Solutions Manual for An Introduction to Mechanics, 2nd Edition by Kleppner

,CONTENTS



1 VECTORS AND KINEMATICS 1

2 NEWTON’S LAWS 21

3 FORCES AND EQUATIONS OF MOTION 33

4 MOMENTUM 54

5 ENERGY 72

6 TOPICS IN DYNAMICS 89

7 ANGULAR MOMENTUM AND FIXED AXIS ROTATION 105

8 RIGID BODY MOTION 138

9 NONINERTIAL SYSTEMS AND FICTITIOUS FORCES 147

10 CENTRAL FORCE MOTION 156

11 THE HARMONIC OSCILLATOR 171

12 THE SPECIAL THEORY OF RELATIVITY 182

13 RELATIVISTIC DYNAMICS 196

14 SPACETIME PHYSICS 206

,1.1 Vector algebra 1
A = (2 î − 3 ĵ + 7 k̂ ) B = (5 î + ĵ + 2 k̂ )
(a) A + B = (2 + 5) î + (−3 + 1) ĵ + (7 + 2) k̂ = 7 î − 2 ĵ + 9 k̂
(b) A − B = (2 − 5) î + (−3 − 1) ĵ(7 − 2) k̂ = −3 î − 4 ĵ + 5 k̂ (c) A
· B = (2)(5) + (−3)(1) + (7)(2) = 21
î ĵ
k̂
(d) A × B = 2 −3 7
5 1 2

= −13 î + 31 ĵ + 17 k̂


1.2 Vector algebra 2
A = (3 î − 2 ĵ + 5 k̂ ) B = (6 î − 7 ĵ + 4 k̂ )
(a) A2 = A · A = 32 + (−2)2 + 52 = 38
(b) B2 = B · B = 62 + (−7)2 + 42 = 101
(c) (A · B)2 = [(3)(6) + (−2)(−7) + (5)(4)]2 = [18 + 14 + 20]2 = 522 = 2704

,2 VECTORS AND KINEMATICS

1.3 Cosine and sine by vector algebra
A = (3 î + ĵ + k̂ ) B = (−2 î + ĵ + k̂ )
(a)

A · B = A B cos (A, B)
A·B
cos (A, B) =
AB
(− 6 + 1 + 1) −4
= √ √ = √ √ ≈ 0.492
(9 + 1 + 1) 4 + 1 + 1) 11 6
(b) method
1:
|A × B| = A B sin (A, B)
|A × B|
sin (A, B) =
AB
î ĵ
k̂
A ×B = 3 1 1
−2 1 1
= (1 − 1) î − (3 + 2) ĵ + (3 + 2) k̂ = −5 ĵ + 5 k̂
√ √
|A × B| = 52 + 52 = 5 2
|A × B| 5 √2
sin (A, B) = =
√ √ ≈ 0.870
AB 11 6
(c) method 2 (simpler) – use:
sin2 θ + cos2 θ = 1
p
sin (A, B) = 1 − cos2 (A, B)
p
= 1 − (0.492)2 from (a) ≈ 0.871



1.4 Direction cosines

Note that here α, β, γ stand
for direction cosines, not for
the angles shown in the
figure: θ x = cos−1 α,
θy = cos−1 β,
θz = cos−1 γ.


continued next page =⇒

, VECTORS AND KINEMATICS 3



A = Ax î + Ay ĵ + Az k̂
Ax = A · î = A cos (A, î) ≡ A α α
= cos (A, î) = cos θ x .

Similarly,

Ay = A cos (A, ĵ) ≡ A β
β = cos (A, ĵ) = cos θy
Az = A cos (A, k̂ ) ≡ A
γ γ = cos (A, k̂ ) = cos
θz

Using these results,
A2 = A2 + A2 + A2
x y z
= A (α + β + γ 2 )
2 2 2



from which it follows that

α2 + β2 + γ2 = 1
Another way to see this is
A2 = ρ2 + A2 = A2 + A2 + A2 = A2 (α2 + β2 + γ2)
z x y z

and it follows as before that

α2 + β2 + γ2 = 1.




1.5 Perpendicular vectors
Given |A−B| = |A+B| with A and B nonzero. Evaluate the magnitudes by squaring.


A2 − 2 A · B + B2 = A2 + 2 A · B + B2
−2 A · B = +2 A · B. A ·
B =0
and it follows that A ⊥ B.

,4 VECTORS AND KINEMATICS

1.6 Diagonals of a parallelogram

The parallelogram
is equilateral, so A
= B.


D1 = A + B
D2 = B − A
D1 · D2 = (A + B) · (B − A) = A2 − B2 = 0.
Hence D1 · D2 = 0 and it follows that D1 ⊥ D2.




1.7 Law of sines

The area A of the triangle is

1 1 1
A = A h = A B sin γ = |A × B|
2 2 2
Similarly,
1 1
A = |B × C| = BC sin α
21 21
A = |C × A| = AC sin β.
2 2
Hence AB sin γ = BC sin α = AC sin β, from which it follows
sin γ sin α sin β
C = A = B
Introducing the cross product makes the notation convenient, and
emphasizes the relation between the cross product and the area of the
triangle, but it is not essential for the proof.

, VECTORS AND KINEMATICS 5

1.8 Vector proof of a trigonometric identity
Given two unit vectors â = cos θ î+sin θ ĵ and b̂ = cos φ î+sin φ ĵ, with a = 1, b = 1.
First evaluate their scalar product using components:



a · b = ab cos θ cos φ + ab sin θ sin φ
= cos θ cos φ + sin θ sin φ

then evaluate their scalar product geometrically.


a · b = ab cos (a, b) = ab cos (φ − θ) = cos (φ − θ)
Equating the two results,

cos (φ − θ) = cos φ cos θ + sin φ sin θ


1.9 Perpendicular unit vector
Given A = (î+ ĵ−k̂ ) and B = (2 î+ ĵ−3 k̂ ), find C such that A · C = 0 and B · C = 0.


C = C x î + Cy ĵ + Cz k̂
= C x (î + (Cy /C x ) ĵ + (Cz /C x ) k̂ )
A · C = Cx(1 + (Cy /Cx) − (Cz/Cx )) = 0
B · C = Cx(2 + (Cy /Cx) − 3(Cz /Cx)) = 0

We have two equations for the two unknowns (Cy/Cx) and

(Cz /Cx). 1 + (Cy/Cx ) − (Cz/Cx) = 0
2 + (Cy /Cx) − 3(Cz/Cx) = 0.
The solutions are (Cy /C x) = − 1 and (Cz /C x ) = 1 , so that C = Cx (î − 1 ĵ + 1 k̂ ). To
2 2 2 2
evaluate Cx, apply the condition that C is a unit vector.

3
C2 = C2 = 1
2p x
Cx = ± (2/3)
1 1
p
Ĉ = ± (2/3) (î − ĵ + k̂ )
2 2
continued next page =⇒

,6 VECTORS AND KINEMATICS

which can be written
1
Ĉ = ± √ (2 î − ĵ + k̂ )
6
Geometrically, C can be perpendicular to both A and B only if C is
perpendicular to the plane determined by A and B. From the standpoint
of vector algebra, this implies that C ∝ A × B. To prove this, evaluate A × B.

î ĵ
k̂
A × B = 1 1 −1
2 1 −3
= −2 î + ĵ − k̂
∝ C.


1.10 Perpendicular unit vectors
Given A = 3î + 4ĵ − 4k̂ , find a unit vector B̂ perpendicular to A.

(a)

B = Bx î + By ĵ = Bx [î + ( By /Bx )ĵ]
A · B = Bx[3 + 4(By/Bx)] = 0
By/Bx = −3/4
3
B = Bx [î − ĵ]
4
To evaluate Bx , note that B is a unit vector, B2 = 1.
!
3 2
2 2
1 = B (1) + !
= 25 B x2
x 4 16
which gives

Bx = ±(4/5)
B̂ = ±(4/5)(î − (3/4)ĵ) = ± (4 î − 3 ĵ)
5
continued next page =⇒

, VECTORS AND KINEMATICS 7

(b)

C = C x î + Cy ĵ + Cz k̂
= C x [î + (Cy /C x ) ĵ + (Cz /C x ) k̂ ]
A · C = 0 ⇒ Cx [3 + 4(Cy/Cx) − 4(Cz/Cx )] = 0
1
B · C = 0 ⇒ 5 Cx [4 − 3(Cy/Cx)] = 0
Cy/Cx = 4/3 Cz/Cx = 25/12
To make C a unit vector,

2 !2 2
C2 = C2 x (1) + 4 ! =1
12
25
3 +
Cx ≈±0.348
(c) The vector B × C is perpendicular (normal) to the plane defined by B and
C, so we want to prove
A ∝B ×C
î ĵ
4 3 k̂
B × C = C x 5 −5 0
25
4 12 !
"1
3
! 100
! #
75 ˆ 25 ˆ
ˆ j+ k
= Cx −! i − 60 15
5 60
= C ( 3 4 ĵ + 4 k̂ ) A.
î
− − ∝
12 x

1.11 Volume of a parallelepiped

With reference to the sketch, the height is A
cos α, so the frontal area is AB cos α. The
depth is
C sin β, so the volume V is
V = (AB cos α)(C sin β) = (A cos α)(BC sin β) = A · (B × C)
The same approach can be used starting with a different
face.

V = C · (A × B) V = B · (C × A)
Note that A, B, C are arbitrary vectors. This proves the vector identity
A · (B × C) = C · (A × B) = B · (C × A)

, 8 VECTORS AND KINEMATICS




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