Solution manual for classical dynamics of particles and systems 5th edition by thornton

Solution Manual for Classical Dynamics of Particles and Systems 5th Edition by Thornton
Full file at .

CHAPTER

1

Matrices, Vectors,
and Vector Calculus

1-1.
x2 = x2′
x1′
45˚
x1
45˚

x3

x3′

Axes x′1 and x′3 lie in the x1 x3 plane.
The transformation equations are:

x1′ = x1 cos 45° − x3 cos 45°
x2′ = x2
x3′ = x3 cos 45° + x1 cos 45°
x1′ =

1
1

x1 −
x3
2
2

x2′ = x2
x3′ =

1
1
x1 −
x3
2
2

So the transformation matrix is:









1
2
0
1
2

0 −
1
0

1 
2

0 
1 

2 

1

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Solution Manual for Classical Dynamics of Particles and Systems 5th Edition by Thornton
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CHAPTER 1

1-2.
a)
x3

D
E

γ
O

β

x2

B

α
A

C

x1

From this diagram, we have

OE cos α = OA
OE cos β = OB

(1)

OE cos γ = OD
Taking the square of each equation in (1) and adding, we find
2

2

2

OE cos 2 α + cos 2 β + cos 2 γ  = OA + OB + OD

2

(2)

But
2

2

2

2

2

2

OA + OB = OC

(3)

and
OC + OD = OE

(4)

Therefore,

2

2

2

OA + OB + OD = OE

2

(5)

Thus,
cos 2 α + cos 2 β + cos 2 γ = 1

(6)

b)
x3
D
E
D′

O
A A′
x1

E′

θ

B′

B
C

x2

C′

First, we have the following trigonometric relation:
OE + OE′ − 2OE OE′ cos θ = EE′
2

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2

2

(7)


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But,
2

2

2

 
 

EE′ = OB′ − OB  + OA′ − OA  + OD′ − OD 

 
 


2

2


 

= OE′ cos β ′ − OE cos β  + OE′ cos α ′ − OE cos α 

 




+ OE′ cos γ ′ − OE cos γ 



2

2

(8)

or,
2
2
2
EE′ = OE′ cos 2 α ′ + cos 2 β ′ + cos 2 γ ′  + OE  cos 2 α + cos 2 β + cos 2 γ 

− 2OE′ OE cos α cos α ′ + cos β cos β ′ + cos γ cos γ ′ 
= OE′ 2 + OE2 − 2OE OE′ cos α cos α ′ + cos β cos β ′ + cos γ cos γ ′ 

(9)

Comparing (9) with (7), we find
cos θ = cos α cos α ′ + cos β cos β ′ + cos γ cos γ ′

(10)

1-3.
x3
e3

O
e1

e2

e2′

e3

A

x2

e2

e1′

e3 ′

e1

x1

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

 cos ( e1′ , e1 ) cos ( e1′ , e2 ) cos ( e1′ , e3 )  0 1 0 



λ =  cos ( e2′ , e1 ) cos ( e2′ , e2 ) cos ( e2′ , e3 )  = 0 0 1 

 1 0 0 

 cos ( e3′ , e1 ) cos ( e3′ , e2 ) cos ( e3′ , e3 )  

1-4.
a) Let C = AB where A, B, and C are matrices. Then,

Cij = ∑ Aik Bkj

(1)

k

(C )
t

Full file at .

ij

= C ji = ∑ Ajk Bki = ∑ Bki Ajk
k

k


Solution Manual for Classical Dynamics of Particles and Systems 5th Edition by Thornton
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CHAPTER 1

( )

Identifying Bki = Bt

ik

( )

and Ajk = At

,

kj

(C ) = ∑ (B ) ( A )
t

t

ij

t

ik

k

(2)

kj

or,
C t = ( AB) = Bt At

(3)

( AB) B−1 A−1 = I = ( B−1 A−1 ) AB

(4)

( AB) B−1 A−1 = AIA−1 = AA−1 = I

(5)

(B

(6)

t

b) To show that ( AB) = B−1 A−1 ,
−1

That is,

1-5.

−1

)

A−1 ( AB) = B−1 IB = B−1B = I

Take λ to be a two-dimensional matrix:

λ =

λ11 λ12
= λ11λ 22 − λ12 λ 21
λ 21 λ 22

(1)

Then,
2
2
2
2 2
2
λ = λ112 λ 22
− 2λ11λ 22 λ12 λ 21 + λ12
λ 21
+ ( λ11
λ 21 + λ122 λ 22
) − ( λ112 λ 212 + λ122 λ 222 )
2

(

)

(

) (

2
2
2
2
2
= λ 22
λ112 + λ122 + λ 21
λ112 + λ122 − λ112 λ 21
+ 2λ11λ 22λ12 λ 21 + λ12
λ 22

(

)(

)

2
2
2
2
= λ11

+ λ12
λ 22
+ λ 21
− ( λ11λ 21 + λ12 λ 22 )

But since λ is an orthogonal transformation matrix,

2

∑λ λ
ij

)
(2)

kj

= δ ik .

j

Thus,
2
2
λ112 + λ122 = λ 21
+ λ 22
=1

(3)

λ11λ 21 + λ12 λ 22 = 0
Therefore, (2) becomes

λ =1
2

1-6.

The lengths of line segments in the x j and x ′j systems are

L=

∑x
j

Full file at .

(4)

2
j

; L′ =

∑ x′
i

i

2

(1)


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If L = L′ , then

∑ x = ∑ x′
2
j

2

i

(2)

xi′ = ∑ λ ij x j

(3)

j

i

The transformation is
j

Then,


∑ x = ∑  ∑ λ
2
j

j

i

ik

k



xk   ∑ λ iA xA 

 A

(4)



= ∑ x k x A  ∑ λ ik λ iA 
 i

k ,A

But this can be true only if

∑λ

ik

λ iA = δ k A

(5)

i

which is the desired result.
1-7.
x3
(0,0,1)

(1,0,1)

(0,0,0)

x1

(0,1,1)

(1,1,1)

(1,0,0)

(0,1,0)

x2

(1,1,0)

There are 4 diagonals:
D1 , from (0,0,0) to (1,1,1), so (1,1,1) – (0,0,0) = (1,1,1) = D1 ;
D 2 , from (1,0,0) to (0,1,1), so (0,1,1) – (1,0,0) = (–1,1,1) = D 2 ;
D 3 , from (0,0,1) to (1,1,0), so (1,1,0) – (0,0,1) = (1,1,–1) = D 3 ; and
D 4 , from (0,1,0) to (1,0,1), so (1,0,1) – (0,1,0) = (1,–1,1) = D 4 .
The magnitudes of the diagonal vectors are

D1 = D 2 = D 3 = D 4 = 3
The angle between any two of these diagonal vectors is, for example,

D1 ⋅ D 2
(1,1,1) ⋅ ( −1,1,1) 1
= cos θ =
=
D1 D 2
3
3

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Solution Manual for Classical Dynamics of Particles and Systems 5th Edition by Thornton
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CHAPTER 1

so that
1
θ = cos−1   = 70.5°
 3

Similarly,
D1 ⋅ D 3
D ⋅D
D ⋅D
D ⋅D
D ⋅D
1
= 1 4 = 2 3 = 2 4 = 3 4 =±
D1 D 3
D1 D 4
D2 D3
D2 D 4
D3 D 4
3
1-8. Let θ be the angle between A and r. Then, A ⋅ r = A2 can be written as

Ar cos θ = A2

or,

r cos θ = A

(1)

This implies
QPO =

π
2

Therefore, the end point of r must be on a plane perpendicular to A and passing through P.
1-9.
a)

A = i + 2j − k

B = −2i + 3j + k

A − B = 3i − j − 2k
2
2
A − B = ( 3) + ( −1) + (−2)2 

12

A − B = 14
b)
B
θ
A

component of B along A

The length of the component of B along A is B cos θ.

A ⋅ B = AB cos θ
B cos θ =

A ⋅ B −2 + 6 − 1 3
6
=
=
or
A
2
6
6

The direction is, of course, along A. A unit vector in the A direction is
1
( i + 2j − k )
6

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So the component of B along A is
1
( i + 2j − k )
2
cos θ =

c)

A⋅B
3
3
3
=
=
; θ = cos −1
AB
6 14 2 7
2 7

θ  71°
i

j

k

2 −1
1 −1
1 2
A × B = 1 2 −1 = i
−j
+k
3 1
−2 1
−2 3
−2 3 1

d)

A × B = 5i + j + 7 k
A − B = 3i − j − 2k

e)

A + B = −i + 5j
i

j

k

( A − B ) × ( A + B ) = 3 −1 − 2
−1

5

0

( A − B) × ( A + B) = 10i + 2j + 14k
1-10.

r = 2b sin ω t i + b cos ω t j

v = r = 2bω cos ω t i − bω sin ω t j
a)

a = v = −2bω 2 sin ω t i − bω 2 cos ω t j = −ω 2 r
speed = v =  4b 2ω 2 cos 2 ω t + b 2ω 2 sin 2 ω t 
= bω  4 cos 2 ω t + sin 2 ω t 

12

speed = bω  3 cos 2 ω t + 1
b)

At t = π 2ω , sin ω t = 1 , cos ω t = 0

So, at this time, v = − bω j , a = −2bω 2 i
So, θ  90°

Full file at .

12

12


Solution Manual for Classical Dynamics of Particles and Systems 5th Edition by Thornton
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CHAPTER 1

1-11.
a) Since ( A × B) i = ∑ ε ijk Aj Bk , we have
jk

(A × B) ⋅ C = ∑∑ ε ijk Aj Bk Ci
j,k

i

= C1 ( A2 B3 − A3 B2 ) − C2 ( A1B3 − A3 B1 ) + C3 ( A1B2 − A2 B1 )
C1
= A1

C2
A2

C3
A1
A3 = − C1

A2
C2

A3 A1
C3 = B1

A2
B2

A3
B3 = A ⋅ ( B × C )

B1

B2

B3

B1

B2

B3

C2

C3

B3

C1

(1)

We can also write
C1

C2

C3

B1

B2

(A × B) ⋅ C = − B1
A1

B2

B3 = C1

C2

C3 = B ⋅ ( C × A )

A2

A3

A2

A3

A1

(2)

We notice from this result that an even number of permutations leaves the determinant
unchanged.
b) Consider vectors A and B in the plane defined by e1 , e2 . Since the figure defined by A, B,
C is a parallelepiped, A × B = e3 × area of the base, but e3 ⋅ C = altitude of the parallelepiped.
Then,

C ⋅ ( A × B) = ( C ⋅ e3 ) × area of the base
= altitude × area of the base
= volume of the parallelepiped

1-12.
O

c
C

a

h

a–c
b

A

b–a

c–b

B

The distance h from the origin O to the plane defined by A, B, C is

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h=

=
=

a ⋅ ( b − a ) × ( c − b)

( b − a ) × ( c − b)

a ⋅ ( b × c − a × c + a × b)
b×c−a×c+a×b
a⋅b× c
a×b+b×c+c×a

(1)

The area of the triangle ABC is:
A=

1-13.

1
1
1
( b − a ) × ( c − b) = ( a − c ) × ( b − a ) = ( c − b) × ( a − c )
2
2
2

Using the Eq. (1.82) in the text, we have
A × B = A × ( A × X ) = ( X ⋅ A ) A − ( A ⋅ A ) X = φ A − A2 X

from which
X=

( B × A ) + φA
A2

1-14.

a)

 1 2 −1  2 1 0   1 −2 1 


 

AB =  0 3 1  0 −1 2  =  1 −2 9 
 2 0 1   1 1 3   5 3 3 

Expand by the first row.
AB = 1

−2 9

3

3

+2

1 9
1 −2
+1
5 3
5 3

AB = −104

b)

 1 2 −1  2 1   9 7 


 

AC = 0 3 1   4 3  = 13 9 
 2 0 1   1 0   5 2 
 9 7


AC = 13 9 
 5 2 

Full file at .

(2)


Solution Manual for Classical Dynamics of Particles and Systems 5th Edition by Thornton
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CHAPTER 1

5
 1 2 −1  8




ABC = A ( BC ) = 0 3 1   −2 −3 
 2 0 1   9 4 

c)

 −5 −5 


ABC =  3 −5 
 25 14 

AB − Bt At = ?

d)

 1 −2 1 


AB =  1 −2 9 
 5 3 3 

(from part a )

 2 0 1  1 0 2  1 1 5


 

B A =  1 −1 1   2 3 0  =  −2 −2 3

 0 2 3   −1 1 1   1 9 3
t

t

 0 −3 −4 


6
AB − B A =  3 0
 4 −6 0 
t

1-15.

t

If A is an orthogonal matrix, then
At A = 1
1 0 0 1 0 0  1 0 0


 

0 a a  0 a − a  = 0 1 0 
 0 − a a  0 a a   0 0 1 
1 0

2
0 2a

0 0
a=

Full file at .

1
2

0  1 0 0
 

0  = 0 1 0 
2a 2  0 0 1 


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1-16.
x3

P
r
θ

r

a

θ a

x2
x1

r cos θ

r ⋅ a = constant
ra cos θ = constant
It is given that a is constant, so we know that
r cos θ = constant
But r cos θ is the magnitude of the component of r along a.
The set of vectors that satisfy r ⋅ a = constant all have the same component along a; however, the
component perpendicular to a is arbitrary.
Thus the surface represented by r ⋅ a = constant
is a plane perpendicular to a.

1-17.
a
B

C

θ

c

b

A

Consider the triangle a, b, c which is formed by the vectors A, B, C. Since

C= A−B
C = ( A − B) ⋅ ( A − B)
2

(1)

= A 2 − 2A ⋅ B + B 2
or,
C = A2 + B2 − 2 AB cos θ
2

which is the cosine law of plane trigonometry.
1-18. Consider the triangle a, b, c which is formed by the vectors A, B, C.
a
B
c

Full file at .

α

C

γ

β
A

b

(2)


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CHAPTER 1

C=A−B

(1)

C × B = ( A − B) × B

(2)

so that

but the left-hand side and the right-hand side of (2) are written as:

C × B = BC sin α e3

(3)

( A − B) × B = A × B − B × B = A × B = AB sin γ e3

(4)

and

where e3 is the unit vector perpendicular to the triangle abc. Therefore,
BC sin α = AB sin γ

(5)

or,
C
A
=
sin γ sin α
Similarly,
C
A
B
=
=
sin γ sin α sin β

(6)

which is the sine law of plane trigonometry.

1-19.
x2
a

a2

b

b2
α

β
a1

b1

x1

a) We begin by noting that

a − b = a 2 + b 2 − 2ab cos (α − β )
2

(1)

We can also write that

a − b = ( a1 − b1 ) + ( a2 − b2 )
2

2

2

= ( a cos α − b cos β ) + ( a sin α − b sin β )
2

(

)

(

2

)

= a 2 sin 2 α + cos 2 α + b 2 sin 2 β + cos 2 β − 2 ab ( cos α cos β + sin α sin β )
= a 2 + b 2 − 2ab ( cos α cos β + sin α sin β )

Full file at .

(2)


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13

Thus, comparing (1) and (2), we conclude that
cos (α − β ) = cos α cos β + sin α sin β
b)

(3)

Using (3), we can find sin (α − β ) :

sin (α − β ) = 1 − cos 2 (α − β )
= 1 − cos 2 α cos 2 β − sin 2 α sin 2 β − 2cos α sin α cos β sin β

(

)

(

)

= 1 − cos 2 α 1 − sin 2 β − sin 2 α 1 − cos 2 β − 2cos α sin α cos β sin β
= sin 2 α cos 2 β − 2sin α sin β cos α cos β + cos 2 α sin 2 β
=

( sin α cos β − cos α sin β )2

(4)

so that
sin (α − β ) = sin α cos β − cos α sin β

(5)

1-20.
a) Consider the following two cases:

When i ≠ j

δ ij = 0

but ε ijk ≠ 0 .

When i = j

δ ij ≠ 0

but ε ijk = 0 .

Therefore,

∑ε

ijk

δ ij = 0

ij

b) We proceed in the following way:

When j = k, ε ijk = ε ijj = 0 .
Terms such as ε j11 ε A11 = 0 . Then,

∑ε

ijk

ε Ajk = ε i12 ε A12 + ε i13 ε A13 + ε i 21 ε A 21 + ε i 31 ε A 31 + ε i 32 ε A 32 + ε i 23 ε A 23

jk

Now, suppose i = A = 1 , then,

∑= ε
jk

Full file at .

123

ε 123 + ε 132 ε 132 = 1 + 1 = 2

(1)


Solution Manual for Classical Dynamics of Particles and Systems 5th Edition by Thornton
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CHAPTER 1

∑= ε

for i = A = 2 ,

213

ε 213 + ε 231 ε 231 = 1 + 1 = 2 . For i = A = 3 ,

jk

A = 2 gives

∑= ε

312

ε 312 + ε 321 ε 321 = 2 . But i = 1,

jk

∑ = 0 . Likewise for i = 2, A = 1 ; i = 1, A = 3 ; i = 3, A = 1 ; i = 2, A = 3 ; i = 3, A = 2 .
jk

Therefore,

∑ε

ijk

ε Ajk = 2δ iA

(2)

j,k

c)

∑ε

ijk

ε ijk = ε 123 ε 123 + ε 312 ε 312 + ε 321 ε 321 + ε 132 ε 132 + ε 213 ε 213 + ε 231 ε 231

ijk

= 1 ⋅ 1 + 1 ⋅ 1 + ( −1) ⋅ ( −1) + ( −1) ⋅ ( −1) + ( −1) ⋅ ( −1) + (1) ⋅ (1)
or,

∑ε

ijk

ε ijk = 6

(3)

ijk

( A × B)i = ∑ ε ijk Aj Bk

1-21.

(1)

jk

( A × B) ⋅ C = ∑
i

∑ε

ijk

Aj Bk Ci

(2)

jk

By an even permutation, we find

ABC = ∑ ε ijk Ai Bj Ck
ijk

1-22.

To evaluate

∑ε

ijk

ε Amk we consider the following cases:

k

a)

i= j:

∑ε

ijk

k

b)

i=A:

∑ε

ε Amk = ∑ ε iik ε Amk = 0 for all i , A , m
k

ijk

k

ε Amk = ∑ ε ijk ε imk = 1 for j = m and k ≠ i , j
k

= 0 for j ≠ m
c)

i = m:

∑ε

ijk

k

ε Amk = ∑ ε ijk ε Aik = 0 for j ≠ A
k

= −1 for j = A and k ≠ i , j
d)

j=A:

∑ε
k

ijk

ε Amk = ∑ ε ijk ε jmk = 0 for m ≠ i
k

= −1 for m = i and k ≠ i , j

Full file at .

(3)


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∑ε

j = m:

e)

15

ε Amk = ∑ ε ijk ε Ajk = 0 for i ≠ A

ijk

k

k

= 1 for i = A and k ≠ i , j
f)

A = m:

∑ε
k

ijk

ε Amk = ∑ ε ijk ε AAk = 0 for all i , j , k
k

i ≠ A or m : This implies that i = k or i = j or m = k.

g)

Then,

∑ε

ijk

ε Amk = 0 for all i , j , A , m

k

j ≠ A or m :

h)

∑ε

ijk

ε Amk = 0 for all i , j , A , m

k

Now, consider δ iA δ jm − δ im δ jA and examine it under the same conditions. If this quantity
behaves in the same way as the sum above, we have verified the equation

∑ε

ijk

ε Amk = δ iA δ jm − δ im δ jA

k

a)

i = j : δ iA δ im − δ im δ iA = 0 for all i , A , m

b)

i = A : δ ii δ jm − δ im δ ji = 1 if j = m, i ≠ j , m

= 0 if j ≠ m
i = m : δ iA δ ji − δ ii δ jA = −1 if j = A , i ≠ j , A

c)

= 0 if j ≠ A
j = A : δ iA δ Am − δ im δ AA = −1 if i = m, i ≠ A

d)

= 0 if i ≠ m
j = m : δ iA δ mm − δ im δ mA = 1 if i = A , m ≠ A

e)

= 0 if i ≠ A
f)

A = m : δ iA δ jA − δ il δ jA = 0 for all i , j , A

g)

i ≠ A , m : δ iA δ jm − δ im δ jA = 0 for all i , j , A , m

h)

j ≠ A , m : δ iA δ jm − δ im δ iA = 0 for all i , j , A , m

Therefore,

∑ε

ijk

ε Amk = δ iA δ jm − δ im δ jA

k

Using this result we can prove that

A × ( B × C ) = ( A ⋅ C ) B − ( A ⋅ B) C

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(1)


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CHAPTER 1

First ( B × C ) i = ∑ ε ijk Bj Ck . Then,
jk

[ A × (B × C) ]A = ∑ ε Amn Am ( B × C ) n = ∑ ε Amn Am ∑ ε njk BjCk
mn

=

mn

∑ε

Amn

njk

ε mn ε jkn Am Bj Ck

Am Bj Ck

jkmn

jk

jkmn



= ∑  ∑ ε lmn ε jkn  Am Bj Ck

jkm  n

(

)

= ∑ δ jlδ km − δ kAδ jm Am Bj Ck

jkm





= ∑ Am BA Cm − ∑ Am BmCA = BA  ∑ AmCm  − CA  ∑ Am Bm 
 m

 m

m
m
= ( A ⋅ C ) BA − ( A ⋅ B) CA
Therefore,

A × ( B × C ) = ( A ⋅ C ) B − ( A ⋅ B) C

1-23.

Write

( A × B) j = ∑ ε jAm AA Bm
Am

( C × D) k = ∑ ε krs Cr Ds
rs

Then,

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17







[ ( A × B) × (C × D)]i = ∑ ε ijk  ∑ ε jAm AA Bm   ∑ ε krs Cr Ds 
Am

jk

=

∑ε

ijk

rs

ε jAm ε krs AA BmCr Ds

jk Amrs

=

∑ε

jAm

jAmrs

=



 ∑ ε ijk ε rsk  AA Bm Cr Ds
k

∑ ε (δ
jAm

ir

)

δ js − δ is δ jr AA BmCr Ds

jAmrs

(

= ∑ ε jAm AA BmCi Dj − AA Bm Di C j
jAm

)





=  ∑ ε jAm Dj AA Bm  Ci −  ∑ ε jAm C j AA Bm  Di
 jAm

 jAm

= (ABD)Ci − (ABC)Di
Therefore,

[( A × B) × (C × D) ] = (ABD)C − (ABC)D
1-24.

Expanding the triple vector product, we have
e × ( A × e) = A ( e ⋅ e) − e ( A ⋅ e)

(1)

A ( e ⋅ e) = A

(2)

A = e ( A ⋅ e) + e × ( A × e)

(3)

But,

Thus,

e(A · e) is the component of A in the e direction, while e × (A × e) is the component of A
perpendicular to e.

Full file at .


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CHAPTER 1

1-25.
er
eφ
θ
φ
eθ

The unit vectors in spherical coordinates are expressed in terms of rectangular coordinates by
eθ = ( cos θ cos φ , cos θ sin φ , − sin θ ) 


eφ = ( − sin φ , cos φ , 0 )


er = ( sin θ cos φ , sin θ sin φ , cos θ ) 

(1)

Thus,

(

e θ = −φ cos θ sin φ − θ sin θ cos φ , φ cos θ cos φ − θ sin θ sin φ , − θ cos θ

)

= −θer + φ cos θ eφ

(2)

Similarly,

(

e φ = −φ cos φ , − φ sin φ , 0

)

= −φ cos θ eθ − φ sin θ er

(3)

e r = φ sin θ eφ + θeθ

(4)

Now, let any position vector be x. Then,
x = rer

(

(5)

)

x = re r + re r = r φ sin θ eφ + θeθ + re r
= rφ sin θ eφ + rθeθ + rer

(

(6)

(

)

)

  cos θ + rφ sin θ e + rφ sin θ e + rθ + rθ e + rθe + 

x = rφ sin θ + rθφ
rer + re r
φ
φ
θ
θ

(

(

)

)

  cos θ + rφ sin θ e + r − rφ 2 sin 2 θ − rθ 2 e
= 2rφ sin θ + 2rθφ
φ
r

(

)

+ 2rθ + rθ − rφ 2 sin θ cos θ eθ
or,

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19

( )

1 d 2   2


x = a = 
r − rθ 2 − rφ 2 sin 2 θ  er + 
r θ − rφ sin θ cos θ  eθ
 r dt

 1

d 2
+
r φ sin 2 θ  eφ

 r sin θ dt


(

1-26.

)

(8)

When a particle moves along the curve
r = k (1 + cos θ )

(1)

we have
r = − kθ sin θ




2



r = − k θ cos θ + θ sin θ  
Now, the velocity vector in polar coordinates is [see Eq. (1.97)]

(2)

v = rer + rθ eθ

(3)

so that
v 2 = v = r 2 + r 2θ 2
2

(

)

= k 2θ 2 sin 2 θ + k 2 1 + 2 cos θ + cos 2 θ θ 2
= k 2θ 2 2 + 2 cos θ 

(4)

and v 2 is, by hypothesis, constant. Therefore,

θ =

v2
2k (1 + cos θ )

(5)

v
2kr

(6)

2

Using (1), we find

θ =

Differentiating (5) and using the expression for r , we obtain

θ =

v 2 sin θ
v 2 sin θ
=
2
4r 2
4 k 2 (1 + cos θ )

(7)

The acceleration vector is [see Eq. (1.98)]

(

)

(

)

a = 
r − rθ 2 er + rθ + 2rθ eθ

so that

Full file at .

(8)


Solution Manual for Classical Dynamics of Particles and Systems 5th Edition by Thornton
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CHAPTER 1

a ⋅ er = 
r − rθ 2

(

)

= − k θ 2 cos θ + θ sin θ − k (1 + cos θ ) θ 2


θ 2 sin 2 θ
= − k θ 2 cos θ +
+ (1 + cos θ ) θ 2 

2 (1 + cos θ )




1 − cos 2 θ
= − kθ 2  2 cos θ +
+ 1
2 (1 + cos θ ) 

3
= − kθ 2 (1 + cos θ )
2

(9)

or,
3 v2
4 k

(10)

3 v 2 sin θ
4 k 1 + cos θ

(11)

( a ⋅ er ) 2 + ( a ⋅ eθ ) 2

(12)

3 v2
4 k

(13)

a ⋅ er = −

In a similar way, we find
a ⋅ eθ = −

From (10) and (11), we have
a =

or,
a =

2
1 + cos θ

1-27. Since

r × (v × r) = (r ⋅ r) v − (r ⋅ v) r

we have
d
d
r × ( v × r )] = [(r ⋅ r ) v − ( r ⋅ v) r ]
[
dt

dt
= (r ⋅ r) a + 2 (r ⋅ v) v − (r ⋅ v) v − ( v ⋅ v) r − (r ⋅ a) r

(

= r 2a + ( r ⋅ v ) v − r v2 + r ⋅ a

)

(1)

Thus,

(

d
[ r × ( v × r )] = r 2a + ( r ⋅ v ) v − r r ⋅ a + v 2
dt

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)

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1-28.

21
grad ( ln r ) = ∑
i

∂
( ln r ) ei
∂x i

(1)

where

∑x

r =

2
i

(2)

i

Therefore,
∂
1

ln r ) =
(
r
∂x i

xi

∑x

2
i

i

=

xi

r

(3)

2

so that

grad ( ln r ) =


1 

xe
2 ∑ i i

r  i

(4)

r
r2

(5)

or,

grad ( ln r ) =

1-29. Let r 2 = 9 describe the surface S1 and x + y + z 2 = 1 describe the surface S2 . The angle θ
between S1 and S2 at the point (2,–2,1) is the angle between the normals to these surfaces at the
point. The normal to S1 is

(

)

(

grad ( S1 ) = grad r 2 − 9 = grad x 2 + y 2 + z 2 − 9
= ( 2xe1 + 2 ye2 + 2ze3 )

x = 2, y = 2, z = 1

)
(1)

= 4e1 − 4e2 + 2e3
In S2 , the normal is:

(

)

grad ( S2 ) = grad x + y + z 2 − 1
= ( e1 + e2 + 2ze3 )
= e1 + e2 + 2e3
Therefore,

Full file at .

x = 2, y =−2, z = 1

(2)


Solution Manual for Classical Dynamics of Particles and Systems 5th Edition by Thornton
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CHAPTER 1

cos θ =

=

grad ( S1 ) ⋅ grad ( S2 )
grad ( S1 ) grad ( S2 )

( 4e1 − 4e2 + 2e3 ) ⋅ (e1 + e2 + 2e3 )

(3)

6 6

or,
cos θ =

4

(4)

6 6

from which

θ = cos −1

3

1-30.

grad ( φψ ) = ∑ ei

i =1

(5)

 ∂ψ ∂ φ 
∂ ( φψ )
= ∑ ei φ
+
ψ
∂x i
i
 ∂x i ∂x i 

= ∑ ei φ
i

6
= 74.2°
9

∂ψ
∂φ
+ ∑ ei
ψ
∂x i
∂x i
i

Thus,
grad ( φψ ) = φ grad ψ + ψ grad φ

1-31.
a)
12
 
∂rn
∂ 
n
2
 ∑ xj  
grad r = ∑ ei
= ∑ ei
∂x i
∂xi  j
 
i =1


n

3

n

2
n
= ∑ ei 2xi  ∑ x 2j 
2 j

i

n


2
= ∑ ei xi n  ∑ x 2j 
 j

i

−1

−1

= ∑ ei x i n r ( n − 2)

(1)

i

Therefore,
grad r n = nr ( n − 2) r

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23

b)
3

grad f ( r ) = ∑ ei
i =1

∂f ( r ) 3 ∂ f ( r ) ∂ r
= ∑ ei
∂x i
∂ r ∂x i
i =1


∂ 
= ∑ ei
x 2j 
∑

∂x i  j

i


= ∑ ei xi  ∑ x 2j 

 j

i
= ∑ ei
i

12

−1 2

∂f ( r )
∂r
∂f ( r )
∂r

x i ∂f
r dr

(3)

Therefore,
grad f ( r ) =

r ∂f ( r )
r ∂r

(4)

c)
12

 
∂ 2 ln r
∂2  
2
∇ ( ln r ) = ∑
= ∑ 2  ln  ∑ x j  
∂xi2
∂x i   j
 
i

2

−1 2
1



 ⋅ 2xi  ∑ x 2j 

2




∂
j
=∑



12


i ∂x i 

2


 ∑ xj 
 j




∂
=∑
i ∂x i

−1
 
 
 xi  ∑ x 2j  
  j
 




= ∑ ( − xi )( 2xi )  ∑ x 2j 
 j


i

(

)( )

= ∑ −2x 2j r 2
i

=−

−2

−2


∂x 
+ ∑ i  ∑ x 2j 

i ∂x i  j

−1

1
+ 3 2 
r 

2r 2 3
1

+ 2 = 2
4
r
r
r

(5)

or,
∇ 2 ( ln r ) =

Full file at .

1
r2

(6)


Solution Manual for Classical Dynamics of Particles and Systems 5th Edition by Thornton
Full file 24
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1-32.

CHAPTER 1

Note that the integrand is a perfect differential:
2ar ⋅ r + 2br ⋅ r = a

d

d
( r ⋅ r ) + b ( r ⋅ r )
dt
dt

(1)

Clearly,

∫ ( 2ar ⋅ r + 2br ⋅ r ) dt = ar
1-33.

2

+ br 2 + const.

(2)

Since

d  r  rr − rr r rr
=
= − 2
dt  r 
r2
r r

(1)

we have

∫

d r 
 r rr 
 r − r 2  dt = ∫ dt  r  dt

(2)

r
 r rr 
 r − r 2  dt = r + C

(3)

from which

∫

where C is the integration constant (a vector).
1-34.

First, we note that

(

)

d
 =A

 ×A
 +A×A

A×A
dt

(1)

But the first term on the right-hand side vanishes. Thus,

∫ ( A × A ) dt = ∫ dt ( A × A ) dt

(2)

∫ ( A × A ) dt = A × A + C

(3)

d

so that

where C is a constant vector.

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25

1-35.
y

x

z

We compute the volume of the intersection of the two cylinders by dividing the intersection
volume into two parts. Part of the common volume is that of one of the cylinders, for example,
the one along the y axis, between y = –a and y = a:

( )

V1 = 2 π a 2 a = 2π a 3

(1)

The rest of the common volume is formed by 8 equal parts from the other cylinder (the one
along the x-axis). One of these parts extends from x = 0 to x = a, y = 0 to y = a2 − x 2 , z = a to

z = a 2 − x 2 . The complementary volume is then
a2 − x 2

a

V2 = 8 ∫ dx ∫
0

0

dy ∫

a2 − x 2
a

dz

= 8 ∫ dx a 2 − x 2  a 2 − x 2 − a 
0


a

a


x 3 a3
x
= 8  a2 x −
− sin −1 
a 0
3
2


=

16 3
a − 2π a 3
3

(2)

Then, from (1) and (2):
V = V1 + V2 =

Full file at .

16 a3
3

(3)