.

Z=

1 M

∫ ρ z dV

The integration is performed using cylindrical coordinates, from 0 to r = R 2 − z 2 over r , from 0 to 2≠ over θ , and from 0 to R over z .

ρ Z= M

R 2π R 2 − z 2

∫∫ ∫ 0 0

zr drdθ dz

0

h

Z=

2πρ R2 − z2 z dz 2 M

∫ 0

Z=

π R4 4V

The volume of a hemisphere is V =

Z=

Z=

14  π R3  .  23 

π R4 4 1 4  π R3  3 2 3R 8

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Classical Mechanics

PROBLEM 3.6 Find the center of mass of a composite solid made of a cone of height h and radius R and a hemisphere of same radius, joined at their base. SOLUTION 3.6 The cone and the hemisphere (Figure 3.4) can be considered as particles, using their h respective center of mass. The center of mass of the cone is Z c = and the one of 4 3R the hemisphere is Z h = , as shown in Problem 3.5. 8 The center of mass of the composite solid is found by computing 2 1 Z= mα zα . 2 mα α =1

α =1

Z=

mc Z c + mh Z h mc + mh

  2 h  h   2 3   −3  ρπ R 3   4  +  3 ρπ R   8 R       Z= 2 2 h 3 ρπ R + ρπ R 3 3

h2 R 2 − Z = 12 4 h 2R + 3 3

Z=

h2 − 3R 2 4(h + 2 R) z

h 0

y R

x

FIGURE 3.4  Cone and half sphere connected to each other from their base.

79

Momentum and Angular Momentum

PROBLEM 3.7

a. A one-­stage rocket consumes 70% of the initial mass as fuel. Find the final speed of the rocket considering that it starts from rest and that the exhaust has the speed vex with respect to the rocket. b. A two-­stage rocket starting from rest consumes 25% of the initial mass as fuel in the first stage. At the end of the first stage, the tank with a mass of 20% of the initial mass is ejected. In the second stage, the rocket consumes another 25% of the initial mass as fuel. Find the speed of the rocket at the end of the stage I and at the end of the stage II.

SOLUTION 3.7 a. The final speed of the one-­stage rocket is v = vex ln

m0 m

where v is the final speed of the rocket, m0 is the initial mass of the rocket plus fuel, and m is the final mass of the rocket. Since a mass of 0.7 m0 is consumed as fuel, the final mass of the rocket is 0.3 m0

v = vex ln

m0 m0 1 = vex ln = vex ln = vex ln 3.33 m m0 − 0.7m0 0.3

b. During the first stage, the rocket will consume fuel with the mass of 0.25 m0 and will gain speed v1 v1 = vex ln

m0 m0 1 4 = vex ln = vex ln = vex ln m m0 − 0.25m0 0.75 3

After the first stage the rocket ejects the tank, having an even smaller mass m0 (1 − 0.25 − 0.2) = 0.55 m0 , and starts this stage with the initial velocity v1 , at the end of the stage is gaining the velocity v2

v2 = v1 + vex ln

0.55m0 4 0.55  4 0.55  = vex ln + vex ln = vex ln  +  = 0.89 vex 0.3m0 3 0.3  3 0.33 

The property of logarithms is used here: ln A + ln B = ln AB .

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R

FIGURE 3.5  A rocket connected to a pole.

PROBLEM 3.8 A rocket is tethered to a pole such that it moves in a circular orbit (Figure 3.5). At t = 0 , the rocket begins burning fuel in a way that decreases its mass in accordance with the equation m(t ) = m0e − tλ . If the initial rope length is R , initial mass is m0 , and initial velocity is v , find an expression for the rope length, as a function of time, that keeps the angular momentum constant. Assume the rocket thrust is such that the rocket’s velocity increases as v(t ) = v + α t . SOLUTION 3.8 The initial angular momentum is given by    l =r×p

Since r ⊥ p

l = l = Rm0 v

At t > 0 , the angular momentum becomes lt > 0 = r (t )m0e − λ t (v + α t )

keeping lt > 0 = l requires

r (t )m0e − λ t (v + α t ) = Rm0 v

Therefore,

r (t ) =

Rm0 veλ t m0 (v + α t )

Note, since the mass is exponentially decreasing, the rope must increase its length exponentially to keep the angular momentum constant.

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PROBLEM 3.9

a. Calculate the center of the mass of a uniform cone. b. How does the center of the mass change if the density depends on the radius as ρ = kr , where k is a constant with appropriate units?

SOLUTION 3.9

a. The cone is positioned such that the apex is at the origin, and the axis of symmetry is in the z direction (Figure 3.6). Due to symmetry, the X and Y coordinates of the center of mass are zero. Only Z remains to be calculated 1 M

Z=

∫ ρ z dV

From the similar triangles it follows that r R = z h

So

r=

R z h

In cylindrical coordinates, dV = rdrdφ dz and dV = π r 2 dz and r will be R substituted by z from the previous equation, knowing that the density per h mass is the inverse of the volume of the cone

ρ 3 = M π R 2h

z

R h

r

z

0

y

x

FIGURE 3.6  Inverted cone of height h and radius of the base R, with the apex centered at the origin of cartesian system.

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1 M

Z= =

ρ z π r 2 dz =

πρ R h2 M

2

h

z 3dz =

0

1 M

ρπ z

R2 2 z dz h2 h

πR 3 z4 3 = h h2 π R 2h 4 0 4 2

Therefore, for a uniformly distributed mass, = X 0= , Y 0,

3 h 4

Z=

b. In this case, the density is not homogeneously distributed, but it depends on the radius as ρ = kr . 1 Z= M =

1 ρ z dV = M

π kR Mh3

3

h

z 4 dz =

0

h

1 R R2 krz dV = k z z π 2 z 2 dz M h h

∫ 0

π kR z Mh3 5 3

5

h

= 0

π kR 3h2 5M

The mass of the cone is also dependent on the density: M=

h

ρ dV = kr π r 2 dz = π k

∫ 0

h

π kR 3 z 4 π kR 3h z3 R3 dz = = 4 h3 h3 4 0

Back to the center of the mass,

Z=

π kR 3h2 π kR 3h2 4 4 = = h 3 5M 5 π kR h 5

Note that with ρ = kr the center of the mass moves closer to the basis of the cone and further from the apex. Also, note that the position of the center of mass depends on how we place the cone in the system of coordinates. In another problem, the cone is placed with the apex up, and the position of the center of mass is inverted, accordingly.

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Momentum and Angular Momentum z

s h

FIGURE 3.7  Uniform cylinder of height h.

PROBLEM 3.10

a. Calculate the moment of inertia of a uniform solid cylinder of radius R, rotating about its axis (Figure 3.7). b. What is the moment of inertia if the density is not uniform, but proportional to the square of the radius ρ = kr 2 , where k is a positive constant with appropriate units? c. What is the moment of inertia if the density is ρ = kr n , with k a constant with appropriate units and n natural number? Check your answer for n = 2.

SOLUTION 3.10 a. The definition of the moment of inertia is N

I=

∑m s

2 α α

α =1

with sα the distances of the mass mα from the axis of rotation. Transform the sum into an integral of the form I=

∫ ρ s dV 2

The element of volume in cylindrical coordinates is dV = r dr dφ dz .

I=

∫ ρ s dV 2

R

h

∫ ∫ ∫

I = ρ r 3dr dφ dz = 2πρ h

0

0

0

r4 4

R

= πρ h 0

R4 2

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Classical Mechanics

The density is represented in terms of mass and the volume as follows: M M = V π R 2h

ρ=

The moment of inertia becomes I = πρ h

Note that this is the moment of inertia of a thin disk of radius R rotating about its axis. b. For the case with the density ρ = kr 2 , the density will remain under the integral sign, because it is not a constant anymore. 2π

R

R 4 π Mh R 4 MR 2 = = 2 π R 2h 2 2

h

r6 I = kr r dr dφ dz = 2π kh 6

∫ ∫

2 3

0

0

0

= π kh 0

R6 3

c. The last case with the density ρ = kr n can be solved in the same manner 2π

R

h

R

∫ ∫

I = kr nr 3dr dφ dz = 2π kh

R

0

0

0

r n+4 Rn+4 = 2π kh n+4 0 n+4

For n = 2, the previous expression from point (b) is obtained. PROBLEM 3.11 Consider the cone in Figure 3.8 with its vertex at the origin and its axis on the z-­axis. Given height h , radius R , and density ρ ( z) = ρ0 sin(λ z) , find its center of mass. z

R h

0

x

FIGURE 3.8  Solid cone of height h and base radius R.

y

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Momentum and Angular Momentum

SOLUTION 3.11 Since this is symmetric about the z-­axis, it is only necessary to determine where the z-­axis center of mass lies. This is given by Z=

1 ρ z dV M

∫ v

where dV = r dφ dr dz

with 0≤r≤

R z, 0 ≤ φ ≤ 2π , 0 ≤ z ≤ h h

Therefore, the Z center of mass is

Z=

=

1 M

R z h h 2π

∫∫ ∫ ρ sin(λ z)rz dφ drdz 0

0 0 0 R z h h 0

2πρ M

∫∫

h

sin(λ z )rz drdz =

0 0

πρ0 R 2 sin(λ z )z 3 dz Mh2

∫ 0

2 2 2 2 πρ R  hλ (6 − h λ ) cos ( hλ ) + 3(h λ − 2)sin(hλ )  = 02   Mh  λ4  2

and the complete center of mass is given by

  πρ0 R 2 R =  0, 0, (hλ (6 − h2λ 2 ) cos(hλ ) + 3(h2λ 2 − 2)sin(hλ ))  2 4 Mh λ  

PROBLEM 3.12 Consider a sphere centered at the origin as in Figure 3.9. Given a radius R and a density ρ (φ ) = ρ0 sin(λφ ) , find the center of mass. SOLUTION 3.12 The center of mass is given by

1 R= M

∫ ρr dV

86

Classical Mechanics z

R

y

0

x

FIGURE 3.9  Solid sphere of radius R.

where r = x x + y y + z z and dV = r 2 sin θ dφ dθ dr . Splitting this into components yields X=

1 M

ρ0 sin(λφ ) x dV =

ρ0 M

R π 2π

∫∫∫ sin(λφ )xr sin(θ ) dφ dθ dr 2

0 0 0

Expressing x in polar coordinates as r sin(θ ) cos(φ ) yields X=

ρ0 M

R π 2π

∫∫∫ sin(λφ )r sin (θ ) cos(φ ) dφ dθ dr 3

2

0 0 0

Similarly, the other coordinates are given by

ρ Y= 0 M ρ Z= 0 M

R π 2π

∫∫∫ 0 0 0

R π 2π

∫∫∫ 0 0 0

ρ sin(λφ ) yr sin(θ ) dφ dθ dr = 0 M 2

ρ sin(λφ )zr sin(θ ) dφ dθ dr = 0 M 2

R π 2π

∫∫∫ sin(λφ )rr sin (θ )sin(φ ) dφ dθ dr 3

2

0 0 0

R π 2π

∫∫∫ sin(λφ )rr sin(θ ) cos(θ ) dφ dθ dr 3

0 0 0

which use y = r sin(θ )sin(φ ) and z = r cos(θ ) . The Z coordinate can be simplified by considering the φ integral only involves sin(λφ ) , which is zero. Therefore,

Z =0

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Momentum and Angular Momentum

The X and Y integrals can be rearranged in the following way:

R

π

0

0

0

R

π

0

0

0

ρ X = 0 r 3 sin 2 (θ ) sin(λφ ) cos(φ ) dφ dθ dr M

∫ ∫

ρ Y = 0 r 3 sin 2 (θ ) sin(λφ )sin(φ ) dφ dθ dr M

∫ ∫

which illustrate they are not terribly complicated. Therefore, the center of mass is given by

 π R 4λρ0 sin 2 (πλ ) π R 4 ρ0 sin 2 (2πλ )  R= , , 0 2 8 M (λ 2 − 1)  4 M (λ − 1) 

PROBLEM 3.13 Consider a mass m moving with velocity v colliding with another mass m attached to a spring with a spring constant k . Determine how much the spring compresses for a. An inelastic collision (Figure 3.10). b. An elastic collision at an angle θ (Figure 3.11) Assume the mass connected to the spring cannot deflect out of the ± x direction.

m

m

v

FIGURE 3.10  Two masses colliding inelastically.

m  v m

FIGURE 3.11  Two masses colliding elastically.

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Classical Mechanics

SOLUTION 3.13 a. The initial momentum is given by  pi = −mvxˆ

Since this is an inelastic collision and the masses are identical, the momentum at the moment when the mass hits the spring is  p f = −2mv f xˆ

where v f is determined via conservation of momentum   pi = p f

so

vf =

v 2

Therefore, the instant the spring starts compressing, the total mass is 2m v and the velocity is . The compression can be determined by considering 2 the initial kinetic energy and final potential energy. KEi = U f

1 1 2mv 2f = kx 2 2 2

= x2

2m v 2 1 m 2 = v k 4 2 k

The compression is thus given by

x=

m v 2k

b. For the elastic collision, the diagram is as in Figure 3.12: The initial momentum is given by  pi = mv(− cos θ x + sin θ y )

89

Momentum and Angular Momentum vf m

f  vi i

m

FIGURE 3.12  Geometry of the elastic collision.

Assuming the mass on the spring cannot deflect out of the ± x direction, the final momentum is given by  p f = −mv f ,s x + mv f (cos θ f x + sin θ f y )

Since this is an elastic collision, the kinetic energy before and after must be the same KEi = KE f

1 2 1 2 1 mv = mv f ,s + mv 2f 2 2 2

The quantity of interest is v f ,s as this will give the initial energy that goes into compressing the spring. Considering conservation of momentum, the following system of equations exists: i. −v cos θ = −v f ,s + v f cos θ f , and yields that v f cos θ f = v f ,s − v cos θ ii. v sin θ = v f sin θ f iii. v 2 = v 2f ,s + v 2f Taking (i) and (ii)

(ii ) (i )

v f sin θ f = v sin θ v f cos θ f = v f ,s − v cos θ

tan θ f =

v sin θ v f ,s − v cos θ

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Classical Mechanics

Equation (ii) now becomes vf =

v sin θ  −1  v sin θ sin  tan    v f ,s − v cos θ 

   

Therefore, using (iii),

v 2f ,s = v 2 − v 2f v f ,s = v 2 −

v 2 sin 2 θ   v sin θ sin 2  tan −1    v f ,s − v cos θ 

v f ,s = v 1 −

   

sin 2 θ   v sin θ sin 2  tan −1    v f ,s − v cos θ 

   

Despite not being able to get a closed-­form solution for v f ,s , this can be solved numerically for a given v and θ . As in part (a), the compression is given by

x2 =

m 2 v f ,s k

x=

m v f ,s k

which can be determined using the numerical solution for v f ,s .

4

Energy

4.1  THEORY This chapters introduces conservative forces, conditions for a force to be conservative, and law of conservation of energy.

4.1.1 Work Kinetic Energy Theorem The variation of kinetic energy of a particle is equal to the work done by the net force acting on the particle T  T2  T1  W

4.1.2 Conservative Forces A force acting on a particle is conservative if it satisfies the following conditions:  i. The force depends only on the particle’s position r and not on the velocity  v , the time t , or any other variables. ii. The work done by the force is path independent. Another condition for a force to be conservative is that the curl of the force should be zero. For example, if the force is represented in cartesian coordinate, the curl is

xˆ    F  x Fx

yˆ  y Fy

zˆ  0 z Fz

It is important to note that the potential energy can be defined only for conservative fields. The relationship between the force and the potential energy is that the force is minus the gradient of the potential energy:  F  U

If the potential energy is known, the force can be obtained.

DOI: 10.1201/9781003365709-4

91

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Classical Mechanics

If the force is known, it is important to verify that the force is conservative (the curl is zero), and then the potential energy is calculated as   U   F· dr

Principle of conservation of mechanical energy: if the forces acting on a particle are conservative, the mechanical energy of the particle is conserved. The total mechanical energy is conserved only when the potential energy is time U independent, that is,  0. t

4.1.3 Obtaining the Equation of the Motion from the Conservation of the Energy In one-­dimensional example, the mechanical energy is the sum of the kinetic energy and potential energy, E  T  U ( x)

The kinetic energy has the form

T=

mx 2 2

The kinetic energy is substituted into the energy equation and expressions for velocity and position can be obtained: E

mx 2  U ( x) 2

x ( x )

2 ( E  U ( x )) m

dx and interestingly, the time is obtained as a function of position, dt and subsequently, the position versus time can be calculated From here, x =

dt =

dx x( x )

By integration, x

t  t0

x0

dx  x ( x )

m 2

x

x0

dx E  U ( x)

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Energy

From here time as a function of position is obtained, and then position as a function of time can be calculated.

4.2  PROBLEMS AND SOLUTIONS PROBLEM 4.1  Check if the following force is conservative F = (ax, bx, cz ). If it is conservative, find  the potential energy and verify that F  U . SOLUTION 4.1 One condition for a force to be conservative is to have the curl equal to zero

x    F  x Fx

y  y Fy

z x    z x Fz ax

y  y by

z  z cz

 (cz ) (by)    (ax ) (cz )    (by) (ax )      z   F  x    y  z  x   x y   z  y    F  (0, 0, 0)    F  0 This force is conservative and the potential energy can be calculated   U   F· dr   Fx dx  Fy dy  Fz dz





 ax dx   by dy   cz dz 

 ax 2 by 2 cz 2        Constant 2 2   2

Since the potential energy is given up to a constant, the constant may be chosen as zero.  Now, it can be verified that the force is the minus gradient of the potential energy, F  U by a simple derivation after writing the gradient in cartesian coordinates

 U  U  U   F  U    x y z   ax x  by y  cz z  (ax, bx, cz)  F y z   x

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Classical Mechanics

PROBLEM 4.2 Check if the following forces are conservative. If they are, find the potential energy.  a. F  ( x 2  x, 2 y  1, 0)  b. F  ( x 2  x, 2 y  z, x 3  y / 2)  c. F  (2 x  y, x  2 z, 2 y  z ) SOLUTION 4.2 If the curl of a force is 0, then the force is conservative. a. x    F   /x x2  x

y  /y 2y 1

z  /z 0

           F   (0)  (2 y  1), ( x 2  x )  (0), (2 y  1)  ( x 2  x )  z z x x y  y 

   F  (0, 0, 0)

   F  0 This force is conservative. The potential energy can be found   U   F· dr

  x3 x2 U     y2  y  3 2  

b. x    F   /x

y  /y

x2  x

2y  z

z  /z y x3  2

    y  y      F    x 3    (2 y  z ), ( x 2  x )   x 3   , 2  z 2 z x   y     (2 y  z )  ( x 2  x )  x y 

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Energy

   F  (3/ 2,  3 x 2 , 0)

This force is not conservative. c. x    F   /x 2x  y

y  /y x  2z

z  /z 2y  z

        F   (2 y  z)  ( x  2 z), (2 x  y)  (2 y  z),     y z z x     ( x  2 z )  (2 x  y )  x y 

   F  (0, 0, 0)

   F  0 This force is conservative. The potential energy can be found   U   F· dr

z2  U    x 2  xy  2 yz   2 

PROBLEM 4.3 Evaluate the line integral for the work done by the following forces on both paths shown. Determine if each force is conservative. Path a is along the line y = x (Figure 4.1a). Path b is along the line y = x 2 (Figure 4.1b).

a. F1 = ( y 2 , x ) b. F2 = (2 x, y)

SOLUTION 4.3   a. Path a : The work done is F1 · dr .

∫ a

Wa  F1x dx  F1y dy

a

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Classical Mechanics y 1

a b

0

x

FIGURE 4.1  Path a (first bisector) and b (parabola).

On path a , x = y . It follows that dx = dy .

Wa  ( F1x  F1y )dx

a 1

Wa  ( y 2  x )dx

0 1

Wa  ( x 2  x )dx

0

5 6

Wa =

  Path b : The work done is F1 · dr

∫ b

Wb  F1x dx  F1y dy

b

On path b , y = x 2 . It follows that dy = 2 xdx

Wb  F1x dx  F1y 2 xdx

b 1

Wb  ( y 2  x  2 x )dx

0

97

Energy 1

Wb  ( x 4  2 x 2 )dx

0

Wb =

13 15

Since Wa ≠ Wb , the force is not conservative. b. Path a : The work done is   Wa  F2  dr  F2 x dx  F2 y dy

a

a

On path a , x = y . It follows that dx = dy

Wa  ( F2 x  F2 y )dx

a 1

Wa  (2 x  y)dx

0 1

Wa  (2 x  x )dx

0

Wa =

3 2

Path b: The work done is   Wb  F2  dr  F2 x dx  F2 y dy

b

b

On path b , y = x 2 . It follows that dy = 2 xdx

Wb  F2 x dx  F2 y 2 xdx

b 1

Wb  (2 x  y  2 x )dx

0

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Classical Mechanics 1

Wb  (2 x  2 x 3 )dx

0

Wb =

3 2

Since Wa = Wb , the force is conservative.

PROBLEM 4.4 k Find the force corresponding to the following potential energy U   , with k a conr stant with appropriate units, then check that the force is conservative (it should be). SOLUTION 4.4 The form of the potential energy suggests that the spherical coordinates may be indicated in this case. The potential energy depends only on the radius, not on azimuthal angle or polar angle.

 k     U ˆ 1 U ˆ 1 U   r   rˆ k   F  U    rˆ   rˆ     sin   r r2 r r r  

The minus sign in the potential indicates that the force is attractive. The force has a potential energy associated with it, so it must be a conservative force, therefore, the curl of the force should be zero. The curl in spherical coordinates is written as    F  rˆ

    1  1    (rF )  (sin  F )  F   ˆ  Fr   sin       r  r  r     1     Fr  ˆ  (rF )  r  r  

     ˆ 1   k  1  (0)  (0)  0    2         r sin    r  r r    1   k     ˆ  (0)  0   r 2   r  r

rˆ

1 r sin 

1 r sin 

Therefore, the force is conservative.

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Energy

H>h

V0 H

h

FIGURE 4.2  Skier going down ski jump ramp.

PROBLEM 4.5 A skier, starting with a velocity v0 , slides down a frictionless ski jump ramp as pictured in Figure 4.2. Using conservation of energy, find a. The maximum speed of the skier. b. The speed of the skier at the end of the ramp. SOLUTION 4.5 a. The mechanical energy is E  T  U . At the top of the ramp, 1 1 2 E1  mv02  mgH . At the lowest point of the ramp, E2 = mvmax . Because 2 2 of the conservation of energy, E1 = E2 . Thus, 1 2 1 mvmax  mv02  mgH 2 2

Solving for vmax , the maximum speed of the skier is vmax  v02  2 gH

b. At the end of the ramp, E3  energy, E1 = E3 .

1 2 mvend  mgh . Because of conservation of 2

1 2 1 mvend  mgh  mv02  mgH 2 2 Solving for vend , the speed of the skier at the end of the ramp is

vend  v02  2 g( H  h)

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Classical Mechanics

0

FIGURE 4.3  Child sitting on a perfectly hemispherical igloo.

PROBLEM 4.6 A child sits on an igloo. Find the vertical height at which the child leaves the surface of the igloo. Consider the igloo a perfect, frictionless hemisphere of radius R . The child’s initial position is on the top of the igloo and consider that the slide is starting with a tiny nudge (Figure 4.3). SOLUTION 4.6 This is a classical problem which can be solved by using the conservation of total mechanical energy and by considering that the motion occurs on a circular trajectory, with the total force being equal to the centripetal force. Also, at the moment the child leaves the igloo, the normal to the surface is zero. The potential energy level is zero at the ground level. The potential energy depends on the angle θ as U ( )  mgR cos  . The total energy is conserved: E  U (0)  mgR

mv 2  mgR cos  2

From here, the kinetic energy is obtained:

T

mv 2  E  U  mgR  mgR cos   mgR(1  cos  ) 2

During the motion on the sphere, Newton’s Second Law is written (Figure 4.4)

N  mg cos

mv 2 R

From the energy conservation equation,

mv 2  2mgR(1  cos  )

101

Energy N

a cp

mg

0

FIGURE 4.4  Child sliding on the hemispherical igloo, with the gravitational force and normal force represented. The centripetal acceleration is keeping the child on the igloo, until the point when the normal force becomes zero, moment in which the child falls from the igloo.

And back in Newton’s Second Law, N  mg cos   2mg(1  cos  )  mg(3 cos   2)

At the moment the child leaves the igloo, the normal force becomes zero, so the angle θ can be obtained as N  mg(3 cos   2)  0

So

cos

2 3

And therefore,

y  R cos

2 R 3

PROBLEM 4.7 A uniform sphere of mass M and radius R is rolling down an incline of height h (Figure 4.5). Find the speed of the sphere at the base of the incline. Solve this problem in two ways. a. First, consider the motion of the sphere about the center of the mass, with the kinetic energy of the center of the mass and the rotational energy about the v center of the mass. Use the non-­slip condition:   . R b. Consider the rotation about the point where the sphere touches the incline. For this, consider Steiner’s theorem I   I  MR 2 , where is the moment of inertia of the sphere with respect to an axis going through the center of the mass 2 MR 2 I= . Calculate the speed of the sphere at the base of the incline. Use the 5 v non-­slip condition:   . R

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Classical Mechanics A

M R

h

B

FIGURE 4.5  Sphere sliding on the incline.

SOLUTION 4.7 a. Here, the total mechanical energy is conserved. E A = EB

On the top of the incline, the sphere starts with zero kinetic energy, but the potential energy is maximum: E A = Mgh

At the bottom of the incline, the total energy is only kinetic – we consider the lowest potential energy at the bottom of the incline EB

Mv 2 I  2  2 2

where I is the moment of inertia of the sphere with respect to the central 2 MR 2 axis, I = , and by using the non-­slip condition and the conservation 5 of the energy,

Mgh

Mv 2 I  2  2 2 2

2 MR 2  v  2 Mv Mv 2 Mv 2 5  R  Mgh     2 2 2 5 By rearranging the terms and dividing by the mass, the speed of the center of the mass of the sphere is obtained:

v=

10 gh 7

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Energy

b. Considering the energy with respect to the contact point, and by using the conservation of the energy,

T  U

2  2 MR 2  v  2  v   MR 2    ( ) I  MR  2   I  R   5   R   7 Mv 2 T   2 2 2 10

2

Note that the moment of inertia I′ is calculated using Steiner’s theorem. As before, the potential energy is

U  0  Mgh And from the conservation of energy,

7 Mv 2 = Mgh 10 The speed of the sphere at the bottom of the incline is, as before,

v=

10 gh 7

PROBLEM 4.8 A block is attached to a horizontal spring of constant k . There is a coefficient of friction µ between the block and the surface. The block is pulled and released at position xs , as in Figure 4.6. Find the velocity of the block when it reaches the equilibrium position the first time, using energy conservation.

Xe

Xs

FIGURE 4.6  Mass connected to a spring released from rest at xs .

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Classical Mechanics

SOLUTION 4.8 The work due to friction is W f   fd    mgd . The change in mechanical energy is due to friction, so E  T  U  W f . At the equilibrium point,

1 1 1 1  mvs2  mve2  kxs2  kxe2    mg( xs  xe ) 2 2 2 2 The origin of the axis is chosen to be the equilibrium point, and recall that vs = 0.

1 2 1 2 mve  kxs    mgxs 2 2 Solving for ve , the speed at equilibrium is

k 2 xs  2  gxs m

ve

PROBLEM 4.9 A ball of mass m is sliding up a frictionless ramp of length L and at an angle θ . Determine the maximum initial velocity of the ball such that it doesn’t slide off the ramp. Consider the following two situations and compare the answers: a. The ball is initially released parallel to the ramp, as in Figure 4.7. b. The ball is initially released parallel to the floor, as in Figure 4.8.

L

V0

FIGURE 4.7  Ball traveling up frictionless ramp.

L

FIGURE 4.8  Ball traveling toward frictionless ramp.

V0

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Energy

Note: This is the same problem text as Problem 1.6, but now should be solved using energy considerations. SOLUTION 4.9 a. Conservation of energy requires Ei = E f

where Ei is pure kinetic energy up the ramp and E f is pure potential at the top of the ramp. Therefore,

1 2 mv0 = mgh 2 By dividing it by the mass and writing the height as h  L sin  ,

1 2 v0  gL sin  2 This can be immediately solved

v0  2 gL sin

b. This will have a similar approach as part a, but care must be taken when the puck first interacts with the ramp. Initially,

E1 =

1 2 mv0 2

but after the ramp, the energy becomes

E2

1 2 mv0 cos2  2

It is this E2 that will be used in the conservation of energy equation, similar to part a, Ei = E f .

1 2 mv0 cos2   mgh 2 As before, after simplification of mass and substitution of height h ,

1 2 v0 cos2   gL sin  2

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Therefore, the velocity is given by

v0

1 cos

2 gL sin

as was found in Chapter 1. PROBLEM 4.10 Consider a block of mass m sliding down a frictionless ramp at an incline θ as in Figure 4.9. Find the velocity of the block at time t if the block is stationary at t = 0 . Note: This is the same problem text as Problem 1.4, but now should be solved using energy considerations. SOLUTION 4.10 Considering conservation of energy, the total energy of each of the following positions must be the same. Specifically, E0 = E ( t )

Since the block is initially at rest, at time t , the difference in potential energy has been converted to kinetic energy (Figure 4.10). Therefore,

mgh

1 2 mv 2

FIGURE 4.9  Block sliding down a ramp. x

h0

ht

FIGURE 4.10  The displacement of the block as it slides down the ramp.

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Energy

The change in height is given by h  x sin  . Considering an infinitesimal displacement, ∆x becomes dx = v dt . From this, the equation from conservation of energy is written as gv sin  dt

1 2 v 2

and the velocity is v  2 g sin  dt

The solution v = 0 is not interesting, being the point up the ramp the object starts moving from. The expression for velocity can be directly integrated, yielding v(t )  2 gt sin   C

where C is a constant. Since the initial velocity at t = 0 is zero, C = 0 and the velocity of the block is v(t )  2 gt sin

Precisely what was found in Chapter 1. PROBLEM 4.11 In Problem 2.8, a mass on a pendulum moved in the presence of linear and quadratic air resistance and the equations of motion were found. For the linear drag case, and assuming small angles, find an expression for total energy as a function of time. Assume the initial position is φ0 and initial velocity is zero. SOLUTION 4.11 From Problem 2.8, the equation of motion was found to be b g     sin   0 m l

In the small angle approximation, this becomes b g       0 m l

This differential equation can be solved considering the characteristic equation

r2

g b r 0 l m

which has roots

r

b2 g b   2 l 2m 4m

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Classical Mechanics

Therefore, the position is given by

(t )  Aer t  Ber t

where A and B are constants. Also,

(t )  Ar er t  Brer t

Using the initial conditions,

r (0)  Ar  Br  0 and B   A  r

The position at t = 0 is

(0)  A  B  0

And by substituting B, A A

r A  0 from where (r  r )  0 r r

The constants are A

0r (r  r )

so B

0r (r  r )

and the position is given by

(t )

0 (rer t  r er t ) (r  r )

The total energy is the sum of the kinetic and potential energies so

E (t )  KE (t )  U (t )

1 2 mv  mgh 2

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Energy

The velocity is given by v  l  l

0 rr (rr er t  rrer t )  l 0   (er t  er t ) (r  r ) (r  r )

and height of the mass is given by h  l (1  cos  )

In the small angle approximation,   2 h  l 1  1   2

 1 2    l  2

Therefore, 1 2 2 1 ml  (t )  mgl  2 (t ) 2 2  02 1  02r2r2  ml  l (er t  er t )2  g (r er t  r er t )2  2 2  (r  r ) 2  (r  r ) 

E (t )

02 ml ( lr2r2 (er t  er t )2  g(rer t  r er t )2 ) 2(r  r )2

To understand this motion, it is necessary to examine er±

e

r t

e

b b2 g     2m m2 l 4 

 t  

e

b2 g b  t t m2 l 4 2m

e

Notice b  2m

b

b2  4m 2

b2 g  4m 2 l

t

Therefore, the e 2 m part of the exponential dominates and the motion of the pendulum exponentially decays. For more insight, consider the other part of the exponential e

t

b2 4 m2

g l

b2 g b2 g b2 g − is real. In this case, damping is > ,   0 , and 2 2 2 l 4m l 4m l 4m large, and the mass essentially swing down to its lowest point and stops. This may When

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Classical Mechanics

b2 g be the situation if the pendulum was in a viscous liquid. When < , it follows 2 4 m l b2 g b2 g that   0 and − is complex (yielding a complex exponential). In 4m 2 l 4m 2 l this case, damping is small, and the mass oscillates around the lowest point with the amplitude of oscillation being exponential damped, in accordance with e

b t 2m

.

PROBLEM 4.12 A ball is placed at the top of a frictionless ramp with a height h and incline θ (Figure 4.11). The ball is also attached to a point 2h above the ground (h above the top of the ramp) with a string of length 2h and negligible mass. Find the initial velocity of the ball must have, parallel to the ramp, in order to reach a maximum height halfway between the fixed point and the top of the ramp. Note: This is the same problem text as Problem 1.15 but now should be solved using energy considerations. SOLUTION 4.12 Considering the two separate motions (down the ramp vs. leaving the ramp and ascending to 2h ), there are several energies involved. There is the initial kinetic energy associated with the initial velocity and the initial potential energy associated with being at the top of a ramp. There is the intermediate energy associated with the moment the ball leaves the ramp where the initial potential energy has been converted to kinetic energy. Then there is the final energy of the ball at its final height where all the kinetic energy has been converted to potential energy. To help keep the goal of the problem in mind, consider working “backward” and looking at the energies in the second phase of the motion. Here, the final energy is purely potential and the initial is purely kinetic. Since the velocity here is the intermediate velocity mentioned above, it is denoted as v1y . Therefore, 1 2 mv1y = mgH 2

fixed point

h

h

FIGURE 4.11  The ball positioned at the top of a ramp, connected to a fixed point.

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Energy

Since the exact spot the ball leaves the ramp is unknown, it can be taken to be a distance y beneath the top of the ramp. This means H  h  y . Therefore, 1 2 v1y  g(h  y) 2

The height h is known, the distance y is unknown but can be determined by geometry, and the velocity v1y is unknown but can be determined by the first phase of the motion. Consider Figure 4.12 to determine y . The equations then follow from the triangle depicted in Figure 4.13. tan

y y cos  so x  x sin

fixed point

h v1y y

v1

h

FIGURE 4.12  The ball moving as far down the ramp as possible before it must leave the surface of the ramp.

h

2h

y

α  x

FIGURE 4.13  Two triangles depicting the angles and heights immediately before the ball leaves the ramp.

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Classical Mechanics

From the Pythagorean theorem in the big triangle, x 2  (h  y )2  (2 h )2

y 2 cos2   h2  2hy  y 2  4h2 sin 2

Rearranging the terms, it follows that y2 (cos2   sin 2  )  2hy  3h2  0 sin 2

And after recalling that cos2   sin 2   1 , y 2  2h sin 2  y  3h2 sin 2   0

From this,

y

2h sin 2   4h2 sin 4   4 3h2 sin 2  2

  h sin 

sin 2   3  sin 

To determine the velocity v1y , consider Figure 4.14. v1y  v1 cos

with    

so     2 2 y

x V1y α

γ

V1 V1x

FIGURE 4.14  Detailed illustration of the forces with respect to the selected axes.



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Energy

  v1y  v1 cos      v1 sin  2 

It also follows that sin(   )

h y 2h

 1  sin  h y 1    sin      sin   2h    1





sin 2   3  sin     2  

Thus,    1  sin  v1y  v1 sin  sin     



 sin 2   3  sin       2    

Using the conservation of energy equation above,

sin 2   3  sin 



sin 2   3  sin 



v1y  2 gh 1  sin 

1 2 v1y  g(h  y) 2

Therefore, 2 gh 1  sin  v1

  sin    sin  sin 1     



 sin 2   3  sin        2    

Now is to obtain v1 from the first phase of the motion. Recall, this is the initial kinetic and potential energies being converted to kinetic energy

1 2 1 mv0  mgy  mv12 2 2

with v0 being the initial velocity (and the objective of the problem). This can be directly solved for v0

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Classical Mechanics

v0  v12  2 gy

sin   3  sin   2gh sin   sin  sin   3  sin         

2 gh 1  sin   2 sin  sin 1     

2

2

2

  

sin 2   3  sin 



 

Slightly different geometry was used here versus that in Problem 1.15, but from the general structure, it is easy to see the similarities with the previous solution. The reorganization can be achieved via trigonometric identities.

5

Oscillations

5.1  THEORY This chapters introduces Hooke’s Law, the simple harmonic motion and different types of oscillations as damped, or driven damped oscillations. The parallelism between mechanical and electrical oscillators is discussed.

5.1.1 Hooke’s Law The force exerted by a spring is F ( x ) = −kx

where k is the spring constant (or force constant, in N/m) and x is the distance with respect to the equilibrium point. The potential energy is U ( x) =

kx 2 2

The potential energy can be expanded in a Taylor series as

U ( x ) = U (0) + U ′(0) x +

U ″(0) x 2 + 2

5.1.2 Simple Harmonic Motion Starting with the force, and expressing the force by using Newton’s Second Law,

F ( x ) = −kx

F ( x ) = mx From this, the equation of motion is

mx = −kx

x=−

k x = −ω 2 x m

where ω is the angular frequency of oscillation. DOI: 10.1201/9781003365709-5

115

116

Classical Mechanics

This is a second-­order, linear, homogeneous differential equation, with two independent solutions, which can be written in different ways, for example: x(t ) = eiωt and x(t ) = e −iωt

The linear combination of the two solutions is also a solution, x(t ) = C1eiωt + C2e −iωt

After writing

e ± iωt = cos ω t ± i sin ω t x(t ) = (C1 + C2 ) cos ω t + i(C1 − C2 ) sin ω t = B1 cos ω t + B2 sin ω t Which is the form of simple harmonic motion.

5.1.3 Energy Potential energy is written as

U=

kx 2 2

T=

mx 2 2

Kinetic energy as

If we consider a third form for the simple harmonic motion, x(t ) = A cos(ω t − δ )

Then the total energy is

E = T +U =

mω 2 A2 kA2 kA2 sin 2 (ω t − δ ) + cos2 (ω t − δ ) = 2 2 2

5.1.4 Particular Types of Oscillations and the Differential Equations Associated with Them 5.1.4.1  Damped Oscillations Consider a system of mass to a spring of spring constant k and subject to  m attached  ˆ a resistive force of type f = −bv = −bxx

mx + bx + kx = 0

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Oscillations

b k = 2 β is called the damping constant, while ω0 = is the natum m ral frequency – the frequency the system would oscillate without the resistive force. The differential equation can be written as The constant

x + 2 β x + ω02 x = 0

with the auxiliary equation

r 2 + 2 β r + ω02 = 0

and the solution

 x(t ) = e − β t  C1e 

β 2 −ω02 t

+ C2 e

− β 2 −ω02 t

  

5.1.4.2  Weak Damping β < ω0 Frequency (smaller than the natural frequency) ω1 = ω02 − β 2 has the solution of the form x(t ) = Ae − β t cos(ω1t − δ )

5.1.4.3  Critical Damping β = ω0 The solution has the form x(t ) = C1e − β t + C2 t e − β t

5.1.4.4  Strong Damping β > ω0 The solution has the form x(t ) = C1e

−  β − β 2 −ω02  t  

+ C2 e

−  β + β 2 −ω02  t  

5.1.4.5  Driven Damped Oscillations In the case of driven damped oscillators, besides Hooke’s law force and the resistive force, now there is another external force which supports the oscillation, called the driving force F (t ) . mx + bx + kx = F (t )

With f (t ) =

F (t ) , the equation of motion is written as m  x + 2 β x + ω02 x = f (t )

Usually, the driving force is of the sine or cosine type.

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Classical Mechanics

5.2  PROBLEMS AND SOLUTIONS PROBLEM 5.1 A cat is playing with a toy connected to a spring fixed to the wall on the other end. The toy has a mass of m = 50 g and the spring constant is k = 20 N/m and is following a simple harmonic motion with the expression x(t ) = A cos(ω t − δ )

a. Calculate the angular frequency ω , the frequency f , and the period τ . b. If the cat stretches the spring by an initial length x0 = 10 cm, and keeps the toy at rest for an instant, identify the amplitude A of the motion and calculate the phase δ . c. In another playful moment, the toy has the speed of 4 m/s at the equilibrium position of the spring x0 = 0 . What are the amplitude A and the phase δ in this case? SOLUTION 5.1 a. The angular frequency is related to the spring constant k and the mass m by the following relationship:

ω2 =

k m

ω=

k = m

N m = 20 s−1 0.05 kg 20

The frequency is obtained from the equation

ω = 2π f

f =

ω 20 s−1 = = 3.18 Hz 2π 2π

The period is immediately following, as the inverse of the frequency

τ=

1 = 0.32 s f

b. The simple harmonic motion has the expression x(t ) = A cos(ω t − δ ) . The velocity is v( t ) =

dx(t ) = x (t ) = −ω A sin(ω t − δ ) dt

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Oscillations

The initial conditions are= x0 x= (0) 10 cm and= v0 v= (0 ) 0 . In the initial expression, for the simple harmonic motion, and with using the properties of the trigonometrical functions cos(−δ ) = cos δ , and sin(−δ ) = − sin δ , x(t = 0) = A cos(−δ ) = A cos δ

The phase can be found from the second condition, v0 = 0

v(t = 0) = −ω A sin(−δ ) = ω A sin δ = 0

δ =0

Back to the first initial condition,

x(t = 0) = A cos δ = A cos 0 = A = 10 cm = 0.1 m

c. In a similar way, starting with the expression for the position and velocity, and, considering the initial condition, the amplitude and phase are calculated as

x(t ) = A cos(ω t − δ ) and v(t ) = −ω A sin(ω t − δ )

So

x0 = 0

x(t = 0) = A cos(−δ ) = A cos δ = 0

δ=

π 2

And from

v0 = 4

m s

v(t = 0) = −ω A sin(−δ ) = ω A sin δ = ω A sin m 4 v0 A= = s = 0.2 m ω 20 1 s

π = ωA 2

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Classical Mechanics

PROBLEM 5.2 a. An oscillation has the solution

x(t ) = β1 cos(ω t ) + β 2 sin(ω t ) Given initial conditions x(0) = 0 and v(0) = 2, solve for β1 and β 2 . b. An oscillation has the solution x(t ) = A cos(ω t − δ )

Given initial conditions x(0) = 3 and v(0) = 1, solve for A and δ . SOLUTION 5.2 a. Since sin(0) = 0 and cos(0) = 1, x(0) = β1 = 0 . The equation can be simplified to x(t ) = β 2 sin(ω t )

v(t ) = x (t ) is obtained by differentiation x (t ) = ωβ 2 cos(ω t )

Thus, v(0) = ωβ 2 = 2 . Solving for β 2 , the result is

β2 =

2 ω

The solution of the oscillation is x(t ) =

b. The derivative of x(t ) is

2 sin(ω t ) . ω

x (t ) = v(t ) = −ω A sin(ω t − δ ) From the given initial conditions, a system of two equations is obtained:

 A cos(−δ ) = 3  ω A sin(−δ ) = 1 After solving the system, values for δ and A are obtained:

 1  δ = arctan  −   3ω 

1 A = 9+  ω 

2

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Oscillations

PROBLEM 5.3 An object of mass m = 0.1 kg is in a simple oscillatory motion defined by equation of π  motion x(t ) = 5 cos  3π t −  , in centimeters. 4  a. Find the amplitude, the angular frequency, the phase δ , the period τ , and the frequency f . b. Calculate the spring constant. c. What are the position and the velocity at t = 0 s ? d. Find the acceleration at t = 0 s . SOLUTION 5.3 a. The general solution in simple harmonic motion is x(t ) = A cos(ω t − δ )

The given equation is

π  x(t ) = 5 cos  3π t −  4 

By identification, A = 5 cm , ω = 3π s−1 , and δ =

π . From here, frequency and 4

period can be easily calculated as follows:

ω = 2π f

f =

ω 3π s−1 = = 1.5 Hz 2π 2π

τ=

1 1 = = 0.67 s f 1.5 Hz

b. The spring constant can be easily calculated from

ω2 =

k m

(

k = mω 2 = 0.1 kg 3π s−1

)

2

= 8.88

N m

c. The initial position is  π π  5 2 x0 = x(t = 0) = 5 cos  −  = 5 cos   = cm = 3.54 cm 4 2   4

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Classical Mechanics

The velocity is

v(t ) = −ω A sin(ω t − δ ) v0 = v(t = 0) = −ω A sin(−δ ) = ω A sin δ = 15π sin = 15π

π cm 4 s

cm 2 cm = 33.3 = 3.33 × 10 −1 m/s 2 s s

d. The acceleration is obtained by differentiating the velocity with respect to time a (t ) =

dv(t ) = −ω 2 A cos(ω t − δ ) = −ω 2 x(t ) dt

The initial acceleration can be easily calculated as

 π  cm  π  cm a0 = a(t = 0) = −ω 2 A cos(−δ ) = −(3π )2 5 cos  −  2 = −(3π )2 5 cos   2  4 s 4 s cm 2 cm = −45π 2 = −314 2 = −3.14 m/s2 2 s2 s

PROBLEM 5.4 An underdamped oscillation has sinusoidal frequency of underdamped oscillator

(

x(t ) = Ae − β t cos t ω 2 − β 2

ω 2 − β 2 . Consider an

)

ω , at what time th is the amplitude of the oscillation half of the 5 initial amplitude? 2π b. The period of oscillation is . What is the amplitude after one period, ω2 − β 2 ω given β = , with a > 1. a a. Given β =

SOLUTION 5.4

a. The amplitude is given by Ae − β t . Thus, the equation Ae − β th = be solved for th .

Ae − β th =

1 A 2

1 A needs to 2

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Oscillations

e

ω − th 5

th =

Therefore, at time th = original amplitude.

=

1 2 5 ln 2 ω

5 ln 2 , the amplitude of the oscillation is half of the ω 2π

b. The amplitude Ap at t =

is the amplitude of oscillation after one

ω2 − β 2

period.

ω − · a

ω2 −

Ap = Ae

ω − · a

Ap = Ae

Ap = Ae

ω2 a2

2π 2

1  ω  1−   a2 

2π a 2 −1

The amplitude of the oscillation after one period is Ap = Ae

2π a 2 −1

PROBLEM 5.5 A critically damped oscillator, where β = ω =

k , has the solution m

x(t ) = C1e − β t + C2te − β t x + 2 β x + ω 2 x = 0 . a. Check that x(t ) satisfies the differential equation  b. Given x(0) = 2 and x (0) = 1 , solve for C1 and C2 .

SOLUTION 5.5 a. The first derivative of x is

x (t ) = −C1β e − β t − C2 β te − β t + C2e − β t

x (t ) = e − β t (−C1β + C2 ) − C2 β te − β t

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The second derivative of x is

x(t ) = C1β 2e − β t + C2 β 2te− β t − 2C2 β e− β t

x(t ) = e − β t (C1β 2 − 2C2 β ) + C2 β 2te − β t Using the derivatives, it follows that  x + 2 β x + ω 2 x = e − β t (C1β 2 − 2C2 β ) + C2 β 2te − β t + e − β t (−2C1β 2 + 2C2 ) −2C2 β 2te − β t + C1β 2e − β t + C2 β 2te − β t = 0

x + 2 β x + ω 2 x = 0 . So x(t ) satisfies  b. Using x(0) = 2 , the value for C1 = 2 . The first derivative of x is x (t ) = e − β t (−C1β + C2 ) − C2 β te − β t

Since x (0) = 1 ,

−C1β + C2 = 1

Thus, C2 = 2 β + 1. Therefore, the solution to the oscillation is

x(t ) = 2e − β t + (2 β + 1)te − β t

PROBLEM 5.6 Consider the case of an overdamped oscillation where β = aω , with a > 1, with solution

x (t ) = e

− β − β 2 −ω 2 t  

+e

− β + β 2 −ω 2 t  

 2π  a. What is the value of x  ?  ω   2π  b. Find x   for a = 2, 5, and 10 .  ω  SOLUTION 5.6 a. First, the equation of motion of the oscillation can be rewritten using the relation β = aω .

x (t ) = e

−  aω − a2ω 2 −ω 2  t  

+e

−  aω + a 2ω 2 −ω 2  t  

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Oscillations

x (t ) = e

x (t ) = e

(

)

  −  aω − ω 2 a 2 −1  t  

−ω  a − a 2 −1  t  

+e

+e

(

)

  −  aω + ω 2 a 2 −1  t  

−ω  a + a 2 −1  t  

 2π  Now, the value x   can be computed:  ω  

−ω  a −  2π  x =e    ω 

 2π x  ω

−2π  a −   =e 

2π a 2 −1  ω

a 2 −1  

+e

+e

2π −ω  a + a 2 −1   ω

−2π  a + a 2 −1   

b. For a = 2 ,

22 −1  

52 −1  

 2π x  ω

−2π  2 −   =e 

 2π x  ω

−2π  5 −   =e 

−2π  2 + 22 −1   

+e

= 0.1857

For a = 5,

+e

−2π  5 + 52 −1   

= 0.5301

For a = 10 , 

−2π  10 −  2π  x =e    ω 

102 −1  

+e

−2π  10 + 102 −1   

= 0.7298

PROBLEM 5.7 A toy car is connected to a spring and moving as a critically damped oscillator. The general solution for a critically damped oscillator is x(t ) = e −ω0 t (C1 + C2t ) , where ω0 is the natural frequency in the absence of any resistive forces. a. The toy is pulled at a position x0 from the equilibrium position, then the toy is released from rest. Find the equation of motion (position vs. time) and the velocity versus time. b. This time the system spring car is passing through the equilibrium position x0 = 0 with a speed v0 . In this case, find the equation of motion x(t ) and the velocity v(t ).

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SOLUTION 5.7 a. The general solution for a critically damped oscillator is x(t ) = e −ω0 t (C1 + C2t )

where C1 and C2 are to be determined from the initial conditions.

x0 = x(t = 0) = e0 (C1 + C2 ⋅ 0) = C1

C1 = x0 The velocity is obtained by,

v( t ) =

dx(t ) = x (t ) = −ω0e −ω0 t (C1 + C2t ) + C2e −ω0 t = e −ω0 t (−ω0C1 − ω0C2t + C2 ) dt

The initial velocity is, considering that C1 = x0

v0 = v(t = 0) = −ω0C1 + C2 = −ω0 x0 + C2 Since v0 = 0 , it yields that C2 = ω0 x0 . With the constants calculated, the equation of motion and the velocity become

x(t ) = e −ω0 t ( x0 + ω0 x0t ) = x0e −ω0 t (1 + ω0t )

v(t ) = e −ω0 t (−ω0 x0 − ω0 2 x0t + ω0 x0 ) = −ω0 2 x0te −ω0 t

b. General solution for a critically damped oscillator is x(t ) = e −ω0 t (C1 + C2t )

In this case, the initial conditions are x0 = 0 and v0

= x0 x= (t 0) = C1

C1 = 0 The velocity is, as calculated before,

v( t ) =

dx(t ) = e −ω0 t (−ω0C1 − ω0C2t + C2 ) dt

v0 = v(t = 0) = −ω0C1 + C2 = C2 C2 = v0

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Oscillations

Consequently, the equation of motion and the velocity are, in this case,

x(t ) = v0t e −ω0 t

v(t ) = v0 (1 − ω0t )e −ω0 t

PROBLEM 5.8 In a simple harmonic motion x(t ) = A cos(ω t − δ )

the motion is fully characterized by the amplitude A , the angular frequency ω, and the phase δ . If the amplitude and angular frequency are not known, they can be found by analyzing the motion and measuring the position and the velocity of the oscillator at two different moments. Find A and ω in two different ways. SOLUTION 5.8 a) First method, by using trigonometry, specifically the equation (sin α )2 + (cos α )2 = 1 . The position and the velocity are x(t ) = A cos(ω t − δ )

and

v(t ) = −ω A sin(ω t − δ )

For two different moments t1 and t2 , the position and the velocities are

x1 = A cos(ω t1 − δ )

v1 = −ω A sin(ω t1 − δ ) and

x2 = A cos(ω t2 − δ )

v2 = −ω A sin(ω t2 − δ ) By substituting sine and cosine from each moment and adding the squares,

x12 v2 + 21 2 = 1 2 A ω A

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Classical Mechanics

1  2 v12   x1 + 2  = 1 A2  ω 

A2 = x12 +

v12 ω2

Similarly, x22 v22 + =1 A2 ω 2 A2

A2 = x22 +

v22 ω2

From both equations, by transitivity, x12 +

v12 v2 = x22 + 22 2 ω ω

x12 − x22 =

v22 − v12 ω2

ω2 =

v22 − v12 x12 − x22

From here, substituting in one of the equations,

A2 = x12 +

2 2 v12 x12 v22 − x12 v12 + x12 v12 − x22 v12 x12 v22 − x22 v12 2 2 x1 − x2 = = x + v = 1 1 v22 − v12 v22 − v12 ω2 v22 − v12

b) Second method, using the conservation of total mechanical energy in the first state, at time t1 , characterized by position x1 and velocity v1 and state 2 at time t2 , characterized by position x2 and velocity v2

E = E1 = T1 + U1 =

m(v12 − v22 ) = k ( x22 − x12 ) Recalling that ω 2 =

mv12 kx12 mv22 kx22 + = E2 = T2 + U 2 = + 2 2 2 2

k it follows that m

ω2 =

v12 − v22 v22 − v12 = x22 − x12 x12 − x22

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Oscillations

Also, when x(t ) = A the potential energy is maximum, and the kinetic energy is zero.

E = E1 = E2 =

kA2 mv12 kx12 = + 2 2 2

m 2 v2 v2 ( x 2 − x 2 ) v1 + x12 = 12 + x12 = 1 2 2 2 1 + x12 ω k v1 − v2 2 2 2 2 2 2 2 2 v x − v x + v x − v2 x1 v12 x22 − v22 x12 x12 v22 − x22 v12 = = = 1 2 1 12 12 1 v1 − v2 v12 − v22 v22 − v12

A2 =

Both methods lead to the same results. The energy method is more physical, based on energies. However, the first method can also be employed. PROBLEM 5.9 Consider a block of mass m connected to a spring with a constant k on a ramp inclined at an angle θ (Figure 5.1). Find an expression for the spring length if the spring’s rest length is d . SOLUTION 5.9 Since the block is stationary, Newton’s Second Law is given by ∑F = 0

Considering the following free body diagram (Figure 5.2). It is clear the gravitational component pulling on the spring is Fg = mg sin θ

The spring length is thus given by L = d + x0 where x0 is the amount the spring stretches. Substituting everything into Newton’s Second Law yields mg sin θ − kx0 = 0

d k m

FIGURE 5.1  Block on a ramp connected to a spring.

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Classical Mechanics

kd m

mg mg sin

FIGURE 5.2  Free body diagram of the block.

x0 =

mg sin θ k

Therefore, the spring’s length is given by

L=d+

mg sin θ k

PROBLEM 5.10 Consider the mass m connected to two springs above and below with spring constants k1 and k2 , respectively, as in Figure 5.3. Find the angular velocity ω after a small displacement in the y direction.

k1

m

k2

FIGURE 5.3  Mass connected to a spring above and a spring below.

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Oscillations

SOLUTION 5.10 To find the mass’s acceleration, consider Newton’s Second Law ∑ F = my

Considering the mass is displaced by an amount y downward, this becomes k1 y + k2 y − mg = my

y=

k1 + k2 y−g m

Noting that y < 0

y=−

k1 + k2 ｜ y ｜− g m

and therefore, the angular velocity is

ω=

k1 + k2 m

It is important to note that assuming the displacement was downward does not impact the solution. If instead the displacement was upward, Newton’s Second Law yields

−k1 y − k2 y − mg = my  y=−

k1 + k2 y−g m

Since y > 0 here, the absolute value of y is not necessary and the angular velocity is exactly what was found assuming the opposite displacement. PROBLEM 5.11 Consider the mass–spring system in Figure 5.4 consisting of two springs with constants k and a mass m. When the mass is in line with the springs, they are at their rest length l0 . The mass is then gently allowed to descend (to prevent oscillations) until

k

m

k

FIGURE 5.4  Mass connected to a spring to the left and a spring to the right.

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Classical Mechanics

the system is in equilibrium. Find the amount the springs stretch, assuming the final position is a small angle away from the initial position. SOLUTION 5.11 After the mass is allowed to come to rest, the system is in the position as shown in Figure 5.5. The amount the springs stretch can be expressed as ∅l , which is illustrated in Figure 5.6. From this triangle, cos θ =

l0 ∆l + l0

To find an expression for θ , Newton’s Second Law can be used, considering the mass is stationary ∑ Fy = 0

k

k

m

FIGURE 5.5  Mass connected to springs on the left and the right in its final position.

l0

Δl

l0

FIGURE 5.6  Triangle illustrating the equilibrium lengths of the system.

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Oscillations

Therefore, 2k ∆l sin θ − mg = 0

where k ∆l sin θ is the amount of force one spring acts on the mass, in the y direction. Assuming small angles, these equations become 1−

θ2 l0 = 2 ∆l + l0

and 2k ∆lθ = mg

Since ∅l is the quantity of interest, the second equation can be rewritten as

θ=

mg 2k ∆l

Substituting this into the first equation yields 1−

8k 2 (∆l )2 − m 2 g 2 =

m2g2 l = 0 8k 2 (∆l )2 ∆l + l0 8k 2 (∆l )2 l0 ∆l + l0

8k 2 (∆l )3 + 8k 2 (∆l )2 l0 − m 2 g 2 ∆l − m 2 g 2l0 = 8k 2 (∆l )2 l0 8k 2 (∆l )3 − m 2 g 2 ∆l − m 2 g 2l0 = 0 (∆l )3 −

m2g2 m2 g2 ∆ l − l0 = 0 8k 2 8k 2

This is a cubic which has the solution 2

∆l =

3

 m 2 g 2l0   m 2 g 2  m 2 g 2l0 +  +− 2 2  2  16k  16k   24k  2

+3

3

 m 2 g 2l0   m 2 g 2  m 2 g 2l0 −  +− 2 2  2  16kk  16k   24k 

3

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Classical Mechanics

l

k

m

FIGURE 5.7  Mass connected to a pendulum and a spring.

PROBLEM 5.12 Consider a mass m attached to a pendulum of length l and a spring of constant k as depicted in Figure 5.7. When the mass is at the lowest point of the pendulum, the spring is at its rest length. Considering a small displacement, find the angular velocity. SOLUTION 5.12 When the mass is displaced, gravity is pulling the mass down toward the equilibrium pendulum position and the spring is pushing/pulling it toward its equilibrium position. If the mass is displaced by an angle φ , the following restoring forces are at play

Fφ = mg sin φ

Fs = kl tan φ

Considering small angles, the φˆ forces due to gravity and the spring are approximately in the same direction. The forces also become

mg sin φ → mgφ

kl tan φ → klφ Newton’s Second Law is then given by

∑ Fφ = mlφ

−klφ − mgφ = mlφ

kl + mg φ = − φ ml

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Oscillations

and the angular velocity is kl + mg ml

ω2 =

kl + mg ml

ω=

PROBLEM 5.13 A small cylinder of mass m is sliding without friction along a thin fixed rod (Figure 5.8). If the spring of spring constant k is elongated a distance h by the force F0 , find the frequency of the oscillations performed by the cylinder. Consider the case of small oscillations. SOLUTION 5.13 The spring elongation is

δ = h2 + x 2 − h

For h  x

2

 x2  x2 x δ = h 1 +   − h = h 1 + 2 − 1 = h  2h  2h Potential energy can be approximated with

U = F0δ = F0

x2 2h

On another side, F0 = kh and recalling that ω 2 = that the frequency of the small oscillations is

ω=

k and substituting k , it yields m

F0 mh

h x m

0

FIGURE 5.8  Small cylinder of mass m oscillating on a thin rod.

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Classical Mechanics

h

M

L m

FIGURE 5.9  Mass m changing position on a rod of mass M fixed on one side and oscillating in a vertical plane.

PROBLEM 5.14 A uniform branch of mass M and length L moves like a simple pendulum in the wind. A ladybug of mass m is climbing the branch and stops to rest in different positions h from the top of the branch (Figure 5.9). Find the angular frequency, the period of the small oscillations, and the value of h for which we obtain the smallest period. What is the minimum period? SOLUTION 5.14 The kinetic energy of the system branch (mass M ) and ladybug (mass m) is T=

I ω 2 mv 2 + 2 2

The moment of inertia I of a thin rod of mass M and length L about the axis ML2 through one end perpendicular to the rod is I = and the angular velocity 3 dφ  ω= = φ . The kinetic energy becomes dt ML2 2 φ  Iω mv mh2φ2 1  ML2 + = 3 + T= =  + mh2  φ2 2 2 2 2 2 3  2

2

The potential energy is, considering that the center of the mass for the uniform rod is at mid-­distance,

U=

M  ML  gL (1 − cos φ ) + mgh(1 − cos φ ) =  + mh  g(1 − cos φ ) 2 2  

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Oscillations

Equilibrium is obtained by deriving the potential energy by the variable φ dU  ML  = + mh  g sin φ dφ  2 

Equilibrium condition

dU = 0 leads to sin φ = 0 , therefore φ = 0 is the equilibdφ

rium position. Stable equilibrium near φ = 0 , and using the approximation of the small oscillaφ2 tions, cos φ ≅ 1 − 2

φ2  ML   ML   + mh  g(1 − cos φ ) ≅  + mh  g  1 − 1 + U = 2  2   2  

 d 2U  ML   ML   φ2 = + mh  g cos φ ≅  + mh  g  1 − +   2 dφ 2 2 2       Or, keeping just the first term,

d 2U  ML  = + mh  g . dφ 2  2 

Back to the kinetic energy, T=

 1  ML2 1 + mh2  φ2 = f φ2  2 3 2 

where  ML2  f = + mh2   3 

So

ω2 =

1  d 2U    f  dφ 2 φ = 0

 ML   2 + mh  g  = ML + 2mh 3g ω2 =  2  ML ML2 + 3mh2 2 2 mh +     3

 1  ML  + mh  gφ 2 =  2 2   

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Classical Mechanics

ω=

ML + 2mh 3g ML2 + 3mh2 2

The period of small oscillations is τ with ω =

τ=

2π and τ

ML2 + 3mh2 2 2π = 2π ω ML + 2mh 3g

The period is minimum for

dτ =0 dh

dτ 2 1  ML2 + 3mh2  = 2π   dh 3g 2  ML + 2mh 

1 2

6mh( ML + 2mh) − ( ML2 + 3mh2 )2m    ( ML + 2mh)2

2 3g

ML + 2mh 2m(3hML + 6mh2 − ML2 − 3mh2 ) ML2 + 3mh2 ( ML + 2mh)2

2 3g

1 2m(3hML + 3mh2 − ML2 ) 1 2 1 ML + 3mh ( ML + 2mh) 2

2

The condition for minimum period is

dτ =0 dh

3mh2 + 3hML − ML2 = 0 After dividing by 3m , the equation is h2 + h

M 2 M L− L =0 3m m

From here, the minimum length (choosing the positive solution) is − hmin =

M L+ m

M 2 2 4 ML2 L + m2 3m = ML  4m + 1 − 1    2 2m  3M 

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Oscillations

The period of small oscillations for the minimum height hmin is

τ min =

2 ML2 + 3mhmin 2π 2 = 2π ω ML + 2mhmin 3g

And after extensive calculations,

 4m  ML h τ min = 2π  + 1 − 1 ≅ 2π min  2mg 3 M g  

PROBLEM 5.15 In damped one-­dimensional oscillations, the object of mass m is moving under the influence of Hooke’s law force −kx and under the influence of the resistive force f = −bv = −bx

leading to Newton’s Second Law equation of the following form: mx + bx + kx = 0

Considering an RLC (resistor–inductor–capacitor) circuit in series in Figure 5.10, write Kirchhoff’s loop equation and, by comparison, find which quantity plays the role of the mass, of the resistive constant b , and of spring constant k . Also, find the resonance frequency ω0 . SOLUTION 5.15 The circuit does not have a battery, so the total voltage on the resistor, on the capacitor, and on the inductance should be zero, VR + VL + VC = 0. L

R

C

I(t)

FIGURE 5.10  A resistor–inductor–capacitor (RLC) circuit in series with no power supply.

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Classical Mechanics

Recalling that the electric current is the derivative of electric charge with respect to time, = I (t )

dq(t ) = q (t ), V= RI = Rq R dt

The voltage on the inductor is dI (t ) d 2 q (t ) = VL L= L = Lq dt dt 2

And the voltage on the capacitor is = VL

q (t ) q = C C

Finally, the sum becomes

Rq + Lq +

q =0 C

Lq + Rq +

q =0 C

or, rearranged

By comparison with the damped oscillator mx + bx + kx = 0, we can identify in the role of mass is the impedance L , in the role of the resistive coefficient b , the resistance R , and in the role of spring constant, the inverse of the capacitance C . b R In mechanics = 2 β , where β is the damping constant, in RLC circuit = 2 β . m L k Recalling that the natural frequency is ω0 = . It follows that for the RLC circuit, m 1 by equivalence, ω0 = , which is the resonance frequency, in the case when the LC capacitive reactance is equal to the inductive reactance

X L = XC

ω0 L =

1 ω0C

ω02 =

1 LC

which is indeed the resonance frequency.

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Oscillations

PROBLEM 5.16 Suppose there exists a spring network of n springs – this looks very similar to networks of resistors and/or capacitors. Considering this, perhaps there is an analogy between springs and circuit elements. Taking this n springs in parallel (Figure 5.11) and in series (Figure 5.12), determine if such an analogy exists. Explain why, intuitively, this makes sense. Assume all springs have the same constant k and rest length l . SOLUTION 5.16 Taking the n springs in parallel first, consider the force required to displace the end of all springs is F = kx + kx + + kx = (nk ) x = Kx

n

k

k

k

k

FIGURE 5.11  Network of parallel springs.

k

k

k

FIGURE 5.12  Network of series springs.

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Classical Mechanics

Therefore, in parallel, an “equivalent spring” would have constant K = nk . Considering the series case next, the amount of force required to displace the end of x the last sprint by x requires all the springs in the chain to extend . The force is then n the amount the last spring in the chain pulls back, specifically

F=k

x n

In an attempt to relate this back to circuit elements and finding an equivalent spring constant, this can be expressed as

k 1 1 = = n 1 1 1 n + ++ k  k  k  k  n

Therefore, an equivalent spring constant is given by

1 1 1 1 = + ++ K  k  k  k  n

Considering these two configurations, it appears springs in parallel behave like capacitors in parallel and springs in series behave like capacitors in series. Intuitively, this makes sense; both springs and capacitor “store” energy so when there are connected in similar ways, it makes sense that they have similar “equivalent element” equations.

6

Lagrangian Formalism

6.1  THEORY This chapters presents the Lagrangian formalism. While this formalism does not appear very useful in solving simple problems, the difficult problems can be solved elegantly in this manner.

6.1.1 The Lagrangian The Lagrangian is   T U

where T is the kinetic energy of the system and U is the potential energy of the system. For a one-­dimensional system with x as generalized coordinate, the Lagrange equation is  d   dx dt dx

6.1.2 Hamilton’s Principle The actual path followed by a particle between point 1, reached at time t1 , and point 2, reached at time t2 , is such that the action integral t2

S   dt

t1

is stationary when taken along the actual path. For a holonomic system of n generalized coordinates q1,…, qn , the Lagrangian is obtained as

(q1,, qn , q1,, q n , t )  T  U

A holonomic system has n degrees of freedom and it can be fully described by exactly n generalized coordinates q1,…, qn .

DOI: 10.1201/9781003365709-6

143

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Classical Mechanics

The condition that the action integral S t2

S  (q1,, qn , q1,, q n , t )dt

t1

is stationary, implies that the n Lagrange–Euler equations hold true,

 d   dq1 dt dq1

 d   dq2 dt dq2

 d   dqn dt dq n

Or, simply called Lagrange equations, with i from 1 to n ,  d   dqi dt dqi

6.2  PROBLEMS AND SOLUTIONS PROBLEM 6.1 Consider two objects of masses m1 and m2 suspended by an inextensible massless string of length l which passes over a massless frictionless pulley of radius R , as in Figure 6.1. a. Obtain the Lagrangian. b. Write the Lagrange equations and obtain the acceleration. c. Compare with the method using Newton’s Second Law. SOLUTION 6.1 a. Consider the distance x of the mass m1 to the level of the center of the pulley as the generalized coordinate. Since the string is inextensible, of length l , the position y of the mass m2 depends on x as y  l   R  x   x  constant .

y

x

m2

FIGURE 6.1  Atwood machine with two masses.

m1

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Lagrangian Formalism

The speeds of the two objects are also interdependent: y   x

The kinetic energy is

T

m1 2 m2 2 (m1  m2 ) 2 x  y  x 2 2 2

The potential energy is U  m1gx  m2 gy  m1gx  m2 g( x  const.)  (m1  m2 )gx  const. The Lagrangian is   T U

(m1  m2 ) 2 x  (m1  m2 )gx 2

b. There is only one generalized coordinate for the generalized coordinate x  d   dx dt dx

By taking the partial derivatives, it yields that

  (m1  m2 )g dx

  (m1  m2 ) x dx

And the derivative with respect to time is

d  d  (m1  m2 ) x  (m1  m2 )  x dt dx dt

The acceleration is easily obtained as expected:

 x

(m1  m2 ) g (m1  m2 )

Note that the Lagrangian formalism allowed solving the problem without considering the tension in the string, which appears in the introductory physics course when using Newton’s Second Law approach.

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Ft Ft m2g

m1g

FIGURE 6.2  Atwood machine with two masses and inextensible string, with gravitational forces and the tensions in the string.

c. Using Newton’s Second Law applied to each of the two objects (Figure 6.2), and by considering the string inextensible, and the fact that the two objects move with the same magnitude of acceleration, it yields

m1a  m1g  Ft

m2 a  m2 g  Ft Adding the two equations, the tension in the string Ft is eliminated and the acceleration is obtained as before

a

(m1  m2 ) g (m1  m2 )

It is always good to check the solution for particular cases. For m1 = m2 , the acceleration is zero. Note that, while acceleration may be zero for equal masses, the system may move with constant speed if a small nudge is applied to the masses to start moving.

PROBLEM 6.2 Consider two objects of masses m1 and m2 suspended by an inextensible massless string of length l which passes over a frictionless pulley of mass M and radius R , as in Figure 6.3. a. Obtain the Lagrangian. b. Write the Lagrange equations and obtain the acceleration.

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Lagrangian Formalism

M

y

x

m2

m1

FIGURE 6.3  Atwood machine with two masses and a pulley of mass M .

SOLUTION 6.2 a. The kinetic energy must include the rotational energy due to the pulley of MR 2 v x mass M and moment of inertia I = , and    2 R R m m I  2 (m1  m2 ) 2 I  2 (m1  m2 ) 2 Mx 2  x   x  T  1 x 2  2 y 2  2 2 2 2 2 2 4 The potential energy is the same U  m1gx  m2 gy  m1gx  m2 g( x  const.)  (m1  m2 )gx  const. The Lagrangian is

M   m1  m2  2    x 2  (m  m )gx  const   T U  1 2 2

b. There is only one generalized coordinate for the generalized coordinate x  d   dx dt dx By taking the partial derivatives, it yields that

  (m1  m2 )g dx

  M  m1  m2   x dx  2 

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And the derivative with respect to time is

d  d  M M  m1  m2   x   m1  m2    x  dt dx dt  2  2  

The acceleration is easily obtained as  x

(m1  m2 ) g M   m1  m2  2   

M at the denomina2 tor, diminishing the acceleration comparing the case with a massless pulley, because some of the energy is used to accelerate the pulley. Note that the disk of mass M is bringing a term of

PROBLEM 6.3 Consider a block of mass m sliding down a frictionless ramp at an incline θ as in Figure 6.4. Find the velocity of the block at time t if the block is stationary at t = 0 . Note: This is the same problem text as Problem 1.4 but now should be solved using Lagrange’s equations. SOLUTION 6.3 Considering the x -axis is parallel to the ramp, the kinetic energy is given by

T=

1 2 mx 2

Without worrying about the true height of the ramp, the potential energy is given by

U  mg(const  x sin  )

θ

FIGURE 6.4  Block placed on a ramp.

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Lagrangian Formalism

Therefore, the Lagrangian is

  T U

1 2 mx  mg(const  x sin  ) 2

Using Lagrange’s equations yields d    dt dx dx

d (mx )  mg sin  dt

 x  g sin

The velocity is then given by

x  gt sin   C

where C is a constant. Since the block is at rest at the top of the ramp,

x (0) = 0

C=0

and the velocity as a function of time is x  gt sin

which is exactly what was found in previous chapters. PROBLEM 6.4 A mass m1 is on an incline attached to a massless string. Another mass m2 is attached to the other end of the string, hanging vertically, as in Figure 6.5. Find the Lagrangian and obtain the equation of motion. SOLUTION 6.4 The Lagrangian is equal to the potential energy subtracted from kinetic energy   T U

The kinetic energy of the system is

T

1 1 m1 x 2  m2 y 2 2 2

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m1

m2

θ

FIGURE 6.5  Masses on incline connected by a string.

x m1 y

m2

θ

FIGURE 6.6  Masses on incline connected by a string.

The potential energy of the system is

U  m2 gy  m1gx sin    m2 gx  m1gx sin   const Using these two equations, the Lagrangian is



1 (m1  m2 ) x 2  (m1 sin   m2 )gx  const 2

Now, the equation of motion can be obtained:

 d   x dt x (m1 sin   m2 )g  (m1  m2 )  x  x

m1 sin   m2 g m1  m2

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Lagrangian Formalism

PROBLEM 6.5 A mass m is attached to a spring of constant k , horizontally on a frictionless plane, as in Figure 6.7.

x. a. Find the Lagrangian and solve for  x. b. Using Newton’s Second Law, solve for 

SOLUTION 6.5 a. The kinetic energy and potential energy of the system are 1 1 T = mx 2 and U = kx 2 2 2 Subtracting potential energy from kinetic energy outputs the Lagrangian:   T U

Using the relation

1 2 1 2 mx  kx 2 2

 d  x is obtained.  , an expression for  x dt x kx

d (mx ) dt

kx  mx  x

k x m

b. According to Newton’s Second Law, Fx = mx . On the block, only the force from the spring is acting in the x direction. Thus, Fx  kx So: kx  mx  x

k x m

The result is the same as the one obtained in part (a).

xe

xs

FIGURE 6.7  Mass attached to a spring extended to point xs . xe is the equilibrium point of the spring.

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PROBLEM 6.6 A particle with initial mass m0 and initial velocity v0 begins losing mass according to the equation m(t )  m0e  t where α is constant. If there are no external forces, find an expression for the velocity. Note: This is the same problem text as Problem 1.7, but now should be solved using Lagrange’s equations. SOLUTION 6.6 Since there is no potential energy in this problem

T

1 1 m(t ) x 2  m0e  t x 2 2 2

Using Lagrange’s equations yields d    dt dx dx

m0

d  t (e x )  0 dt

m0 (  xe  t   e  t x )  0

m0e  t (  x   x )  0

 x   x  0 This is a separable differential equation.

dx   x dt

dx   dt x

ln x   t  C

x  Ae t

where C and A are constants. Since the initial velocity is v0 ,

x (0= ) v= A 0

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Lagrangian Formalism

Therefore, the velocity is given by x (t )  v0e t

which is exactly what was found in Chapter 1. PROBLEM 6.7 A small bead of mass m can slide without friction on a parabolic wire of equation z = ax 2 rotating with angular velocity ω , as in Figure 6.8. a. Find the velocity of the bead. b. Find the Lagrangian. c. Use Lagrange equation to find the equation of motion. d. Discuss the equilibrium position. SOLUTION 6.7 a. The cylindrical polar coordinates are the most appropriate in this case, the parabola rotating around the z-­axis with the angular velocity    . Consider the equation of the parabola as z = ax 2 . The position of the bead is ( x, 0, ax 2 ) and the velocity is calculated following the steps:

z = ax 2

z = 2axx

v 2  x 2  ( x )2  z 2  x 2  x 2 2  (2axx )2  x 2  x 2 2  4a 2 x 2 x 2 b. The Lagrangian is   T U

m 2 ( x  x 2 2  4a 2 x 2 x 2 )  mgax 2 2

z ω

m x 0

FIGURE 6.8  A bead is sliding without friction on a parabolic wire rotating about the vertical axis with angular velocity ω .

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c. Lagrange equation for x

 d   dx dt dx

  mx 2  4ma 2 xx 2  2mgax dx

  mx  4ma 2 x 2 x  dx d     x 2   mx  4ma 2 (2 xxx x )  mx  4ma 2 (2 xx 2  x 2  x)  dt dx

And by considering both terms of the Lagrangian equation, it yields

mx 2  4ma 2 xx 2  2mgax  mx  4ma 2 (2 xx 2  x 2  x)

And with a bit of rearrangement and after dividing by mass,

 x  4a 2 x 2  x  4a 2 xx 2  x 2  2 gax

From here, the equation of motion is  x(1  4a 2 x 2 )  4a 2 xx 2  x( 2  2 ga)

d. If the bead positions itself at height H = ax 2 and x is an equilibrium position, x = 0. it means that x = 0 and  The equation of motion yields

x( 2  2 ga)  0

This can happen if x = 0 , at the apex of the parabola, or when  2  2 ga  0 , which means that the angular velocity is   2ga . In this particular case, the bead can position itself at any height z , and these constitute unstable positions of equilibrium. Looking at the case when the bead is at x = 0 and checking the stability of the bead, if we consider a position ε very small, close to zero, and ε very small as well, with  2  0 , then the equation of motion becomes

 x  x( 2  2 ga)

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Lagrangian Formalism

For angular velocities ω smaller than 2ga , the bead may go up a bit and then slide back toward the equilibrium position x = 0. For high angular velocities, that is,   2ga , the bead may go further and further from the equilibrium position. PROBLEM 6.8 A mass m is moving on a frictionless sphere of radius R . The force on the sphere is F  mg r . Find the Lagrangian and solve in polar coordinates. SOLUTION 6.8 1 The kinetic energy of the mass is T = mv 2 , which converted in polar coordinates 2 is written as:

T

1 2 1 mv  m((r sin  )2  (r)2 ) 2 2

The potential energy of the mass is U = mgr . The Lagrangian,   T  U , is obtained, after substituting r by R (the surface of the sphere).



1 m( R 22 sin 2   R 2 2 )  mgR 2

Solving for φ is first done by using the relation  d    dt 

so

0

d (mR 2 sin 2  ) dt

0  mR 2 sin 2 

and   0 . Now, an expression for θ is obtained the same way.

 d    dt 

mR 22 cos  sin   mR 2

  2 cos  sin

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PROBLEM 6.9 A mass m is suspended vertically on a spring of constant k , as in Figure 6.9. Find x. the Lagrangian and solve for  SOLUTION 6.9 The sum of the forces acting on the mass is F  mg  kx

The potential energy is U   F dx . So: U  mgx

1 2 kx 2

The kinetic energy of the mass is T=

1 2 mx 2

Thus, the Lagrangian is

  T U

1 2 1 mx  mgx  kx 2 2 2

x is straightforward since Using the Lagrangian, solving for 

mg  kx

d (mx ) dt

 x  g

k x m

X

m

FIGURE 6.9  Mass hanging vertically, attached to a spring.

 d   . x dt x

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Lagrangian Formalism

PROBLEM 6.10 Consider a bead of mass m on the spinning wire as in Figure 6.10. If the wire is in 2 1 2 the shape of an upside-­down Gaussian, specifically f (  )   e , find any 2 equilibrium positions and comment on their stability. SOLUTION 6.10 This type of analysis can be performed using Lagrange’s equations which means the kinetic energy T and potential energy U must be found. This requires the position of the bead, which is given by  1  r     f (  ) z     e 2

2 2 z

with the velocity given by 2

1 v   ˆ   e 2 z   2

Thus, the kinetic energy is

T

1 2 1  2 1 2 2 2  mv  m      e   2 2  2 2  2 

The potential energy is given by

2  1 1 2 U  mg ｜ f (0) ｜ ｜ f (  ) ｜  mg   e  2 2 

f(ρ) ρ

ω

FIGURE 6.10  Bead on a spinning wire.

m

2  mg    1  e 2  2  

   

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The Lagrangian is therefore

2  1  2 1 2 2 2 mg  2 2      T U  m     e     1 e 2 2  2 2  

   

which will be used in Lagrange’s equations d    dt d  d

Now, to calculate each element

2   1  1 1 2 2    e   m  2    2  e      m  1  d  2     2 

d  1 2 2  d 1 2 2    e    m  1   e   m  1  dt d  dt  2  2   2 2 1 2 2    1   m  1   e    m  (2  e     2e   (2  ))   2   2 

1 2   2   2 m   2   m  1   e  e (1   2 )  2    2 2  1  1 2  mg    m e  (2  e     2e    2  )  2  2   d  2  2 2 

2 2 1 1 2 mg    3e   )  2  2   m   (  e     e 2  2 

2 2

2 2

Combining everything yields 1 2   2   2 m   2  m  1   e  e (1   2 )   2 

2 2 1 1  mg   e  m   2 (  e     3e   )  2  2   2  2 

2 2

Since the goal is to analyze the stability, the bead cannot be moving in the ρ direction; therefore,

  0,   0

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Lagrangian Formalism

and the equation reduces to m  2

mg   e 2

2 2

2  mg  2 2   m  e  2 

0

 0  

From this, it is clear   0 is an equilibrium position. Intuitively, this makes sense; when the mass is at   0 , it doesn’t experience any force due to the rotation. Considering equilibrium when   0 yields m 2

mg  e 2

2 2

0

Solving for ω

g  e 2

2 4

Due to the shape of the wire, the angular velocity must depend on the position. For ρ close to zero,

g 2

As ρ increases, the slope of wire decreases (exponentially); so, in order to keep the bead from moving, the velocity must decrease (exponentially). Very far from   0 , where the wire is very flat, an extremely small ω is required. PROBLEM 6.11 Consider a mass m attached to a pendulum of length l and a spring of constant k as depicted in Figure 6.11. When the mass is at the lowest point of the pendulum, the spring is at its rest length. Considering a small displacement, find the angular velocity. Note: This is the same problem text as Problem 5.12, but now should be solved using Lagrange’s equations. SOLUTION 6.11 While it may be clear at this point that the velocity of a mass on a pendulum is lφ but to get more practice in deriving it (for cases which aren’t so easy), consider the position

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l

k m

FIGURE 6.11  Mass connected to a pendulum and a spring.

r  l sin  x  l cos  y  l (sin  x  cos  y )

with velocity

v  l (cos   x  sin   y )

so

T

1 2 1   1 2 2 mv  mv  v  ml  2 2 2

The potential energy is due to a combination of gravity and the spring U  mgl (1  cos  )

1 k (l tan  )2 2

Considering small angles, this becomes U

mgl 2 1 2 2   kl  2 2

Therefore, the Lagrangian is given by

  T U

1 2 2 mgl 2 1 2 2 1 2 2 l ml     kl   ml   (mg  kl ) 2 2 2 2 2 2

Using Lagrange’s equations yields

d    dt 

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Lagrangian Formalism

ml 2  l (mg  kl )

mg  kl    ml Therefore, the angular velocity is

mg  kl ml

PROBLEM 6.12 Consider a block of mass m on ramp at an incline θ and connected to a spring of constant k , as shown in Figure 6.12. Using Lagrange’s equations, find the angular frequency of oscillations following a displacement of the block. SOLUTION 6.12 Considering the x -axis is parallel to the ramp, the kinetic energy is given by T=

1 2 mx 2

Without worrying about the true height of the ramp, the potential energy is given by

U

1 k ( x  d )2  mg(const  x sin  ) 2

where d is the “pre-­stretched” displacement of the spring due to the mass but not due to the oscillations. This quantity is found by considering Figure 6.13 Therefore, kd  mg sin

k

X m

FIGURE 6.12  Block on a ramp connected to a spring.

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kd m

mg mg sin

FIGURE 6.13  Free body diagram of the block.

The Lagrangian is given by

  T U

1 2 1 mx  k ( x  d )2  mg(const  x sin  ) 2 2

Using Lagrange’s equations yields

d    dt dx dx

mx  (k ( x  d )  mg sin  )

mx  (kx  kd  mg sin  ) Notice kd  mg sin   0, so this reduces to

 x

k x m

k m

and the angular velocity is

An interesting note: The angular frequency of oscillation does not depend on θ at all. PROBLEM 6.13 Consider the pendulum–spring system from Figure 6.14 consisting of a mass m connected to a pendulum of length l and a spring with constant k and rest length l . Assuming a small angular displacement, find the angular frequency of the oscillations.

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Lagrangian Formalism

l

m

k

FIGURE 6.14  Mass connected to a pendulum above and a spring below.

SOLUTION 6.13 This can be solved using Lagrange’s equations, with kinetic energy given by T

1 2 2 ml  2

In order to find the potential energy associated with the spring, consider Figures 6.15 and 6.16, where d is the amount the spring stretches. Using the law of cosines yields

(l  d )2  (2l )2  l 2  2(2l )(l ) cos

(l  d )2  4l 2  l 2  4l 2 cos

(l  d )2  l 2 (1  4(1  cos  ))

l

m l

k

FIGURE 6.15  Projection of the mass, pendulum, and spring upon displacement.

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l

2l l+d

FIGURE 6.16  Triangle illustrating distances involved in the displacement.

l  d  l 1  4(1  cos  )

d l



1  4(1  cos  )  1

Therefore, the potential energy is given by 1 U  mgl (1  cos )  kl 2 2



1  4(1  cos  )  1

2

and the Lagrangian is

  T U

1 2 2 1 ml   mgl (1  cos  )  kl 2 ( 1  4(1  cos  )  1)2 2 2

Considering a small angle approximation 1  cos

2 2

so 

1 2 2 mgl 2 1 2 ml     kl ( 1  2 2  1)2 2 2 2

Also notice

1  2 2  1  1   2  1   2

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Lagrangian Formalism

Therefore,



1 2 2 mgl 2 1 2 4 ml     kl  2 2 2

Since φ  1, φ 4  φ 2 and, provided the spring constant isn’t massive, the last term in the Lagrangian can be omitted; essentially, the spring does not stretch enough in the small angle approximation to affect the angular frequency. The Lagrangian then becomes



1 2 2 mgl 2 ml    2 2

Using Lagrange’s equations yields

d    dt 

ml 2  lmg

g    l

and the angular frequency of oscillation is

g l

Note: In the case of a massive spring constants, it is possible φ 2 and kφ 4 are of similar order, and the last term in the Lagrangian could no longer be omitted. This would have resulted in the differential equation

ml 2  lmg  2kl 2 3

which is a nonlinear differential equation and could be solved numerically.

7

Hamiltonian Formalism

7.1  THEORY This chapter presents the Hamiltonian formalism. Many problems solved in Chapter 6 using Lagrangian formalism are solved here by using the Hamiltonian formalism. It is important to note that the Hamiltonian formalism is based on Lagrangian formalism.

7.1.1 The Hamiltonian The Lagrangian is obtained as

(q1,, qn , q1,, q n , t )  T  U The Hamiltonian is defined as n

H

p q  L i i

i 1

where the generalized momenta are pi

 qi

The next step is to express the Hamiltonian function as a function of the variables qi and pi . For this, the generalized velocities qi are written in terms of the coordinates qi and pi and the Hamiltonian is written in terms of qi and pi , that is, (q1,…, qn , p1,…, pn , t ) . Hamilton’s equations are, with i from 1 to n , qi

 pi

p i

 qi

7.1.2 Example – One-Dimensional Systems

  (q, q )  T  U

DOI: 10.1201/9781003365709-7

167

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The generalized momentum is p

 q

and it is substituted in the Hamiltonian, H  pq  L

Hamilton’s equations for a one-­dimensional system are q

 p

and p

 q

7.2  PROBLEMS AND SOLUTIONS PROBLEM 7.1 A mass m is moving in xyz coordinates. A force F is acting on the mass. For the following cases, get the Hamiltonian and find Hamilton’s equations.  a. F = ax x  b. F  ay x  ax y  c. F  a x  by y  cz z SOLUTION 7.1 a. First, the Lagrangian is determined:

1 U   F d x   ax 2 2

  T U

1 1 m( x 2  y 2  z 2 )  ax 2 2 2

The momentum in each direction is obtained:

px

L  mx x

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py

L  my y

pz

L  mz z

Using the expressions of momentum and the Lagrangian, the Hamiltonian is

  px x  py y  pz z

px2 py2 pz2 1  px2 py2 pz2  1 2    m     ax m m m 2  m2 m2 m2  2

1 1 ( px2  py2  pz2 )  ax 2 2m 2

1 1 m( x 2  y 2  z 2 )  ax 2 2 2

Deriving the Hamiltonian leads to Hamilton’s equations.

   p  x   ax x  p x  px m x

p    0 y   y p y  y py m

p    z  p z  0  z pz m z

b. The process is the same as in part (a). First, the Lagrangian is determined: U  axy

Checking that the potential energy U is correctly calculated can be achieved by taking its gradient:

U  U  U  F  U   x y z  ay x  ax y x y z 

1 m ( x 2  y 2  z 2 )  axy 2

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The momentum in each direction is obtained:

px = mx

py = my

pz = mz Using the expressions of momentum and the Lagrangian, the Hamiltonian is 

1 ( px2  py2  pz2 )  axy 2m

Deriving the Hamiltonian leads to Hamilton’s equations:

= x

px = p x ay m

= y

py = p y ax m

z =

pz p z 0 = m

c. The process is the same as in parts (a) and (b). First, the Lagrangian is determined: U  ax  

1 2 1 2 by  cz 2 2

1 1 1 m ( x 2  y 2  z 2 )  ax  by 2  cz 2 2 2 2

The momentum in each direction is obtained:

px = mx

py = my

pz = mz Using the expressions of momentum and the Lagrangian, the Hamiltonian is



1 1 1 ( px2  py2  pz2 )  ax  by 2  cz 2 2m 2 2

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Hamiltonian Formalism

Deriving the Hamiltonian leads to Hamilton’s equations: px m py y  m pz z  m x

p x  a p y  by p x  cz

PROBLEM 7.2 A mass is attached to a spring of constant k , horizontally on a frictionless plane a. Find the Hamiltonian and Hamilton’s equations. b. Find the equation of motion. SOLUTION 7.2 a. The Lagrangian   T U

1 2 1 2 mx  kx 2 2

is used to find the Hamiltonian: px = mx



px2 1 2  kx 2m 2

Now, Hamilton’s equations can be solved by derivation:

x

  kx x

p x

 px  px m

b. From Hamilton’s equations obtained in part (a), the momentum of the mass is px = mx . After derivation,

p x = mx

Again, from results obtained in part (a), px  kx . So

mx  kx

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PROBLEM 7.3 Consider the Atwood machine in Figure 7.1: two masses m1 and m2 suspended by an inextensible string over a frictionless pulley. a. Obtain the Lagrangian. b. Find the generalized momentum by using p

 and obtain the Hamiltonian x

H  pq  L for one degree of freedom.   c. Use Hamilton’s equations x  and p   to obtain the acceleration of x p the system.

Note: This is the same problem as 6.1, but now it is solved using Hamilton’s equations. SOLUTION 7.3 a. As in the similar problem from Chapter 6 (Problem 6.1), the length of the string is constant, so x  y   R  constant , so y   x   R , and y   x and y 2 = x 2 . The kinetic energy is T

m1  m2 2 x 2

The potential energy is

U  m1gx  m2 gy  (m1  m2 )gx  constant

The Lagrangian is

  T U

m1  m2 2 x  (m1  m2 )gx 2

R

y

x

m1

FIGURE 7.1  Atwood machine with two masses.

m2

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Hamiltonian Formalism

b. The generalized momentum is   (m1  m2 ) x x

p

The next step is to write x in terms of the generalized momentum: x

p (m1  m2 )

The Lagrangian in terms of x and p becomes 2

p  (m1  m2 )   m  m   (m1  m2 )gx 2 2   1 p2   (m1  m2 )gx 2(m1  m2 )

( x, p)

The Hamiltonian is obtained by using the expression H  pq  L .  p

p p2   (m1  m2 )gx m1  m2 2(m1  m2 )

p2  (m1  m2 )gx 2(m1  m2 )

c. Hamilton’s equations are

x

p

 p  (7.1) p m1  m2   (m1  m2 )g (7.2) x

By taking the derivative of x with respect to time in the first Hamilton Equation (7.1) and by substituting p from the second Hamilton Equation (7.2), the acceleration is obtained

 x

p (m  m2 )  1 g m1  m2 m1  m2

Which is the same result from Chapter 6, as expected.

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PROBLEM 7.4 Consider a block of mass m sliding down a frictionless ramp at an incline θ as in Figure 7.2. Find the velocity of the block at time t if the block is stationary at t = 0 and x is measured parallel to the incline. Note: This is the same problem text as Problem 6.3 but now should be solved using Hamilton’s equations. SOLUTION 7.4 In order to use Hamilton’s equations, the Lagrangian is required. The kinetic energy is simply T=

1 2 mx 2

with the potential given by:

U  mg(const  x sin  ) Therefore, the Lagrangian is

  T U

1 2 mx  mg(const  x sin  ) 2

To use this in Hamilton’s equations, the velocity term must be substituted for the generalized momentum. So

p

  mx x

x =

p m

The Lagrangian can now be rewritten in terms of x and p , yielding

θ

FIGURE 7.2  Block placed on a ramp.

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Hamiltonian Formalism



p2  mg(const  x sin  ) 2m

The Hamiltonian is then given by H  px  L

p2 p2 p2   mg(const  x sin  )   mg(const  x sin  ) m 2m 2m

Hamilton’s equations yield x

 p  p m

and p

  mg sin  x

Therefore, p

  mgt sin   C  pdt

where C is a constant. This can be plugged into the equation for x x

p  gt sin   C m

Since x (0) = 0 , C = 0 and x (t )  gt sin

as was found before.

PROBLEM 7.5 A  mass is moving on a frictionless sphere of radius R . The force on the sphere is F  mgrˆ. Find the Hamiltonian and Hamilton’s equations. SOLUTION 7.5 As shown in Problem 6.8, the Lagrangian is



1 m( R 22 sin 2   R 22 )  mgR 2

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The momentum and first derivative for each variable are formulated:

p

L  mR 2 sin 2  



p mR sin 2  2

and p

L  mR 2 

p    2 mR

Now, the Hamiltonian can be found 

p2 p2 p2 1  2 p2  2 2  2   R 2 4 4 sin   R 2 4   mgR 2 mR sin  mR 2m  m R sin  m R 



p2 p2   mgR 2mR 2 sin 2  2mR 2

2

Deriving the Hamiltonian gives Hamilton’s equations for φ and θ .

p     2 p mR sin 2  p

 0

and

 p  2   p mR p

p2 cos    2 3  mR sin

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Hamiltonian Formalism

PROBLEM 7.6 Consider the following system of masses m1 and m2 , connected by an inextensible massless string as in Figure 7.3. The mass m1 is on a table and connected to the wall by a spring of spring constant k . a. Obtain the Lagrangian of the system. b. Use Lagrange equation to obtain the equation of motion. c. Find the Hamiltonian, write the Hamilton’s equation, and compare the equation obtained for acceleration. SOLUTION 7.6 a. As discussed in previous problems involving the Atwood machine, since the masses, m1 and m2 , are connected with an inextensible string, the kinetic energy is T

m1  m2 2 x 2

and the potential energy is U  m2 gx

kx 2 2

so the Lagrangian is

  T U

m1  m2 2 kx 2 x  m2 gx  2 2

b. Lagrange equation for x is  d   dx dt dx m1

x m2

FIGURE 7.3  System of two masses connected by an inextensible string. The system is connected to the wall by a spring of spring constant k . The spring below is simply showing the equilibrium position.

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  m2 g  kx dx

  (m1  m2 ) x dx

d   (m1  m2 )  x dt dx so (m1  m2 )  x  m2 g  kx

(m1  m2 )  x  kx  m2 g  0

c. The generalized momentum is p

  (m1  m2 ) x x

so x

p (m1  m2 )

By substituting this into the Lagrangian,

( x, p)

p2 kx 2  m2 gx  2(m1  m2 ) 2

The Hamiltonian is in this case defined as p p2 kx 2 p2 kx 2 H  pq  L  p   m2 gx    m2 gx  2 2(m1  m2 ) 2 m1  m2 2(m1  m2 ) Hamilton’s equations are

x

 p  p m1  m2

p

  m2 g  kx x

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Hamiltonian Formalism

Deriving again the velocity with respect to time, the acceleration is  x

p m g  kx  2 m1  m2 m1  m2

Which is the same expression as before. (m1  m2 )  x  kx  m2 g  0

PROBLEM 7.7 Consider a block of mass m on ramp at an incline θ and connected to a spring of constant k , as shown in Figure 7.4. Using Hamilton’s equations, find the angular frequency of oscillations following a displacement of the block. Note: This is the same problem text as Problem 6.12 but now should be solved using Hamilton’s equations. SOLUTION 7.7 In order to use Hamilton’s equations, the Lagrangian is required. The kinetic energy is given by T=

1 2 mx 2

and the potential energy (without worrying about the true height of the ramp) is given by

U

1 k ( x  d )2  mg(const  x sin  ) 2

where d is the “pre-­stretched” displacement of the spring due to the mass but not due to the oscillations. This was found to be d

mg sin  k

k

x

m

θ

FIGURE 7.4  Block on a ramp connected to a spring.

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in Chapter 6. Therefore, the Lagrangian is given by   T U

1 2 1 mx  k ( x  d )2  mg(const  x sin  ) 2 2

To use this in Hamilton’s equations, the velocity term must be substituted for the generalized momentum. So

p

  mx x

x =

p m

Now, rewriting the Lagrangian in terms of x and p yields 

p2 1  k ( x  d )2  mg(const  x sin  ) 2m 2

The Hamiltonian is thus given by H  px  L



p2 p2 1   k ( x  d )2  mg(const  x sin  ) m 2m 2

p2 1  k ( x  d )2  mg(const  x sin  ) 2m 2

Considering Hamilton’s equations, the angular velocity can be found via

x

 p  p m

 x=

p m

where p comes from the other of Hamilton’s equations

p

 mg sin    k ( x  d )  mg sin   k  x  x k 

Therefore,

 x

p k  x m m

   mg sin   kx 

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Hamiltonian Formalism

and the angular velocity is given by

k m

as was found in Chapter 6. PROBLEM 7.8 Consider an object of mass m connected to the ceiling by a spring, as in Figure 7.5. The system sphere-spring can only move in a vertical plane.

a. Choose the appropriate coordinates and write the kinetic and the potential energies. b. Find the Lagrangian. c. Obtain the equation of motion from Lagrange equations. d. Write the Hamiltonian and Hamilton’s equations and compare with part (c).

SOLUTION 7.8 a. The spring is moving in a vertical plane and forming the angle θ with the vertical as in Figure 7.6. The mass m is positioned at the length l + x from the fixed point on the ceiling, where x is the distance from the equilibrium position. The position of the mass m is ((l  x )sin  ,(l  x ) cos  ) . The velocity is v 2  x 2  [(l  x )]2 and the kinetic energy is

T

mv 2 m 2  ( x  (l  x )22 ) 2 2

The potential energy is both gravitational and elastic,

U  mg(l  x ) cos

kx 2 2

m

FIGURE 7.5  A sphere of mass m connected to the ceiling by a spring and moving only in a vertical plane.

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θ

l+x

m

FIGURE 7.6  A sphere of mass m connected to the ceiling by a spring and moving only in a vertical plane, with the angle with respect to the vertical being θ , and the length of the spring length l + x .

b. The Lagrangian is as follows:   T U

m 2 kx 2 ( x  (l  x )22 )  mg(l  x ) cos   2 2

c. The next step is to write the Lagrange equations for x and θ and to calculate them. We start with x :

 d   dx dt dx

  m(l  x )2  mg cos   kx dx

  mx dx

d   mx dt dx

mx  m(l  x )2  mg cos   kx (7.3)  x  (l  x )2  g cos

k x m

Now, the equation for θ :

 d   d dt d

  mg(l  x )sin  d

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Hamiltonian Formalism

  m(l  x )2 d

d   m(l  x )2  2m(l  x ) x dt d

mg(l  x )sin   m(l  x )2  2m(l  x ) x

(l  x )  2 x  g sin   0

d. The Hamiltonian is obtained by n

H

p q  L  p q  p q i i

x x

 

L

i 1

The generalized momenta are px

x =

  mx x px m

and

p

  m(l  x )2 



p m(l  x )2

By substituting the generalized momenta, the new form of the Lagrangian is obtained as  m  px2 p2 kx 2 2  mg ( l  x ) cos    2  (l  x ) 2  2 m 2 m (l  x )4  2 2 2 kx p p  mg(l  x ) cos    x  2 2m 2m(l  x ) 2



The Hamiltonian is obtained by subsequent substitutions in the formula

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px2 p2 p2 p2   x  2 m m(l  x ) 2m 2m(l  x )2

H  px q x  p q  L  mg(l  x ) coos

kx 2 kx 2 p2 p2 mg l x  x   (  ) cos   2 2m 2m(l  x )2 2

x

 px  px m

 x=

p x m

p x

 x

p2    mg cos   kx x m(l  x )3

p2 k  g cos   x m 2 (l  x )3 m

Recalling that



p m(l  x )2

Equation (7.3) from part (c)

mx  m(l  x )2  mg cos   kx

becomes

 x

(l  x ) p2 k  g cos   x m 2 (l  x )4 m

which is the same one as using Hamiltonians

 x

p2 k  g cos   x m (l  x )3 m 2

Now, checking the equation in θ :

 p    p m(l  x )2

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Hamiltonian Formalism



p 2 p x  2 m(l  x ) m(l  x )3

p

  mg(l  x )sin

Therefore,

mg(l  x )sin  2 p  x   m(l  x )2 m(l  x )3

g sin  2p p  2  x 3   lx m (l  x ) p p and x = x and this is how Equation (7.3) from m m(l  x )2 part (c) is obtained, similarly as by Lagrangian formalism Recalling that 

(l  x )  2 x  g sin   0

PROBLEM 7.9 A mass m is moving inside and against  the side of a half sphere of radius R as in Figure 7.7. The force on the mass is F  kr r , directed toward the bottom of the sphere. Find the Hamiltonian and Hamilton’s equations. SOLUTION 7.9 The kinetic energy of the mass is (Figure 7.8)

y

m

F x 0 z

FIGURE 7.7  Mass moving inside a half sphere, with force F acting on it.

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r

R x 0

z

FIGURE 7.8  Mass moving inside a half sphere, with force F acting on it.

T

 1 2 1 mv  m( R 2  2 sin 2   R 2 2 ) 2 2

The potential energy is U=

1 2 kr 2

Using geometric relations and the properties of isosceles triangles, r is expressed as     r  2 R sin    2 

Thus,

U

1     k 4 R 2 sin 2   2  2 

The Lagrangian of the system is obtained:

  T U

1     m( R 22 sin 2   R 22 )  2kR 2 sin 2   2  2 

The momentum corresponding to each variable is obtained:

p

L  mR 2 sin 2 

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Hamiltonian Formalism

p

L  mR 2

Now, it is possible to find the Hamiltonian , and Hamilton’s equations:



p2 p2       2kR 2 sin 2   2 2 2 2mR sin  2mR  2 

p



H 0

p mR 2 sin 2

p  

p cos       2 2  2kR 2 cos   mR sin   2

      sin  2    

p    2 mR

PROBLEM 7.10 A small bead of mass m can slide without friction on a parabolic wire of equation z = ax 2 rotating with angular velocity ω as in Figure 7.9. a. Write the Lagrangian (similarly with the problem in Chapter 6). b. Write the Hamiltonian. c. Find the highest point zmax = Z at which the bead can position itself if the initial velocity at the origin of the coordinates is v0 . Note: This is the same problem text as Problem 6.7 but now should be solved using Hamilton’s equations.

z ω

x 0

FIGURE 7.9  A bead is sliding without friction on a parabolic wire rotating about the vertical axis with angular velocity ω .

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SOLUTION 7.10 a. Similarly with Chapter 6 (Problem 6.7), the Lagrangian is written by recalling that z = ax 2 and starting with the velocity as v 2  x 2  ( x )2  z 2  x 2  ( x )2  4a 2 x 2 x 2 . m The Lagrangian is   T  U  ( x 2  x 2 2  4a 2 x 2 x 2 )  mgax 2 . 2 b. The Hamiltonian is defined as n

H

p q  L i i

i 1

Here, there is a one degree of freedom problem, so H  pq  L , with q = x and

p

  mx  4ma 2 x 2 x x

Then the Hamiltonian is m H  pq  L  mx 2  4ma 2 x 2 x 2  ( x 2  x 2 2  4a 2 x 2 x 2 )  mgax 2 2  ( x 2  x 2 2  4a 2 x 2 x 2 )  mgax 2

Since the Lagrangian is time independent, the Hamiltonian is conserved and equal to the total energy, from the initial condition, at= x 0= , x v0 , the Hamiltonian is written as m m mv02 (1  4a 2 x 2 ) x 2  ( 2  2 ga) x 2  (7.4) 2 2 2

c. The bead moves to the highest point zmax = Z given by the condition x = 0 . Equation (7.4) becomes

(2 ga   2 ) x 2  v02

Multiplying on both sides by a and recalling that z = ax 2 , it follows that the maximum height is

Z

av02 2 ga   2

It is easy to see that, in order to have a positive value for Z , the condition 2 ga   2  0 needs to be fulfilled, so  2  2 ga , or   2ga . In the case   2ga , the bead goes up to infinity on the parabola.

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Hamiltonian Formalism

PROBLEM 7.11 Consider a bead of mass m on the spinning wire as in Figure 7.10. If the wire is in 2 1 2 the shape of an upside-­down Gaussian, specifically f (  )   e , find any 2 equilibrium positions and comment on their stability. Note: This is the same problem text as Problem 6.10 but now should be solved using Hamilton’s equations. SOLUTION 7.11 In order to use Hamilton’s equations, the Lagrangian is required. Recalling the position can be used to find velocity  1  r     f    z     e 2

2 2 z

with 2

1 v      e 2 z  ˆ 2

the kinetic energy is given by

T

1 2 1  2 1 2 2 2  mv  m      e   2 2  2 2  2 

and the potential energy is given by

2  1 1 2 U  mg ｜ f (0) ｜ ｜ f (  ) ｜  mg   e  2 2 

f(ρ) ρ

ω

FIGURE 7.10  Bead on a spinning wire.

m

2  mg    1  e 2  2  

   

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Classical Mechanics

Therefore, the Lagrangian is

2  1  2 1 2 2 2 mg  2 2      T U  m     e     1 e 2 2  2 2  

2   mg  1  2 1 2 2  1  e 2 m    1   e    2 2   2   2 2   

   

   

To use this in Hamilton’s equations, the velocity term must be substituted for the generalized momentum. Thus, p

2   1 2 2  e  m   1   2  

p 1 2     1   e m  2 

2 2

   

   

1

This can now be substituted into the Lagrangian

2 1  p2  1 2 2 1   e  m 2  m 2  2

2 2 1 1 2 2 p  e  m  2 1  2  m  2

p2  1 2  1   e  2m  2

2 2

2  2    1 2 2  mg  2 2  1  1  e 2  e        2  2       1  2   mg     2 2   1  e 2   2    

1

2    2 2 m mg    1  e 2   2 2  

   

   

The Hamiltonian can now be found via H  p   L p2  1 2  1   e  m  2

2 2

1

2  p2  1 2 2   1   e  2m  2 

1

2    2 2 m mg    1  e 2   2 2  

   

   

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Hamiltonian Formalism

1 2  p2  1   e  2m  2

2 2

1

2    2 2 m mg    1  e 2    2 2  

   

For stability analysis, consider Hamilton’s equations



 p

p



 will not have ω in its expression. This means the p derivative is p not useful for the analysis and will be ignored. Also, the ρ momentum must be zero so the first term in the Hamiltonian can be ignored as well. Therefore, consider Notice 

p



0

2     2 2 m mg  1  e 2     2 2  

mg 0   m  2 2

2 m

mg   e 2

   

2   e 2 (   )     

2 2

As was discussed previously,   0 is an uninteresting equilibrium point so it can be ignored. Therefore,

g  e 2

2 4

which is exactly what was found in Chapter 6.

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PROBLEM 7.12 Consider a mass m attached to a pendulum of length l and a spring of constant k as depicted in Figure 7.11. When the mass is at the lowest point of the pendulum, the spring is at its rest length. Considering a small displacement, find the angular velocity. Note: This is the same problem text as Problem 6.11 but now should be solved using Hamilton’s equations. SOLUTION 7.12 In order to use Hamilton’s equations, the Lagrangian is required. The kinetic energy of a pendulum should be familiar T

1 2 1 2 2 mv  ml  2 2

with the potential energy a combination of gravity and the spring U  mgl (1  cos  )

1 k (l sin  )2 2

Considering small angles, the Lagrangian is given by

  T U

1 2 2 mgl 2 1 2 2 1 2 2 l ml     kl   ml   (mg  kl ) 2 2 2 2 2 2

To use this in Hamilton’s equations, the velocity term must be substituted for the generalized momentum. Thus,

p

  ml 2 

ϕ

l

k m

FIGURE 7.11  Mass connected to a pendulum and a spring.

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Hamiltonian Formalism

p   2 ml

Using this in the Lagrangian yields 

l p2  (mg  kl ) 2 2 2ml 2

The Hamiltonian is then given by p2 p2 l   (mg  kl ) 2 2 ml 2ml 2 2 p2 l   (mg  kl ) 2 2ml 2 2

H  p  L

Considering Hamilton’s equations, the angular velocity can be found via

 p    p ml 2

p   2 ml

where p comes from the other of Hamilton’s equations p

  l (mg  kl )

Therefore,

p l (mg  kl ) mg  kl   2    2 ml ml ml

and the angular velocity is given by

as was found in Chapter 6.

mg  kl ml

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PROBLEM 7.13 A particle with initial mass m0 and initial velocity v0 begins losing mass according to the equation m(t )  m0e  t where α is constant. If there are no external forces, find an expression for the velocity. Note: This is the same problem text as Problem 6.6 but now should be solved using Hamilton’s equations. SOLUTION 7.13 In order to use Hamilton’s equations, the Lagrangian is required. The only energy in the problem is kinetic, so

T

1 1 m(t ) x 2  m0e  t x 2 2 2

To use this in Hamilton’s equations, the velocity term must be substituted for the generalized momentum. So

p

  m0e  t x x

x

p t e m0

The Lagrangian can now be rewritten as 

p2  1 p2  t m0e  t  2 e2 t   e 2  m0  2m0

Therefore, the Hamiltonian is given by

H  px  L

p2  t p2  t p2  t e  e  e 2m0 2m0 m0

An expression for velocity can be found by taking Hamilton’s equations p

 0 x

p=C

where C is a constant. Also,

x

 p t C t e  e  p m0 m0

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Hamiltonian Formalism

Considering the initial velocity is v0 ,

) v= x (0= 0

C m0

C = m0 v0

and exactly as was found before.

x (t )  v0e t

8

Coupled Oscillators and Normal Modes

8.1  THEORY A variety of systems as atoms in molecules or molecules in certain types of polymers may oscillate as coupled oscillators, not independent oscillators. In this case, a different method shall be employed. For a system of n degrees of freedom, the n generalized coordinates q1, q2 ,⊃ , qn can be written as a column matrix q. With  and  the mass and the spring constant matrices, respectively, the equation of motion for small oscillation about the equilibrium position (with q = 0 ) is Mq = −Kq

The two matrices are obtained from the kinetic and potential energies of the system, T= U=

1 2

∑ q q

1 2

∑ q q

jk

j k

j ,k

jk

j k

j ,k

The motion in which all coordinates oscillate with the same frequency ω is called a normal mode and is written as q(t ) = Re( aeiωt )

The column matrix a satisfies the eigenvalue equation

(K − ω 2M) a = 0

If the matrix has a non-­zero determinant, then the only solution is the trivial solution a = 0. If the determinant is zero, DOI: 10.1201/9781003365709-8

det(K − ω 2M) = 0 197

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Classical Mechanics

Then there is a non-­trivial solution, and the frequencies ω at which the system is oscillating can be determined. For a system with n degrees of freedom, there are n normal frequencies, each with a corresponding eigenvector.

8.2  PROBLEMS AND SOLUTIONS PROBLEM 8.1 A system of oscillators with generalized coordinates q1, q2 , q3 has the following kinetic and potential energy (consider the masses and the spring constants equal to one):

T=

1 2 (q1 + q22 + q32 ) 2

U=

1 (4q12 + 5q22 + 4q32 − 2q1q2 − 2q2q3 ) 2

a. Show that the position with q= q= q3 = 0 is an equilibrium position. 1 2 b. Find the three normal frequencies. SOLUTION 8.1 a. Condition for equilibrium ∂U =0 ∂qi

For q1 ,

∂U =0 ∂q1

which yields to

4q1 − q2 = 0

For q2 ,

∂U =0 ∂q2

which yields to

Coupled Oscillators and Normal Modes

199

5q2 − q1 − q3 = 0

Lastly, for q3 ,

∂U =0 ∂q3

with the equation

4q3 − q2 = 0

The determinant of the system is

4  det  −1 0 

−1 5 −1

0  −1  = 80 − 4 − 4 = 72 ≠ 0 4 

Since the determinant is not zero, the only solution is q= q= q3 = 0 1 2

Note that the potential energy can be rewritten as

1 U = [3(q12 + q22 + q32 ) + (q12 − 2q1q2 + q22 ) + (q22 − 2q2q3 + q32 )] 2 1 = [3(q12 + q22 + q32 ) + (q1 + q2 )2 + (q2 + q3 )2 ] 2 It is easy to see that the total potential energy is positive, therefore the origin is a position of stable equilibrium. b. The Lagrangian is  = T −U =

1 2 1 (q1 + q22 + q32 ) − (4q12 + 5q22 + 4q32 − 2q1q2 − 2q2q3 ) 2 2

Lagrange equations for q1, q2 , q3 are For q1 ,

d ∂ ∂ = dt ∂q1 ∂q1 q1 = −4q1 + q2

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For q2 , d ∂ ∂ = dt ∂q2 ∂q2

q2 = −5q2 + q1 + q3

For q3 ,

d ∂ ∂ = dt ∂q3 ∂q3

q3 = −4q3 + q2

The three coupled equations can be written in the compact form as

−1 0   q1  1 0 0   q1  4         −1 q2  5 0 1 0  q2  = −  −1  0 0 0 1   q3  4   q3  −1       M

det(K − ω 2M) = 0

K

 4  det   −1   0 

−1 5 −1

 4 − ω 2  det   −1  0 

0 1   −1 − ω 2 0 0 4  −1 5 − ω2 −1

0 1 0

0   0  = 0 1  

0   −1   = 0 4 − ω 2   

(4 − ω 2 )(5 − ω 2 )(4 − ω 2 ) − (4 − ω 2 ) − (4 − ω 2 ) = 0

which can be written as

(ω 2 − 4)2 (ω 2 − 5) − 2(ω 2 − 4) = 0

or

(ω 2 − 4)2 [(ω 2 − 5)(ω 2 − 4) − 2] = 0

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Coupled Oscillators and Normal Modes

From here, ω1 = 2 (note that we choose the positive solutions only). The second equation becomes

ω 4 − 9ω 2 + 18 = 0

And the other two frequencies are calculated as ω2 = 6 and ω3 = 3 .

PROBLEM 8.2 Consider a system of two masses and three springs, as pictured in Figure 8.1. For the following cases, find the two normal frequencies ω1 and ω2 . k and m = m= m; 1 2 3 b. = m1 2= m2 m and k= k= k3 = k 1 2 a. k= k= k ; k2 = 1 3

SOLUTION 8.2 a. Using Newton’s Second Law, F = mx, two equations M x = −Kx are obtained

k k 4k mx1 = −k1 x1 + k2 ( x2 − x1 ) = −kx1 + ( x2 − x1 ) = − x1 + x2 3 3 3

4 k k k k  mx2 = −k3 x2 + k2 ( x1 − x2 ) = −kx2 + ( x1 − x2 ) = x1 −  + k  x2 = x1 − x2 3 3 3 3 3  The matrices  and  are found: m = 0

 4k  3 = − k  3

k1

m1

0  m k −  3  4k  3 

k2

FIGURE 8.1  System of three springs and two masses.

m2

k3

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Classical Mechanics

From these, the matrix K − ω 2M is written:

 4k − mω 2  3 K − ω 2M =   −k 3 

   4k − mω 2   3 −

k 3

Using det(K − ω 2M) = 0 , the two normal frequencies are determined 2

 4k  5k 2 k  2   2k 2  3 − mω  −  3  =  3 − mω   3 − mω        

ω1 =

5k 3m

ω2 =

2k 3m

b. The process is similar to part (a). Using Newton’s Second Law, F = mx, two equations M x = −Kx are obtained

2mx1 = −k1 x1 + k2 ( x2 − x1 ) = −kx1 + k ( x2 − x1 ) = −2kx1 + kx2

mx2 = −k3 x2 + k2 ( x1 − x2 ) = −kx2 + kx1 − kx2 = kx1 − 2kx2 The matrix K − ω 2M is written:

2k − 2mω 2 K − ω 2M =  −k 

−k   2k − mω 2 

Using det(K − ω 2M) = 0 , the two normal frequencies are determined

(2k − 2mω 2 )(2k − mω 2 ) − k 2 = 0

3k 2 − 64mω 2 + 2m 2ω 4 = 0 Letting W = ω 2 , substitution is used to solve the equation

3k 2 − 64mW + 2m 2W 2 = 0

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Coupled Oscillators and Normal Modes

W=

3 −3 3+ 3 k, − k 2m 2m

So, the two normal frequencies are

3+ 3 k 2m

ω1 =

ω2 = −

3 −3 k 2m

PROBLEM 8.3 Consider a system of two identical springs of spring constant k and two objects of mass m connected as in Figure 8.2. Ignoring the gravitational field, determine the frequencies ω at which the two objects oscillate. SOLUTION 8.3 Considering y1 and y2 the positions of the two masses from the equilibrium positions, the kinetic energy is

T=

my12 my22 + 2 2

and the potential energy is written as

k

1

m

k

2

m

FIGURE 8.2  A system of two objects of identical mass m connected by two identical springs of spring constant k .

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Classical Mechanics

U=

ky12 k ( y2 − y1 )2 ky 2 + = ky12 + 2 − ky1 y2 2 2 2

The Lagrangian is

 = T −U =

my12 my22 ky 2 + − ky12 − 2 + ky1 y2 2 2 2

Lagrange equations for y1 and y2 become d ∂ ∂ = dt ∂y1 ∂y1

my1 = −2ky1 + ky2

d ∂ ∂ = dt ∂y2 ∂y2

my2 = ky1 − ky2

Note that the two coupled equations can be written in the compact matrix form

0    y1  −k   y1  m  2k      = −    m   y2  −k k y 0     2  M

K

where the two square matrices are the mass matrix  and the spring constant matrix y1     y1   with the column matrices  y =   and y =   y2     y2  M y = − Ky

To find the normal modes, consider the determinant det(K − ω 2M) = 0

  2k det     −k

−k  2 m  −ω  k 0

 2k − ω 2 m det      −k

0  = 0 m  

−k    = 0 k − ω 2 m  

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Coupled Oscillators and Normal Modes

2k 2 − 2kmω 2 − kmω 2 + m 2ω 4 − k 2 = 0

k 2 − 3kmω 2 + m 2ω 4 = 0 After dividing by m 2 and by recalling that ωe =

k the equation becomes m

ω 4 − 3ωe2ω 2 + ωe4 = 0 ω2 =

3ωe2 ± 9ωe4 − 4ωe4 3± 5 = ωe2 2 2

ω = ωe

3± 5 2

These two normal frequencies are the frequencies at which the two spheres can oscillate in purely sinusoidal motion. PROBLEM 8.4 Consider a system of three masses and three springs hanging vertically, as shown in Figure 8.3. All masses are equal, and all spring constants are equal. Find the matrix K − ω 2M . SOLUTION 8.4 The kinetic energy of the system is T=

1 2 1 2 1 2 my1 + my2 + my3 2 2 2

The potential energy of the system is

U=

=

1 2 1 1 ky1 + k ( y2 − y1 )2 + k ( y3 − y2 )2 2 2 2 1 2 1 2 1 2 1 1 ky1 + ky2 + ky1 − ky1 y2 + ky32 + ky22 − ky2 y3 2 2 2 2 2

So the Lagrangian is

 = T −U =

1 2 1 2 1 2  2 1  my1 + my2 + my3 −  ky1 + ky22 − ky1 y2 + ky32 − ky2 y3  2 2 2 2  

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Classical Mechanics

m

m

m

FIGURE 8.3  System of three masses and three springs.

By derivation, a system of equations is obtained, M y = −Ky. d ∂ ∂ = dt ∂y1 ∂y1

my1 = −2ky1 + ky2

d ∂ ∂ = dt ∂y2 ∂y2

my2 = k (−2 y2 + y1 + y3 )

d ∂ ∂ = dt ∂y3 ∂y3

my3 = −ky3 + ky2

Thus, the matrix  is

m   = 0  0

0 m 0

0  0 m 

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Coupled Oscillators and Normal Modes

And the matrix  is  2k   =  −k  0

−k 2k −k

0  −k  k 

So the matrix K − ω 2M is the following:  −mω 2 + 2k  K −ω M =  −k  0  2

 0  −k  −mω 2 + k 

−k −mω 2 + 2k −k

PROBLEM 8.5 Consider two pendulums with strings of same length L , coupled by a spring of constant k as in Figure 8.4. Find the normal frequencies, if m = m= m , and in the 1 2 small angle approximation. SOLUTION 8.5 The kinetic energy of the system is

T=

1 2 2 2 mL (φ1 + φ2 ) 2

The potential energy of the system is

U = mgL (1 − cos φ1 ) + mgL (1 − cos φ2 ) +

1

2

m1

FIGURE 8.4  Coupled oscillators.

1 2 kL (φ2 − φ1 )2 2

m2

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Classical Mechanics

Using small angles, cos φ1 is approximated: 1 cos φ1 ≈ 1 − φ12 2

Rewriting the potential energy, the following expression is obtained: U=

1 mgL 2 (φ1 + φ22 ) + kL2 (φ22 + φ12 − 2φ1φ2 ) 2 2

The Lagrangian is  = T −U =

1 2 2 2 mgL 2 1 mL (φ1 + φ2 ) − (φ1 + φ22 ) − kL2 (φ22 + φ12 − 2φ1φ2 ) 2 2 2

By derivation, a system of equations is obtained, Mφ = Kφ .

d ∂ ∂ = dt ∂φ1 ∂φ1

mL2φ1 = −mgLφ1 − kL2φ1 + kL2φ2 = −(mgL + kL2 )φ1 + kL2φ2

and

d ∂ ∂ = dt ∂φ2 ∂φ2 mL2φ2 = −mgLφ2 − kL2φ2 + kL2φ1 = kL2φ1 − (mgL + kL2 )φ2 The matrices  and  are

 mL2 =  0

0   mL2 

 mgL + kL2 = 2  −kL

−kL2   mgL + kL2 

The matrix K − ω 2M is written:

 mgL + kL2 − ω 2 mL2 K − ω 2M =  −kL2 

 −kL2  mgL + kL2 − ω 2 mL2 

Coupled Oscillators and Normal Modes

209

Using det(K − ω 2M) = 0 , the two normal frequencies are determined.

det(K − ω 2M) = (mgL + kL2 − ω 2 mL2 )2 − (kL2 )2 = 0

(mgL + kL2 − ω 2 mL2 − kL2 )(mgL + kL2 − ω 2 mL2 + kL2 ) = 0 The first normal frequency is

mgL − ω 2 mL2 = 0

ω2 =

ω1 =

mgL g = mL2 L g L

The second normal frequency is

mgL + 2kL2 − ω 2 mL2 = 0

ω2 = =

ω2 =

mgL + 2kL2 mL2 mg + 2kL mL g 2k + L m

PROBLEM 8.6 Consider the system of coupled pendulums as in Figure 8.5. If each pendulum has length L , spring constant k , and mass m, find the normal modes. SOLUTION 8.6 Since the kinetic energy is the sum of energies due to the individual pendulums, it is given by

T=

1 2 2 2 mL (φ1 + φ2 ) 2

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Classical Mechanics

1

2

L

k

L

k m1

m2

FIGURE 8.5  Two pendulums connected by springs.

The potential energy is due to the pendulums (gravitational) and the springs (elastic), U g = mgL (1 − cos φ1 ) + mgL (1 − cos φ2 ) ≈

mgL 2 (φ1 + φ22 ) 2

and 1 1 k ( L sin φ1 )2 + k ( L sin φ2 − L sin φ1 )2 2 2 1 2 2 1 ≈ kL (φ1 + (φ2 − φ1 )2 ) = kL2 (2φ12 − 2φ1φ2 + φ22 ) 2 2

Us =

where small angles were used to approximate the trigonometric functions. Therefore, U=

1 mgL 2 (φ1 + φ22 ) + kL2 (2φ12 − 2φ1φ2 + φ22 ) 2 2

The Lagrangian is now given by

 = T −U =

1 2 2 2 mgL 2 1 mL (φ1 + φ2 ) − (φ1 + φ22 ) − kL2 (2φ12 − 2φ1φ2 + φ22 ) 2 2 2

and the normal frequencies can be found by considering Lagrange’s equations

d ∂ ∂ = dt ∂φ1 ∂φ1

and

d ∂ ∂ = dt ∂φ2 ∂φ2

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Coupled Oscillators and Normal Modes

Thus, Lagrange equation for φ1 d ∂ ∂ = dt ∂φ1 ∂φ1

yields the equation

d (mL2φ1 ) = −(mgLφ1 + kL2 (2φ1 − φ2 )) dt

mL2φ1 = −((mgL + 2kL2 )φ1 − kL2φ2 )

and Lagrange equation for φ2 d ∂ ∂ = dt ∂φ2 ∂φ2

yields

d (mL2φ2 ) = −(mgLφ2 + kL2 (φ2 − φ1 )) dt

mL2φ2 = −(−kL2φ1 + (mgL + kL2 )φ2 )

Rewriting these equations as a matrix equation yields

 mL2  mgL + 2kL2 −kL2  φ1  0  φ1  = −    2     2 0 mL  φ1  −kL mgL + kL2  φ2        M

K

Therefore, the normal frequencies can be found by considering the determinant det(K − ω 2M) = 0

  mgL + 2kL2 det   2    −kL

2 −kL2  2  mL − ω   mgL + kL2   0

  mgL + 2kL2 − ω 2 mL2 det    −kL2 

0   = 0 mL2  

 −kL2 =0 2 2 2 mgL + kL − ω mL  

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Classical Mechanics

(mgL + 2kL2 − ω 2 mL2 )(mgL + kL2 − ω 2 mL2 ) − k 2 L4 = 0

2 2 2 2 2 2 4 2 4 2 4 (mgL + 2kL )(mgL + kL ) − ω mL (mgL + kL + mgL + 2kL ) + ω m L − k L = 0

(mgL + 2kL2 )(mgL + kL2 ) − ω 2 mL2 (2mgL + 3kL2 ) + ω 4 m 2 L4 − k 2 L4 = 0

ω 4 m 2 L4 − ω 2 mL2 (2mgL + 3kL2 ) + m 2 g 2 L2 + 3mgL3k + 2k 2 L4 − k 2 L4 = 0

ω 4 m 2 L2 − ω 2 m(2mgL + 3kL2 ) + m 2 g 2 + 3mgLk + k 2 L2 = 0

ω2 =

m(2mgL + 3kL2 ) ± m 2 (2mgL + 3kL2 )2 − 4m 2 L2 (m 2 g 2 + 3mgLk + k 2 L2 ) 2m 2 L2

ω2 =

ω2 =

2mg + 3kL ± (2mg + 3kL )2 − 4(m 2 g 2 + 3mgLk + k 2 L2 ) 2mL

2mg + 3kL ± 4m 2 g 2 + 12mgkL + 9k 2 L2 − 4m 2 g 2 − 12mgLk − 4k 2 L2 2mL

ω2 =

2mg + 3kL ± 5k 2 L2 2mL

ω2 =

g k + (3 ± 5 ) L m

ω=

g k + (3 ± 5 ) L m

Notice, when ω =

g k + ( 3 + 5 ) the pendulums move toward/away from each L m

other and when ω =

g k + ( 3 − 5 ) the pendulums move together to the left/right. L m

PROBLEM 8.7 Consider the cart from Figure 8.6, which is attached to the wall with a spring and has pendulum hanging below it. For a spring of constant k , a cart of mass m , and a pendulum of length L and mass M , find the normal frequencies. Assume small φ .

213

Coupled Oscillators and Normal Modes x m

k

L

M

FIGURE 8.6  Cart connected to a spring with a pendulum connect beneath it.

SOLUTION 8.7 The normal frequencies can be found using the Lagrangian. In order to find expres  sions for energy the positions of the cart, rm , and pendulum, rM , must be found. These are given by

rm = x x

rM = ( x + L sin φ ) x − L cos φ y Considering small angles,  φ2   rM ≈ ( x + Lφ ) x − L  1 −  y 2  

The velocities are

vm = x x

vM = ( x + Lφ) x + Lφφ y Therefore, the kinetic energy expressions are

Tm =

1 2 mx 2

TM =

1 1 M (( x + Lφ)2 + ( Lφφ)2 ) = M ( x 2 + 2 Lxφ + L2φ2 + L2φ 2φ2 ) 2 2

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Classical Mechanics

Since φ is small, L2φ 2φ2 is very small and can be ignored. Thus,

T = Tm + TM =

1 2 1 1 1   mx + M ( x 2 + 2 Lxφ + L2φ2 ) = x 2 (m + M ) + ML  xφ + Lφ2  2 2 2 2  

The potential energy has contributions from the spring and the pendulum. From the pendulum U M = MgL (1 − cos φ )

Considering small angles, this becomes UM ≈

MgL 2 φ 2

The potential energy is then given by U=

1 (kx 2 + MgLφ 2 ) 2

The Lagrangian is then given by  = T −U =

1 2 1   1 x (m + M ) + ML  xφ + Lφ2  − (kx 2 + MgLφ 2 ) 2 2   2

In order to find the normal frequencies, Lagrange’s equations must be considered: d ∂ ∂ = dt ∂x ∂x

and

d ∂ ∂ = dt ∂φ ∂φ Therefore, from the Lagrange equation in x,

d ((m + M ) x + MLφ) = −kx dt

(m + M )  x + MLφ = −kx

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Coupled Oscillators and Normal Modes

and from the Lagrange equation in φ :

d ( MLx + ML2φ) = − MgLφ dt

MLx + ML2φ = − MgLφ  x + Lφ = − gφ

These can be re-­expressed as the matrix equation ML    x k 0   x  m + M     = −    1 0 g  φ  L φ     

M

K

To find the normal modes, consider the determinant det(K − ω 2M) = 0

 k det    0

0 2 m + M  −ω  g  1

  k − ω 2 (m + M ) det    −ω 2 

ML    = 0 L  

−ω 2 ML    = 0 g − ω 2 L  

By calculating the determinant, the following equation yields

(k − ω 2 (m + M ))( g − ω 2 L ) − ω 4 ML = 0

kg − (kL + g(m + M ))ω 2 + L (m + M )ω 4 − MLω 4 = 0

kg − (kL + g(m + M ))ω 2 + mLω 4 = 0 From this equation, ω 2 is obtained as

ω2 =

kL + g(m + M ) ± (kL + g(m + M ))2 − 4mLkg 2mL

And by taking the radical, the frequencies of the normal modes are obtained

ω=

kL + g(m + M ) ± (kL + g(m + M ))2 − 4mLkg 2mL

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Classical Mechanics

Following a suggestion from Taylor, let m = M = L= g= 1 and k = 2 (in appropriate units). Therefore,

ω=

2 + 2 ± (2 + 2)2 − 4(2) = 2

4± 8 = 2± 2 2

The two normal modes are then the cart and pendulum oscillating in the same direction, ω = 2 − 2 , and the cart and pendulum oscillating in opposite directions,

ω = 2+ 2 . PROBLEM 8.8 Consider the cart on a ramp (Figure 8.7), cart which is attached to the wall with a spring and has pendulum hanging below it. For a spring of constant k and an equilibrium position a distance l from the bottom of the ramp, a cart of mass m, a pendulum of length L and mass M , and a ramp at angle θ , find the normal frequencies. Assume small x and φ . SOLUTION 8.8 Since the cart is on a ramp, the spring is already displaced from its equilibrium position, d , before the system oscillates. This distance is given by ∑ Fx = 0 mg sin θ = kd

d=

mg sin θ k

Expressions for energy can now be found by considering the positions of the cart,   rm , and the pendulum, rM

rm = x(cos θ x − sin θ y )

with y

k

r m

x

L M

FIGURE 8.7  Cart on a ramp connected to a spring with a pendulum connect beneath it.

217

Coupled Oscillators and Normal Modes

vm = r = x (cos θ x − sin θ y )

and

L   rM = ( x cos θ + L sin φ ) x + ( x sin θ − L cos φ ) y ≈ ( x cos θ + Lφ ) x +  x sin θ − φ 2  y 2   where a small angle approximation was taken. Also  vM = ( x cos θ + Lφ) x + ( x sin θ − Lφφ) y

The energies can now be found. Starting with the cart, the kinetic energy is given by Tm =

1   1 1 mvm ⋅ vm = mx 2 (cos2 θ + sin 2 θ ) = mx 2 2 2 2

and the potential energy is given by U m = mg(l − x cos θ ) +

1 1  mg  k ( x + d )2 = mg(l − x cos θ ) + k  x + sin θ  2 2  k 

Since the oscillations are small, the energy U m ( x ) can be expressed as Um =

1 ∂ 2U 2 x 2 ∂x 2

Observe

∂U mg   = −mg cos θ + k  x + sin θ  k ∂x  

and

∂ 2U =k ∂x 2

so

Um =

1 2 kx 2

2

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Classical Mechanics

For the pendulum, the kinetic energy is 1   1 MvM ⋅ vM = M (( x cos θ + Lφ)2 + ( x sin θ − Lφφ)2 ) 2 2 1 = M ( x 2 cos2 θ + 2 x cos θ Lφ + L2φ2 + x 2 sin 2 θ + 2 x sin θ Lφφ + L2φ 2φ2 ) 2 1 ≈ M ( x 2 + 2 x cos θ Lφ + L2φ2 ) 2

TM =

where the last two terms in the parenthesis can be ignored; xφφ and φ 2φ2 are very small compared to the other terms. The potential energy is given by 1   U M = Mg((l − x cos θ ) + L (1 − cos θ )) ≈ Mg  l − x cos θ + Lφ 2  2  

Since the oscillations are small, U M ( x, φ ) can be rewritten as UM =

∂ 2U ∂ 2U 2  1  ∂ 2U 2 x x φ φ  + + 2  ∂x∂φ ∂φ 2  2  ∂x 2

with

∂ 2U =0 ∂x 2

∂ 2U =0 ∂x∂φ ∂ 2U = MgL ∂φ 2

Therefore,

UM =

1 MgLφ 2 2

The total energies are then 1 2 1 mx + M ( x 2 + 2 x cos θ Lφ + L2φ2 ) 2 2 1 1 2  φ cos θ + ML2φ2 = (m + M ) x + xML 2 2

T = Tm + TM =

219

Coupled Oscillators and Normal Modes

and U = Um + U M =

1 2 1 kx + MgLφ 2 2 2

Putting everything together, the Lagrangian is given by  = T −U =

1 1 1 1  φ cos θ + ML2φ2 − kx 2 − MgLφ 2 (m + M ) x 2 + xML 2 2 2 2

In order to find the normal frequencies, Lagrange’s equations must be considered: d ∂ ∂ = dt ∂x ∂x

and

d ∂ ∂ = dt ∂φ ∂φ

Therefore,

d ∂ ∂ = dt ∂x ∂x

d ((m + M ) x + MLφ cos θ ) = −kx dt

(m + M )  x + MLφ cos θ = −kx

and d ∂ ∂ = dt ∂φ ∂φ

d  ( xML cos θ + ML2φ) = − MgLφ dt

xML cos θ + ML2φ = − MgLφ Expressing this as a matrix equation yields

ML cos θ    x 0  x k  m+M    = −     2 cos θ ML  φ  0 MgL  φ    ML     M

K

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Classical Mechanics

and the normal modes can be found by considering the determinant. det(K − ω 2M) = 0

 k det    0

0  2 m+M  −ω  MgL   ML cos θ

  k − ω 2 (m + M ) det    −ω 2 ML cos θ 

ML cos θ    = 0 ML2  

−ω 2 ML cos θ    = 0 MgL − ω 2 ML2  

ω 4 M 2 L2 cos2 θ − ( MgL − ω 2 ML2 )(k − ω 2 (m + M )) = 0

ω 4 M 2 L2 cos2 θ − (kMgL − ω 2 ((m + M ) MgL + kML2 ) + ω 4 (m + M ) ML2 ) = 0

ω 4 ML cos2 θ − kg + ω 2 ((m + M )g + kL ) − ω 4 (m + M )L = 0

ω 4 ( ML cos2 θ − mL − ML ) + ((m + M )g + kL )ω 2 − kg = 0

ω 4 (−mL − ML sin 2 θ ) + ((m + M )g + kL )ω 2 − kg = 0

ω 4 (mL + ML sin 2 θ ) − ((m + M )g + kL )ω 2 + kg = 0

ω2 =

(m + M )g + kL ± ((m + M )g + kL )2 − 4(mL + ML sin 2 θ )(kg) 2(mL + ML sin 2 θ )

ω=

(m + M )g + kL ± ((m + M )g + kL )2 − 4kg(mL + ML sin 2 θ ) 2(mL + ML sin 2 θ )

Note: When θ = 0 , this reduces to what was found in the previous problem.

9

Nonlinear Dynamics and Chaos

9.1  THEORY The systems encountered thus far have, for the most part, been linear in position, velocity, etc. While anything deviating from this may seem to a “special case”, the reality is most systems are not so well behaved, that is, nonlinear. The simplest example of such a system is one which is governed by the logistics equation, nt +1 = rnt

where the “next value”, nt +1 , is dependent on the “current value”, nt . Nonlinear systems have interesting properties which require a full course in nonlinear dynamics to explore. This chapter serves as a brief sampling of this incredibly rich topic. In fact, the equations governing the system must be nonlinear to get chaos (however, not all nonlinear systems exhibit chaos). Trajectories of chaotic systems are not typically periodic, and these systems are very sensitive to initial conditions, that is, a slight difference in initial conditions can lead to drastically different trajectories. Numerical software is generally required to fully explore the chaos of a system and while numerically generated solutions are provided here, the authors recommend further education in numerical analysis to gain confidence in manipulating the equations describing such systems.

9.2  PROBLEMS AND SOLUTIONS PROBLEM 9.1 Exponential population growth. By applying the simple example of a growth equation, calculate how many rabbits would populate Australia from 1850 to 1860, considering that two rabbits were brought to Australia and that the rabbits were reproducing at a rate of 20 rabbits per year (the rate could range between 18 and 30 rabbits per female rabbit). SOLUTION 9.1 The number of rabbits in year t +1 is based on the number of rabbits in the previous year as

nt +1 = f (nt ) = rnt The function f (n) = rn , where r is the growth rate (here r = 10 per one female).

DOI: 10.1201/9781003365709-9

221

222

Classical Mechanics

Starting with year 1850,

n1850 +1 = f (n1850 ) = rn1850

n1850 + 2 = f (n1851 ) = r 2 n1850 And so on, n1850 + k = f (n1850 + k −1 ) = r k n1850

And in ten years,

n1860 = n1850 +10 = f (n1859 ) = r10 n1850 With our initial conditions n1850 = 1 and growth rate r = 10 n1860 = r10 n1850 = 1010 ⋅ 1 = 1010

After this model, in ten years the rabbit population increased to ten billion rabbits. The rabbits determined overgrazing, leading to decrease in plant diversity and land degradation and erosion. The migration of rabbits is thought to reach at times 80 miles in one year. PROBLEM 9.2 The logistic growth model. Let N (t ) be the population dependent on time, based on maximum population growth rate r . The limiting factor is the carrying capacity K , which represents the total population that the environment could support based on the existent resources. Then the rate of change of population over time is dN rN ( K − N ) = dt K

N , the fraction of population alive K divided by the total population that the environment could support. Calculate the rate of population x =

SOLUTION 9.2 Starting with the formula

dN rN ( K − N ) = dt K

N After the substitution of rate of alive population by the x = , and by considering K dN dx that =K , the equation becomes dt dt

223

Nonlinear Dynamics and Chaos

dx = rx(1 − x ) dt

Integrating x

x0

t

dx′ = r dt ′ x′(1 − x′)

∫ 0

x

(ln x′ − ln (1 − x′)) x0 = rt

ln

ln

= rt x0

x(1 − x0 ) = rt x0 (1 − x )

ert =

x

x x − ln 0 = rt 1− x 1 − x0 ln

x 1− x

x(1 − x0 ) x0 (1 − x )

x0ert − x0 xert = x − xx0

x[1 + x0 (ert − 1)] = x0ert x (t ) =

x0ert = 1 + x0 (ert − 1)

1 1 = 1 + 1 − e −rt 1 +  1 − 1  e −rt x0ert  x0 

The fraction of the population x can be calculated, and this is called a sigmoid curve. When solved numerically, the method is to take small intervals steps (logistic map, Markov chain), following:

xn +1 = rxn (1 − xn )

where xn+1 is the population fraction for the next generation, xn is the population fraction for present generation, and r is the growth rate.

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Classical Mechanics

PROBLEM 9.3 Considering that the Feigenbaum relation ( yn +1 − yn ) =

1 ( yn − yn −1 ) δ

is correct, confirm that the limit is yc = 1.0829 . SOLUTION 9.3 The Feigenbaum relation is

( yn +1 − yn ) =

1 ( yn − yn −1 ) δ

with y1 = 1.0663 , y2 = 1.0793, and δ = 4.6692016 . Rewriting the relation, the following is obtained:

1 1  yn +1 − yn  1 +  + yn −1 = 0  δ δ Using the characteristic root technique for sequences, the roots can be found:

1 1  x2 − x 1 −  + = 0  δ δ

1  ( x − 1)  x −  = 0 δ  So the roots are r1 = 1 and r2 =

1 . The recurrence is of the form δ n

n

1 1 yn = C1   + C21n = C1   + C2 δ  δ 

where C1, C2 are constants. Using the given values of y1 , y2 , and δ solving for C1 and C2 is possible

1 C1   + C2 = 1.0663 δ 

 1  C1  2  + C2 = 1.0793 δ 

225

Nonlinear Dynamics and Chaos

1 Thus, C2 = 1.0663 − C1   . Plugging this value in the second equation, solving δ  for C1 is easy.

 1 C1  2 δ

 1  + 1.0663 − C1  δ  = 1.0793     1 1 C1  2 −  = 0.013 δ δ

C1 = C2 = 1.0663 −

0.013 = −0.07724  1 1 − δ2 δ   

0.013  1  = 1.0829  1 1   δ  − δ2 δ   

  1 n  So yc = lim yn = lim  C1   + C2  = C2 . Thus, yc = 1.0829 .  n →∞ n →∞   δ  

PROBLEM 9.4 Consider a driven damped pendulum with small drive strength γ  1. Find a solution to equation of motion, using small angle approximation. SOLUTION 9.4 A driven damped pendulum is governed by the equation

φ + 2 βφ + ω02 sin φ = γω02 cos ω t Using small angle approximation, sin φ = φ :

φ + 2 βφ + ω02φ = γω02 cos ω t

To solve the differential equation, the general solution φn and a particular solution φt need to be found.

φ = φn + φt

First, φn is determined:

φ + 2 βφ + ω02φ = 0

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Classical Mechanics

To solve the equation, it is rewritten as x 2 + 2 β x + ω 2 x = 0 . The roots of the equation are x = ± β 2 − ω2 − β

The solution for φn is

φn = C1et (

β 2 −ω 2 − β )

+ C2 e t ( −

β 2 −ω 2 − β )

Now, φt is found:

φ + 2 βφ + ω02φ = γω02 cos ω t

The following φt is a solution to the differential equation:

φt =

γω sin ω t 2β

Using small angle approximation, the equation of motion of a damped pendulum is

φ = C1et (

β 2 −ω 2 − β )

+ C2 e t ( −

β 2 −ω 2 − β )

+

γω sin ω t 2β

PROBLEM 9.5 Numerically solve and graph, for t = 0 to 100 , the nonlinear differential equation φ + 2 βφ + ω02 sin φ = γω02 cos ω t for ω = 1, ω0 = 2ω , 2 β = ω / 3 , φ (0) = φ(0) = 0 and a. γ = 0.3 b. γ = 1.06 c. γ = 2 SOLUTION 9.5 a. The plot is represented in Figure 9.1. b. The plot is represented in Figure 9.2. c. The plot is represented in Figure 9.3.

Nonlinear Dynamics and Chaos

FIGURE 9.1  Driven damped oscillator with γ = 0.3 .

FIGURE 9.2  Driven damped oscillator with γ = 1.06 .

227

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Classical Mechanics

FIGURE 9.3  Driven damped oscillator with γ = 2 .

PROBLEM 9.6 Consider the double pendulum in Figure 9.4 with masses m1 and m2 , and lengths L1 and L2 . Without assuming small angles, investigate the chaos of the system using numerical software. SOLUTION 9.6 The equations of motion will be derived using Lagrange’s equations. This requires the kinetic and potential energies. The kinetic energy can be found by considering the positions of the masses

1

L1

m1

2 L2

m2

FIGURE 9.4  Double pendulum.

Nonlinear Dynamics and Chaos

r1 = L1 (sin φ1 x + cos φ1 y )

and

r2 = r1 + L2 (sin φ2 x + cos φ2 y) = ( L1 sin φ1 + L2 sin φ2 ) x + ( L1 cos φ1 + L2 cos φ2 ) y The velocities are then given by   v1 = r1 = L1φ1 (cos φ1 x − sin φ1 y )

and

v2 = r2 = ( L1φ1 cos φ1 + L2φ2 cos φ2 ) x − ( L1φ1 sin φ1 + L2φ2 sin φ2 ) y

The kinetic energy of each mass is T1 =

1   1 m1v1 ⋅ v1 = m1L21φ12 2 2

and 1   1 m2 v2 ⋅ v2 = m2 (( L1φ1 cos φ1 + L2φ2 cos φ2 )2 + ( L1φ1 sin φ1 + L2φ2 sin φ2 )2 ) 2 2 1 2 2 = m2 ( L1φ1 cos2 φ1 + 2 L1L2φ1φ2 cos φ1 cos φ2 + L22φ22 cos2 φ2 + L21φ12 sin 2 φ1 2 + 2 L1L2φ1φ2 sin φ1 sin φ2 + L22φ22 sin 2 φ2 ) 1 = m2 ( L21φ12 + 2 L1L2φ1φ2 (sin φ1 sin φ2 + cos φ1 cos φ2 ) + L22φ22 ) 2 1 = m2 ( L21φ12 + 2 L1L2φ1φ2 cos(φ1 − φ2 ) + L22φ22 ) 2

T2 =

Therefore, T=

1 1 (m1 + m2 )L21φ12 + m2 L22φ22 + m2 L1L2φ1φ2 cos(φ1 − φ2 ) 2 2

The potential energies are given by

U1 = m1gL1 (1 − cos φ1 )

and

U 2 = m2 g( L1 (1 − cos φ1 ) + L2 (1 − cos φ2 ))

229

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Classical Mechanics

Therefore, U = gL1 (m1 + m2 )(1 − cos φ1 ) + m2 gL2 (1 − cos φ2 )

The Lagrangian is then given by  = T −U =

1 1 (m1 + m2 )L21φ12 + m2 L22φ22 + m2 L1L2φ1φ2 cos(φ1 − φ2 ) 2 2 − gL1 ( m1 + m2 ) (1 − cos φ1 ) − m2 gL2 (1 − cos φ2 )

The equations of motion are thus d ∂ ∂ = dt ∂φ1 ∂φ1

and

d ∂ ∂ = dt ∂φ2 ∂φ2

Since the objective is to investigate chaos numerically, and for the sake of simplicity, all constants are set to 1. Therefore,

1  = φ12 + φ22 + φ1φ2 cos(φ1 − φ2 ) − 2(1 − cos φ1 ) − (1 − cos φ2 ) 2 1 = φ12 + φ22 + φ1φ2 cos(φ1 − φ2 ) − 3 + 2 cos φ1 + cos φ2 2

with

d ∂ ∂ = dt ∂φ1 ∂φ1

which yields

sin(φ1 − φ2 )φ22 + 2(sin φ1 + φ1 ) + cos(φ1 − φ2 )φ22 = 0

and

d ∂ ∂ = dt ∂φ2 ∂φ2

Nonlinear Dynamics and Chaos

231

which yields

sin φ2 + cos(φ1 − φ2 )φ12 + φ2 − sin(φ1 − φ2 )φ12 = 0

Since these differential equations involve trigonometric functions mixed with nonlinear second-­order derivates, this cannot be solved analytically. Luckily, this is not required to investigate the chaos of the system; the differential equations can be solved numerically, given initial conditions. Consider a time range between 0 and 40, this system is solved for four different initial (φ1,i , φ2,i ) pairs (in units of degrees): (45, 45) , (46, 45) , (45, 46) , and (46, 46) . Notice all these initial conditions are quite close. Plotting the results for φ1 and φ2 yields Figures 9.5 and 9.6. Notice that despite the very close initial conditions, the trajectories of the pendulums are quite different; this is a hallmark of chaotic systems. Another way to look at this is the state-­space orbit. Solving this system for (φ1,i , φ2,i ) = (45, 45) over the time range of 800 yields the following φ1 versus φ1 state-­space orbit (Figure 9.7). Observe the trajectory never repeats, another clear indication of chaos. Therefore, the double pendulum is clearly a chaotic system.

FIGURE 9.5  Plot of φ1 (t ) for various initial positions.

232

FIGURE 9.6  Plot of φ2 (t ) for various initial positions.

FIGURE 9.7  State-space plot of φ1 (t ) versus φ1 (t ) .

Classical Mechanics

233

Nonlinear Dynamics and Chaos

PROBLEM 9.7 Consider the two-­pendulum system in Figure 9.8, each with mass m and length L , connected with a sprint of constant k . Without assuming small angles, investigate the chaos of the system using numerical software. SOLUTION 9.7 The equations of motion will be obtained via Lagrange’s equations. For this system, the kinetic energy is simply given by T=

1 2 2 2 mL (φ1 + φ2 ) 2

and the potential energy is U = mgL ((1 − cos φ1 ) + (1 − cos φ2 )) +

= mgL (2 − cos φ1 − cos φ2 ) +

1 2 kL (sin φ2 − sin φ1 )2 2

1 2 kL (sin φ2 − sin φ1 )2 2

where the last term in U is due to the spring. Therefore, the Lagrangian is given by

 = T −U =

1 2 2 2 1 mL (φ1 + φ2 ) − mgL (2 − cos φ1 − cos φ2 ) − kL2 (sin φ2 − sin φ1 )2 2 2

Since this will be analyzed numerically, the constants are set to 1. Thus,

=

1 2 2 1 (φ1 + φ2 ) − 2 + cos φ1 + cos φ2 − (sin φ2 − sin φ1 )2 2 2

1

2

L

L

k m

FIGURE 9.8  Two pendulums connected by a spring.

m

234

Classical Mechanics

The equations of motion are thus d ∂ ∂ = dt ∂φ1 ∂φ1

and

d ∂ ∂ = dt ∂φ2 ∂φ2

With the objective of investigating chaos numerically, numerical software can be used from this point out to calculate derivatives and plot results. Therefore, Lagrange equation for φ1 d ∂ ∂ = dt ∂φ1 ∂φ1

has the form

(1 + cos φ1 )sin φ1 + φ1 − cos φ1 sin φ2 = 0

and Lagrange equation for φ2

d ∂ ∂ = dt ∂φ2 ∂φ2

has the form

sin φ2 + φ2 − cos φ2 (sin φ1 − sin φ2 ) = 0

Given the mixture of derivatives and trigonometric functions, this cannot be solved analytically. Luckily, this is not required to investigate the chaos of the system; this can be solved numerical, given initial conditions. Consider a time range between 0 and 40, this system is solved for four different initial (φ1,i , φ2,i ) pairs (in units of degrees): (25, 45) , (26, 45) , (25, 46) , and (26, 46) . Notice all these initial conditions are quite close. Plotting the results for φ1 and φ2 yields Figures 9.9 and 9.10. It appears the trajectories differ more and more each period, however this pattern does not clearly indicate chaos. To dig into this a bit more, the state-­space orbit of φ1 versus φ1 can be calculated using the initial conditions (φ1,i , φ2,i ) = (25, 45) over the time range of 800 (Figure 9.11). While the individual trajectories did not show wildly diverging paths for very similar initial conditions, the state-­space orbit clearly indicates this is a chaotic system. This suggests that the trajectories would have diverged more significantly given more time.

Nonlinear Dynamics and Chaos

FIGURE 9.9  Plot of φ1 (t ) for various initial positions.

FIGURE 9.10  Plot of φ2 (t ) for various initial positions.

235

236

Classical Mechanics

FIGURE 9.11  State-space plot of φ1 (t ) versus φ1 (t ) .

PROBLEM 9.8 Consider the cart below, which is attached to the wall with a spring and has pendulum hanging below it. For a spring of constant k , a cart of mass m, and a pendulum of length L and mass M , determine if this system is complicated enough to exhibit chaos. Do not assume small angles (Figure 9.12). Note: This system was analyzed in Problem 8.7 using small angle approximation. SOLUTION 9.8 The equations of motion will be obtained via Lagrange’s equations. The kinetic energy of the cart is simply x m

k

L

M

FIGURE 9.12  Cart connected to a spring with a pendulum oscillating beneath it.

237

Nonlinear Dynamics and Chaos

Tm =

1 2 mx 2

The position of the pendulum is  rM = ( x + L sin φ ) x − L cos φ y

so the velocity is

vM = ( x + Lφ cos φ ) x + Lφ sin φ y

The kinetic energy of the pendulum is then given by TM =

1   1 MvM ⋅ vM = M ( x 2 + 2 Lxφ cos φ + L2φ2 ) 2 2

and the total kinetic energy is T=

1 1 (m + M ) x 2 + ML2φ2 + MLxφ cos φ 2 2

The potential is simply that of a spring and a pendulum. U=

1 2 kx + MgL (1 − cos φ ) 2

Therefore, the Lagrangian is

 = T −U =

1 1 1 (m + M ) x 2 + ML2φ2 + MLxφ cos φ − kx 2 − MgL (1 − cos φ ) 2 2 2

Since this will be analyzed numerically, the constants are set to 1, except M which is set to 2. Thus,

=

3 2 2 1 x + φ + 2 xφ cos φ − x 2 − 2 + 2 cos φ 2 2

The equations of motion are thus

d ∂ ∂ = dt ∂x ∂x

238

Classical Mechanics

and d ∂ ∂ = dt ∂φ ∂φ

With the objective of investigating chaos numerically, numerical software can be used from this point out to calculate derivatives and plot results. Therefore, Lagrange equation for x d ∂ ∂ = dt ∂x ∂x

yields to the equation

x + 3 x + 2φ cos φ − 2φ2 sin φ = 0

and Lagrange equation for φ

d ∂ ∂ = dt ∂φ ∂φ

yields to the equation

sin φ +  x cos φ + φ = 0

Given the mixture of derivatives and trigonometric functions, this cannot be solved analytically. Luckily, this is not required to investigate the chaos of the system. Consider a time range between 0 and 40, this system is solved for three different initial ( xi , φi ) pairs (angles are in degrees): (2, 45) , (2, 46) , and (2, 44) . Notice all these initial conditions are quite close. Plotting the results for x and φ yields Figures 9.13 and 9.14. There are clearly deviations of the trajectories which are larger than the initial conditions differences. This can be seen even more clearly by considering the state-­space orbits x versus x and φ versus φ over a time range of 800 and initial conditions ( xi , φi ) = (2, 45) (Figures 9.15 and 9.16). It is clear these orbits do not repeat themselves indicating this system is in fact complex enough to exhibit chaos.

Nonlinear Dynamics and Chaos

FIGURE 9.13  Plot of x(t ) for various initial positions.

FIGURE 9.14  Plot of φ (t ) for various initial positions.

239

240

FIGURE 9.15  State-space plot of x (t ) versus x(t ) .

FIGURE 9.16  State-space plot of φ(t ) versus φ (t ) .

Classical Mechanics

10

Special Relativity

10.1  THEORY This chapter presents a brief discussion of special relativity and contains problems concerning length contraction, time dilation, as well as relativistic dynamics and redshift.

10.1.1 Galileo’s Transformations Consider two reference frames S and S′, with S′ moving with a velocity V relative to S along x-­axis. In Newtonian mechanics, it is an assumption that there is a universal time t . The origins of the two reference frames are chosen such that they coincide at time t = 0 .  An event occurs in frame S at position r = ( x, y, z ) at time t , and it is measured in  frame S′ such that it occurs at r′ = ( x′, y′, z′) at time t ’ . The following equations constitute Galilean transformations:

x′ = x − Vt

y′ = y

z′ = z

t′ = t

Furthermore, the vector position r ′ in frame S′ can be written as r ′ = r − Vt and     the vector velocity v′ in frame S′ can be written as v′ = v − V , showing that in classical mechanics the velocity addition formula follows the vector addition rules (Figure 10.1). y

y′

S

y

y′

0

0′

S′ P(x, y, z, t)

V

x′

x′

z z

z′

z′

FIGURE 10.1  A system of two reference frames S and S′ , with frame S fixed and frame S′ moving along the x direction with constant velocity. DOI: 10.1201/9781003365709-10

241

242

Classical Mechanics

Newton’s Laws are invariant under the Galilean transformation. Electromagnetism laws, however, are not invariant under the Galilean transformations. For example, Maxwell claims that electromagnetic wave propagates in vacuum 1 with a speed c = = 3 × 108 m/s . If light is traveling with speed c with respect ε 0 µ0 to reference frame S , and if the reference frame S′ moves with a speed V along the x direction with respect to the reference frame S , then the speed in reference frame S′ is given by v′ = c + V , which would be greater than the speed of light. This obviously contradicts Maxwell’s equations. This dilemma was solved by Einstein in his theory on special relativity. An inertial reference frame is a reference frame (system of coordinates ( x, y, z) and time t ) in which all laws of physics hold.

10.1.2 Postulates of the Theory of Relativity First postulate of relativity: All laws of physics are the same in all inertial reference frames. Second postulate of relativity: The speed of light in vacuum is the same in all inertial reference systems and in all directions. In relativity, the time is not an invariant, as in Galilean transformations. Lorentz transformations are correct in relativistic mechanics and Maxwell’s equations are invariant under Lorentz’s transformations.

10.1.3 Lorentz Transformations Consider two reference frames S and S′, with S′ moving with a velocity V relative to S along the x-­axis (Figure 10.2). An event which occurs in frame S occurs at posi  tion r = ( x, y, z) at time t , and measured in frame S′ it occurs at r ′ = ( x′, y′, z′) at time t ′ . The following equations constitute Lorentz transformations: x′ =

x − Vt 1−

y

y′

y

y′

V2 c2

P(x, y, z, t) S′

S

0′

0

V

x′

x′

z z

z′

z′

FIGURE 10.2  Two frames S and S′ , with frame S fixed and frame S′ moving along the x  direction with velocity V .

243

Special Relativity

y′ = y

z′ = z

x 2 c t′ = V2 1− 2 c t −V

Similarly, the equations for the coordinates in reference frame S as a function of coordinates in reference frame S′ are x=

x′ + Vt ′ 1−

V2 c2

y = y′

z = z′

x′ 2 c t= 2 V 1− 2 c t′ + V

V 1 and Gamma factor is γ = . c 1− β 2 Note that Galileo’s transformations are obtained for c → ∞ , or as approximation V for V  c , β =  1. c All physics transformations are Lorentz transformation invariant, or L-­invariant. Classical mechanics is invariant to Galileo’s transformations, not to Lorentz transformations. For speeds close to speed of light, that is, in relativistic mechanics, Lorentz transformations need to be employed. Some usual notations are β =

10.1.4 Length Contraction, Time Dilation The length l measured in the reference system in motion S′ with velocity V in the x direction is smaller, compared to l0 , the length in the reference system at rest S (length contraction):

    V2 V2 l = x2′ − x1′ =  x2 1 − 2 − Vt ′  −  x1 1 − 2 − Vt ′      c c    

244

Classical Mechanics

= ( x2 − x1 ) 1 −

V2 V 2 l0 = l 1 − = < l0 0 γ c2 c2

The length of a ruler is maximum in the reference frame at rest. Note: Only the longitudinal dimensions in the x direction (the direction of the velocity V ) are changed, the dimensions in y and z directions are not changed. For example, a cube in reference frame S′ becomes a parallelepiped with a smallest side in the x direction, and with sides in y and z directions equal to each other. This is why the volume change is due to the change in dimension along the x direction, and has the formula indicated below. Please note that dV and dV0 refer to the elements of volume in reference frames S′ and S , while V refers to the velocity along the x direction.

dV = dV0 1 −

V2 dV = dV0 1 − β 2 = 0 2 c γ

The time τ a process takes in a reference system in motion S′ with velocity V in x direction is larger, compared to the time τ 0 it takes for the same process in the reference system at rest S (time dilation). Vx Vx t1 − 2 2 c c = t2 − t1 = τ 0 = γτ 0 > τ 0 τ = t2′ − t1′ = − 2 V V2 V2 V2 1− 2 1− 2 1− 2 1− 2 c c c c t2 −

The time of a process observed from a reference frame in motion is larger compared to the time of a process observed from a reference frame at rest. The time of a process observed from the reference frame at rest is minimum. On simultaneity: Simultaneity is relative, that is, two events simultaneous observed from a reference frame at rest S may not appear simultaneous while observed from a reference frame in motion S′. The speed of light or the maximum speed of light in vacuum cannot be overcome. Speed of light in vacuum is the maximum attainable speed. For the limit speed V close to speed of light, the dimensions of the objects go to zero limit, and the periods of time move to an infinity limit.

10.1.5 Composing Velocities By differentiating the Lorentz transformations, it yields that dx =

dx′ + Vdt ′ 1−

V2 c2

245

Special Relativity

dy = dy′

dt =

dz = dz′

Vdx′ c2 V2 1− 2 c

dt ′ +

The velocities in reference frame S are obtained by dividing the first three equations by dt ,

dx′ +V v′ + V dx dx′ + Vdt ′ vx = = = x ′ = dt ′ Vv dt dt ′ + Vdx′ 1 + Vdx′ 1 + 2x c 2 dt ′ c2 c

V2 V2 ′ 1 v − y v′y dy c2 = c2 = vy = = dx′ Vv′  Vv′  dt dt ′ + V 2 1 + 2x γ  1 + 2x  c c c  

V2 V2 vz′ 1 − 2 2 dz vz′ c = c = vz = = ′ dx′ Vv  Vv′  dt dt ′ + V 2 1 + 2x γ  1 + 2x  c c c  

dy′ 1 −

dz′ 1 −

and the equivalent equations for the velocities in reference frame S′ v′x =

vx − V Vv 1 − 2x c

V2 vy c2 = v′y = Vvx  Vv′  1− 2 γ  1 − 2x  c c  

V2 vz c2 = v′z = ′ Vv  Vv′  1 − 2x γ  1 − 2x  c c  

vy 1 −

vz 1 −

246

Classical Mechanics

Minkowski space (spacetime) is a space with the three Euclidian special dimensions and one time dimension ( x, y, z, ct ) . Note that the time t is multiplied by speed of light c for dimensional purposes.

10.1.6 Relativistic Dynamics The relativistic (or variable) mass mrel is dependent of velocity of the reference frame. If rest mass m is the mass measured in the reference frame at rest, the mass in a reference frame of the particle moving with velocity v in the x direction is mrel = m

1

m

=

2

v 1− 2 c

1− β 2

=γm

Linear momentum of a particle with rest mass m and velocity v is    p = mrel v = mv

1 1−

v2 c2

=

mv 1− β 2

Note that when the speed v is close to speed of light, the momentum increases to infinity. Relativistic kinetic energy K has the expression K = E − E0 =

mc 2 1−

2

v c2

− mc 2 = (γ − 1)mc 2

The rest energy E0 = mc 2 . The total energy of the particle is (Einstein equation) E = mrelc 2 =

mc 2 1−

2

v c2

= γ mc 2

Relativistic linear momentum is related to the total energy of the particle.

E 2 = p2c 2 + (mc 2 )2

Note: For photons, with m = 0 , the equation becomes E = pc .

247

Special Relativity

10.1.7 Doppler Shift 10.1.7.1  Redshift When an object moves away from the Earth, the velocity of the object can be determined from the following equation, in which λ is the observed wavelength, λ0 is the initial wavelength, and v is the speed of the galaxy, or object moving away (redshift) or closer (blueshift)

λ = λ0

c+v c

∆λ v = λ0 c

A redshift has been observed from different galaxies, which implied that they move away from each other, and therefore, the universe is expanding. 10.1.7.2  Blueshift This has the same expression, but the velocity v has negative sign

λ = λ0

c−v c

∆λ v =− λ0 c

10.2  PROBLEMS AND SOLUTIONS PROBLEM 10.1  An object is moving with velocity v = (vx , 0, 0) with respect to the system S , and  with velocity  v′ = (v′x , 0, 0) with respect to the system S′, which is moving with a velocity V = (V , 0, 0) relative to S along the x-­axis. Compose the speeds in both relativistic and classical way, and show that, in classical view, the absolute speed 3 becomes greater than speed of light c. Consider v′x = V = c . 4 SOLUTION 10.1 Relativistic calculation

3c 3c 6c + 6c 16 24 v′x + V 4 4 = cc 4 4 4 2

Note: This large speed requires, as expected, relativistic calculations, and this is why the classical method led to a speed greater than the speed of light, which is contradicting the relativistic expectation that the speed of light is the upper limit. PROBLEM 10.2 A trip taking two years in intergalactic space takes one year observed from the Earth. Assuming the rocket moves in the x direction, find the speed V . SOLUTION 10.2 The time is dilated in the reference frame in motion,

τ = t2′ − t1′ =

t2 − t1 1−

2

V c2

=

τ0 1−

V2 c2

so

τ = τ0

1 1−

V2 c2

By taking the square and rearranging,

τ 02 V2 = 1− 2 2 τ c

V2 τ2 = 1 − 02 2 c τ

V = c 1−

τ 02 τ2 2

 1 year  1 3c V = c 1−  = 0.866 c  = c 1− = years 2 4 2  

249

Special Relativity

PROBLEM 10.3 Muons are unstable elementary particles with a charge equal to the charge of an electron, but the mass about 207 times larger than the mass of an electron. They are created by cosmic radiation colliding with atoms in the atmosphere. The lifetime of stationary muons is 30 times smaller than the lifetime of moving muons. Calculate the velocity of the muons. (The slow-­moving muons have a lifetime of about 2.2 ∝s.) SOLUTION 10.3 As before,

τ0

τ=

1−

V2 c2

Here,

τ0 =

1 τ 30

Following the calculations as before: 2

V = c 1−

τ 02 900 − 1 899  1  = c 1−   = c= c = 0.99944 c τ2 30 30  30 

PROBLEM 10.4 The life span of a giant Pacific octopus may be of about five years (the common octopus lives one to two years). a. Find the life span of the octopus when measured by an observer moving at a speed of 0.9c relative to the octopus. b. If the speed is increased by 10%, what happens to the time interval in the reference frame in motion compared to the one from point (a)? SOLUTION 10.4 a. The proper time interval measured in the rest reference frame of the octopus is five years. The time interval measured in the frame moving with speed V is

τ = γτ 0 =

τ0 1−

2

V c2

=

5 years 1−

( 0.9c ) c

2

2

=

5 years 1 − ( 0.9 )

2

=

5 years = 11.46 years 0.436

250

Classical Mechanics

b. Now, the velocity is increased by 10%, which means that V2 = 0.9 c + 10%(0.9 c) = 1.1 × 0.9 c = 0.99 c 5 years

τ 2 = γ 2τ 0 =

1−

( 0.99 c ) c

2

=

5 years 1 − ( 0.99 )

2

=

5 years = 35.44 years 0.141

2

The time interval increases as

τ 2 − τ 35.44 years − 11.46 years = = 2.09 = 209% τ 11.46 years

The 10% increase in velocity leads to an increase of 209% in the dilated time interval.

PROBLEM 10.5 Twins Alpha and Beta do an experiment when they are 20 years old. Alpha remains on the Earth (reference frame), and Beta goes on a trip with a speed of V = 0.87c . If the proper time interval τ 0 is 10 years, calculate the relativistic time, assuming that Beta arrives at the destination and goes promptly back with the same speed. What is the age of the twin brothers when Beta arrives back to Earth? SOLUTION 10.5 Using the formula for the time dilation, it follows that

τ = γτ 0 =

τ0 1−

2

V c2

=

10 years 1−

( 0.87c )

2

=

10 years 1 − ( 0.87 )

2

=

10 years ≅ 20 years 0.5

c2

Therefore, now, Alpha is 40 years old, and Beta is only 30 years old. Also, note that there is no paradox in fact – Beta is moving with respect to an observer at rest with respect to the Earth, while Alpha remains in the reference frame at rest with respect to the observer. Also, the positions of the twins cannot be interchanged, because Beta will feel that he is decelerating and stopping momentarily at the destination, and then promptly coming back with the same velocity V , while Alpha does not perceive any acceleration or deceleration in the reference frame at rest.

251

Special Relativity

PROBLEM 10.6 A snake (“relativistic snake”) is measured as having the length of one meter in a reference frame at rest with respect to the snake. a. Calculate the length of the snake in a reference frame traveling on x direction (along the snake) with velocity V = 0.95c ? b. What is the length of the snake if the reference frame would move perpendicular to the snake? SOLUTION 10.6 a. The length of the snake measured in the reference frame moving with velocity V is

l=

l0 V2 (0.95c)2 = l0 1 − 2 = 1 m 1 − = 1 m 1 − (0.95)2 = 0.31 m c c2 γ

b. If the reference frame moves perpendicular to the snake, the length of the snake is observed to be still one meter, it is not changed. However, the diameter of the snake will be measured in the moving reference system and become smaller following the same formula d=

d0 V2 = d0 1 − 2 = 0.31 d0 γ c

PROBLEM 10.7 The energy of an electron in fast motion is four times the rest energy. a. Calculate the rest energy. b. Calculate the speed V of the electron. c. Calculate the kinetic energy of the electron in eV. d. Calculate the linear momentum of the electron in eV/c. SOLUTION 10.7 a. The rest energy is 2

m  E0 = mc 2 = 9.1 × 10 −31 kg  3 × 108  = 8.19 × 10 −14 J s   −14 8.19 × 10 J = 5.12 × 105 eV = 0.512 MeV = J 1.6 × 10 −19 eV

252

Classical Mechanics

b. The relativistic energy is four times the rest energy = E 4= E0 4 mc 2

The relativistic mass is E = mrelc 2 = mc 2 1−

2

v c2

mc 2 v2 1− 2 c

= γ mc 2

= 4 mc 2

From here,

1−

v2 1 = c2 4

1−

v2 1 = c 2 16 v2 1 = 1− c2 16

which leads to v = 1−

1 15 15 c= c= c = 0.97 c 16 16 4

c. Kinetic energy K is, using point (a),

K = E − E0 = 4 mc 2 − mc 2 = 3mc 2 = 3 × 0.512 MeV = 1.536 MeV d. Linear momentum is E 2 = p2c 2 + (mc 2 )2 = (4mc 2 )2

p2c 2 = 16(mc 2 )2 − (mc 2 )2 = 15(mc 2 )2

Simplifying by c 2 and taking the square root,

= p

mc 2 = 15 c

15

0.512 MeV = 1.98 MeV/c c

253

Special Relativity

PROBLEM 10.8 The spectroscopic measurement of light at wavelength λ = 656 nm coming from the (fictional) galaxy X has a redshift of ∆λ = 15 nm. Find the recessional speed of the galaxy. SOLUTION 10.8 The fractional redshift is given by ∆λ v = λ0 c

The recessional speed is

v=

∆λ 15 nm m m c= c = 0.0228 c = 2.28 × 3 × 106 = 6.86 × 106 λ0 656 nm s s

The galaxy X is moving away from Earth with a speed of 2.28% of speed of light. Note that for objects in the solar system, the wavelength shift may be very small, less than 10 −3 of an Angstrom, and there is a need for a very sensible spectrometer.

PROBLEM 10.9 Knowing the expressions for energy E = γ mc 2 and for linear momentum p = γ mv , find the relationship between the energy and linear momentum. SOLUTION 10.9 Method I

E c2 From energy and momentum, by dividing the two relationships it yields that = p v p c2 and the speed is v = E 1 By substituting γ = in the energy relationship for linear momentum v2 1− 2 c p = γ mv , p2c 4 2 2 4 mv E2 = m p c p2 = = 2 4 2 2 v pc E − p2c 2 1− 2 2 c 1 − E2 c 2 2

m2

254

Classical Mechanics

By cross-­multiplying, p2 ( E 2 − p2c 2 ) = m 2 p2c 4

After dividing both sides by the squared momentum (which is not zero), the relationship energy momentum is found as E 2 = p2c 2 + m 2c 4

Method II

Note that the difference can be calculated in a more elegant way:

E 2 − p2c 2 = (γ mc 2 )2 − (γ mv)2 c 2 = γ 2 m 2c 4 − γ 2 m 2 v 2c 2  v2  1 = γ 2 m 2c 4  1 − 2  = γ 2 m 2c 4 2 = m 2c 4 γ  c  From this it follows that

E 2 = p2c 2 + m 2c 4

Note that, when considering the photon, with mass m = 0 , the energy of the photon is simply

E = pc

Appendix Differential Equations In Classical Mechanics, many problems come down to solving an equation of motion. These equations can be comprised of positions, velocities, and accelerations; noting the last two are simply derivates of position. Therefore, the equations of motion are typically differential equations. This appendix serves as a refresher on how to solve a few different types of differential equations which are encountered in this book. This is by no means a complete representation of the field of differential equations and is more of an appendix which can be referred to as problems in the various chapters are worked. For a complete treatment of differential equations, the authors recommend the books Fundamentals of Differential Equations and Boundary Value Problems by Nagle, Saff, and Snider, or Elementary Differential Equations and Boundary Value Problems by Boyce and DiPrima.

SEPARABLE EQUATIONS Consider the first-­order ordinary differential equation (ODE) of the form dy = f ( x, y ) dx

where f ( x, y) is some general function of x and y . For an equation to be separable, the function f can be rewritten as f ( x, y ) =

g( x ) h( y)

Therefore, the ODE becomes dy g( x ) = dx h( y)

This can be solved by rewriting the equation as

h( y)dy = g( x )dx

and integrating

∫ h( y) dy = ∫ g( x) dx

Note: Do not forget the constant of integration! 255

256Appendix

FIRST-ORDER EQUATIONS WITH AN INTEGRATING FACTOR Consider the first-­order ODE of the form dy + p( x ) y = q( x ) dx

where p( x ) and q( x ) are general functions of x . Such an equation can be solved by considering an integrating factor ∝( x ) given by

µ ( x ) = exp

( ∫ p(x) dx )

To understand how this transforms the original ODE, consider multiplying the original equation by ∝( x )

µ ( x)

dy + µ ( x ) p( x ) y = µ ( x )q( x ) dx

Noting that

d µ ( x ) = p( x )µ ( x ) dx

the original ODE can be rewritten as

d ( µ ( x ) y) = µ ( x )q( x ) dx Therefore, the solution is given by

µ ( x ) y = µ ( x )q( x ) dx y=

1 µ ( x )q( x ) dx µ ( x)

Note: Do not forget the constant of integration!

SECOND-ORDER HOMOGENEOUS EQUATIONS Consider the second-­order homogeneous differential equation of the form

a

d2y dy + b + cy = 0 dx 2 dx

257

Appendix

where a , b , and c are constants. This type of equation can be solved by considering a solution of the form y = erx

Note the following derivatives

dy rx = re = ry dx

d2y 2 rx = r= e r2y dx 2 Substituting these into the original equation yields

ar 2 y + bry + cy = 0

and since y ↑ 0 , this reduces to

ar 2 + br + c = 0

Therefore, finding the solution to the second-­order homogeneous ODE amounts to finding the values of r that satisfy the above equation. Specially,

r± =

−b ± b2 − 4ac 2a

Since there are two solutions for r , there are two solutions to the ODE. Therefore, the full solution is a linear combination of all solutions, specifically

y = c1er+ x + c2er− x

where c1 , c2 are constants which can be found via the initial conditions of the problem.

Bibliography Burlacu, Lucian, Culegere de probleme de mecanica analitica, David, Dorin Gheorghe, Universitatea din Bucuresti, Facultatea de Fizica, Bucuresti, Romania, 1988. Byron, Frederick, Mathematics of Classical and Quantum Physics, Robert Fuller, Dover Publication, Inc., New York, 1992. Goldstein, Herbert, Classical Mechanics, Charles Poole, John Safko, Addison Wesley, San Francisco, CA, 2002. Hristev, Anatolie, Mecanica si Acustica, Editura Didactica si Pedagogica, Bucuresti, 1984. Morin, David, Introduction to Classical Mechanics: With Problems and Solutions, 1st edition, Cambridge University Press, Cambridge, UK, 2008. Nagle, Kent, Edward Saff, David Snider, Fundamentals of Differential Equations and Boundary Value Problems, Addison WesleyPearson, Boston, 2012. Sauer, Timothy, Numerical Analysis, Pearson, Boston, 2012. Taylor, John R., Classical Mechanics, University Science Books, Sausalito, CA, 2005.

259

Index β, damping constant, 117, 140 A acceleration, 1, 5, 8, 14, 17, 22, 28–30, 32, 46, 52–3, 101, 121–22, 131, 144–46, 148, 172–73, 177, 179, 250, 255 centripetal, 28–9, 101 in 2D polar coordinates, 8 in cartesian coordinates, 29, 46, 48 addition of velocities, 244 air resistance comparison of linear and quadratic, 43, 60, 66, 107 linear, 56–7, 60, 66, 62 quadratic, 43, 50, 52, 54, 56, 57, 60, 62, 66 angular momentum, 69, 70–1 conservation of, 70 angular velocity, 8, 28, 130, 134, 136, 153, 159, 161, 180, 187, 192 approach to chaos, 221, 228, 230–31, 233–34, 236–39 Attwood machine, 144 by Hamiltonian formalism, 188, 193 by Lagrange equations, 145–46 including pulley, 147 auxiliary equation, 117 axial symmetry, 76–7, 81 B bead on rotating rod, 153–5 on wire, using Lagrangean, 153–55, 157 on wire, using Hamiltonian, 187–89 block on the incline, 11, 57, 103, 129–30, 148–49, 161–62, 174, 179 on horizontal plane, 72, 103, 151 C center of mass, see CM centripetal acceleration, 28–9, 101 chaos, 221, 228, 230–31, 233–34, 236, 238 characteristic equation, 107 classical definition of force, 5–8 definition of momentum, 69 CM, center of mass, 69, 71, 75, 78, 81–2, 84–5, 87 collision elastic, 87–9 perfectly inelastic, 72–4, 87–9

conservation of angular momentum, 70 of energy, 92, 99, 100, 102–03, 107, 113, 128 of momentum, 72–3, 88–9 conservative force, 91 constrained system, 52 coupled oscillators, 197, 200, 204, 207 coupled pendulums, 207 critical damping, 117 cross product, 1 curl, of vector, 91–4, 98 cylindrical polar coordinates, 2, 6, 76–7, 81, 83, 153 D damped oscillations, 115–57, 123–36, 139 damping constant, β, 117, 140 degrees of freedom, 143, 197–98 differential equations, 116, 231, 255 general solution of, 121, 125–26, 225 Doppler shift, 247 dot product (scalar product), 1 double pendulum, 228, 231 drag, 43–4, 46, 50, 52–3, 56, 58, 60, 62–4, 66, 107 driven damped oscillations, 117, 225, 227 E elastic collision, 87–9 energy conservation of, 100, 103 kinetic, 73, 88, 91–2, 100, 102, 105, 108, 110, 113, 116, 129, 136, 143, 147–48, 151, 155, 157, 161, 163, 172, 174, 177, 179, 181, 185, 192, 194, 197, 203, 217, 228, 233, 236, 246, 251 mechanical, 108, 115–16, 118, 120, 144 of two particles, 73 potential, 88, 91–5, 98, 100, 102, 103, 105, 106, 110, 113, 115, 129, 135–37, 143, 147–48, 150–52, 155, 157, 160–61, 169, 172, 177, 181, 186, 192, 197–99, 207–08, 210, 214, 217, 228, 233 relativistic, 246 rest, 252, 251–52 equation of motion, 24, 45, 51, 54, 107, 115, 117, 121, 124, 127, 149–50, 153, 171, 197, 225–26 equilibrium, 25, 103, 115, 118, 125, 132, 134, 137, 151, 153–55, 157, 159, 177, 181, 189, 191, 197, 199, 203, 216

261

262Index F Feigenbaum relation, 224 force, 5–8 as gradient of potential energy, 91–5, 98 conservative, 91–2 frequency, 117, 127, 162 natural, 125 resonance, 140 G Galilean relativity, 241–42 general solution of second–order differential equation, 121, 125, 225 generalized coordinate, 143 generalized momentum, 168, 172–74, 178, 180, 190, 192, 194 gradient, 18, 107, 109, 185 H Hamiltonian, 167–73 definition of, 167 Hamilton’s equations, 167–76 compared with Lagrange’s equations, 181 homogeneous equation, 256–57 homogeneous solution, 256–57 Hooke’s law, 115, 117, 139 I incline block on, 11, 57, 103, 129–30, 148–49, 161–62, 174, 179 inelastic collision, 72–4, 87–9 invariance of Newton’s laws under Galilean transformation, 242 inverse Lorentz transformation, 243 K kinetic energy T, 73, 88, 91–2, 100, 102, 105, 108, 110, 113, 116, 129, 136, 143, 147–48, 151, 155, 157, 161, 163, 172, 174, 177, 179, 181, 185, 192, 194, 197, 203, 217, 228, 233, 236, 246, 251 relativistic, 246 Kronecker delta symbol, 3 L l angular momentum, 69–71 L total angular momentum, 70–1

Lagrange’s equations, 143 compared with Hamilton’s equations, 181 Law of Inertia, 5 length contraction, 241, 243 line integral, 2, 95 linear drag, 56–7, 60, 66, 62 compared to quadratic, 43, 60, 66, 107 horizontal motion with, 44 vertical motion with, 46 logistic map, 223 Lorentz transformation, 242–4 LRC circuit, 139 M mass energy, 252 invariant, 5,6 in relativity, 246, 252 variable, 69 matrix in coupled oscillations, 197–98, 202, 204–48 mechanical energy, 92, 99, 102, 128 mechanics classical, 6 Hamiltonian, 167 Lagrangian, 143 nonlinear, 221 Minkowski, 246 moment of inertia, 70 momentum angular, 70–1 conservation of, 70 generalized, 168, 172–74, 178, 180, 190, 192, 194 relativistic, 246 momentum–energy, 246 multistage rocket, 69, 79 N natural frequency, 117, 125, 140 Newton’s First Law, 5–6 validity of, 5 Newton’s Laws, 6–7 Newton’s Second Law, 6 in 2D polar coordinates, 8 in Cartesian coordinates, 22 Newton’s Third Law, 6 invalid in relativity, 6 validity of, 6 nonlinear mechanics, 221 normal frequencies, 198, 201–03, 205, 207, 209–14, 216, 219 numeric solution, 221, 223, 226, 228, 230–31, 233–34, 237–38

263

Index O

S

one–dimensional systems, 167 oscillations, 115, 197 coupled, 197 damped, 115–17, 123–26, 139 driven, 117 of bead on spinning hoop overdamped, 117 underdamped, 117

scalar, 1, 2, 3 scalar product, 1 sensitivity to initial conditions, 221–22, 231, 234, 238, 257 separation of variables, 20, 25, 35, 44 simple harmonic motion, 115, 118–19, 121, 127 energy, 116 simple pendulum, 56, 136 simultaneity, 6, 244 spacetime, 246 special relativity postulates of, 242 speed of light, 242 as speed limit for inertial frames, 243–44 invariance of, 242–44 non–invariance under Galilean transformation, 248 string, 8, 31, 110, 144, 146, 149, 172, 177, 207

P partial derivatives, 145, 147 particle, 5, 20, 30, 70, 71, 73, 91, 152 particular solution, 38, 225 pendulum, 31, 54, 55–6, 107, 109, 134, 159, 163, 209 double, 228, 231 driven damped, 225 simple, 56, 136 photon, 246, 254 polar coordinates, 4, 34, 86 cylindrical, 4, 153 spherical, 4 potential energy, 88, 91–5, 98, 100, 102, 103, 105, 106, 110, 113, 115, 129, 135–37, 143, 147–48, 150–52, 155, 157, 160–61, 169, 172, 177, 181, 186, 192, 197–99, 207–08, 210, 214, 217, 228, 233 product cross, 1 dot, 1 proper time, 249–50

T terminal speed, 62, 65 with linear drag, 46 with quadratic drag, 53 time in relativity, 249–50 proper, 249 torque, 70–1 twin paradox, 250 U underdamped oscillations, 122

Q quadratic drag compared to linear, 43, 60, 66, 107 horizontal and vertical motion, 50 vertical motion with, 52 R reference frame, 241, 244–45 inertial, 241–42 rest energy, 246, 251–52 RLC circuit, 139 rockets, 69 multistage, 69, 79 single stage, 74 thrust of, 69, 80 rolling, 101 rotation, 83, 101 about a fixed axis, 147

V variable mass, 246 vector cross product, 1 curl of, 91–4, 98 dot product, 1 scalar product, 1 velocity, 1, 5 velocity addition formula relativistic, 241 viscosity, 43 W weak damping, 117 work done by force, 95–7, 104 work – kinetic energy theorem, 91