Solution manual for analytical mechanics 7th edition by fowles

u x=Brx Cr = ByCz is the area of the parallelogram formed by the vectors Br , Cr... The figure above shows the unit vectors and nˆ τ which are normal and tangent to the ˆ curve.. The vec

CHAPTER 1 FUNDAMENTAL CONCEPTS: VECTORS

1.1 (a) A Bv v+ = + + +(iˆ ˆj) (ˆj kˆ)= +iˆ 2ˆj k+ ˆ

1 2

(1 4 1) 6

A Bv v+ = + + = (b) 3Av−2Bv=3(iˆ+ −ˆj) 2(ˆj k+ ˆ)= + −3iˆ ˆj 2k ˆ

(c) A Bv v⋅ =(1)(0) (1)(1) (0)(1)+ + =1

(d)

ˆ

ˆ ˆ

1 1 0 (1 0) (0 1) (1 0)

0 1 1

i j k

A Bv v× = =i − + j − +k − = − +i j k

1 2

(1 1 1) 3

A Bv v× = + + =

1.2 (a) A B Cv⋅(v+ v)=(2iˆ+ ⋅ +ˆj) (iˆ 4ˆj k+ ˆ)=(2)(1) (1)(4) (0)(1)+ + =6 (A B Cv+ v)⋅ =v (3iˆ+ + ⋅ˆj kˆ) 4ˆj=(3)(0) (1)(4) (1)(0)+ + =4

(b) ( ) 12 10 01 8

0 4 0

(A B Cv× v)⋅ = ⋅v A B Cv (v× v)= −8 (c) Av×(B Cv× v) (= A C Bv v⋅ ) (v− A B Cv⋅ v) v =4( )i kˆ+ ˆ −2 4( )ˆj =4iˆ−8ˆj+4 kˆ

( ) ( ) ( ) ( )

( ˆ ˆ) ( )ˆ ˆ ˆ ˆ

× × = − × × = −⎣ ⋅ − ⋅ ⎦

= −⎣ + − + ⎦= +

1.3

2 2

( )( ) (2 )(2 ) (0)(3 ) 5 cos

5 14

5 14

θ = ⋅v = + + =

cos 1 5 53

14

1.4

(a) A iv= + +ˆ ˆj kˆ : body diagonal

A= ⋅ =A Av v i iˆ ˆ⋅ + ⋅ + ⋅ =ˆ ˆj j k kˆ ˆ 3 (b) B iv= +ˆ ˆj : face diagonal

B= ⋅ =B Bv v 2

(c)

ˆ ˆ ˆ 1 1 1

1 1 0

k j i

Cv = × =A Bv v

(d) cos 1 1 0

3 2

A B AB

θ = v v⋅ = − = ∴ =θ 90o

1.5

sin

B= Bv = × =A Cv v AC θ y sin B

A

θ

cos

A Cv v⋅ = AC θ =u x cos u

A

θ

2

A

v v

v

B A

v

v v

u2 12

A A

= v+ v× v

1.6 ˆ d ( ) ˆ d ( )2 ˆ d ( )3 ˆ ˆ2 ˆ3 2

dt dt

dt dt

dAv = α + β + γ = α+ β + γ

2

d A

dtv = β+ γ

v

1.7 ( )( ) ( )( ) ( )( ) 2

0= ⋅ =A Bv v q q + 3 − +q 1 2 =q −3q+2

0 (q−2) (q− =1) , q= or 2 1

2

A Bv+ v = A Bv+ v ⋅ A Bv+ v = A +B + A Bv⋅v

⎡⎣Av + Bv⎤⎦2 = A2+B2+2AB

Since A B ABv v⋅ = cosθ ≤AB , A Bv+ ≤v Av + Bv

A Bv⋅ =v ABcosθ = A Bv v cosθ ≤ A Bv v

cos

1.9 Show Av×(B Cv× v) (= A C Bv v⋅ ) (v− A B Cv⋅ v) v

ˆ ˆ ˆ

x y z x x y y z z x x y y z z

x y z

k j i

)

B B A C A C A C B A B A B A B C

x x x y x y z x z x x x y y x z z x

A B C A B C A B C A B C A B C A B C i

x y x y y y z y z x x y y y y z z y

A B C A B C A B C A B C A B C A B C j

x z x y z y z z z x x z y y z z z z

A B C A B C A B C A B C A B C A B C k

ˆ ˆ ˆ

y x y z x z y y x z z x

x y x z y z x x y z z y

x z x y z y x x z y y z

A B C A B C A B C A B C i

A B C A B C A B C A B C j

A B C A B C A B C A B C k

ˆ ˆ

ˆ

z y

x

y z z y z x x z x y y x

k j

i

A A

A

B C −B C B C −B C B C −B C

ˆ ˆ ˆ

y x y y y x z z x z x z

z y z z z y x x y x y x

x z x x x z y y z y z y

i A B C A B C A B C A B C

j A B C A B C A B C A B C

k A B C A B C A B C A B C

1.10

sin

y A= θ

2xy y B x xy yB xy AB inθ

Α = A Bv v×

1.11

( )

ˆ

ˆ ˆ

x y z

x y z x y z x x y y z z ( )

x y z

v v v v

z

A

=

=

1.12

Let Av = (Ax,Ay,Az),

Bv = (0,By,0) and Cr

= (0,Cy,Cz)

Cz is the perpendicular distance between the plane Av

, Bv

and its opposite u B is directed along the x-axis since the vectors

xC

= r r r

Br , Cr are in the y,z plane u x=Brx Cr = ByCz

is the area of the parallelogram formed by the vectors Br

, Cr Multiply that area times the height of plane v

,

A Bv

= Ax to get the volume of the parallopiped

V =A u x x =A B C x y z= •A B x Cr (r r)

Br

Ar

Cr

ur

y

1.13 For rotation about the z axis:

i iˆ ˆ⋅ =′ cosφ = ⋅ˆ ˆj j′, k k′ˆ ˆ⋅ =1

ˆ ˆi j⋅ = −′ sinφ

ˆ ˆj i⋅ =′ sinφ

For rotation about the y′ axis:

i iˆ ˆ⋅ =′ cosθ = ⋅k kˆ ˆ′, ˆ ˆj j′⋅ = 1

i kˆ⋅ =ˆ′ sinθ

k iˆ ˆ⋅ = −′ sinθ

cos 0 sin cos sin 0 cos cos cos sin sin

0 cos

1 0

1 sin cos sin sin cos 0

T

φ

⎟

⎟

⎟

⎠

1.14

3

ˆ ˆ cos 30

2 1

ˆ ˆ sin 30

2 ˆ

i i

i j

i k

′

′

⋅ = − = −

′

⋅ =

o

o

1

ˆ ˆ sin 30

2 3

ˆ ˆ cos 30

2

ˆ

j i

j j

j k

′

′

′

⋅ =

o

o

ˆ ˆ 0

ˆ ˆ 0

ˆ ˆ 1

k i

k j

k k

′

⋅ =

′

⋅ =

′

⋅ =

2

1

x y z

A A A

′

′

′

⎢ ⎥= −⎢ ⎥⎢ ⎥ ⎢= − ⎥

Av =3.232iˆ′+1.598ˆj′−kˆ′

1.15 1 Rotate thruφ about z-axis φ =45o Rφ

Rotate thru

2 θ about x’-axis θ =45o Rθ

Rotate thru

3 ψ about z’-axis ψ =45o Rψ

0

1 1

0

Rφ

⎟

⎜

⎟

⎜

1 1 0

1 1 0

Rθ

⎟

⎜

⎟

⎜

⎟

0

1 1

0

Rψ

⎟

⎜

⎟

⎜

( )

2 2 2 2 2 2

, ,

2 2 2 2 2 2

R ψ θ φ R R Rψ θ φ

⎟

⎜

2

2 1

1

, , 0

0

R

α

ψ θ φ β

γ

Condition is: xv′ =Rxv where and

1 0 0

x

⎛ ⎞

⎜ ⎟

′ = ⎜ ⎟

⎜ ⎟

⎝ ⎠

β γ

⎛ ⎞

⎜ ⎟

= ⎜ ⎟

⎜ ⎟

⎝ ⎠ v

Since x xv v⋅ =1 we have: 2 2 2

1

ψ +β +α =

After a lot of algebra: 1 2,

4 2

α = − 1 2,

4 2

β = + 1

2

γ =

1.16 v vv= τˆ=ctτˆ

ˆ v ˆ ˆ c

b

ρ

ˆ

t

at b

t c

= , vv=τˆ b c and a cv= τˆ+cnˆ

2

1 cos

2 2

v a c bc

va bc c

θ = v v⋅ = =

45

θ = o

1.17 v tv( )= −ibˆωsin( )ωt + ˆj b2 ωcos( )ωt

( ) ˆ 2cos ˆ2 2sin

a tv = −ibω ωt− j bω ωt

1 3sin

at t=0, vv =2bω; at

2

t π ω

= , vv =bω

1.18 v tv( )=ibˆ ωcosωt− ˆjbωsinωt k ct+ ˆ2

a tv = −ibω ωt− jbω ωt k c+

1.19 v rev= &ˆr+r eθ&ˆθ =bke e ktˆr +bce eˆ kt θ

av= r r&&− θ& e + rθ&&+ r e&θ& θ =b k −c e e + bcke eˆθ

1 1

2 cos

4

v a va

φ = ⋅ = − +

v v

, a constant

av= R R&&− φ& e + R&φ&+R eφ&& φ +ze&&ˆ

2

ˆR 2 ˆz

av= −b eω + ce

4

av = bω + c

1.21 r tv( )=iˆ(1−e−kt)+ ˆje kt

( ) ˆ kt ˆ kt

r tv =ike− + jke

( ) ˆ 2 kt ˆ 2 kt

r tv = −ik e− + jk e

1.22 v e r e rv= ˆr&+ˆφ φsinθ +e rˆθ θ&

1

v e bφ ω ⎧π ⎡ ωt ⎤⎫ e bθ πω t

= ⎨⎩ ⎢⎣ + ⎥⎦⎬⎭−

ˆ cos cos 4 ˆ sin 4

2 8

v e b= φ ω ⎡π ωt ⎤−e bθ ωπ t

⎥

⎣

1

cos cos 4 sin 4

4 8

⎠

⎣

Path is sinusoidal oscillation about the equator

1.23 v v vv v⋅ = 2

2

dt

dtv⋅ + ⋅v v v = &

2v av v⋅ =2vv&

v a vvv v⋅ = &

1.24 d r v a( ) dr (v a) r d (v a)

dt⎡⎣v v v⋅ × ⎤⎦= dtv⋅ ×v v + ⋅v dt v v×

v v a( ) r dv a v da

dt dt

× + ⋅ ⎜ × ⎟ ⎜+ × ⎟

⎠

⎠ ⎝

v

v v v v

0 r ⎡0 v a ⎤

= +v⋅ + ×⎣ v v& ⎦

d

r v a r v a

dt⎡⎣v v v⋅ × ⎤⎦= ⋅ ×v v v&

1.25 v vv= τˆ and a av= ττˆ+a n nˆ

v a vav v⋅ = τ, so v a

a v

τ

⋅

= v v

n

a =aτ +a2

, so ( )1

n

1.26 For 1.14,

1

a

τ

+

⋅ +

⋅

−

=

c t

2 1

4 4

c t a

τ

ω

=

+

1

16 4

4

n

c t

ω

ω

⎞

⎛

⎝

For 1.15, ( )

1

2

kt kt

b k k c e b c ke

be k c

τ

+

ke k +c

n

1.27

ˆ

n

ˆ

τ

ˆ ˆ

dτ= −n dθ

dθ

ρ

The figure above shows the unit vectors and nˆ τ which are normal and tangent to the ˆ curve The vectors are shown at 2 different points along the trajectory Since the particle

is moving tangentially to the curve and in a direction perpendicular to ρ , the local radius

of curvature, we have …

ˆ

v

= v

v τ

2

ρ

v & & &

a τ τ n n (since v&=0and from the above figure)

2 2

ˆ

v v

3

v

ρ ρ ρ

⎞

⎛

=

=

=

v

1.28

ˆ sin ˆ cos

P

rvo =ib θ + jb θ

ˆ cos ˆ sin

rel

vv =ibθ& θ − jbθ& θ

rel

av =ib θ&& θ θ− & θ − jb θ&& θ θ+ & θ

at the point θ π= , vvrel = −vv

So, vvrel =bθ& v=

v b

θ&= v a

b b

θ&&= =& o

Now,

ˆ rel ˆ ˆ

rel rel

b

ρ

1

2 2

rel

v

b

=⎜ + ⎟

⎝ o ⎠ v

P rel

vv = +v vv v and avP = +avo avrel

P

o

⎠

1

2 2

2

P

v v

⎛

o

v

is a maximum at

P

av θ =0, i.e., at the top of the wheel

2

2 sin v cos 0

a b

o 2 1

tan v

a b

⎝ o ⎠

Comments : Note that a point on the bottom of the wheel is instantaneously at rest, i.e.,

there is no relative motion between the ground and the bottom of the wheel assuming no slipping

2 2

0 0 1

⎜

= ⎜

2

x=

The transformation represents a rotation of 45o about the z-axis (see Example 1.8.2)

1.30 (a) a = iˆcosθ + ˆjsinθ

a

b

θ

φ

ˆcosϕ ˆsinϕ

cos θ ϕ cosθ sinθ cosϕ sin

cos θ ϕ− =cos cosθ ϕ+sin sinθ ϕ

(b) b a× = kˆsin(θ ϕ− )= (iˆcosθ + ˆjsinθ) (× iˆcosϕ+ ˆjsinϕ )

sin θ ϕ− =sin cosθ ϕ−cos sinθ ϕ -