Title: Solution Manual for Classical Dynamics of Particles and Systems – 5th Edition by Stephen T. Thornton and Jerry B. Marion | Complete Step-by-Step Solutions for Newtonian Mechanics, Lagrangian and Hamiltonian Dynamics, Rigid Body Motion, Oscillations

All 14 Chapters Covered




SOLUTION MANUAL

, Contents



Preface v
Problems Solved in Student Solutions Manual vii

1 Matrices, Vectors, and Vector Calculus 1

2 Newtonian Mechanics—Single Particle 29

3 Oscillations 79

4 Nonlinear Oscillations and Chaos 127

5 Gravitation 149

6 Some Methods in The Calculus of Variations 165

7 Hamilton’s Principle—Lagrangian and Hamiltonian Dynamics 181

8 Central-Force Motion 233

9 Dynamics of a System of Particles 277

10 Motion in a Noninertial Reference Frame 333

11 Dynamics of Rigid Bodies 353

12 Coupled Oscillations 397

13 Continuous Systems; Waves 435

14 Special Theory of Relativity 461


iii

,iv CONTENTS

, Preface



This Instructor’s Manual contains the solutions to all the end-of-chapter problems (but not the
appendices) from Classical Dynamics of Particles and Systems, Fifth Edition, by Stephen T.
Thornton and Jerry B. Marion. It is intended for use only by instructors using Classical Dynamics
as a textbook, and it is not available to students in any form. A Student Solutions Manual
containing solutions to about 25% of the end-of-chapter problems is available for sale to
students. The problem numbers of those solutions in the Student Solutions Manual are listed on
the next page.
As a result of surveys received from users, I continue to add more worked out examples in
the text and add additional problems. There are now 509 problems, a significant number over
the 4th edition.
The instructor will find a large array of problems ranging in difficulty from the simple
“plug and chug” to the type worthy of the Ph.D. qualifying examinations in classical mechanics.
A few of the problems are quite challenging. Many of them require numerical methods. Having
this solutions manual should provide a greater appreciation of what the authors intended to
accomplish by the statement of the problem in those cases where the problem statement is not
completely clear. Please inform me when either the problem statement or solutions can be
improved. Specific help is encouraged. The instructor will also be able to pick and choose
different levels of difficulty when assigning homework problems. And since students may
occasionally need hints to work some problems, this manual will allow the instructor to take a
quick peek to see how the students can be helped.
It is absolutely forbidden for the students to have access to this manual. Please do not
give students solutions from this manual. Posting these solutions on the Internet will result in
widespread distribution of the solutions and will ultimately result in the decrease of the
usefulness of the text.
The author would like to acknowledge the assistance of Tran ngoc Khanh (5th edition),
Warren Griffith (4th edition), and Brian Giambattista (3rd edition), who checked the solutions of
previous versions, went over user comments, and worked out solutions for new problems.
Without their help, this manual would not be possible. The author would appreciate receiving
reports of suggested improvements and suspected errors. Comments can be sent by email to
, the more detailed the better.
Stephen T. Thornton
Charlottesville, Virginia


v

,vi PREFACE

, CHAPTER 1
Matrices, Vectors,
and Vector
Calculus


1-1.

x1

45˚
x1

45˚
x3 x3

Axes x1 and x3 lie in the x1x3 plane.
The transformation equations are:
x1 x1 cos 45 x3 cos 45
x2 x2
x3 x3 cos 45 x1 cos 45
1 1
x x x
1
2 1 2 3
x2 x2
1 1
x x x
3 1 3
2 2
So the transformation matrix is:

 1 1 
 0 
2 2 

 0 1 0 
 1 1 
 0 
 2 2 


1

,2 CHAPTER 3


1-2.
a)
x3
D
E

 x2
O
 B
A C
x1

From this diagram, we have
OE cos

OA OE cos (1)

OB OE cos

OD

Taking the square of each equation in (1) and adding, we find
2 2 2 (2)
OE cos2 cos2 cos2 OA OB
2
OD
But
2 2 2
OA  OB  OC (3)
and
2 2 2
OC  OD  OE (4)
Therefore,
2 2 2 2
OA  OB  OD  OE (5)
Thus,

cos2   cos2   cos2   1 (6)
b)
x3
D
E
D
E
 B B x2
O

A A C C
x1

First, we have the following trigonometric relation:
2 2
OE OE 2OE OE cos (7)
2
EE

,MATRICES, VECTORS, AND VECTOR CALCULUS 3


But,
2 2 2 2
EE OB OA OD OD
OB
O
A

2 2
OE cos OE cos OE cos OE cos

2
OE cos OE (8)
cos

or,
2 2 2
EE OE cos2 cos2 cos2 OE cos2 cos2
cos2

2OE OE cos cos cos cos cos cos

OE 2 OE2 2OE OE cos cos cos cos cos cos (9)
Comparing (9) with (7), we find

cos   cos  cos    cos  cos    cos  cos   (10)


1-3.
3
3



e2
1




Denote the original axes by x1 , x2 , x3 , and the corresponding unit vectors by e1 , e2 , e3 . Denote
the new axes by x1 , x2 , x3 and the corresponding unit vectors by e1 , e 2 , e 3 . The effect
of the rotation is e1 e 3 , e2 e1 , e3 e 2 . Therefore, the transformation matrix is written
as:
cos e , e cos e , e cos e , e 0 1 0
cos e , e 1
cos e ,1 e
1 2
cos 1 e3 , e 0 0 1
2 1 2 2 2 3
cos e ,e  cos e ,e cos e ,e 1 0 0
3 1 3 2 3 3




1-4.
a) Let C = AB where A, B, and C are matrices. Then,
Cij Aik Bkj (1)
k

Ct C A B B A
ij ji jk ki ki jk
k k

,4 CHAPTER 3


Identifying B and A At ,
Bt
ki ik jk kj


C
ij  Bt  At  ik kj
(2)
k
t
or,
t
Ct AB Bt At (3)
1
b) To show that AB B 1A 1 ,

AB B 1A 1 I B 1A 1 (4)
That is, AB
(5)

AB B 1A 1 AIA 1 (6)
AA 1 I

B 1A 1 AB B 1IB
B 1B I


1-5. Take to be a two-dimensional matrix:

11 12 (1)
  11 22 12 21
21 22

Then,
2 2 2 2 2
2 2 2 2
2 2 2 2 2
11 22 11 22 12 21 12 21 11 21 12 22 11 21 12 22


2 2 2
2
22
2
11 12 21 11 12 11 21
2 11 22 12 21
2
12 22
2

2 2 2



2 2 2 2
(2)
2

11 12 22 21 11 21 12 22


But since is an orthogonal transformation matrix, ij kj ik .
j

Thus,
2 2 2 2 1
11 12 21 22
(3)
11 21 12 22

Therefore, (2) becomes 0

(4)
2 1

, MATRICES, VECTORS, AND VECTOR CALCULUS 5

1-6. The lengths of line segments in the xj and systems are
x j
x 2
j  x i
2

L j ; L i (1)