Fowles G.R., Cassiday G.L. - Analytical mechanics, 7ed., Solutions manual-Brooks (2004)

Instructor’s Solutions Manual
To Accompany
Analytical Mechanics
7th Edition
Grant R. Fowles - University of Utah
George L. Cassiday - University of Utah
ISBN-10: 0-534-49493-5
ISBN-13: 978-0-534-49493-3
Contents
1. Fundamental Concepts: Vectors.
2. Newtonian Mechanics: Rectilinear Motion of a Particle.
3. Oscillations.
4. General Motion of a Particle in Three Dimensions.
5. Noninertial Reference Systems.
6. Gravitation and Central Forces.
7. Dynamics of Systems of Particles.
8. Mechanics of Rigid Bodies: Planar Motion.
9. Motion of Rigid Bodies in Three Dimensions.
10. Lagrangian Mechanics.
11. Dynamics of Oscillating Systems.
CHAPTER 1
FUNDAMENTAL CONCEPTS: VECTORS
1.1
K K
(a) A + B = (iˆ + ˆj ) + ( ˆj + kˆ) = iˆ + 2 ˆj + kˆ
1
K K
A + B = (1 + 4 + 1) 2 = 6
K
K
(b) 3 A − 2 B = 3(iˆ + ˆj ) − 2( ˆj + kˆ) = 3iˆ + ˆj − 2kˆ
K K
(c) A ⋅ B = (1)(0) + (1)(1) + (0)(1) = 1
iˆ ˆj kˆ
K K
(d) A × B = 1 1 0 = iˆ(1 − 0) + ˆj (0 − 1) + kˆ(1 − 0) = iˆ − ˆj + kˆ
0 1 1
1
K K
A × B = (1 + 1 + 1) 2 = 3
K K K
1.2 (a) A ⋅ ( B + C ) = 2iˆ + ˆj ⋅ iˆ + 4 ˆj + kˆ = (2)(1) + (1)(4) + (0)(1) = 6
)(
(
K
K
)
K
( A + B ) ⋅ C = ( 3iˆ + ˆj + kˆ ) ⋅ 4 ˆj = (3)(0) + (1)(4) + (1)(0) = 4
2 1 0
K K K
(b) A ⋅ ( B × C ) = 1 0 1 = −8
0 4 0
K
K
K
K
K
K
( A × B ) ⋅ C = A ⋅ ( B × C ) = −8
K K K
K K K
K K K
(c) A × ( B × C ) = ( A ⋅ C ) B − ( A ⋅ B ) C = 4 iˆ + kˆ − 2 4 ˆj = 4iˆ − 8 ˆj + 4kˆ
K
K
K
K
K
K
(
) ( )
K K K
K K K
( A × B ) × C = −C × ( A × B ) = − ( C ⋅ B ) A − ( C ⋅ A) B 
= − 0 ( 2iˆ + ˆj ) − 4 ( iˆ + kˆ )  = 4iˆ + 4kˆ


1
K K
5a 2
A ⋅ B (a )(a ) + (2a )(2a) + (0)(3a)
1.3 cos θ =
=
= 2
AB
a 5 14
5a 2 14a 2
θ = cos −1
5
≈ 53°
14
1.4
K
(a) A = iˆ + ˆj + kˆ
K K
A = A⋅ A
K
(b) B = iˆ + ˆj
K K
B = B⋅B
: body diagonal
= iˆ ⋅ iˆ + ˆj ⋅ ˆj + kˆ ⋅ kˆ = 3
: face diagonal
= 2
iˆ ˆj kˆ
K K K
(c) C = A × B = 1 1 1
1 1 0
K K
A⋅ B
1−1
(d) cos θ =
=
=0
AB
3 2
1.5
∴θ = 90D
K K
K
B = B = A × C = AC sin θ
K K
A ⋅ C = AC cos θ = u
K
K A
C = Cx +
A
u K
= 2 A+
A
1.6
B
A
u
∴ Cx = C cos θ =
A
∴ C y = C sin θ =
K K
K K
B× A
u K B× A B 
K K Cy = 2 A +
 
A
AB  A 
B× A
1 K K
B× A
A2
K
d
d
dA ˆ d
= i (α t ) + ˆj ( β t 2 ) + kˆ (γ t 3 ) = iˆα + ˆj 2 β t + kˆ3γ t 2
dt
dt
dt
dt
K
d2A ˆ
= j 2 β + kˆ6γ t
dt 2
2
K K
0 = A ⋅ B = ( q )( q ) + ( 3)( − q ) + (1)( 2 ) = q 2 − 3q + 2
1.7
( q − 2 ) ( q − 1) = 0 ,
q = 1 or 2
K K2
K K
K K
K K
A + B = ( A + B ) ⋅ ( A + B ) = A2 + B 2 + 2 A ⋅ B
1.8
K K 2
 A + B  = A2 + B 2 + 2 AB


K K
K K
A+ B ≤ A + B
K K
Since A ⋅ B = AB cosθ ≤ AB ,
K K
K K
K K
A ⋅ B = AB cosθ = A B cosθ ≤ A B
B cos θ ≤ B
K K K
K K K
K K K
Show A × ( B × C ) = ( A ⋅ C ) B − ( A ⋅ B ) C
1.9
iˆ
K
A × Bx
Cx
or
kˆ
K
K
Bz = ( Ax Cx + Ay C y + Az Cz ) B − ( Ax Bx + Ay By + Az Bz ) C
Cz
ˆj
By
Cy
= ( Ax BxCx + Ay Bx C y + Az Bx Cz − Ax Bx Cx − Ay By Cx − Az Bz Cx ) iˆ
+ ( Ax By Cx + Ay By C y + Az By Cz − Ax Bx C y − Ay By C y − Az Bz C y ) ˆj
+ ( Ax Bz Cx + Ay Bz C y + Az Bz C z − Ax BxCz − Ay By Cz − Az Bz Cz ) kˆ
( A B C + A B C − A B C − A B C ) iˆ
= + ( A B C + A B C − A B C − A B C ) ˆj
+ ( A B C + A B C − A B C − A B C ) kˆ
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
3
iˆ
Ax
By Cz − Bz C y
iˆ ( Ay Bx C y − Ay By Cx − Az Bz Cx + Az Bx Cz )
kˆ
Az
= + ˆj ( Az By Cz − Az Bz C y − Ax Bx C y + Ax By Cx )
Bx C y − By Cx + kˆ A B C − A B C − A B C + A B C
( x z x x x z y y z y z y)
ˆj
Ay
Bz Cx − Bx Cz
1.10
y = A sin θ
1 
Α = 2  xy  + y ( B − x ) = xy + yB − xy = AB sin θ
2 
K K
Α = A× B
1.11
iˆ
K K K
K
A ⋅ ( B × C ) = A ⋅ Bx
Cx
1.12
x
G
u
kˆ
Ax
Bz = Bx
Cz Cx
ˆj
By
Cy
Ay
By
Cy
Bx
Az
Bz = − Ax
Cz
Cx
By
Ay
Cy
Bz
K K K
Az = − B ⋅ ( A × C )
Cz
z
G
A
G
C
x
G
B
G
K
K
Let A = (Ax,Ay,Az), B = (0,By,0) and C = (0,Cy,Cz)
K K
G G G
Cz is the perpendicular distance between the plane A , B and its opposite. u = B x C is
G G
G G
directed along the x-axis since the vectors B , C are in the y,z plane. u x = B x C = ByCz
G G
is the area of the parallelogram formed by the vectors B , C . Multiply that area times the
K K
height of plane A , B = Ax to get the volume of the parallopiped
G G G
V = Ax u x = Ax By Cz = A • B x C
(
)
4
1.13
For rotation about the z axis:
iˆ ⋅ iˆ′ = cos φ = ˆj ⋅ ˆj ′, kˆ ⋅ kˆ′ = 1
iˆ ⋅ ˆj ′ = − sin φ
ˆj ⋅ iˆ′ = sin φ
For rotation about the y′ axis:
iˆ ⋅ iˆ′ = cos θ = kˆ ⋅ kˆ′,
iˆ ⋅ k̂ ′ = sin θ
kˆ ⋅ iˆ′ = − sin θ
 cos θ
I 
T = 0
 sin θ

ˆj ⋅ ˆj ′ = 1
0 − sin θ   cos φ

1
0   − sin φ
0 cosθ   0
sin φ
cos φ
0
0   cos θ cos φ
 
0  =  − sin φ
1   sin θ cos φ
cosθ sin φ
cos φ
sin θ sin φ
− sin θ 

0 
cosθ 
1.14
3
iˆ ⋅ iˆ′ = cos 30D =
2
ˆj ⋅ iˆ′ = sin 30D = 1
2
kˆ ⋅ iˆ′ = 0
1
iˆ ⋅ ˆj ′ = − sin 30D = −
2
ˆj ⋅ ˆj ′ = cos 30D = 3
2
kˆ ⋅ ˆj ′ = 0
ˆj ⋅ kˆ′ = 0
iˆ ⋅ kˆ′ = 0
 3 1

3

0

3+ 

2
2
2
 2  
 Ax′  

   3
3
 A  = − 1

 y′   2 2 0   3  =  2 3 − 1
  −1
 Az ′  
0 1     −1 
 0






K
A = 3.232iˆ′ + 1.598 ˆj ′ − kˆ′
kˆ ⋅ kˆ′ = 1
5
1.15
1.
Rotate thru φ about z-axis
φ = 45D
Rφ
2.
Rotate thru θ about x’-axis
Rθ
3.
Rotate thru ψ about z’-axis
θ = 45
ψ = 45D
 1
 2

 1
Rφ =  −
2

 0


1
2
1
2
0

1
0

1

Rθ =  0
2

1

0 −
2

D

0


0

1



0 

1 
2 
1 

2
1
 1

0
 2
2


1
 1

Rψ =  −
0
2
2


0 1
 0




1
1 1
 1
+
 2−
2
2
2
2

1
1
1
 1
R (ψ ,θ , φ ) = Rψ Rθ Rφ =  − −
− +
2 2 2
2 2 2

1
1

−

2
2

1
K  
where x′ =  0  and
0
 
2
2
we have: ψ + β + α 2 = 1
K
K
Condition is: x′ = Rx
K K
Since x ⋅ x = 1
After a lot of algebra: α =
1.16
Rψ
1 
2 

1
α 
1 
 
 
or  0  = R (ψ ,θ , φ )  β 

2
0
γ 

 
 
1 

2
α 
K  
x =β 
γ 
 
1
2
1
2
1
−
, β= +
, γ =
2 4
2 4
2
K
v = vτˆ = ctτˆ
v2
c 2t 2
K
a = vτˆ + nˆ = cτˆ +
nˆ
b
ρ
6
b K
, v = τˆ bc and
c
K K
v ⋅a
c bc
cos θ =
=
=
va
bc 2c 2
K
a = cτˆ + cnˆ
at t =
1
2
θ = 45D
1.17
K
ˆ ω sin (ω t ) + ˆj 2bω cos (ω t )
v ( t ) = −ib
1
1
K
v = ( b 2ω 2 sin 2 ω t + 4b 2ω 2 cos 2 ω t ) 2 = bω (1 + 3cos 2 ω t ) 2
K
ˆ ω 2 cos ω t − ˆj 2bω 2 sin ω t
a ( t ) = −ib
1
K
a = bω 2 (1 + 3sin 2 ω t ) 2
1.18
K
v = 2bω ;
t = 0,
at
at
t=
π
,
2ω
K
v = bω
K
ˆ ω cos ω t − ˆjbω sin ω t + kˆ 2ct
v ( t ) = ib
K
ˆ ω 2 sin ω t − ˆjbω 2 cos ω t + kˆ 2c
a ( t ) = −ib
1
1
K
a = ( b 2ω 4 sin 2 ω t + b 2ω 4 cos 2 ω t + 4c 2 ) 2 = ( b 2ω 4 + 4c 2 ) 2
1.19
K
ˆr + rθeˆθ = bke kt eˆr + bce kt êθ
v = re
K
a = r − rθ 2 eˆr + rθ + 2rθ eˆθ = b ( k 2 − c 2 ) e kt eˆr + 2bcke kt êθ
K K
b 2 k ( k 2 − c 2 ) e 2 kt + 2b 2 c 2 ke 2 kt
v ⋅a
cos φ =
=
1
1
va
kt
2
2 2
kt 
2
2 2
2 2 2
be ( k + c ) be ( k − c ) + 4c k


(
cos φ =
1.20
)
(
)
k ( k 2 + c2 )
(k
2
+c
1
2 2
) (k
2
+c
k
=
2
) (k
2
+c
1
2 2
)
, a constant
K
− Rφ ) eˆ + ( 2 Rφ + Rφ ) eˆ + a = (R
zeˆz
R
φ
K
a = −bω 2 eˆR + 2ceˆz
1
K
a = ( b 2ω 4 + 4c 2 ) 2
7
1.21
1.22
K
r ( t ) = iˆ (1 − e − kt ) + ˆje kt
K
ˆ − kt + ˆjkekt
r ( t ) = ike
K
ˆ 2 e − kt + ˆjk 2 ekt
r ( t ) = −ik
K
v = eˆr r + eˆφ rφ sin θ + eˆθ rθ
π  1
π
K

v = eˆφ bω sin  1 + cos ( 4ω t )   − eˆθ b ω sin ( 4ω t )
2

2  4
π
K
π

v = eˆφ bω cos  cos ( 4ω t )  − eˆθ bω sin ( 4ω t )
2
8

1
2

2
K
π
 π
v = bω cos 2  cos 4ω t  + sin 2 4ω t 
8
 4


Path is sinusoidal oscillation about the equator.
1.23
K K
v ⋅ v = v2
K
K
dv K K dv
⋅v + v ⋅
= 2vv
dt
dt
K K
2v ⋅ a = 2vv
K K
v ⋅ a = vv
8
1.24
1.25
K
d K K K
dr K K K d K K
 r ⋅ ( v × a )  =
⋅ (v × a ) + r ⋅ (v × a )
dt
dt
dt
K
K
K K K K  dv K   K da  
= v ⋅ ( v × a ) + r ⋅  × a  +  v ×  
dt  
 
 dt
K 
K K 
= 0 + r ⋅ 0 + ( v × a )
d K K K
K K K
 r ⋅ ( v × a )  = r ⋅ ( v × a )
dt
K
K
v = vτˆ and a = aτ τˆ + an nˆ
K K
v ⋅a
K K
v ⋅ a = vaτ , so aτ =
v
1
a 2 = aτ2 + an2 , so an = ( a 2 − aτ2 ) 2
1.26
For 1.14, aτ =
−b 2ω 3 cos ω t ⋅ sin ω t + b 2ω 3 sin ω t ⋅ cos ω t + 4c 2t
(b ω
2
aτ =
2
1
cos 2 ω t + b 2ω 2 sin 2 ω t + 4c 2t 2 ) 2
4c 2t
(b ω
2
2
1
2 2 2
+ 4c t
)
1
 2 2
2
16c 4t 2
2
an =  b ω + 4c − 2 2

b ω + 4c 2 t 2 

1
b 2 k ( k 2 − c 2 ) e 2 kt + 2b 2 c 2 ke 2 kt
2
2 2
kt
= bke ( k + c )
For 1.15, aτ =
1
kt
2
2 2
be ( k + c )
1
1
2
2
an = b 2 e 2 kt ( k 2 + c 2 ) − b 2 k 2e 2 kt ( k 2 + c 2 )  = bce kt ( k 2 + c 2 ) 2


2
1.27
v
K
K
v = vτˆ , a = vτˆ + nˆ
ρ
v 2 v3
K K
v × a = v ⋅ an = v =
ρ
ρ
9
1.28
K
ˆ sin θ + ˆjb cos θ
rD P = ib
K
ˆ θ cos θ − ˆjbθ sin θ
vrel = ib
K
ˆ θ cosθ − θ 2 sin θ − ˆjb θsin θ + θ 2 cosθ
arel = ib
(
)
at the point θ =
π
2
K
So, vrel = bθ = v
(
)
K
K
, vrel = −v
v aD
=
b b
v2
v2
K
Now, arel = vrelτˆ + rel nˆ = aDτˆ + n̂
b
ρ
θ =
v
b
θ =
1
 2 v4  2
K
arel =  aD + 2 
b 

K
K K
K
K K
vP = v + vrel and aP = aD + arel

a

a

v2
v2
K
aP = iˆ  aD + b  D cos θ − 2 sin θ   − ˆjb  D sin θ + 2 cosθ 
b
b
b

b


1

2
v4
2v 2
K
aP = aD  2 + 2 cos θ + 2 2 −
sin θ 
aD b
aDb


K
aP is a maximum at θ = 0 , i.e., at the top of the wheel.
2v 2
−2sin θ −
cos θ = 0
aDb

v2 

 aDb 
θ = tan −1  −
2
0 0
 x -x 0   x x 0   2 x





= x x 0 − x x 0 = 0 2 x 2 0  Therefore, x = 1
1.29 RR



 
2
0 0 1 0 0 1  0

0 1


 
The transformation represents a rotation of 45o about the z-axis (see Example 1.8.2)
10
1.30
(a)
a
θ
b
φ
a = iˆ cos θ + ˆj sin θ
b = iˆ cos ϕ + ˆj sin ϕ
(
)(
a ⋅ b = cos (θ − ϕ ) = iˆ cos θ + ˆj sin θ ⋅ iˆ cos ϕ + ˆj sin ϕ
)
cos (θ − ϕ ) = cos θ cos ϕ + sin θ sin ϕ
(b)
(
) (
b × a = kˆ sin (θ − ϕ ) = iˆ cos θ + ˆj sin θ × iˆ cos ϕ + ˆj sin ϕ
)
sin (θ − ϕ ) = sin θ cos ϕ − cos θ sin ϕ
--------------------------------------------------------------------------------------------------------
11
CHAPTER 2
NEWTONIAN MECHANICS:
RECTILINEAR MOTION OF A PARTICLE
2.1
1
( F + ct )
m
t 1
F
c 2
x = ∫ ( F + ct ) dt = t +
t
0 m
m
2m
t F
c 2
F
c 3
x=∫  t+
t  dt = t 2 +
t
0 m
m
2m 
6m

(a) x =
F
sin ct
m
t F
F
F
t
x=∫
sin ct dt = −
cos ct 0 =
(1 − cos ct )
0 m
cm
cm
t F
F
1
x=∫
(1 − cos ct ) dt = 1 − sin ct 
0 cm
cm  c

(b) x =
F ct
e
m
F ct t F ct
x=
e =
( e − 1)
cm 0 cm
F  1 ct 1  F
ct
x=
 e − − t  = 2 ( e − 1 − ct )
cm  c
c  cm
(c) x =
2.2
dx dx dx
dx
= ⋅ =x
dt dx dt
dx
dx 1
x = ( F + cx )
dx m
1
xdx = ( F + cx ) dx
m
1 2 1
cx 2 
x = F x+

m
2
2 
(a) x =
1
x
2
x =  ( 2 F + cx ) 
m

(b) x = x
dx 1
= F e − cx
dx m
12
1
F e − cx dx
m
1 2
F − cx
F
x =−
e − 1) =
(
(1 − e−cx )
2
cm
cm
xdx =
1
 2F
2
x=
1 − e − cx ) 
(
 cm

dx 1
(c) x = x = ( F cos cx )
dx m
F
xdx = cos cx dx
m
1 2 F
x =
sin cx
cm
2
1
 2F
2
sin cx 
x=
 cm

2.3
(a) V ( x ) = − ∫
x
( F + cx ) dx = − F x −
x
cx 2
+C
2
F − cx
e +C
c
x
F
(c) V ( x ) = − ∫ F cos cx dx = − sin cx + C
x
c
x
(b) V ( x ) = − ∫ F e − cx dx =
x
2.4
dV ( x )
= −kx
dx
x
1
V ( x ) = ∫ kx dx = kx 2
0
2
(b) T = T ( x ) + V ( x )
(a) F ( x ) = −
T ( x) = T −V ( x) =
1
k ( A − x2 )
2
1 2
kA
2
(d) turning points @ T ( x1 ) → 0 ∴ x1 = ± A
(c) E = T =
2.5
x
kx 3 
1
1 kx 4
V ( x ) = ∫  kx − 2 dx = kx 2 −
0
2
4 A2
A 

1 2 1 kx 4
(b) T ( x ) = T − V ( x ) = T − kx +
2
4 A2
(c) E = T
(a) F ( x ) − kx +
kx 3
A2
so
13
(d) V ( x ) has maximum at F ( xm ) → 0
3
kx
xm = ± A
kxm − m2 = 0
A
1
1 kA4 1 2
V ( xm ) = kA2 −
= kA
2
4 A2 4
If E < V ( xm ) turning points exist.
Turning points @ T ( x1 ) → 0 let u = x1
2
1
1 ku 2
E − ku +
=0
2
4 A2
solving for u , we obtain
1


4E  2 

2 
u = A 1 ± 1 − 2 
  kA  


or
1

4E   2

x1 = ± A 1 − 1 − 2  
 kA  

2.6
x = v ( x) =
α
x=−
x
α
x2
x=−
α2
x3
mα 2
F ( x ) = mx = − 3
x
2.7
2.8
F ≥ Mg sin θ
F = mx = mx
dx
dx
x = bx −3
dx
= −3bx −4
dx
F = m ( bx −3 )( −3bx −4 )
F = −3mb 2 x −7
2.9

m
m

(a) V = mgx = (.145kg )  9.8 2  (1250 ft )  .3048  = 541J
s 
ft 


14
(b) T =
T=
1 2 1
1  mg  1 m 2 g
2
mv = mvt = m 
=
2
2
2  c2  2 .22 D 2
(.145kg )
2
m

 9.8 2 
s 

kg
( 2 )(.22 ) ( 2 )(.0366 )
m
2
= 87 J
3

 t 
2
3
∫ Fdx = ∫ −cv dx = −c ∫ v dt = −c ∫  −vt tanh  τ   dt
t
 t   t 
3  1
= cvt τ  − tanh 2   + ∫ tanh   d   
τ 
 τ   τ 
 2
t
 t 
3  1
= cvt τ  − tanh 2   + ln cosh   
τ 
 τ 
 2
t
Now tanh 2 
τ

 ≅ 1 for t τ


 t 
t
Meanwhile x = ∫ vdt = ∫  −vt tanh    dt = vtτ ln cosh  
τ 
τ 

t x
ln cosh   =
 τ  vtτ

m
x = (1250 ft )  .3048  = 381 m
ft 

1

m 2

kg
.145
9.8
(
)



 mg 
m
s2  


 = 34.72
vt = 
 =
kg
2
s
 (.22 )(.0732 )

 c2 


m 
1
2
1

2
1


 m 2
(.145kg )

 = 3.543s
τ =
=

 (.22 )(.0732 )2 kg  9.8 m  
 c2 g 



m
s 2  


3.81
2
3
Fdx
=
.22
.0732
34.72
3.543
−
.5
+
(
)(
)
(
)
(
)

 = 454 J
∫
( 34.72 )( 3.54 ) 

V − T = 541J − 87 J = 454 J
2.10
F
1F 2
t
t, x=
2m
m
F
F 2
t1 , t = t1
For t1 ≤ t ≤ 2t1 : v = t1 , x =
2m
m
For 0 ≤ t ≤ t1 : v =
15
F 2 F
1 2F
2
t1 + t1 ( t − t1 ) +
( t − t1 )
2m
2 m
m
F 2 F 2 F 2 5F 2
t1 + t1 + t1 =
t1
At t = 2t1 : x =
2m
2m
m
m
dv dv dx
dv
c 3
a=
= ⋅ = v ⋅ = − v2
dt dx dt
dx
m
1
−
c
v 2 dv = − dx
m
1
v −
xmax
c
∫v v 2 dv = ∫0 − mdx
1
c
−2v 2 = − xmax
m
x=
2.11
1
xmax
2.12
2mv 2
=
c
Going up: Fx = − mg sin 30 − µ mg cos 30
x = − g ( sin 30 + 0.1cos 30 ) = −5.749
v = v + at
at the highest point v = 0 so tup = −
m
s2
v
= 0.174v s
a
1
2
2
2
xup = v tup + atup2 = 0.174v − .087v = 0.087v m
2
2
Going down: x ′ = 0.087v , v ′ = 0 , a′ = −9.8 ( 0.5 − 0.0866 )
ttotal
2.13
1
2
2
xdown = 0 = 0.087v − 4.0513tdown
2
tdown = 0.207v s
= tup + tdown = 0.381v s
At the top v = 0 so e −2 kxmax =
Coming down x = xmax
g
k
g
2
+v
k
and at the bottom x = 0
g 2
 v
1
g g
k
2
1) =  
v = − 
(
g
2
k k g
2
+v
 +v 
k
k

2
16
v=
vt v
(v
t
2.14
2
+v
1
2 2
)
, vt =
g
mg
=
k
c2
Going up: Fx = −mg − c2 v 2
dv
c2
2
a = v = − g − kv , k =
m
dx
v
x
vdv
∫v − g − kv 2 = ∫0 dx
−
v
1
ln ( − g − kv 2 ) = x
v
2k
g + kv 2
= e −2 kx
2
g + kv
g
g
2
v 2 =  + v  e −2kx −
k
k

2
Going down: Fx = −mg + c2 v
dv
v = − g + kv 2
dx
v
x
vdv
=
∫0 − g + kv2 ∫0 dx
v
1
ln ( − g + kv 2 ) = x − x
0
2k
k
1 − v 2 = e 2 kx e −2 kx
g
g g

v 2 = −  e −2 kx  e 2 kx
k k

2.15
Using
dv
= mg − c1v − c2 v 2
dt
t dt
v
dv
=
∫0 m ∫0 mg − c1v − c2v 2
m
dx
1
2cx + b − b 2 − 4ac
=
ln
∫ a + bx + cx 2 b2 − 4ac 2cx + b + b2 − 4ac ,
17
2
−2c2 v − c1 − c1 + 4mgc2
t
1
=
ln
2
2
m
c1 + 4mgc2 −2c2 v − c1 + c1 + 4mgc2
(
t
2
c1 + 4mgc2
m
)
1
2
2c v + c +
(
= ln
( 2c v + c −
2
0
)( c −
+ 4mgc )( c +
2
1
c1 + 4mgc2
2
1
c1
2
v
2
2
)
+ 4mgc )
1
c1 + 4mgc2
1
c1
2
2
2
as t → ∞ , 2c2 vt + c1 − c1 + 4mgc2 = 0
c1  c1  mg 

+ 
 +
2c2  2c2 
c2 


Alternatively, when v = vt ,
dv
2
m = 0 = mg − c1vt − c2 vt
dt
2
1
2
vt = −
c1  c1  mg 

+ 
 +
2c2  2c2 
c2 


2
vt = −
2.16
1
2
dv
k
= − x −2
dx
m
v
x
kdx
∫0 vdv = ∫b − mx 2
1 2 k 1 1
v =  − 
2
m x b
a=v
1
1
dx  2k  1 1   2  2k  b − x   2
v=
=
 −  =


dt  m  x b  
 mb  x  
1
 x 2
3
 x
t
0  mb 
 mb  0 
x 
∫0 dt = ∫b  2k  b − x  dx =  2k  ∫1  b x  d  b 
 1 − 
 b
x
Since x ≤ b , say = sin 2 θ
b
1
2
1
1
2
1
 mb3  2 0 sin θ ( 2sin θ cos θ dθ )  2mb3  2 0
2
t =
=
 ∫− π
 ∫− π sin θ dθ
k
k
2
cos
θ

 2

 2
1
 mb3  2
t =
 π
 8k 
18
2.17
dv
dv
= mv = f ( x )i g ( v )
dt
dx
mvdv
= f ( x ) dx
g (v)
m
By integration, get v = v ( x ) =
If F ( x,t ) = f ( x )i g ( t ) :
dx
dt
d 2x
d  dx 
= m   = f ( x )i g ( t )
2
dt
dt  dt 
This cannot, in general, be solved by integration.
If F ( v,t ) = f ( v )i g ( t ) :
m
dv
= f ( v )i g ( t )
dt
mdv
= g ( t ) dt
f (v)
m
Integration gives v = v ( t )
dx
= v (t )
dt
dx = v ( t ) dt
A second integration gives x = x ( t )
2.18
c1 = (1.55 ×10−4 )(10−2 ) = 1.55 × 10−6
c2 = ( 0.22 ) (10−2 ) = 2.2 × 10−5
2
kg
s
kg
s
1
 1.55 ×10−6  2 (10−7 ) ( 9.8 )  2
1.55 × 10

vt = −
+ 
 +
2 × 2.2 ×10−5  2 × 2.2 ×10−5 
2.2 × 10−5 


m
vt = 0.179
s
−6
(10 ) ( 9.8) = 0.211 m
−7
Using equation 2.29, vt =
2.2 ×10−5
s
2.19
F ( x ) = − Aeα x = mx
Let u = eα v
or
du = α eα v dv
F ( v ) = − Aeα v = mv
dv =
du
du
=
αv
αe
αu
dv
A
= − dt
αv
e
m
du
αA
∴ 2 =−
dt
u
m
19
Integrating
1 1 A
− = αt
u u m
and substituting eα v = u
 A

ln 1 + eα v α t 
α  m

(b) t = T @ v = 0
 A

α v = ln 1 + eα v αT 
 m

A
m
1 − e −α v 
eα v = 1 + eα v αT
T=
αA
m
dv
A
vdv
A
= − dx
(c) v = v = − eα v
αv
dx
m
e
m
(a) v = v −
1
again, let u = eα v
du = α udv
or
dv =
du
αu
v=
1
α
ln u
1
 du
α ln u  α u
A
= − dx
Integrating and solving
u
m
m
x = 2 1 − (1 + α v ) e −α v 
α A
2.20
d ( mv )
= mv + vm = mg
dt
4
but m = ρ π r 3
m = ρ1π r 2 v
3
4
4
4
so (1) π ρ r 3v + π ρ1r 2 v 2 = π 2 = π ρ r 3 g
3
3
3
F=
Now
ρ1
≈ 10−3
ρ
so, second term is negligible-small
hence v ≈ g and v ≈ gt speed ∝ t but
m = ρ 4π r 2 r = ρ1 π r 2 v or r ≅
1 ρ1
v
4ρ
Hence r ≈
1 ρ1
gt and rate of
4ρ
growth ∝ t
The exact differential equation from (1) above is:
2
4
4ρ
4ρ r
4
r + πρ1
πρ r
= π ρ rg
3
3
ρ1
ρ1
20
3r 2
ρ
= 1 g
4ρ
r
Using Mathcad, solve the above non-linear d.e. letting
which reduces to: r +
ρ1
≈ 10−3 and R ≈ 0.01mm (small raindrop). Graphs
ρ
show that
v ∝ r ∝ t and r ∝ t 2
21
CHAPTER 3
OSCILLATIONS
3.1
x = 0.002sin  2π ( 512 s −1 ) t  [ m ]
m
m
xmax = ( 0.002 )( 2π )( 512 )   = 6.43  
s
s
2
2 m
m
xmax = ( 0.002 )( 2π ) ( 512 )  2  = 2.07 × 104  2 
s 
s 
3.2
x = 0.1sin ω t [m]
When t = 0, x = 0
and
ω = 5 s −1
3.3
m
x = 0.1ω cos ω t  
s
m
x = 0.5   = 0.1ω
s
2π
T=
= 1.26 s
ω
x ( t ) = x cos ω t +
x
ω
sin ω t and ω = 2π f
x = 0.25cos ( 20π t ) + 0.00159sin ( 20π t ) [ m ]
3.4
cos (α − β ) = cos α cos β + sin α sin β
x = A cos (ω t − φ ) = A cos φ cos ω t + A sin φ sin ω t
x = Α cos ω t + Β sin ω t , Α = A cos φ , Β = A sin φ
3.5
1 2 1 2 1 2 1 2
mx1 + kx1 = mx2 + kx2
2
2
2
2
2
2
2
2
k ( x1 − x2 ) = m ( x2 − x1 )
1
k  x22 − x12  2
=
ω =

m  x12 − x22 
1 2 1 2 1 2
kA = mx1 + kx1
2
2
2
m
x2 x2 − x2 x2
A2 = x12 + x12 = 1 12 22 1 + x12
k
x2 − x1
x x −x x 
A=

 x −x 
2 2
1 2
2
2
2 2
2 1
2
1
1
2
22
3.6
3.7
l
1
1
T =π
s ≈ 2.5 s
=π
9.8
g
2
6
For springs tied in parallel:
Fs ( x ) = − k1 x − k2 x = − ( k1 + k2 ) x
1
(k + k ) 2
ω= 1 2 
 m 
For springs tied in series:
The upward force m is keq x .
Therefore, the downward force on spring k2 is keq x .
The upward force on the spring k2 is k1 x′ where x′ is the
displacement of P, the point at which the springs are tied.
Since the spring k2 is in equilibrium, k1 x′ = keq x .
Meanwhile,
The upward force at P is k1 x′ .
The downward force at P is k2 ( x − x′ ) .
Therefore, k1 x′ = k2 ( x − x′ )
x′ =
k2 x
k1 + k2
 k x 
And keq x = k1  2 
 k1 + k2 
1
ω=
3.8
 k1k2
2
=

m  ( k1 + k2 ) m 
keq
For the system ( M + m ) , − kX = ( M + m ) X
The position and acceleration of m are the same as for ( M + m ) :
k
xm
M +m


k
k
xm = A cos 
t + δ  = d cos
t
M
m
M
m
+
+


The total force on m, Fm = mxm = mg − Fr
xm = −
23
Fr = mg +
mk
mkd
k
xm = mg +
t
cos
M +m
M +m
M +m
For the block to just begin to leave the bottom of the box at
the top of the vertical oscillations, Fr = 0 at xm = −d :
mkd
0 = mg −
M +m
g ( M + m)
d=
k
3.9
x = e −γ t A cos (ω d t − φ )
dx
= −e −γ t Aω d sin (ω d t − φ ) − γ e −γ t A cos (ω d t − φ )
dt
dx
= 0 = ω d sin (ω d t − φ ) + γ cos (ω d t − φ )
maxima at
dt
tan (ω d t − φ ) = −
γ
ωd
thus the condition of relative maximum occurs every time that t increases by
ti +1 = ti +
2π
2π
ωd
:
ωd
For the i th maximum: xi = e −γ ti A cos (ω d ti − φ )
xi +1 = e
−γ ti +1
A cos (ω d ti +1 − φ ) = e
−γ
2π
ωd
xi
2π
−γ
xi
= e ωd = eγ Td
xi +1
3.10
(a)
(b)
(c)
c
= 3 s −1
2m
2
ω d = ω 2 − γ 2 = 16 s −2
γ=
k
= 25 s −2
m
2
ω r = ω d 2 − γ 2 = 7 s −2
ω2 =
∴ω r = 7 s −1
F
48
=
Amax =
m = 0.2 m
Cωd 60.4
tan φ =
2γω r
2γω r ω r
7
=
=
=
2
2
2
(ω − ω r ) 2γ γ 3
∴φ ≈ 41.4
24
3.11
17 2
β mx = 0
2
3
17
γ = β and ω 2 = β 2
2
2
2
2
2
2
ω r = ω − 2γ = 4 β
mx + 3β mx +
(a)
Amax =
(b)
=
ω d2 = ω 2 − γ 2 =
25 2
β
4
∴ω d =
5
β
2
2A
15β 2
e −γ Td =
3.12
F
2mγω d
∴ω r = 2 β
γ=
1
2
1
ln 2 = f d ln 2
Td
1
ω d = (ω 2 − γ 2 ) 2
(a)
1
2 2
So, ω = (ω + γ )
2
d
1
1
1
1
 2  γ 2  2
  ln 2  2  2
f =  fd + 
  = f d 1 + 
 
 2π  
  2π  

f = 100.6 Hz
1
(b)
ω r = (ω d2 − γ 2 ) 2
 2  γ 2  2
  ln 2 2  2
fr =  fd − 
  = f d 1 − 
 
 2π  
  2π  

f r = 99.4 Hz
3.13
Since the amplitude diminishes by e −γ Td in each complete period,
n
1
e −γ Td = = e −1
e
γ Td n = 1
ω
1
γ=
= d
Td n 2π n
(
Now
So
)
(
ωd = ω 2 − γ 2
ω = (ω d2 + γ
)
1
2 2
)
1
2
1
1 2

= ω d 1 + 2 2 
 4π n 
25
2π
1
Td ω d ω 
1 2
=
=
= 1 + 2 2 
2π ω d  4π n 
T
ω
For large n,
Td
1
≈ 1+ 2 2
T
8π n
1
3.14
(
(a) ω r = ω − 2γ
2
1
2 2
)
2 2

ω  
= ω 2 − 2    = 0.707ω
 2  

1
 1 2
2
ω
1 − 
ω −γ
ωd
4
=
(b) Q =
= 
= 0.866
2γ
2γ
ω 
2 
 2 
ω 
2   ( 2ω )
2γω
2
2
(c) tan φ = 2
=  2  2 =−
2
3
ω −ω
ω − 4ω
1
2 2
(
)
 2


φ = tan −1  −  = 146.3
3
1
2
 2
2
ω 
2 2
(d) D (ω ) = (ω − 4ω ) + 4   ( 4ω 2 )  = 3.606 ω 2
 2 


F
F
A (ω ) = m = 0.277
D (ω )
mω 2
3.15
A (ω ) ≈
Amaxγ
1
 (ω − ω ) 2 + γ 2  2


1
1
γ
for A (ω ) = Amax ,
=
1
2
2
 (ω − ω ) 2 + γ 2  2


2
(ω − ω ) + γ 2 = 4γ 2
ω − ω = ±γ 3
ω =ω ±γ 3
26
3.16
(b) Q =
ω2 −γ 2
ωd
=
2γ
2γ
ω2 =
1
,
LC
γ=
R
2L
2
 1   R 

− 2 
 LC   4 L 
 L  1
=  2 − 
Q=
 R 
 R C  4
2

 2L 
L
ω 
R
(c) Q =   = C =
R
R
 2γ 
3.17
Fext = F sin ω t = Im  F eiω t 
and x ( t ) is the imaginary part of the solution to:
mx + cx + kx = F eiω t
i.e.
i ω t −φ
x ( t ) = Im  Ae ( )  = A sin (ω t − φ )
where, as derived in the text,
F
A=
1
( k − mω 2 )2 + c 2ω 2  2


and
2γω
tan φ = 2
ω −ω2
3.18
Using the hint, Fext = Re ( F e β t ) ,
where β = −α + iω ,
and x(t) is the real part of the solution to:
mx + cx + kx = F e β t .
x = Ae β t −iφ
Assuming a solution of the form:
( mβ 2 + cβ + k ) x =  FA  xeiφ
F
mα 2 − 2imαω − mω 2 − cα + icω + k = ( cos φ + i sin φ )
A
F
m (α 2 − ω 2 ) − cα + k = cos φ
A
F
ω ( −2mα + c ) = sin φ
A
27
φ = tan −1
ω ( c − 2mα )
m (α 2 − ω 2 ) − cα + k
Using sin 2 φ + cos 2 φ = 1 ,
2
F2
2
=  m (α 2 − ω 2 ) − cα + k  + ω 2 ( c − 2mα )
2
A
F
A=
{m (α − ω ) − cα + k  + ω ( c − 2mα ) }
2
2
2
2
1
2 2
and x ( t ) = Ae −α t cos (ω t − φ ) + the transient term.
3.19
l  A2 
T ≈ 2π
1 −

g
8 
(a)
for A =
(b)
π
4
−
1
2
l
×1.041
g
, T ≈ 2π
4π 2l
× 1.084
T2
l
4π 2l
gives g = 2 , approximately 8% too small.
Using T = 2π
g
T
(c)
g=
λ A3
ω2
and λ =
B=−
32ω 2
6
B
A2
=
A 192
for A =
3.20
π
4
,
f ( t ) = ∑ cn einω t
B
= 0.0032
A
n = 0, ±1, ±2, . . .
n
f ( t ) = ∑ cn cos nω t + ∑ cni sin nω t , n = 0, ±1, ±2, . . .
n
T
2
T
−
2
T
2
T
−
2
n
and cn =
1
T∫
f ( t ) e − inωt dt ,
cn =
1
T∫
f ( t ) cos ( nω t ) dt −
n = 0, ±1, ±2, . . .
i
T
T
2
T
−
2
∫
f ( t ) sin ( nω t ) dt
The first term on cn is the same for n and − n ; the second term changes sign for
n vs. − n . The same holds true for the trigonometric terms in f ( t ) . Therefore, when
terms that cancel in the summations are discarded:
28
1 T

f ( t ) = c + ∑  ∫ 2T f ( t ) cos ( nω t ) dt  cos nω t
−
n T
2

T
1

+ ∑  ∫ 2T f ( t ) sin ( nω t ) dt  sin nω t ,
−
n T
2

1 T2
n = ±1, ±2, . . ., and c = ∫ T f ( t ) dt
T −2
Now, due to the equality of terms in ± n :
 2 T2

f ( t ) = c + ∑  ∫ T f ( t ) cos ( nω t ) dt  cos nω t
−
n T
2

T
2

+ ∑  ∫ 2T f ( t ) sin ( nω t ) dt  sin nω t ,
−
n T
2

n = 1, 2,3, . . .
Equations 3.9.9 and 3.9.10 follow directly.
3.21
f ( t ) = ∑ cn e
inω t
n
T=
2π
ω
so
,
1 T2
cn = ∫ T f ( t ) e− inωt dt ,
T −2
ω
cn =
2π
=
π
ω
π
−
ω
∫
and n = 0, ±1, ±2, …
f ( t ) e− inω t dt
π

ω  0
− inω t
ω − inω t
−
+
e
dt
e dt 
(
)
π
 ∫−
∫
0
2π  ω


ω  1 −inω t
=
e
2π  inω

0
−
π
ω
π

1 − inω t ω 
−
e

inω
0

1
1 − e + inπ − e− inπ + 1
2π in
For n even, einπ = e − inπ = 1 and the term in brackets is zero.
For n odd, einπ = e − inπ = −1
4
cn =
,
n = ±1, ±3, . . .
2π in
4 inω t
f (t ) = ∑
e , n = ±1, ±3, . . .
n 2π in
4 1 1 inω t − inω t
=∑
( e − e ) , n = 1,3,5, . . .
n π n 2i
41
=∑
sin ( nω t ) , n = 1,3,5, . . .
n π n
4
1
1

f ( t ) = sin ω t + sin 3ω t + sin 5ω t + …
π
3
5

=
29
3.22
In steady state, x ( t ) = ∑ An e (
i nω t −φn )
n
An =
Fn
m
1
(ω 2 − n 2ω 2 )2 + 4γ 2 n 2ω 2  2


4F
,
n = 1,3,5, . . .
and
Now Fn =
nπ
ω
9ω 2
Q = 100 ≈
so γ 2 ≈
40, 000
2γ
4F
1
A1 =
⋅
1
mπ
2
2 2
2

9ω ⋅ ω 
2
2
( 9ω − ω ) + 4 40000 


F
A1 ≈
2mπω 2
4F
1
A3 =
⋅
1
3mπ
2 2 2

2
 9ω 
( 9ω 2 − 9ω 2 ) + 4 
 

 200  
400 F
A3 ≈
27 mπω 2
4F
1
A5 =
⋅
1
5mπ
2
2


2
2
 3ω 
2
2
−
+
ω
ω
ω
9
25
4
5
)  200  ( ) 
(


F
A5 ≈
20mπω 2
i.e., A1 : A3 : A5 = 1 : 29.6 : 0.1
3.23
(a)
x +ω2x = 0
y=x
ω = 3ω
Thus y = −ω 2 x
x= y
y dy
ω x
=
=−
x dx
y
2
divide these two equations:
(b)
Solving
ydy
ω2
+ xdx = 0 and Integrating
Let 2C = A2
y2
x2
+
=1
ω 2 A2 A 2
→
y2
x2
+
=C
2ω 2 2
an ellipse
30
3.24
The equation of motion is F ( x ) = x − x 3 = mx . For simplicity, let m=1. Then
(a)
(b)
x = x − x 3 . This is equivalent to the two first order equations …
x = y and
y = x − x3
The equilibrium points are defined by
x − x 3 = x (1 − x )(1 + x ) = 0
Thus, the points are: (-1,0), (0,0) and (+1,0). We can tell whether or not
the points represent stable or unstable points of equilibrium by examining
the phase space plots in the neighborhood of the equilibrium points. We’ll
do this in part (c).
dy y x − x 3
The energy can be found by integrating
= =
or
dx x
y
∫ y dy = ∫ ( x − x ) dx + C
3
or
y 2 x2 x4
= − +C
2
2 4
In other words … E = T + V =
`
(c)
y 2  x4 x2 
+  −  = C . The total energy C is
2 4 2
constant.
The phase space trajectories are given by solutions to the above equation
1

2
x4
y = ±  x 2 − + 2C  .
2


The upper right quadrant of the trajectories is shown in the figure below.
The trajectories are symmetrically disposed about the x and y axes. They
form closed paths for energies C<0 about the two points (-1,0) and (+1,0).
Thus, these are points of stable equilibrium for small excursions away
from these points. The trajectory passes thru the point (0,0) for C=0 and is
a saddle point. Trajectories never pass thru the point (0,0) for positive
energies C>0. Thus, (0,0) is a point of unstable equilibrium.
1
0.5
0
0
0.5
1
1.5
2
31
3.25
θ + sin θ = 0

d θ 2
 − cos θ  = 0
dt  2

⇒
θ2
Integrating:
2
θ
θ
= cos θ θ
0
)
0
θ
∴T = 4 ∫
θ 2 = 2 ( cosθ − cosθ
or
dθ
1
 2 ( cos θ − cos θ )  2
Time for pendulum to swing from θ = 0 to θ = θ is
Now—substitute sin φ =
θ
sin
sin
2
so φ =
θ
and after some algebra …
2
at θ = θ
2
cos θ = 1 − 2sin
and use the identity
π
T
4
2
θ
2
dφ

2θ 
1 − sin 2 
1
2
=
θ
dθ
0
2
  2θ
2 θ 
−
4
sin
sin

 
2
2  
 
∴T = 4 ∫
dθ
  2θ
2 θ 
 4  sin 2 − sin 2  

 
1
1
2
or
π
dφ
2
T = 4∫
(a)
0
1 − α sin 2 φ 
(1 − α sin 2 φ )
(b)
−
1
2
where α = sin 2
1
2
θ
2
1
3
≈ 1 + α sin 2 φ + α 2 sin 4 φ + …
2
8
π
2
3
 1

T = 4 ∫ dφ 1 + α sin 2 φ + α 2 sin 4 φ + …
8
 2

0
2
θ θ 3
 θ2
...
(c)
α = sin
≈
− + … ∼
2  2 48
4

 θ2

T = 2π 1 + + …
 16

-----------------------------------------------------------------------------------------------------2
θ
 α 9

T = 2π 1 + + α 2 + …
 4 64

32
CHAPTER 4
GENERAL MOTION OF A PARTICLE
IN THREE DIMENSIONS
--------------------------------------------------------------------------------------------------------Note to instructors … there is a typo in equation 4.3.14. The range of the projectile is …
v 2 sin 2α
v 2 sin 2 2α
R=x= 0
… NOT ... 0
g
g
--------------------------------------------------------------------------------------------------------4.1
∂V ˆ ∂V ˆ ∂V
−j
−k
∂x
∂y
∂z
ˆ
ˆ + ˆjxz + kxy
F = −c iyz
(a) F = −∇V = −iˆ
(
)
(b) F = −∇V = −iˆ2α x − ˆj 2 β y − kˆ 2γ z
(c) F = −∇V = ce−(α x + β y +γ z ) iˆα + ˆj β + kˆγ
(
(d) F = −∇V = −eˆr
)
∂V
1 ∂V
1 ∂V
− eˆθ
− eˆφ
∂r
r ∂θ
r sin θ ∂φ
F = −eˆr cnr n −1
4.2
(a)
∇× F =
iˆ
ˆj
∂
∂x
x
∂
∂y
y
iˆ
ˆj
∂
∂x
y
∂
∂y
−x
iˆ
ˆj
∂
∂x
y
∂
∂y
x
kˆ
∂
=0
∂z
z
conservative
(b)
∇× F =
kˆ
∂
= kˆ ( −1 − 1) ≠ 0
∂z
z2
conservative
(c)
∇× F =
kˆ
∂
= kˆ (1 − 1) = 0
∂z
z3
conservative
(d)
33
eˆr
∂
1
∇× F = 2
r sin θ ∂r
− kr − n
4.3
eˆθ r
∂
∂θ
0
eˆφ r sin θ
∂
= 0 conservative
∂φ
0
(a)
∇× F =
iˆ
ˆj
∂
∂x
xy
∂
∂y
cx 2
kˆ
∂
= k ( 2cx − x )
∂z
z3
2cx − x = 0
1
c=
2
(b)
∇× F =
also
4.4
(a)
iˆ
ˆj
kˆ
∂
∂x
z
y
∂
∂y
cxz
y2
∂
∂z
x
y
 x cx 
= iˆ  − 2 − 2  +
y 
 y
x cx
− 2 − 2 =0
y
y
cz z
+
=0
y2 y2
ˆj  1 − 1  + kˆ  cz + z 
2
y2 
 y y
y
c = −1
implies that
c = −1 as it must
1
E = constant = V ( x, y, z ) + mv 2
2
1 2
at the origin E = 0 + mv
2
1
1
at (1,1,1)
E = α + β + γ + mv 2 = mv 2
2
2
34
v2 = v2 −
2
(α + β + γ )
m
1
2

2
v = v 2 − (α + β + γ ) 
m


(b)
v2 −
2
(α + β + γ ) = 0
m
1
2
2
v =  (α + β + γ ) 
m

(c)
mx = Fx = −
∂V
∂x
mx = −α
∂V
my = −
= −2 β y
∂y
∂V
mz = −
= −3γ z 2
∂z
4.5
(a)
ˆ + ˆjy
F = ix
ˆ + ˆjdy
dr = idx
on the path x = y :
(1,1)
∫(
0,0 )
1
1
1
1
0
0
0
0
F ⋅ dr = ∫ Fx dx + ∫ Fy dy = ∫ xdx + ∫ ydy = 1
ˆ
dr = idx
dr = ˆjdy
on the path along the x-axis:
and on the line x = 1 :
(1,1)
∫(
0,0 )
1
1
0
0
F ⋅ dr = ∫ Fx dx + ∫ Fy dy = 1
F is conservative.
(b)
ˆ − ˆjx
F = iy
on the path x = y :
(1,1)
∫(
0,0 )
1
1
0
0
1
1
0
0
F ⋅ dr = ∫ Fx dx + ∫ Fy dy = ∫ ydx − ∫ xdy
(1,1)
and, with x = y
∫(
on the x-axis:
∫(
and, with y = 0 on the x-axis
∫(
on the line x = 1 :
∫(
0,0 )
(1,0 )
0,0 )
(1,0 )
0,0 )
(1,1)
1,0 )
1
1
0
0
F ⋅ dr = ∫ xdx − ∫ ydy = 0
1
1
0
0
F ⋅ dr = ∫ Fx dx = ∫ ydx
F ⋅ dr = 0
1
1
0
0
F ⋅ dr = ∫ Fy dy = ∫ xdy
35
(1,1)
∫(
and, with x = 1
on this path
1,0 )
(1,1)
∫(
0,0 )
1
F ⋅ dr = ∫ dy = 1
0
F ⋅ dr = 0 + 1 = 1
F is not conservative.
4.6
From Example 2.3.2, V ( z ) = −mg
re2
( re + z )

z
V ( z ) = − mgre 1 + 
 re 
From Appendix D,
(1 + x )
−1
−1
= 1 − x + x 2 +…


z z2
V ( z ) = − mgre 1 − + 2 + … 
 re re

2
mgz
V ( z ) = −mgre + mgz −
+…
re
With − mgre an additive constant,

z
V ( z ) ≈ mgz 1 − 
 re 
∂
F = −∇V = − kˆ V ( z )
∂z
ˆ 1 − z + z  − 1  
= − kmg


 re  
 re
ˆ 1 − 2 z 
F = − kmg


re 

mx = Fx = 0 ,
mz
my = Fy = 0
 2z 
mz = − mg 1 − 
re 

 2z 
dz
= − mg 1 − 
dz
re 

h
2z 
zdz = − g ∫ 1 −  dz
0
z
re 


1
h2 
− v 2z = − g  h − 
re 
2

∫
0
v
re v 2z
=0
h − re h +
2g
2
36
h=
re 1 2 2re v 2z
−
re −
g
2 2
h=
re re
2v 2
1− z
−
2 2
gre
From Appendix D,
( h, z
1
(1 + x ) 2 = 1 +
h=
re re v 2z
v4
− +
+ z +…
2 2 2 g 4 gre
h≈
v 2z 
v 2z 
+
1

2 g  2 gre 
re )
x x2
− +…
2 8
−1
v2 
v2 
From Example 2.3.2, h =
1 −

2 g  2 gre 
v2 
v2 
−1
And with (1 − x ) ≈ 1 + x ,
+
1
h≈


2 g  2 gre 
4.7
For a point on the rim measured from the center of the wheel:
ˆ cos θ − ˆjb sin θ
r = ib
vt
θ =ωt = ,
so r = −iˆv sin θ − ˆjv cos θ
b
Relative to the ground, v = iˆv (1 − sin θ ) − ˆjv cosθ
For a particle of mud leaving the rim:
y = −b sin θ and v y = −v cosθ
So
v y = v y − gt = −v cos θ − gt
1
y = −b sin θ − v t cosθ − gt 2
2
At maximum height, v y = 0 :
and
t=−
v cos θ
g
 v cos θ
h = −b sin θ − v  −
g

2
2
v cos θ
h = −b sin θ +
2g
Maximum h occurs for

1  v cos θ 
 cos θ − g  −

g 
2 

2
dh
2v 2 cos θ sin θ
= 0 = −b cos θ −
2g
dθ
37
sin θ = −
gb
v2
cos 2 θ = 1 − sin 2 θ =
v 4 − g 2b 2
v4
gb 2 v 4 − g 2b 2 gb 2 v 2
+
= 2+
v2
2 gv 2
2v
2g
Measured from the ground,
gb 2 v 2
′ =b+ 2 +
hmax
2v
2g
hmax =
 gb 
The mud leaves the wheel at θ = sin −1  − 2 
 v 
4.8
x = R cos φ
so t =
and
x = v x t = ( v cos α ) t
R cos φ
v cos α
1
1
and y = v y t − gt 2 = ( v sin α ) t − gt 2
2
2
y = R sin φ
R cos φ 1  R cos φ 
R sin φ = ( v sin α )
− g

v cos α 2  v cos α 
gR cos 2 φ
sin φ = tan α cos φ − 2
2v cos 2 α
α
2
φ
2v 2 cos 2 α
2v 2 cos α
tan
cos
sin
α
φ
−
φ
=
(
)
( sin α cos φ − cos α sin φ )
g cos 2 φ
g cos 2 φ
From Appendix B, sin (θ + φ ) = sin θ cos φ + cos θ sin φ
R=
2v 2 cos α
sin (α − φ )
R=
g cos 2 φ
dR
2v 2
=0=
 − sin α sin (α − φ ) + cos α cos (α − φ ) 
dα
g cos 2 φ 
Implies that cos α cos (α − φ ) − sin α sin (α − φ ) = 0
R is a maximum for
From appendix B, cos (θ + φ ) = cos θ cos φ − sin θ sin φ
so cos ( 2α − φ ) = 0
2α − φ =
Rmax =
π
2
α=
π
4
+
φ
2
2v
π φ  π φ 
cos  +  sin  − 
2
g cos φ
 4 2  4 2
2
38
π  π φ 
π φ 
π φ 
Now sin  −  = cos  −  −   = cos  + 
 4 2
 4 2
 2  4 2 
2v 2
π φ 
cos 2  + 
Rmax =
2
g cos φ
 4 2
Again using Appendix B, cos 2θ = cos 2 θ − sin 2 θ = 2 cos 2 θ − 1
v2   π
2v 2  1
π
 1
 
cos  + φ  + 1
Rmax =
cos  + φ  +  =


2
2
g cos φ  2
 
2
 2  g cos φ   2
π

Using cos  + θ  = − sin θ ,
2

2
v
Rmax =
(1 − sin φ )
g (1 − sin 2 φ )
2
Rmax =
v
g (1 + sin φ )
4.9
(a) Here we note that the projectile is launched
“downhill” towards the target, which is located a
distance h below the cannon along a line at an angle
φ below the horizon. α is the angle of projection
that yields maximum range, Rmax. We can use the
α
results from problem 4.8 for this problem. We
simply have to replace the angle φ in the above
φ
result with the angle -φ, to account for the downhill
h
slope. Thus, we get for the downhill range …
Rmax
2v0 2 cos α sin (α + ϕ )
R=
cos 2 ϕ
g
The maximum range and the angle is α are obtained from the problem above again by
v 2 (1 + sin ϕ )
π
and … 2α = − ϕ .
replacing φ with the angle -φ … Rmax = 0
2
2
g cos ϕ
2
2
v (1 + sin ϕ )
v0
h
We can now calculate α … Rmax =
= 0
=
2
sin ϕ
g cos ϕ
g (1 − sin ϕ )

gh 
1 + 2 
 v0 
π

But, from the above … sin ϕ = sin  − 2α  = cos 2α = 1 − 2sin 2 α
2

gh 
gh 
Thus … 1 − 2 sin 2 α = 2  1 + 2 
v0  v0 
Solving for sin ϕ … sin ϕ =
gh
v0 2
39
2sin 2 α =
2
gh
= 1− 2
2
csc α
v0

1
gh 
1 + 2  =
 v0  1 + gh
v0 2

gh 
Finally … csc 2 α = 2 1 + 2 
 v0 
(b)
Solving for Rmax … Rmax =
h
h
h
=
=
2
sin ϕ 1 − 2sin α 1 − 2 csc 2 α
Substituting for csc 2 α and solving …
v2
gh 
Rmax = 0 1 + 2 
g  v0 
4.10 We can again use the results of problem 4.8. The maximum slope range from
problem 4.8 is given by …
v0 2
h
=
Rmax =
g (1 + sin ϕ ) sin ϕ
Solving for sin ϕ …
sin ϕ =
gh
v0 2
 gh 
1 − 2 
 v0 
Thus …
cos ϕ
sin ϕ
We can calculate cos ϕ from the above relation for sin ϕ …
xmax = Rmax cos ϕ = h
1
2

gh   gh 
cos ϕ = (1 − sin 2 ϕ ) = 1 − 2 2   1 − 2 
v0   v0 

Inserting the results for sin ϕ and cos ϕ into the above …
1
2
1
xmax
cos ϕ v0 2 
gh  2
1
2
=h
=
−


sin ϕ
g 
v0 2 
4.11 We can simplify this problems somewhat by noting that the trajectory is
symmetric about a vertical line that passes through the highest point of the trajectory.
Thus we have the following picture …
40
z
zmax
v0
h0
h1
α
x0
δ
x
x1
R
We have “reversed the trajectory so that h0 (= 9.8 ft), and x0 , the height and range within
which Mickey can catch the ball represent the starting point of the trajectory. h1 (=3.28
ft) is the height of the ball when Mickey strikes it at home plate. δ is the distance behind
home plate where the ball would be hypothetically launched at some angle α to achieve
the total range R. x1 (=328 ft) is the distance the ball actually would travel from home
plate if not caught by Mickey. (Note, because of the symmetry, v0 is the speed of the ball
when it strikes the ground … also at the same angle α at which was launched. We will
calculate the value of x0 assuming a time-reversed trajectory!)
v 2 sin 2α 2v0 2 sin α cos α
=
(1) The range of the ball … R = 0
g
g
2
R
g
R
tan α −
 
2
2
2
2v0 cos α  2 
g
2
h1 = x1 tan α −
(3) The height at x1 …
( x1 )
2
2
2v0 cos α
g
tan α
=
and inserting this into (2) gives …
From (1) …
2
2
R
2v0 cos α
R
R
R
zmax = tan α − tan α = tan α
2
4
4
4z
Thus, R = max and inserting this expression and the first previously derived into (3)
tan α
(2) The maximum height … zmax =
(4)
h1 = x1
( x tan α )
tan α − 1
2
4 zmax
Let u = x1 tan α and we obtain the following quadratic …
u 2 − 4 zmax u + 4 zmax h1 = 0 and solving for u …
1
h


u = 2 zmax 1 ± (1 − h1 zmax ) 2  and letting ε = 1 , we get …
zmax


41
u ≈ zmax ε = h1
or
u ≈ 2 zmax ( 2 − ε ) = 2 zmax ( 2 − .0475 ) = 3.9 zmax . This result is the correct one …
3.9 zmax
= 0.821 ∴ α = 39.40
x1
Now solve for x0 using a relation identical to (4) …
Thus, tan α =
h0 = x0
( x tan α )
tan α − 0
2
4 zmax
Again we obtain a quadratic expression for u = x0 tan α which we solve as before. This
time, though, the first result for u is the correct one to use …
u = zmax ε ≈ h0 and we obtain …
h
x0 = 0 = 11.9 ft
tan α
4.12
The x and z positions of the ball vs. time are
1
1
1
x = v t cos θ
z = v t cos θ sin θ − gt 2
2
2
2
1
Since vx = v cos θ
2
v2
1
cos 2 θ sin 2θ
R=
The horizontal range is
2
g
dR
The maximum range occurs @
=0
dθ
1
1
dR v 2 

2 1
=
 2 cos θ cos 2θ − cos θ sin θ sin 2θ  = 0
dθ
g 
2
2
2

1
1
1
Thus, 2 cos 2 θ cos 2θ = cos θ sin θ sin 2θ
2
2
2
2
Using the identities: 2 cos θ = 1 + cos 2θ and sin 2θ = 2sin θ cosθ
We get:
(1 + cosθ ) ( 2 cos2 θ − 1) = sin θ sin θ cosθ = (1 − cos 2 θ ) cosθ
or
(1 + cosθ ) ( 3cos 2 θ
Thus
cos θ = −1 ,
− cosθ − 1) = 0
cosθ =
(
1
1 ± 13
6
)
Only the positive root applies for the θ -range:
(
)
1
cosθ = 1 + 13 = 0.7676
6
Thus (b) for v = 25 m s −1 Rmax = 55.4 m
0 ≤θ ≤
π
2
θ = 39 51′
@ θ = 39 51′
42
(a) The maximum height occurs at
or
1
v cos θ sin θ = gT
2
v2
1
H=
cos 2 θ sin 2 θ
2g
2
or at
dz
=0
dt
1
v cos θ sin θ
2
T=
g
maximum at fixed θ
dH
=0
dα
dH v 2 
1
1

2 1
2
=
 2 cos θ sin θ cosθ − cos θ sin θ sin θ  = 0
dα 2 g 
2
2
2

Using the above trigonometric identities, we get
1
1
(1 + cosθ ) sin θ cosθ = sin θ sin 2 θ = sin θ (1 − cos 2 θ )
2
2
or
sin θ (1 + cosθ )( 3cos θ − 1) = 0
The maximum possible height occurs @
1
3
The first two roots give minimum heights; the last gives the maximum
1
Thus, H max = 18.9m @θ = cos −1 = 70 32′
3
There are 3-roots:
4.13
sin θ = 0 ,
cos θ = −1 ,
cos θ =
The trajectory of the shell is given by Eq. 4.3.11 with r replacing x
z
g
z = r − 2 r2
where r = v cos θ
z = v sin θ
r
2r
Thus, z = r tan θ −
g r2
sec 2 θ
2v 2
Since sec2 θ = 1 + tan 2 θ
We have:
g r2
g r2
2
tan
θ
−
tan
θ
+
+
=0
r
z
2v 2
2v 2
(r,z) are target coordinates.
The above equation yields two possible roots:
1
1  2
4
2
2 2 2
±
−
−
tan θ =
v
v
2
gzv
g
r
(
)

gr 

The roots are only real if
v 4 − 2 gzv 2 − g 2 r 2 ≥ 0
The critical surface is therefore:
v 4 − 2 gzv 2 − g 2 r 2 = 0
4.14
If the velocity vector, of magnitude s , makes an angle θ with the z-axis, and its
43
projection on the xy-plane make an angle φ with the x-axis:
x = s sin θ cos φ ,
and Fx = Fr sin θ cos φ = mx
y = s sin θ sin φ ,
and Fy = Fr sin θ sin φ = my
z = s cos θ ,
and Fz = − mg + Fr cos θ = mz
Since Fr = −c2 s = −c2 ( x + y 2 + z 2 ) , the differential equations of motion are not
2
2
separable.
mx = −c2 s 2 sin θ cos φ = −c2 sx
dx
dx ds
dx
m = m ⋅ = ms = −c2 sx
dt
ds dt
ds
dx
c2
c
= − ds = −γ ds , where γ = 2
m
x
m
x
ln x − ln x = ln = −γ s
x
θ
s
y
x = x e −γ s
y = y e −γ s
Similarly
4.15
z
x
φ
z
g γ x
g  γx 
From eqn 4.3.16,  + 2  max + 2 ln 1 − max  = 0
γ
x 
γ γ  x

2
3
u u
From Appendix D: ln (1 − u ) = −u − − − …
for u < 1
2 3
3
 γx 
γx
γ 2 xmax 2 γ 3 xmax
−
+ terms in γ 4
ln  1 − max  = − max −
2
3
x 
x
2x
3x

2
3
z xmax gxmax gxmax gxmax
gγ xmax
+
−
−
−
+ terms in γ 2 = 0
2
3
x
2x
3x
γx
γx
2
xmax
+
3x
3x 2 z
xmax −
≈0
gγ
2γ
1
xmax
3x  9 x 2 3x 2 z  2
≈−
±
+

gγ 
4γ  16γ 2
1
3 x 3 x  16γ z  2
±
xmax ≈ −
1 +

4γ 4γ 
3g 
Since xmax > 0 , the + sign is used.
From Appendix D:
1
2
 16γ z  2
8γ z 1  16γ z 
3
− 
1 +
 = 1+
 + terms in γ
3g 
3g 8  3g 

3x 3x 2 x z 8 x γ z 2
−
+ terms in γ 2
xmax = −
+
+
2
4γ 4γ
g
3g
44
2 x z 8x z 2
−
γ +…
3g 2
g
z = v sin α and 2 x z = v 2 sin 2α :
xmax =
For
xmax =
v 2 sin 2α 4v 3 sin 2α sin α
−
γ +…
3g 2
g
4.16
y
x
x = A cos (ω t + α ) ,
x = − Aω sin (ω t + α )
from x = 0 ,
from x = A ,
y = B cos (ω t + β ) ,
α =0
x = A cos ω t
y = −ω B sin (ω t + β )
1 2 1 2 1 2
kB = ky + my
2
2
2
with
B 2 = 16 A2 +
Then
1
ω2
( 9ω
2
y = 4 A , y = 3ω A
and ω =
k
:
m
A2 ) = 25 A2
B = 5A
4 A = 5 A cos β and 3ω A = −5ω A sin β
4
 3
β = cos −1   = sin −1  −  = −36.9
5
 5
y = 5 A cos (ω t − 36.9
)
Since maximum x and y displacements are ± A and ±5A ,
respectively, the motion takes place entirely within a
rectangle of dimension 2 A and 10A .
∆ = β − α = −36.9 − 0 = −36.9
2 AB cos ∆
From eqn 4.4.15, tan 2ψ = 2
A − B2
4
( 2 A)( 5 A) cos ( −36.9 ) 10  5  1
=
tan 2ψ =
=−
2
−24
3
A2 − ( 5 A )
1
 1
ψ = tan −1  −  = −9.2
2
 3
4.17
mx = Fx = −

x = A cos 

∂V
= − kx = −π 2 mx
∂x

k
t + α  = A cos (π t + α )
m

45
∂V
= −4π 2 mx
∂y
y = B cos ( 2π t + β )
my = −
∂V
= −9π 2 mz
∂z
z = C cos ( 3π t + γ )
mz = −
Since x = y = z = 0 at t = 0 ,
π
α = β =γ = −
π
2

x = A cos  π t −  = A sin π t
2

x = Aπ cos π t
Since v 2 = x 2 + y 2 + z 2 and x = y = z ,
v
x =
= Aπ
3
v
A=
π 3
v
x=
sin π t
π 3
y = B sin 2π t ,
y = 2 Bπ cos 2π t
v
= 2π B
y =
3
v
B=
2π 3
v
sin 2π t
y=
2π 3
z = C sin 3π t ,
z = 3Cπ cos 3π t
v
= 3Cπ
z =
3
v
C=
3π 3
v
sin 3π t
z=
3π 3
Since ω x = π , ω y = 2π , and ω z = 3π the ball does retrace its path.
tmin =
2π n1
ωx
=
2π n2
ωy
=
2π n3
ωz
The minimum time occurs at n1 = 1 , n2 = 2 , n3 = 3 .
46
tmin =
4.18
2π
π
=2
Equation 4.4.15 is
tan 2ψ =
2 AB cos ∆
A2 − B 2
Transforming the coordinate axes xyz to the new axes x′y′z ′ by a rotation about
the z-axis through an angleψ given, from Section 1.8:
x′ = x cosψ + y sin ψ ,
y′ = − x sinψ + y cosψ
or,
x = x′ cosψ − y′ sinψ , and
y = x′ sinψ + y′ cosψ
From eqn. 4.4.10:
x2
2 cos ∆ y 2
x
−
+ 2 = sin 2 ∆
y
2
A
AB
B
Substituting:
1
x′2 cos 2 ψ − 2 x′y′ cosψ sinψ + y′2 sin 2 ψ )
2 (
A
2 cos ∆ 2
 x′ cosψ sinψ + x′y′ ( cos 2 ψ − sin 2 ψ ) − y′2 cosψ sin ψ 
−

AB 
1
+ 2 ( x′2 sin 2 ψ + 2 x′y′ cosψ sin ψ + y′2 cos 2 ψ ) = sin 2 ∆
B
For x′ to be a major or minor axis of the ellipse, the coefficient of x′y′ must
vanish.
2 cosψ sinψ 2 cos ∆
2 cosψ sinψ
cos 2 ψ − sin 2 ψ ) +
−
−
=0
(
2
A
AB
B2
From Appendix B, 2 cosψ sin ψ = sin 2ψ and cos 2 ψ − sin 2 ψ = cos 2ψ
sin 2ψ 2 cos ∆ cos 2ψ sin 2ψ
=0
−
−
+
A2
AB
B2
1  2 cos ∆
 1
tan 2ψ  2 − 2  =
A 
AB
B
2 AB cos ∆
tan 2ψ = 2
A − B2
4.19
Shown below is a face-centered cubic lattice. Each atom in the lattice is centered
within a cube on whose 6 faces lies another adjacent atom. Thus each atom is
surrounded by 6 nearest neighbors at a distance d. We neglect the influence of
atoms that lie at further distances. Thus, the potential energy of the central atom
can be approximated as
6
V = ∑ cri −α
i =1
2d
47
1
r1 = ( d − x ) + y 2 + z 2  2


2
−
α
α
 2x x2 + y2 + z 2  2
+
r = ( d − 2dx + x + y + z ) = d 1 −

d
d2


1
n
From Appendix D, (1 + x ) = 1 + nx + n ( n − 1) x 2 + …
2
2
 α  2 x x + y 2 + z 2  1  α   α 
r1−α = d −α 1 −  − +
 +  −   − − 1
d2
 2  d
 2  2  2 
−α
1
2
2
2
2 −2
−α
2
2
2
2
2
2 2
 2 x  2
x3 
 2x   x + y + z   x + y + z  
+
term
s
in
+
 −  − 2   





d2
d2
d3 
 d 
 d 
 
 
x3  
α  α   4 x2
 αx α
− 2 ( x 2 + y 2 + z 2 ) +  + 1  2 + terms in 3  
r1−α = d −α 1 +
d  
d 2d
4  2  d

 αx α
α α  
− 2 ( x 2 + y 2 + z 2 ) + 2  + 1 x 2 
r1−α ≈ d −α 1 +
d 2d
d 2  

1
1
r2 = ( − d − x ) + y 2 + z 2  2 =  d 2 + 2dx + x 2 + y 2 + z 2  2


2
−
−α
2
r
r2−α
α
 2 x x2 + y 2 + z 2  2
= d 1 +
+

d
d2


 αx α
α α  
≈ d −α 1 −
− 2 ( x 2 + y 2 + z 2 ) + 2  + 1 x 2 
d 2d
d 2  

−α
α
α


r1−α + r2−α ≈ d −α  2 − 2 ( x 2 + y 2 + z 2 ) + 2 (α + 2 ) x 2 
d
 d

Similarly:
α
α


r3−α + r4−α ≈ d −α  2 − 2 ( x 2 + y 2 + z 2 ) + 2 (α + 2 ) y 2 
d
 d

α
α


r5−α + r6−α ≈ d −α  2 − 2 ( x 2 + y 2 + z 2 ) + 2 (α + 2 ) z 2 
d
 d

 3α

 α 2 2α 
V ≈ cd −α 6 − 2 ( x 2 + y 2 + z 2 ) +  2 + 2  ( x 2 + y 2 + z 2 ) 
d
d 
d


−α
−α − 2
2
2
2
2
≈ 6cd + cd
(α − α )( x + y + z )
V ≈ A + B ( x2 + y 2 + z 2 )
4.20
48
z
k̂B
(
F = q E+v×B
)
ˆ × kB
ˆ = iyB
ˆ + ˆjy + kz
ˆ − ˆjxB
v × B = ix
ĵE
ˆ0
iv
(
)
ˆ
F = iqyB
+ ˆjq ( E − xB )
y
mx = Fx = qyB
qB
x−x =
y
m
x
qB

my = Fy = qE − qxB = qE − qB  x +
m


y

2
2
qE qBx  qB 
eE eBx  eB 
−
−
+
−  y
y=
 y=−
m
m  m
m
m m
eE
eB
+ωx ,
ω=
y +ω2 y = −
m
m
1  eE

+ ω x  + A cos (ω t + θ )
y = 2 −
ω  m

y = − Aω sin (ω t + θ )
y = 0 , so θ = 0
1  eE

−
+ωx 
2 
ω  m

1  eE

y = a (1 − cos ω t ) ,
+ωx 
a = 2 −
ω  m

qB
x=x +
y = x − ω y = x − ω a (1 − cos ω t )
m
x = ( x − ω a ) + ω a cos ω t
y = 0 , so A = −
x = ( x − ω a ) t + a sin ω t
x = a sin ω t + bt ,
mz = Fz = 0
z=zt+z =0
b = x −ωa
4.21
1 2
b
mv + mqh = mg
2
2
2
v = g ( b − 2h )
y
b
h
x
mv 2
= −mg cos θ + R
Fr = −
b
49
h
b
h mv 2 mg
mg
R = mg −
=
 h − ( b − 2h )  =
( 3h − b )
b
b
b
b
the particle leaves the side of the sphere when R = 0
b
b
above the central plane
h = , i.e.,
3
3
cos θ =
4.22
1 2
mv + mgh = 0
2
at the bottom of the loop, h = −b
1 2
mv = mgb ,
2
v = 2gb
so
h
b
mv 2
Fr = −mg + R =
b
2
mv
R = mg +
= mg + 2mg = 3mg
b
v
4.23
From the equation for the energy as a function of s in Example 4.6.2,
1
1  mg  2
E = ms 2 + 
s ,
2
2  4A 
s is undergoing harmonic motion with:
"k "
g
1 g
ω=
=
=
m
4A 2 A
Since s = 4 A sin φ , φ increases by 2π radians during the time interval:
 A
= 2π  2

ω
 g
For cycloidal motion, x and z are functions of 2φ so they undergo a
complete cycle every time φ changes by π . Therefore, the period for the
cycloidal motion is one-half the period for s .
1
A
T = T ′ = 2π
2
g
------------------------------------------------------------------------------------------------------
T′ =
2π
50
CHAPTER 5
NONINERTIAL REFERENCE SYSTEMS
5.1 (a) The non-inertial observer believes that he is in equilibrium and that the net force
G
acting on him is zero. The scale exerts an upward force, N , whose value is equal to the
scale reading --- the “weight,” W’, of the observer in the accelerated frame. Thus
G
G
G
N + mg − mA0 = 0
G
N
N − mg − mA0 = N − mg − m
G
A0
g
5
= N − mg = 0
4
4
5
5
mg = W
4
4
W ′ = 150lb.
W′ = N =
(a)
(b)
G
mg
(b) The acceleration is downward, in the same
G
direction as g
W 3
g
N − mg + m   = 0 W ′ = W − = W
4 4
4
W ′ = 90lb.
G
G G G
5.2 (a) Fcent = −mω × (ω × r ′ )
G G G
For ω ⊥ r ′ , Fcent = mω 2 r ′ eˆr ′
ω = 500 s −1 = 1000π s −1
G
2
Fcent = 10−6 × (1000π ) × 5 eˆr = 5π 2 dynes outward
Fcent mω 2 r ′ (1000π ) 5
=
=
= 5.04 × 104
Fg
mg
980
2
(b)
5.3
G
G G
mg + T − mAD = 0
(See Figure 5.1.2)
g
− mg ˆj + T cos θ ˆj + T sin θ iˆ − m   iˆ = 0
 10 
mg
T cos θ = mg , and T sin θ =
10
1
tan θ = , θ = 5.71D
10
mg
= 1.005mg
T=
cos θ
G
5.4
The non-inertial observer thinks that g ′ points downward in the direction of the
hanging plumb bob… Thus
51
g
G G G
g ′ = g − AD = g ˆj − iˆ
10
For small oscillations of a simple pendulum:
1
T = 2π
g′
2
g
g ′ = g +   = 1.005 g
 10 
2
T = 2π
5.5 (a)
1
1
= 1.995π
1.005g
g
f = − µ mg is the frictional force acting on the
G
A0
G
f
box, so
G
G
G
f − mA0 = ma′
(b)
(a)
G
( a′ is the acceleration of the box relative to the truck. See
G
Equation 5.1.4b.) Now, f the only real force acting horizontally, so the
acceration relative to the road is
f
µ mg
g
a= =−
= −µ g = −
m
m
3
(For + in the direction of the moving truck, the – indicates that friction opposes
the forward sliding of the box.)
g
AD = −
(The truck is decelerating.)
2
from above, ma − mA0 = ma′ so
g g g
a′ = a − AD = − + =
3 2 6
K
5.6 (a) r = iˆ ( xD + R cos Ωt ) + ˆjR sin Ωt
K
r = −iˆΩR sin Ωt + ˆjΩR cos Ωt
K K
r ⋅ r = v 2 = Ω 2 R 2
∴ v = ΩR circular motion of radius R
K K K K
K ˆ
′ + ˆjy′
(b) r′ = r − ω × r ′ where r ′ = ix
ˆ ′ + ˆjy′
= −iˆΩR sin Ωt + ˆjΩR cos Ωt − ω kˆ × ix
(
)
52
= −iˆΩR sin Ωt + ˆjΩR cos Ωt − ˆjω x′ + iˆω y′
x ′ = ω y′ − ΩR sin Ωt
y ′ = −ω x′ + ΩR cos Ωt
(c) Let u′ = x′ + iy′
here i = −1 !
u ′ = x ′ + iy ′ = ω y′ − ΩR sin Ωt − iω x′ + iΩR cos Ωt
iω u ′ =
+iω x′
= −ω y′
∴ u ′ + iω u′ = −ΩR sin Ωt + iΩR cos Ωt = iΩ ReiΩt
Try a solution of the form
u′ = Ae − iω t + BeiΩt
u ′ = −iω Ae− iω t + iΩBeiΩt
iω u′ = iω Ae− iω t + iω BeiΩt
∴ u ′ + iω u ′ = i (ω + Ω ) Beiω t
so B =
ΩR
ω +Ω
Also at t = 0 the coordinate systems coincide so
u ′ = A + B = x′ ( 0 ) + iy′ ( 0 ) = xD + R
ΩR
ω +Ω
ΩR iΩt
e
+
ω +Ω
∴ A = xD + R − B = xD + R −
ω R  − iω t

Thus, u′ =  xD +
e
ω + Ω 

so, A = xD +
ωR
ω +Ω
5.7
The x, y frame of reference is attached to the Earth, but the x-axis always points
away from the Sun. Thus, it rotates once every year relative to the fixed stars. The X,Y
frame of reference is fixed inertial frame attached to the Sun.
(a) In the x, y rotating frame of reference
x ( t ) = R cos ( Ω − ω ) t − Rε
y ( t ) = − R sin ( Ω − ω ) t
where R is the radius of the asteroid’s orbit
and RE is the radius of the Earth’s orbit. Ω
is the angular frequency of the Earth’s
revolution about the Sun and ω is the
angular frequency of the asteroid’s orbit.
(b)
x ( t ) = − ( Ω − ω ) R sin ( Ω − ω ) t → 0 at t = 0
y ( t ) = − ( Ω − ω ) R cos ( Ω − ω ) t → − ( Ω − ω ) R at t = 0
(c)
K K K K K
K K K K K
a = A − Aε − Ω
× r − 2Ω × r − Ω × Ω × r
K
Where a is the acceleration of the asteroid in the x, y frame of reference,
53
K K
A, Aε are the accelerations of the asteroid and the Earth in the fixed, inertial frame
of reference.
K K K K K
1st : examine: A − Aε − Ω × Ω × r
K K
K K K K K K K K
= ω × ω × R − Ω × Ω × Rε − Ω × Ω × ( R − Rε )
K
K
K
K K K K
K
= (ω × ω − Ω × Ω ) × R = − (ω 2 − Ω 2 ) R
note: ω = ω k̂ , Ω = Ωk̂
Thus:
K
K K
K
a = ( Ω 2 − ω 2 ) R − 2Ω × v
Therefore:
ˆ + ˆjy = ( Ω 2 − ω 2 ) iR
ˆ
ˆ
ˆ
ˆ

ix
 cos ( Ω − ω ) t − jR sin ( Ω − ω ) t  −2 jΩx + 2i Ωy
(
)
Thus:
x = ( Ω 2 − ω 2 ) R cos ( Ω − ω ) t + 2Ωy
y = − ( Ω 2 − ω 2 ) R sin ( Ω − ω ) t − 2Ωx
Let x = ( Ω − ω ) y
and
Then, we have
y = ( Ω − ω ) R cos ( Ω − ω ) t +
y = ( Ω − ω ) x
2Ω
y
(Ω − ω )
which reduces to
y = − ( Ω − ω ) R cos ( Ω − ω ) t
Integrating … y = − R sin ( Ω − ω ) t → 0 at t = 0
Also,
or
− x ( Ω − ω ) = − ( Ω 2 − ω 2 ) R sin ( Ω − ω ) t − 2Ωx
x = ( Ω + ω ) R sin ( Ω − ω ) t + 2Ωx
x = − ( Ω − ω ) R sin ( Ω − ω ) t
Integrating … x = R cos ( Ω − ω ) t + const
x = R cos ( Ω − ω ) t − Rε → R − Rε at t = 0
5.8
Relative to a reference frame fixed to the turntable the cockroach travels at a
constant speed v’ in a circle. Thus
y′
v ′2
G
G
′
a
eˆr ′ .
=
−
ω
b
Since the center of the turntable is fixed.
G
′
x
b
AD = 0
The angular velocity, ω, of the turntable is
constant, so
54
G
G
ω = ω k̂ ′ , with ω = 0
G
G G G
r ′ = beˆr ′ , so ω × (ω × r ′ ) = −bω 2 eˆr ′
G G
G
v′ = v′êθ ′ , so ω × v′ = −ω v′eˆr ′
G G
G G G G G
From eqn 5.2.14,
a = a′ + 2ω × v′ + ω × (ω × r ′) and putting in terms from above
v′ 2
− 2ω v′ − bω 2
ar ′ = −
b
G
G
For no slipping F ≤ µ s mg , so a ≤ µ s g
v′ 2
+ 2ω v′ + bω 2 ≤ µ s g
b
2
vm′ + 2ω bvm′ + b 2ω 2 − bµ s g = 0
vm′ = −ω b ± ω 2b 2 − b 2ω 2 + bµ s g
Since v′ was defined positive, the +square root is used.
vm′ = −ω b + bµ s g
(b)
5.9
G
v′ = −v′êθ ′
G G
ω × v′ = +ω v′eˆr ′
v′ 2
+ 2ω v′ − bω 2
ar ′ = −
b
2
′
v
− 2ω v′ + bω 2 ≤ µ s g
b
vm′ = ω b + bµ s g
G VD 2
VD ˆ
iˆ′
As in Example 5.2.2, ω =
j ′ and AD =
G
ρ
ρ
For the point at the front of the wheel:
2
G
G V
r ′ = D ˆj ′ and v′ = −VD k̂ ′
b
G
ω =0
Vb
G G V
ω × r ′ = D kˆ′ × −bjˆ′ = D iˆ′
ρ
G
G G
ω × (ω × r ′ ) =
G G
ω × v′ =
(
)
ρ
2
VD ˆ VDb ˆ VD b ˆ
j′
k′×
i′ =
ρ
ρ
(
)
ρ
VD ˆ
k ′ × −VD kˆ′ = 0
ρ
G VD 2
 VD 2 VD 2b 
G G G G G
ˆ
+ 2  ˆj ′
a = r ′ + ω × (ω × r ′ ) + AD =
i′ + 
ρ
ρ 
b

55
5.10
(See Example 5.3.3)
mω 2 x′ = mx′
x′ ( t ) = Aeω t + Be −ω t
x ′ ( t ) = ω Aeω t − ω Be −ω t
Boundary Conditions:
l
x2 ( 0) = = A + B
2
x ′ ( 0 ) = 0 = ω ( A − B )
∴A=
l
4
B=
l
cosh ω t
2
l l l
x′ (T ) = + = cosh ωT
2 2 2
l
x ′ ( t ) = ω sinh ω t
2
x′ ( t ) =
(a)
(b)
∴ cosh ωT = 2
when the bead reaches the end of the rod
T=
or
1
ω
cosh −1 2 =
1.317
ω
l
x ′ (T ) = ω sinh ωT
2
l
l
= ω sinh cosh −1 2  = ω (1.732 ) = 0.866ω l
2
2
1
l
ωl
or ω  cosh 2 ωT − 1 2 = 3
= 0.866ω l
2
2
(c)
5.11
l
4
G
v′ = 400 ˆj ′ mph = 586.67 ˆj ′ ft ⋅ s −1
G
ω = 7.27 ×10−5 cos 41D ˆj ′ + sin 41D kˆ′ s −1
G G
ω × v′ = − ( 7.27 ×10−5 ) ( 586.67 ) ( sin 41D ) iˆ′ ft ⋅ s −2
G G
−2mω × v′
Fcor
=
Fgrav
mg
(
=
)
2 ( 7.27 ×10−5 ) ( 586.67 ) ( sin 41D )
32
Fcor
= 0.0017
Fgrav
G G
The Coriolis force is in the −ω × v′ direction, i.e., +iˆ′ or east.
5.12
(See Figure 5.4.3)
G
ω = ω y′ ˆj′ + ω z′ kˆ′
G
v′ = vx′iˆ′ + v y′ ˆj ′
56
G G
ω × v′ = −ω z′v y′iˆ′ + ω z′vx′ ˆj′ − ω y′vx′ kˆ′
G G
(ω × v′ )horiz = −ω z′v′y′iˆ′ + ω z′vx′ ˆj′
1
G G
1
= (ω z2′v y2′ + ω z2′vx2′ ) 2 = ω z′ ( vx2′ + v y2′ ) 2 = ω z ′v′
G
G G
Fcor = −2mω × v′
G
G
G G
= 2m (ω × v′ )horiz = 2mω z′ v′ , independent of the direction of v′ .
Fcor
(ω × v′ )horiz
( )
5.13
horiz
From Example 5.4.1 …
1
1  8h 3  2
′
xh = ω 
 cos λ
3  g 
and yh′ = 0 .
1
3
3 2
1
−5 −1  8 × 1250 ft 
D
′
xh = ( 7.27 × 10 s ) 
 cos 41
−2
3
 32 ft ⋅ s

xh′ = 0.404 ft to the east.
5.14
From Example 5.4.2:
ωH 2
∆≈
sin λ is the deflection of the baseball towards the south since it
vD
was struck due East at Yankee Stadium at latitude λ = 41D N (problem 5.13). v0 is the
initial speed of the baseball whose range is H. From eqn 4.3.18b, without air resistance
in an inertial reference frame, the horizontal range is …
vD2 sin 2α
H=
g
1
Solving for v0 …
 gH  2
vD = 

 sin 2α 
1
 32 ft ⋅ s −2 × 200 ft  2
−1
vD = 
 = 113 ft ⋅ s
D
sin 30


∴
( 7.27 ×10
∆≈
−5
s −1 )( 2002 ft 2 )
sin 41D = 0.0169 ft = 0.2 in
113 ft ⋅ s −1
A deflection of 0.2 inches should not cause the outfielder any difficulty.
5.15 Equation 5.2.10 gives the relationship between the time derivative of any vector
in a fixed and rotating frame of reference. Thus …
G
G
G  da 
G G
 da 
=   +ω × a
r = 
 dt  fixed  dt  rot
G G G G
G G G G G
a = r ′ + ω × r ′ + 2ω × r′ + ω × (ω × r ′ )
57
G
G G G G G
G G
G G
 da 
  = r ′ + ω × r ′ + ω × r ′ + 2ω × r ′ + 2ω × r ′
 dt rot
G G G
G G G
G G G
+ω × (ω × r ′ ) + ω × ω × r ′ + ω × ω × r′
( ) ( )
G G G G G G G
G G G
G
G G G
ω × a = ω × r ′ + ω × (ω × r ′ ) + 2ω × (ω × r′ ) + ω × ω × (ω × r ′ ) 
G
G
is ⊥ to ω and r ′ . Let this define a direction n̂ :
G G
G G
ω × r ′ = ω × r ′ nˆ
G
G
G
G G
Since ω ⊥ n̂ , ω × (ω × r ′ ) is in the plane defined by ω and r ′ and
G G G
G
G G
G G
ω × (ω × r ′) = ω × nˆ ω × r ′ = ω ω × r ′ .
G G G G
Since ω ⊥ ω × (ω × r ′ ) …
G
G G G
G G
ω × ω × (ω × r ′ )  = ω 2 ω × r ′
G
G G G
And ω × ω × (ω × r ′ )  is in the direction − n̂
G
G G G
G G
Thus ω × ω × (ω × r ′ )  = −ω 2 (ω × r ′ )
G G G G G G G
G G
G G
ω × a = ω × r ′ + ω × ω × r ′ + 2ω × ω × r′ − ω 2 (ω × r ′ )
Now
G G
(ω × r ′ )
(
)
(
)
G G G G
G G
G G G G G
r = r ′ + ω × r ′ + 3ω × r′ + 3ω × r ′ + ω × (ω × r ′ )
G G G
G G G
G G
+2ω × ω × r ′ + 3ω × ω × r′ − ω 2 (ω × r ′ )
(
5.16
)
(
)
With xD′ = yD′ = zD′ = xD′ = yD′ = 0 , and zD′ = vD′ Equations 5.4.15a – 5.4.15c become:
1
x′ ( t ) = ω gt 3 cos λ − ω t 2 vD′ cos λ
3
y′ ( t ) = 0
1
z′ ( t ) = − gt 2 + vD′t
2
When the bullet hits the ground z ′ ( t ) = 0 , so
2v′
t= D
g
 8v′ 3 
 4v ′ 2 
1
x′ = ω g  D3  cos λ − ω  D2  vD′ cos λ
3
 g 
 g 
4ω vD′ 3
x′ = −
cos λ
3g 2
x′ is negative and therefore is the distance the bullet lands to the west.
5.17
With xD′ = yD′ = zD′ = 0 and xD′ = vD cos α yD′ = 0 zD′ = vD sin α
we can solve equation 5.4.15c to find the time it takes the projectile
to strike the ground …
58
1
z ′ ( t ) = − gt 2 + vD′ t sin α + ω vDt 2 cos α cos λ = 0
2
2vD′ sin α
2v′ sin α
or
t=
≈ D
g − 2ω vD′ cos α cos λ
g
We have ignored the second term in the denominator—since vD′ would have to be
impossibly large for the value of that term to approach the magnitude g
g − 2ω vD′ cos α cos λ ≈ g − ω vD′
For example, for λ = 41o and α = 45o
or
vD′ ≈
g
ω
≈ 144
km
!
s
Substituting t into equation 5.4.15b to find the lateral deflection gives
4ω vD′ 3
y ( t ) = − [ω vD cos α sin λ ] t 2 = −
sin λ sin 2 α cos α
g2
5.18
Let …
G
r
y
x
G
R0
G
aD = acceleration of object relative to Earth
G
ωD = ωD k̂ = its angular speed
G
AD = acceleration of satellite
G
ω = ω k̂ = its angular speed
G
G G
G G G G G
aD = a + 2ω × v + ω × (ω × r ) + AD (Equation 5.2.14)
G G G
G G G G G
a = aD − AD − 2ω × v − ω × ω × r
G
R
As in problem 5.7 Evaluate the term …
G
G G G G G G G G G G G G G G G
∆ a = aD − AD − ω × ω × r = ω D × ω D × RD − ω × ω × R − ω × ω × RD − R
G
G
∆ a = (ω D2 − ω 2 ) RD
(
… given the condition that
but
ωD2 RD3 = ω 2 R 3
G
G  R3 
∆ a = −ω 2 RD  1 − 3 
 RD 
G G
G G G G
RD ⋅ RD = R + r ⋅ R + r = R 2 + r 2 + 2rR cos θ
(
Letting x = cos θ
)(
)
G G
 2x 
RD ⋅ RD = R 2 + r 2 + 2 Rx ≈ R 2  1 +  for small r
R

3
−
3
R3  2 x  2
= 1 + 
or
and
RD3 
R
3
G
− 

G
G
2
R
x
2


∆ a = −ω 2 RD 1 −  1 +   ≈ −3ω 2 x D ≈ −3ω 2 x iˆ for small r
 
R 
R


 2x 2
RD3 ≈ R 3 1 + 
R

59
)
G G
G G
ˆ + ĵy
Hence: a = ∆ a − 2ω × v = −3ω 2 x iˆ − 2ω kˆ × ix
G
ˆ − 2ω xĵ
a = iˆ
x + ˆjy = −3ω 2 xiˆ + 2ω yi
So
x − 2ω y − 3ω 2 x = 0
y + 2ω x = 0
(
5.19
G
G
G G
mr = qE + q v × B
(
Equation 5.2.14
)
)
G G G G
G G G G G
r = r ′ + ω × r ′ + 2ω × v′ + ω × (ω × r ′ )
G G G G
v = v′ + ω × r ′
Equation 5.2.13
q G
G
G
ω =−
B so ω = 0
2m
G G
G
G
q G G G
G
G G G
mr′ − q B × v′ − B × (ω × r ′ ) = qE + q ( v′ + ω × r ′ ) × B 
2
G
G
G
G
G G q G G
G G
G G
mr ′ + q v′ × B + (ω × r ′ ) × B = qE + q v′ × B + q (ω × r ′ ) × B
2
G q G G
G
G
mr ′ = qE + (ω × r ′ ) × B
2
G q qB
q G G
(ω × r ′) × B =   ( r ′)( sin θ )( B ) ∝ B 2
2
2  2m 
G
G
Neglecting terms in B 2 , mr′ = qE
5.20
(
)
(
)
(
)
For x′ = x cos ω ′t + y sin ω ′t
y′ = − x sin ω ′t + y cos ω ′t
x ′ = x cos ω ′t − xω ′ sin ω ′t + y sin ω ′t + yω ′ cos ω ′t
y ′ = − x sin ω ′t − xω ′ cos ω ′t + y cos ω ′t − yω ′ sin ω ′t
x ′ = x cos ω ′t + y sin ω ′t + ω ′ y′
y ′ = − x sin ω ′t + y cos ω ′t − ω ′x′
x′ = x cos ω ′t − xω ′ sin ω ′t + y sin ω ′t + yω ′ cos ω ′t + ω ′ y ′
y′ = − x sin ω ′t − xω ′ cos ω ′t + y cos ω ′t − yω ′ sin ω ′t − ω ′x ′
x′ = x cos ω ′t + y sin ω ′t + 2ω ′ y ′ + ω ′2 x′
y′ = − x sin ω ′t + y cos ω ′t − 2ω ′x ′ + ω ′2 y′
Substituting into Eqns 5.6.3:
x cos ω ′t + y sin ω ′t + 2ω ′ y ′ + ω ′2 x′
g
g
= − x cos ω ′t − y sin ω ′t + 2ω ′ y ′
l
l
′
′
′
x sin ω t + y cos ω t − 2ω x ′ + ω ′2 y′
g
g
= + x sin ω ′t − y cos ω ′t − 2ω ′x ′
l
l
60
Collecting terms and neglecting terms in ω ′2 :
g 
g 


x + x  cos ω ′t +  y + y  sin ω ′t = 0
 l 
l 


g 
g 


x + x  sin ω ′t −  y + y  cos ω ′t = 0
 l 
l 


24
hours
sin λ
24
T=
= 73.7 hours
sin19D
5.21
T=
5.22
Choose a coordinate system with the origin at the center of the wheel, the x′ and
y′ axes pointing toward fixed points on the rim of the wheel, and the z ′ axis
pointing toward the center of curvature of the track. Take the initial position of
G
the x′ axis to be horizontal in the −VD direction, so the initial position of the y′
axis is vertical.
V
The bicycle wheel is rotating with angular velocity D about its axis, so …
b
V
G
ωl = kˆ′ D
b
A unit vector in the vertical direction is:
Vt
Vt
nˆ = iˆ′ sin D + ˆj ′ cos D
b
b
At the instant a point on the rim of the wheel reaches its highest point:
Vt
Vt
G

r ′ = bnˆ = b  iˆ′ sin D + ˆj ′ cos D 
b
b 

Since the coordinate system is moving with the wheel, every point on the rim is
fixed in that coordinate system.
G
G
r′ = 0 and r′ = 0
The x′y′z ′ coordinate system also rotates as the bicycle wheel completes a circle
around the track:
V V 
Vt
Vt
G
ω 2 = nˆ D = D  iˆ′ sin D + ˆj ′ cos D 
ρ ρ
b
b 
The total rotation of the coordinate axes is represented by:
V
Vt
Vt
G G G V 
ω = ω1 + ω 2 = D  iˆ′ sin D + ˆj ′ cos D  + kˆ′ D
ρ
b
b
b 
2
Vt
Vt
G V 
ω = D  iˆ′ cos D − ˆj′ sin D 
b
ρb 
b 
2
Vt
Vt  V2
G G V 
ω × r ′ = D  kˆ′ cos 2 D + kˆ′ sin 2 D  = D k̂ ′
b
b  ρ
ρ 
61
G G
ω × r′ = 0
VDb  ˆ
VDt
VDt ˆ
Vt
Vt
Vt
Vt

− k ′ sin D cos D  + VD  ˆj ′ sin D − iˆ′ cos D 
 k ′ sin cos
ρ 
b
b
b
b 
b
b 

2
2
V 
Vt
Vt V 
Vt
Vt
G G G
ω × (ω × r ′ ) = D  kˆ′ sin 2 D + kˆ′ cos 2 D  + D  −iˆ′ sin D − ˆj ′ cos D 
b
b  b 
b
ρ 
b 
2
2
V
V
G G G
ω × (ω × r ′ ) = kˆ′ D − nˆ D
b
ρ
Since the origin of the coordinate system is traveling in a circle of radius ρ :
G
V2
AD = kˆ′ D
G G
ω × r′ =
ρ
G
G G G G
G G
G G G
r = r ′ + ω × r ′ + 2ω × r′ + +ω × (ω × r ′ ) + AD
V2
V2
V2
V2
G
r = kˆ′ D + kˆ′ D − nˆ D + kˆ′ D
b
ρ
ρ
ρ
V2
V2
G
r = 3 D kˆ′ − D nˆ
b
ρ
With appropriate change in coordinate notation, this is the same result as obtained
in Example 5.2.2.
--------------------------------------------------------------------------------------------------
62
CHAPTER 6
GRAVITATIONAL AND CENTRAL FORCES
6.1
4
m = ρ V = ρ π rs3
3
1
 3m  3
rs = 

 4πρ 
F=
Gmm
( 2rs )
2
2
2
Gm 2  4πρ  3 G  4πρ  3 43
=

 = 
 m
4  3m 
4 3 
2
F
F
Gm 2  4πρ  3 3
=
=

 m
W mg
4g  3 
1
2
1
F 6.672 × 10−11 N ⋅ m 2 ⋅ kg −2  4π × 11.35 g ⋅ cm −3 1 kg 106 cm3  3
kg
=
×
1
×
×
)3

 (
W
4 × 9.8 m ⋅ s −2
3
103 g
1 m3 

F
= 2.23 × 10−9
W
6.2
(a) The derivation of the force is identical to that in Section 6.2 except here r < R.
This means that in the last integral equation, (6.2.7), the limits on u are R – r to R + r.
GmM R + r  r 2 − R 2 
F=
Q
1 +
 ds
s2 
4 Rr 2 ∫R − r 
R
s
GmM 
R2 − r 2 R2 − r 2 
R
r
R
r
=
+
−
−
+
−
(
)


θ ψ
4 Rr 2 
R+r
R−r 
P
r
GmM
 2r + R − r − ( R + r )  = 0
F=
4 Rr 2 
(b) Again the derivation of the gravitational potential
energy is identical to that in Example 6.7.1,
except that the limits of integration on s are ( R − r ) → ( R + r ) .
2πρ R 2 R + r
ds
rR ∫R − r
2πρ R 2
= −G
 R + r − ( R − r ) 
rR 
4π R 2 ρ
M
= −G
φ = −G
R
R
For r < R , φ is independent of r. It is constant inside the spherical shell.
φ = −G
6.3
F =−
GMm
eˆr
r2
63
The gravitational force on the particle is due only to the mass of the earth that is
inside the particle’s instantaneous displacement from the center of the earth, r.
The net effect of the mass of the earth outside r is zero (See Problem 6.2).
4
M = π r3ρ
3
4
F = − Gπρ mreˆr = − kreˆr
3
r
F
The force is a linear restoring force and induces simple
harmonic motion.
T=
2π
ω
= 2π
3
m
= 2π
4Gπρ
k
The period depends on the earth’s density but is independent of its size.
At the surface of the earth,
GMm Gm 4 3
= 2 ⋅ π Re ρ
Re2
Re 3
4Gπρ g
=
3
Re
mg =
T = 2π
6.4
Re
6.38 × 106 m 1 hr
= 2π
×
≈ 1.4 hr
9.8 m ⋅ s −2
3600 s
g
Fg = −
GMm
4
eˆr , where M = π r 3 ρ
2
r
3
The component of the gravitational force perpendicular to the tube
is balanced by the normal force arising from the side of the tube.
The component of force along the tube is
Fx = Fg cos θ
The net force on the particle is …
4
F = −iˆ Gπρ mr cos θ
3
r cos θ = x
4
ˆx
F = −iˆ Gπρ mx = −ik
3
As in problem 6.3, the motion is simple harmonic with a period of 1.4 hours.
64
6.5
6.6
GMm mv 2
GM
v2 =
=
so
2
r
r
r
for a circular orbit r, v is constant.
2π r
T=
v
4π 2 r 2 4π 2 3
T2 =
r ∝ r3
=
2
v
GM
2π r
v
From Example 6.5.3, the speed of a satellite in circular orbit is …
(a) T =
1
 gR 2  2
v= e 
 r 
3
T=
2π r 2
1
g 2 Re
1
 T 2 gRe2  3
r =

2
 4π 
1
1
r  T 2 g  3  242 hr 2 × 36002 s 2 ⋅ hr −2 × 9.8 m ⋅ s −2  3
=
 =

Re  4π 2 Re  
4π 2 6.38 × 106 m

r
= 6.62 ≈ 7
Re
3
(b)
T=
2π r 2
1
2
g Re
3
=
2π ( 60 Re ) 2
1
2
g Re
= 2π
603 Re
g
1

2
603 × 6.38 × 106 m
= 2π 
−2
−2
−2 
2 2
2
2
 9.8 m ⋅ s × 3600 s ⋅ hr × 24 hr ⋅ day 
T = 27.27 day ≈ 27 day
6.7
From Example 6.5.3, the speed of a satellite in a circular orbit just above the
earth’s surface is …
v = gRe
T=
2π Re
Re
= 2π
v
g
65
This is the same expression as derived in Problem 6.3 for a particle dropped into a
hole drilled through the earth. T ≈ 1.4 hours.
6.8
The Earth’s orbit about the Sun is counter-clockwise as seen from, say, the north
star. It’s coordinates on approach at the latus rectum are ( x, y ) = ( ε a, −α ) .
The easiest way to solve this problem is to note that ε =
orbit is almost circular!
GM S m mv 2
GM S
∴
=
and v 2 =
2
r
r
r
with r = α ≈ a ≈ b when ε ≈ 0
 GM S 
4 m
v≈
 = 3 ⋅10
s
 α 
More exactly
ml 2
r × v = α v cos β = l , but α =
k
2
l
α=
,
Since k = GM S m
GM S
1
is small. The
60
(equation 6.5.19)
1
hence l = α v cos β = (α GM S ) 2
1
 GM S  2 1
v=
Or

 α  cos β
The angle β can be calculated as follows:
y 2 x2
+
=1
b2 a 2
dy
b2 x
∴ =− 2
dx
a y
(see appendix C)
and at ( x, y ) = ( ε a, −α )
so
dy b 2 ε a b 2
ε
b2
2
= 2
= 2
=
ε
since
1
−
ε
=
dx a α a (1 − ε 2 )
a2
here
dy
= tan β = ε
dx
and
 GM S  2 1
 GM S  2
v=
≈
 as before.

 α  cos ε  α 
6.9
F ( r ) = Fs + Fd
1
or β ≈ ε (small ε)
1
GMm
r2
GM d m
Fd = −
r2
The net effect of the dust outside the planet’s radius is zero (Problem 6.2). The mass of
the dust inside the planet’s radius is:
Fs = −
66
4
M d = π r3ρ
3
GMm 4
F ( r ) = − 2 − πρ mGr
3
r
6.10
1 1 − kθ
= e
r r
du
k
= − e − kθ
dθ
r
u=
d 2u k 2 − kθ
= e = k 2u
dθ 2 r
From equation 6.5.10 …
d 2u
1
+ u = k 2u + u = − 2 2 f ( u −1 )
2
du
ml u
−1
2
2
f ( u ) = − ml ( k + 1) u 3
f (r ) = −
ml 2 ( k 2 + 1)
r3
The force varies as the inverse cube of r.
From equation 6.5.4, r 2θ = l
dθ
l
= 2 e −2 kθ
dt r
l
e 2 kθ dθ = 2 dt
r
1 2 kθ lt
e = 2 +C
r
2k

1  2klt
ln  2 + C ′ 
2k  r

θ varies logarithmically with t.
θ=
k
= ku 3
3
r
From equation 6.5.10 …
d 2u
1
ku
+ u = − 2 2 ⋅ ku 3 = − 2
2
dθ
ml u
ml
2
d u 
k 
+ 1 + 2  u = 0
2
dθ
 ml 
6.11
f (r ) =
67
k 

If 1 + 2  < 0 ,
 ml 
k 

If 1 + 2  = 0 ,
 ml 
du
= C1
dθ
u = c1 θ + c2
1
r=
c1 θ + c2
k 

If 1 + 2  > 0 ,
 ml 
u = A cos
(
cθ + δ
d 2u
− cu = 0 ,
dθ 2
d 2u
=0
dθ 2
)
d 2u
+ cu = 0 ,
dθ 2



k
r =  A cos  1 + 2 θ + δ  
ml


 
6.12
c > 0 , for which u = aebθ is a solution.
c>0
−1
1
1
=
r r cos θ
du
sin θ
=
dθ r cos 2 θ
u=
d 2u 1  1
2sin 2 θ
=
+

dθ 2 r  cosθ cos3 θ

1
=
 r cos θ
 2 − 2 cos 2 θ
1 +
cos 2 θ


1  2

− 1
=

2

 r cosθ  cos θ
d 2u
= u ( 2r 2u 2 − 1) = 2r 2u 3 − u
2
dθ
Substituting into equation 6.5.10 …
1
2r 2u 3 − u + u = − 2 2 f ( u −1 )
ml u
−1
2
2 5
f ( u ) = −2r ml u
2r 2 ml 2
f (r ) = −
r5
6.13 From Chapter 1, the transverse component of the acceleration is … aθ = rθ + 2rθ
If this term is nonzero, then there must be a transverse force given by …
f (θ ) = m(rθ + 2rθ )
r = aθ , and θ = bt
For
f (θ ) = 2mab 2 ≠ 0
Since
f (θ ) ≠ 0 , the force is not a central field.
68
For r = aθ , and the force to be central, try θ = bt n
f (θ ) = m  2ab 2 n 2t 2 n − 2 + ab 2 n ( n − 1) t 2 n − 2 
For a central field … f (θ ) = 0
2n + ( n − 1) = 0
n=
1
3
1
θ = bt 3
(a)
6.14
Calculating the potential energy
 4 a2 
dv
− = f ( r ) = −k  3 + 5 
dr
r 
r
 2 a2 
Thus, V = −k  2 + 4 
4r 
r
The total energy is …
1
1  1  9k  9k
 2
E = T + V = v2 − k  2 + 2  =  2  − 2 = 0
2
 a 4 a  2  2a  4a
Its angular momentum is …
9k
l 2 = a 2v 2 =
= constant = r 4θ 2
2
Its KE is …
2
 2 1  dr  2 2  l 2
1
1  dr 
2
T = r 2 + r 2θ 2 = 
r
+
 θ = 

 +r  4
2
2  dθ 
2  dθ 

 r
The energy equation of the orbit is …
2
 2
 2 a2 
1  dr 
2 l
+
−
T + V = 0 = 
r
k

 2+ 4

4
2  dθ 
4r 
r
 r
(
)
 dr  2 2  9k
 2 a2 
r
k
= 
+
−

 2+ 4

4
4r 
 dθ 
r
 4r
2
or
 dr  1 2 2

 = (a − r )
 dθ  9
r = a cos φ
Letting
2
So
 dφ  1

 =
 dθ  9
then
dφ
dr
= −a sin φ
dθ
dθ
1
∴φ = θ
3
69
1
r = a cos θ
( r = a @θ = 0 )
3
3π
(b) at θ =
r → 0 the origin of the force. To find how long it takes …
2
av
v
l
θ= 2=
=
1
1
r
a 2 cos 2 θ a cos 2 θ
3
3
1
a
dt = cos 2 θ dθ
3
v
Thus
T=
3π
2
∫
0
π
3a 2
3π a
a
2 1
cos θ dθ =
cos 2 φ dφ =
∫
3
4v
v
v 0
1
 9k  2
Since v =  2 
 2a 
1
1
3 2  2  2 π a2  2  2
T = πa   =
 
4
4 k
 9k 
(c) Since the particle falls into the center of the force
v→∞
(since l = vr⊥ = const )
6.15
From Example 6.5.4 …
v  2r1 
=

vc  r1 + r 
1
2
1
2


 2 
v

we have V = 
Letting V =
vc
 1+ r 

r1 

−2
dV
1   r   r 
 −2 1 +   −  
So:
=
dr1 2V   r1   r12  


dV
1
1
r 1
1
r
V =
Thus
=
2
 dr1  
  r  r1 2  r  r1
   2  1 + 
1 + 
 r1  
 r1 
  r1 
r
 1+ 

r1 

70
 dV 


 V  ≈ 1 r (b)  dr1  = 2r1  dV  = 2 60 1% = 120%!
(a)
( )
 


 dr1  2 r1
 r1  r  V 
 
 r1 
The approximation of a differential has broken down – a correct result can be obtained by
calculating finite differences, but the implication is clear – a 1% error in boost causes
rocket to miss the moon by a huge factor --- ∼ 2!
From section 6.5, ε = 0.967 and r = 55 ×106 mi.
1+ ε
From equations 6.5.21a&b, r1 = r
1− ε
r
1
55 × 10 6 mi
1 AU
a = ( r + r1 ) =
=
×
2
1− ε
1 − 0.967 93 × 10 6 mi
a = 17.92 AU
6.16
3
From equation 6.6.5, τ = ca 2
τ = 1yr ⋅ AU
τ = 75.9 yr
−
3
2
3
3
×17.92 2 AU 2
From equation 6.5.21a and 6.5.19 …
ml 2
α
−1
ε = − 1=
r0
kr0
ε=
mr v 2
− 1 and
k
k = GMm
1
 GM
2
v =
( ε + 1) 
 r

From Example 6.5.3 we can translate the factor GM into the more convenient
GM = ae ve2
… with ae the radius of a circular orbit and ve the orbital speed …
1
2
1
a v

 93 ×106 mi
2
v =
1.967
(ε + 1)  = 
(
)
 ve
6
 55 × 10 mi

 r

v0 = 1.824 ve
Since l is constant … r1v1 = r v
r
1− ε
1 − .967
× 1.824 ve = 0.0306 ve
v1 = v =
v =
1+ ε
1.967
r1
2
e e
ve ≈
2π ae
τ
=
2π × 93 ×106 mi
= 66, 705 mph
1 yr × 365 day ⋅ yr −1 × 24 hr ⋅ day −1
v = 1.22 × 105 mph
and
v1 = 2.04 × 103 mph
71
6.17
From Example 6.10.1 …
1
 
2
r
v
2 2
ε = 1 +  q 2 −  ( qd sin φ ) 
where q =
and d =
d
ve
ae
 

are dimensionless ratios of the comet’s speed and distance from the Sun in terms of the
Earth’s orbital speed and radius, respectively (q and d are the same as the factors V and R
in Example 6.10.1). φ is the angle between the comet’s orbital velocity and direction
vector towards the Sun (see Figure 6.10.1).
The orbit is hyperbolic, parabolic, or elliptic as ε is > , = , or < 1 …
2

i.e., as  q 2 −  is > , = , or < 0.
R

 2 2
2
 q −  is > , = , or < 0 as q d is > , = , or < 2.
R

Since l is constant, vmax occurs at r and vmin occurs at r1 , i.e. vmax = v and
vmin = v1 and form the constancy of l … v1r1 = v r
r
vmin vmax = v1v = v 2
r1
k
r v 2 = ( ε + 1)
(See Example 6.5.4)
m
2
k
 2π a 
= GM = 
From equation 6.6.5 …
 a
m
 τ 
6.18
 2π a  a ( ε + 1)
vmin vmax = 
 ⋅
r1
 τ 
2
1+ ε
. With 2a = r + r1 :
1− ε
a ( ε + 1) 1 ( r + r1 )( ε + 1) 1  r

1  1 − ε
=
=  + 1 ( ε + 1) = 
r1
r1
2
2  r1 
2  1 + ε
From equation 6.5.21a&b … r1 = r
vmin vmax
 2π a 
=

 τ 
 
 + 1 (ε + 1) = 1
 
2
6.19 As a result of the impulse, the speed of the planet instantaneously changes; its
orbital radius does not, so there is no change in its potential energy V. The instantaneous
change in its total orbital energy E is due to the change in its kinetic energy, T, only, so
δv
δv
1

= 2T
δ E = δ T = δ  mv 2  = mvδ v = mv 2
v
v
2

δE
δv
=2
T
v
72
But the total orbital energy is
k
k
δE = 2 δa
E=−
So
2a
2a
Since planetary orbits are nearly circular
k
k
V ~−
and
T~
a
2a
δa
δE δa
and
=
Thus, δ E ≅ T
T
a
a
δa
δv
=2
We obtain
a
v
6.20 (a) V =
1
τ
Vdt
τ∫
0
k
r
From equation 6.5.4, l = r 2θ
dθ
l
r 2 dθ
= 2 or dt =
dt r
l
τ
2π kr
∫0 Vdt = −∫0 l dθ
From equation 6.5.18a …
a (1 − ε 2 )
r=
1 + ε cos θ
ka (1 − ε 2 ) 2π
τ
dθ
∫0 Vdt = − l ∫0 1 + ε cosθ
2π a 2
From equation 6.6.4 … τ =
1− ε 2
l
V (r ) = −
k 1 − ε 2 2π
dθ
∫
0 1 + ε cos θ
2π a
2π
dθ
2π
k
2
∫0 1 + ε cosθ = 1 − ε 2 , ε < 1 ∴V = − a
V =−
(b)
This problem is an example of the virial theorem which, for a bounded, periodic
τ
system, relates the time average of the quantity
∫∑ p⋅r
i
0
to its kinetic energy T. We will
i
derive it for planetary motion as follows:
τ
τ
τ
1
1
1
p
⋅
rdt
=
mr
⋅
rdt
=
∫
∫
∫ F ⋅ rdt
τ
0
τ
0
τ
0
73
Integrate LHS by parts
τ
τ
1
1
1
 mr ⋅ r  τ0 − ∫ mr 2 dt = ∫ F ⋅ rdt
τ
τ
τ
0
0
The first term is zero – since the quantity has the same value at 0 and τ .
denote time average of the quantity within brackets.
Thus 2 T = − F ⋅ r where
but
− r ⋅ F = r ⋅∇V = r
dV
k
=
=− V
dr
r
hence 2 T = − V
but
hence
and
Thus:
V
V
+ V =
2
2
k
V =2 E
but
E=−
= constant
2a
τ
k
1
k
so
2E = −
E = ∫ Edt = E = −
a
2a
τ0
k
1
k
V = − as before and therefore T = − V =
2
2a
a
E = T + V =−
The energy of the initial orbit is
1 2 k
k
mv − = E = −
2
2a
r
k 2 1
v2 =  − 
(1)
m r a
Since ra = a (1 + ε ) at apogee, the speed v1 , at apogee is
6.21
v12 =
k
2
1  k (1 − ε )
−  =

m  a (1 + ε ) a  ma (1 + ε )
To place satellite in circular orbit, we need to boost its speed to vc such that
1 2 k
k
mvc − = −
since the radius of the orbit is ra
2
2ra
ra
k
k
vc2 =
=
mra ma (1 + ε )
Thus, the boost in speed ∆v1 = vc − v1
1
(2)
1

2 
k

∆v1 = 
 1 − (1 − ε ) 2 

 ma (1 + ε )  
Now we solve for the semi-major axis a and the eccentricity ε of the first orbit. From (1)
above, at launch v = v at r = RE , so
74
k  2 1
− 

m  RE a 
and solving for a
RE
noting that …
a=
2 RE
2 − mv
k
k
GM E
(3)
=
= gRE
mRE
RE
RE
R
a=
= E = 4.49 ⋅103 km
2

v  1.426
2−

gRE 

v2 =
The eccentricity ε can be found from the angular momentum per unit mass, l, equation
6.5.19, and the data on ellipses defined in figure 6.5.1 …
1
 ka (1 − ε 2 )  2
 kα 
2

l = r θ = v ( RE sin θ ) =   = 
m
m


where v , θ are the launch velocity, angle
1
2
Solving for ε (using (3) above)
v2 
v2  2
ε = 1−
2−
 sin θ = 0.795
gRE 
gRE 
∴ ε = 0.892
2
Inserting these values for a, ε into (2) and using (3) gives
1
(a)
(b)
6.22
1

 R a  2 

∆v1 =  gRE  E   1 − (1 − ε ) 2  = 4.61 ⋅103 km ⋅ s −1
 1 + ε  


3
h = a (1 + ε ) − RE = 2.09 ⋅10 km
{altitude above the Earth … at perigee}
 −be − br
e −br 
e − br
−
=
f ′ ( r ) = −k 
2
k

2
r3 
r2
 r
f ′(a)
2

= −b + 
f (a)
a


f ′(a) 
From equation 6.14.3, ψ = π 3 + a

f (a) 

ψ=
−
2

b + 
r

1
2
= π 3 − ( ab + 2 ) 
−
1
2
π
1 − ab
75
6.23
From Problem 6.9, f ( r ) = −
GMm 4
− πρ mGr
3
r2
2GMm 4
− πρ mG
3
r3
4
4πρ a 3
−3
−
−
+
2
πρ
2
GMma
mG
f ′(a)
3
3M
=
=
4
f ( a ) −GMma −2 − πρ mGa
 4πρ a 3 
a 1 +

3
3M 

f ′(r ) =

f ′(a) 
From equation 6.14.3, ψ = π 3 + a

f (a) 


4πρ a
 −2 + 3M
ψ = π 3 +
4πρ a 3

1+

3M
3





−
1
2
1
 1+ c 2
ψ =π 
 ,
 1 + 4c 
6.24
c=
−
1
2

 4πρ a 3  
1
4
+



 3M  
=π 

4πρ a 3 
+
1


3M


−
1
2
4πρ a 3
3M
We differentiate equation 6.11.1b to obtain mr = −
For a circular orbit at r = a , r = 0 so
dU
r =a = 0
dr
For small displacements x from r = a ,
r=x
r = x+a
and
From Appendix D …
x2
f ( x + a ) = f ( a ) + xf ′ ( a ) +
f ′′ ( a ) + …
2
dU
d 2U
Taking f ( r ) to be
, f ′(r ) = 2
dr
dr
Near r = a …
dU dU
d 2U
=
+
x
r =a
r =a + …
dr
dr
dr 2
d 2U
mx = − x 2 r = a
dr
dU ( r )
dr
This represents a “restoring force,” i.e., stable motion, so long as
d 2U
> 0 at r = a .
dr 2
76
6.25
f ′(r ) =
2k 4ε
+
r3 r5
From equation 6.13.7, the condition for stability is f ( a ) +
k ε a  2k 4ε 
− +  + <0
a 2 a 4 3  a3 a5 
k
ε
− 2 + 4 <0
3a 3a
a
f ′(a) < 0
3
−
ε
a2
<k
1
 ε 2
a> 
k
6.26 (a)
e − br
r2
e − br
 b 2
f ′ ( r ) = − ke −br  − 2 − 3  = k 2
r 
r
 r
f ( r ) = −k
2

b + 
r

From equation 6.13.7, the condition for stability is f ( a ) +
a
f ′(a) < 0
3
e − ba a e −ba 
2
−k 2 + k 2  b +  < 0
a
3 a 
a
− ba
− ba
e
be
−k 2 + k
<0
3a
3a
1
1
b<
a<
a
b
(b)
Since
k
r3
3k
f ′(r ) = 4
r
a
k a  3k 
f (a) + f ′(a) = − 3 +  4  = 0
3
a 3a 
a
f ( a ) + f ′ ( a ) is not less than zero, the orbit is not stable.
3
f (r ) = −
77
a (1 − ε 2 )
1 + ε cos θ
and the data on ellipses in Figure 6.5.1 p = a (1 − ε ) so
(1 + ε )
r=p
1 + ε cos θ
For a parabolic orbit, ε = 1
The comet intersects earth’s orbit at r = a .
2p
a=
1 + cos θ
2p
cos θ = −1 +
a
6.27
(See Figure 6.10.1) From equation 6.5.18a r =
6.28
(See Figure 6.10.1) T = ∫ dt along the comet’s trajectory inside earth’s orbit
From equation 6.5.4, r 2θ = r 2
dθ
=l
dt
so
dt =
r 2 dθ
l
r 2 dθ
T =∫
l
a (1 − ε 2 )
1 + ε cos θ
and the data on ellipses in Figure 6.5.1 p = a (1 − ε ) so
(1 + ε )
r=p
1 + ε cos θ
From equation 6.5.18b, with ε = 1 for a parabolic orbit:
2p
r=
1 + cos θ
π
2p
At θ = the distance to the comet is r = α =
= 2p
π
2
1 + cos
2
2
l2
ml
, where k = GMm , so p =
From equation 6.5.19, α =
2GM
k
2
As shown in Example 6.5.3, GM = ave
2π a
For a circular orbit, ve =
1 yr
From equation 6.5.18a r =
1
1
3
1
l = ( 2GMp ) 2 = ( 2ap ) 2 ve = ( 2a ) 2 p 2π yr −1
T =∫
+θ
−θ
1
3
+θ
−
r 2 dθ
4 p2
−
−1
2 p 2π dθ yr
=∫
⋅
2
a
(
)
2
θ
−
l
(1 + cosθ )
2p 

where θ = cos −1  −1 +
 from Problem 6.27
a 

78
3
T=
2 p2
πa
3
2
dθ
+θ
∫ θ (1 + cosθ )
−
From a table of integrals,
2
yr
dx
∫ (1 + cos x )
2
=
1
x 1
x
tan + tan 3
2
2
2 6
3
2  p 2  θ 1 3 θ 
T=
yr
tan + tan
2 3
2 
π  a  
1
x  1 − cos x  2
tan = 

2  1 + cos x 
1
2 p 2
1

2−
2




θ
−
a
p
a
tan = 
=


2  2p   p 


 a 
1
3
3


2  p  2  a − p  2 1  a − p  2 
T=
yr
+
π  a   p  3  p  


1
3
2  p 2  a − p 2  a − p 
=
1+
 yr
3p 
π  a   p  
1
p 2
2  2 p 
T=
+ 1  1 −  yr

3π  a
 a 
1
T is a maximum when ( 2 p + a )( a − p ) 2 is a maximum.
d 
2
2
2 p + a ) ( a − p )  = 2 ( 2 p + a )( 2 )( a − p ) + ( 2 p + a ) ( −1)
(

dp 
= ( 2 p + a )( 3a − 6 p )
T is a maximum when p =
a
.
2
1
1 2
2
T=
( 2 )   = yr = 77.5 day
3π
3π
2
When p = 0.6a
2
T=
6.29
2
3π
( 2.2 )
.04 = 0.2088 yr = 76.2 day
k kε
V (r ) = − − 3
r r
79
dV k 3kε k 2
= +
= 4 ( r + 3ε )
dr r 2 r 4
r
2k 12kε
2k 2
f ′ ( r ) = − 3 − 5 = − 5 ( r + 6ε )
r
r
r
2
f ′(a)
2  r + 6ε 
=−  2

f (a)
a  r + 3ε 
f (r ) = −

f ′(a) 
From equation 6.14.3, ψ = π 3 + a

f (a) 

−
For
For
6.30
−
1
2
1

 r 2 + 6ε   2
r 2 + 3ε
ψ = π 3 − 2  2
 = π 2
r − 3ε
 r + 3ε  

2
ε = R∆R , R = 4000 mi , ∆R = 13 mi
5
2
ε = ( 4000 )(13) = 2.08 ×104 mi 2
5
r ≈ R , r 2 = 1.6 ×107 mi 2
ψ = 1.0039π = 180.7
k
1 
k
Vrel = − −
E+ 
2 
r 2m c 
r
f (r ) = −
2
dV
k
2 
k  k 
=− 2 +
E +  − 2 
2 
dr
r
r  r 
2m c 

1 
k 
E + 
1 +
2 
r 
 mc 
2k 
1 
k  k  1   k 
f ′ ( r ) = 3 1 +
E +  − 2 
 − 
2 
r  mc 
r  r  m c2   r 2 
f (r ) = −
k
r2

1 
3k  
2E + 
2 +
2 
mc 
r 

1 
3k  

2+
2E +  
2 

f ′(a)
1
mc 
a 

=− 
1 
k 
f (a)
a
 1 + m c 2  E + a  



3 
1 
3k  
k
3+
2E + 
E + −2−

2 
2 
f ′(a)  m c 
a
mc 
a 
3+ a
=
f (a)

1 
k 
1 + m c 2  E + a  



f ′(r ) =
k
r3
80

E 
1 + m c 2 


=

1 
k 
1 + m c 2  E + a  





 m c2 + E 
=
3+ a

f ( a )  m c2 + E + k 
a

f ′(a)

f ′(a) 
ψ = π 3 + a

f (a) 

−
1
2
1
k

2


a
ψ = π 1 +

2
 mc +E


a (1 − ε 2 )
…
(Here θ is the polar angle of conic
6.31 From equation 6.5.18a r =
1 + ε cos θ
section trajectories as illustrated by the coordinates in Figure 6.5.1)
… and the data on ellipses in Figure 6.5.1 r0 = a(1 − ε )
so
1+ ε
rcom = r
1 + ε cos θ
From equation 6.5.18b r =
α
α
and at θ = 0 r0 =
1 + ε cos θ
1+ ε
And from equation 6.5.19 α =
ml 2
k
m
m
1
=
=
k GMm GM
From Example 6.5.3, GM = ae ve2
rcom
2
2
rcom
vcom
sin 2 φ (1 + ε )
=
ae ve2 (1 + ε )(1 + ε cos θ )
1 = RV 2 sin 2 φ
cos θ =
1
( RV
ε
so
r0 =
ml 2
k (1 + ε )
2
2
and l 2 = rcom
vcom
sin 2 φ
1
1 + ε cos θ
2
sin 2 φ − 1)
sin θ = (1 − cos 2 θ )
1
2
1
2 2
1

= 1 − 2 ( RV 2 sin 2 φ − 1) 
 ε

81
1
2
1
2
sin θ = ε 2 − ( RV 2 sin 2 φ ) + 2 ( RV 2 sin 2 φ ) − 1


ε
Again from Example 6.5.3 …
1
 
2
2 2
ε = 1 +  V 2 −  ( RV sin φ ) 
R
 

ε 2 = 1 + ( RV 2 sin φ ) − 2 RV 2 sin 2 φ
2
1
2
2 2
1
sin θ = ( RV 2 sin φ ) − ( RV 2 sin 2 φ ) 

ε
1
sin θ = RV 2 sin φ cos φ
ε
cos θ
RV 2 sin 2 φ − 1
=
sin θ RV 2 sin φ cos φ
2
cot θ = tan φ −
2
RV sin 2φ
2


θ = cot −1  tan φ −
csc 2φ 
2
RV


V = 0.5 , R = 4 , φ = 30 :
For


csc 60 


4 (.5 )




 1
2 
 = cot −1 − 3
= cot −1 
−
3
 3


2 

θ = −30
θ = cot −1  tan 30 −
2
2
(
)
6.32
The picture at left shows the orbital transfer and the position of
the two satellites at the moment the transfer is initiated.
Satellite B is “ahead” of satellite A by the angle θ0
r +r
a = 1 2 is the semi-major axis of the elliptical transfer orbit.
2
r1
From Kepler’s 3rd law (Equation 6.6.5) applied to objects in
orbit about Earth …
θ0
4π 2 3
2
a
τ =
GM E 2
The time to intercept is …
3
3
GM E
τ
1
π
Tt = = π
a2 =
a2
since
=g
RE 2
2
GM E
RE g
r2
82
Letting r1 = RE + h1 and r2 = RE + h 2 where h1 and h2 are the heights of the 2 satellites
above the ground. Inserting these into the above gives …
3
3
h1 + h 2  2
h1 + h 2  2

π 
2
Tt =
RE  2 +
=
+



RE 
RE 
RE g
RE g 

From Example 6.6.2, RE = 6371 km, h1 = 200 mi = 324 km and h2 = r2 - RE = 42,400 km 36,029 km. Putting in the numbers …
Tt = 4.79 hr
 T 
(b) Thus, θ 0 = 1800  1 − t  = 1080
 12 
π
3
2
6.33
The potential for the inverse-cube force law
k
is … V ( r ) = 2
2r
Letting u = r −1 , we have (Equation 6.9.3)
2

1 2  du 
2
−1
ml 
 + u  + V (u ) = E
2
 dθ 

θS
θ
θ0
θ0
2( E −V ) 2
du
=
−u
dθ
ml 2
r
b
dθ =
2( E −V ) 2
−u
ml 2
du =
ml
2m ( E − V ) − m 2 l 2 u 2
Now, integrating from r = ∞ ( u = 0 ) up to r = rmin ( u = umax ) …
θ0 =
umax
∫
0
ml
2m ( E − V ) − m 2 l 2 u 2
du
1
mv0 2 , l = bv0 , so …
2
umax
mbv0
du
θ0 = ∫
1 2
1
0
2
2 2 2 2
2m  mv0 − ku  − m b v0 u
2
2

But E =
θ0 =
umax
∫
0
mbv0
1
1

2m  mv0 2 − ku 2  − m 2b 2 v0 2u 2
2
2


du
83
du
Before evaluating this integral, we need to find umax ( = rmin −1 ) , in other words, the
distance of closest approach to the scattering center.
1
1 k
1
E = T ( rmin ) + V ( rmin ) = mv 2 +
= mv0 2
2
2
2 rmin
2
But, the angular momentum per unit mass l is …
l = bv0 = rmin v and substituting for v into the above gives …
mv0 2
ml 2
k
1
2
so
…
+
=
mv
=
= umax 2
0
rmin 2 rmin 2
rmin 2 ml 2 + k
Solving for umax …
1
umax =
k
b2 +
mv0 2
Now we evaluate the integral for θ0 …
θ0 =
umax
b
∫

k  2
1 −  b2 +
u
mv0 2 

0
Solving for b …
b (θ 0 ) =
k
mv0 2
du =
b
 2
k 
b +

mv0 2 

sin
−1
u
umax
umax
=
0
b
π
 2
k 2
b +

mv0 2 

2θ 0
π 2 − 4θ 0 2
1
(π − θ S ) . Thus, we have …
2
π − θS
k
b (θ S ) =
2
mv0 θ S ( 2π − θ S )
But θ 0 =
We can now compute the differential cross section …
kπ 2 ( π − θ S )
b
db
=
σ (θ S ) =
sin θ S dθ S mv0 2θ S 2 ( 2π − θ S )2 sin θ S
Since d Ω = 2π sin θ s dθ s we get …
 (π − θ S ) 

 dθ S
2
2
 ( 2π − θ S ) θ S 
---------------------------------------------------------------------------------------------------
σ (θ S ) d Ω = 2π bdb =
kπ 3
E
84
Chapter 7
Dynamics of Systems of Particles
7.1
z
1
∑ mi ri
m i
From eqn. 7.1.1, rcm =
(
3
)
1
1
( r1 + r2 + r3 ) = iˆ + ˆj + ˆj + kˆ + kˆ
3
3
1 ˆ
rcm = i + 2 ˆj + 2kˆ
3
d
1
1
vcm = rcm = ( v1 + v2 + v3 ) = 2iˆ + ˆj + iˆ + ˆj + kˆ
dt
3
3
1
vcm = 3iˆ + 2 ˆj + kˆ
3
From eqn 7.1.3, p = ∑ mi vi = v1 + v2 + v3
rcm =
(
2
)
(
(
)
)
y
1
x
i
p = 3 iˆ + 2 ˆj + kˆ
1
7.2 (a) From eqn. 7.2.15, T = ∑ mi vi2
i 2
1
T =  22 + 12 + (12 + 12 + 12 )  = 4
2
1
(b) From Prob. 7.1, vcm = 3 iˆ + 2 ˆj + kˆ
3
1 2 1
1 2
1
mvcm = × 3 × ( 3 + 22 + 12 ) = 2
2
2
9
3
(c) From eqn.7.2.8, L = ∑ ri × mvi
(
(
)
(
i
)
)
(
)
L =  iˆ + ˆj × 2iˆ  +  ˆj + kˆ × ˆj  +  kˆ × iˆ + ˆj + kˆ 

 
 

ˆ
ˆ
L = −2k + −iˆ + ˆj − iˆ = −2iˆ + ˆj − 2k
( ) ( ) (
7.3
)
v = vb − vg
Since momentum is conserved and the bullet and gun were initially at rest:
0 = mvb + Mvg
vg = −γ vb ,
v = (1 + γ ) vb
vb =
γ=
m
M
v
1+ γ
85
vg = −
7.4
γv
1+ γ
v 
mv = m   + Mvblk
2
1
m
vblk = γ v
γ=
2
M
2
2
1 2 1  v  1  γ v  
Ti − T f = mv −  m   + M 
 
2
 2  2  2  2  
Momentum is conserved:
1 2 1 1 γ2 
= mv 1 − − M

2
4 
 4 m
Ti − T f 3 γ
= −
4 4
Ti
7.5
v0 / 2
At the top of the trajectory:
ˆ cos 60 = iˆ v
v = iv
2
Momentum is conserved:
ˆ v = ˆj  m   v  + m v2
im
2
 2  2  2
ˆ − ˆj v
v2 = iv
2
 v 
− 
Direction: θ = tan −1  2  = 26.6 below the horizontal.
 v 


v0
θ
60 o
1
 2  v 2  2
Speed: v2 =  v +    = 1.118 v
 2  

7.6
When a ball reaches the floor,
As a result of the bounce,
1 2
mv = mgh .
2
v′
=ε .
v
The height of the first bounce: mgh′ =
1
mv′2
2
v ′2 ε 2 v 2
=
= ε 2h
2g
2g
Similarly, the height of the second bounce, h′′ = ε 2 h′ = ε 4 h
h′ =
86
∞


Total distance = h + 2ε 2 h + 2ε 4 h + … = h  −1 + ∑ 2ε 2 n 
n=0


∞
a
, r <1.
Now ∑ ar n =
1− r
n=0
∞
2
1+ ε 2

2n 
−
+
=
−
+
=
ε
1
2
1
∑


1− ε 2 1− ε 2
n=0


 1+ ε 2 
total distance = h 
2 
 1− ε 
For the first fall,
1 2
2h
gt = h , so t =
2
g
2h′
2h
=ε
g
g
Accounting for equal rise and fall times:
∞
2h
2h 

−
+
1 + 2ε + 2ε 2 + …) =
1
2ε n 
Total time =
(
∑

g
g 
n=0

For the fall from height h′ : t1 =
Total time =
7.7
2h  1 + ε 


g  1− ε 
m
v0
-v0/2
vc
vt
4m
From eqn. 7.5.5:
( m − ε m2 ) x1 + ( m2 + ε m2 ) x2
x1′ = 1
m1 + m2
x2′ =
( m1 + ε m1 ) x1 + ( m2 − ε m1 ) x2
m1 + m2
1
1



 v 
 v 
 m − 4m  v +  4m + 4m   −  0 + 5m  − 
4
4


 2  =
 2  =−v
vc = 
2
m + 4m
5m
1 
1  v 


 m + m  v +  4m − m   − 
4 
4  2 

vt = 
m + 4m
5
15  v 
mv + m  − 
v
4
4  2
=
=−
5m
8
Both car and truck are traveling in the initial direction of the truck
v
v
and , respectively.
with speeds
2
8
87
7.8
1
1
1
From eqn. 7.2.15, T = ∑ mi vi2 = m1v12 + m2 v22
2
2
i 2
Meanwhile:
2
1 2 1 2 1  m1v1 + m2 v2  1 m1m2
2
mvcm + µ v = m 
( v1 − v2 )
 +
m
2
2
2 
 2 m1 + m2
1
 m12 v12 + m22 v22 + 2m1m2 v1 ⋅ v2 + m1m2 ( v12 + v22 − 2v1 ⋅ v2 ) 
=

2m 
1
1
= m1v12 + m2 v22
2
2
1 2 1 2
Therefore, T = mvcm + µ v
2
2
7.9
7.10
1 2 1 2
mvcm + µ v
2
2
′ :
Q = T − T ′ and since vcm = vcm
1
1
Q = µ v 2 − µ v ′2
2
2
v′
From eqn. 7.5.4, ε =
v
1
Q = µ v 2 (1 − ε 2 )
2
From Prob. 7.8, T =
Conservation of momentum:
m1v1 = m1v1′ + m2 v2′
m
v1′ = v1 − 2 v2′
m1
2m2
m2
v1v2′ + 22 v2′2
m1
m1
Conservation of energy:
1
1
1
m1v12 = m1v1′2 + m2 v2′2
2
2
2
1
1
m2
1
m1v12 = m1v12 − m2 v1v2′ + 2 v2′2 + m2 v2′2
2
2
2m1
2
v1′2 = v12 −
m2  m2  2
+ 1 v2′ − m2 v1v2′ = 0

2  m1

2m1v1
v2′ =
m1 + m2
T1 − T1′ =
1
1
1
2m12 m2
m1v12 − m1v1′2 = m2 v2′2 =
v2
2 1
2
2
2
( m1 + m2 )
88
2m1µ 2
v1
T1 − T1′
4µ
m
=
=
1
T1
m
m1v12
2
7.11
From eqn. 7.2.14, L = rcm × mvcm + ∑ ri × mi vi
∑r ×m v
i
i i
i
= r1 × m1v1 + r2 × m2v2
i
 m 
 m + m2  m1
From eqn. 7.3.2, R = r1 1 + 1  = r1  1
r1
=
 m2 
 m2  µ
m
Since from eqn. 7.3.1, r1 = − 2 r2
m1
m
R = − 2 r2
µ
∑r ×m v
i
i
i i
 µ 
R × m1v1 +  −
 R × m2v2
m1
m
2 

= µ R × ( v1 − v2 ) = R × µ v
=
µ
L = rcm × mvcm + R × µ v
7.12
Let ms = mass of Sun and ae = semi-major axis of Earth’s orbit
then from eqn. 7.3.9c,
−
1
3
 m m  2  a 2
τ =  1 + 2    yr
 ms ms   ae 
1
2
 m m 3
a
=τ 3  1 + 2 
ae
 ms ms 
16 ms
20 ms
2
2
1
−

1 yr  3
3
3
=  5.6 da ×
 ( 20 + 16 ) yr
da
365


1
a = 0.20 ae = ae
5
7.13
(Ignore primes in notation)
a)
The coordinates of the two primaries, P1
and P2, are shown at left – along with the
coordinates of L4 and L5 .
89
(1 − α )
b) V ( x, y ) = −
1
2
α
−
( x − α ) + y 
( x + 1 − α ) + y 




∂V (1 − α )( x − α ) α ( x + 1 − α )
−x
=
+
3
3
∂x
2
2
2
2
[]
2
2
1
2
x2 + y 2
−
2
(7.4.13)
[]
Now x = α − 0.5 at L4 and L5
also, each bracket term in the denominator equals 1 at L4 , L5
∂V
= (1 − α )(α − 0.5 − α ) + α (α − 0.5 + 1 − α ) − (α − 0.5 )
∂x
= −0.5 + 0.5α + 0.5α − α + 0.5 ≡ 0
∂V (1 − α ) y α y
=
+ 3 −y
3
∂y
2
2
[]
[]
Again, the denominator in brackets equals 1 @ L4 , L5
So,


3
3 
∂V
3
= (1 − α )  ±
 + α  ±
 −  ±

∂y
 2 
 2   2 
3
3
∓α
±α
2
2
∂V
∇V ( x, y ) = iˆ
+
∂x
=±
Thus
7.14
3
3
∓
≡0
2
2
ˆj ∂V ≡ 0 at L4 , L5 .
∂y
v’p
Conservation of momentum:
m p v = m p v′p + 4m p vα′
v = v′p cos 45 + 4vα′ cos φ
4vα′ cos φ = v −
v′p
2
0 = v′p sin 45 − 4vα′ sin φ
4vα′ sin φ =
45 o
v′p
φ
vo
v’
2
16vα′ = v − 2 v v′p + v′p2
2
2
Conservation of energy:
1
1
1
m p v 2 = m p v′p2 + 4m p vα′ 2
2
2
2
16vα′2 = 4v 2 − 4v′p
Subtracting:
0 = −3v 2 − 2 v v′p + 5v′p 2
90
2v ± 2v 2 + 60v 2 v
=
2 ± 62
10
10
v′p > 0 , so the positive square root is used.
(
v′p =
)
v′p = 0.9288 v
v′px = v′py =
v′p
2
= 0.657 v
1
1
v
1 2
v − v′p2 ) 2 = (1 − .9288 2 ) 2
(
2
2
vα′ = 0.1853v
v′p
v′p
.9288
2 =
=
tan φ =
v′p
2v − v′p
2 − .9288
v −
2
−1
φ = tan 1.9134 = 62.41
vα′ x = vα′ cos φ = 0.086 v
vα′ y = −vα′ sin φ = −0.164 v
vα′ =
7.15
Conservation of energy:
1
1
1
11

m p v 2 = m p v′p2 + 4m p vα′ 2 +  m p v 2 
2
2
2
42

2
2
2
16vα′ = 3v − 4v′p
From the conservation of momentum eqn of Prob. 7.14:
16vα′ 2 = v 2 − 2v v′p + v′p2
Subtracting:
0 = −2v 2 − 2v v′p + 5v′p2
2v ± 2v 2 + 40v 2 v
=
2 ± 42
10
10
Using the positive square root, since v′p > 0 :
(
v′p =
)
v′p = 0.7895 v
v′px = v′py =
v′p
2
= 0.558 v
1
1
1
 3
2 v
vα′ =  v 2 − v′p2  = (.75 − .7895 2 ) 2
4 
2
 16
vα′ = 0.1780v
From the conservation of momentum eqns of Prob. 7.14:
v′p
.7895
=
tan φ =
2v − v′p
2 − .7895
91
φ = tan −1 1.2638 = 51.65
vα′ x = vα′ cos φ = 0.110 v
vα′ y = −vα′ sin φ = −0.140 v
7.16
sin θ
φ1 and θ are the scattering angles in
γ + cosθ
the Lab and C.M. frames respectively.
m
From eqn. 7.6.16, for Q = 0, γ = 1
m2
sin θ
tan 45 =1 =
1
+ cos θ
4
1
+ cos θ = sin θ
and squaring …
4
1 1
+ cos θ + cos 2 θ = 1 − cos 2 θ
16 2
1
15
2 cos 2 θ + cos θ − = 0
2
16
1
1 15
− ±
+
2
4
2 = −.125 ± .696
cosθ =
4
From eqn. 7.6.14,
Since 0 < θ <
7.17
π
2
,
tan φ 1 =
θ = cos −1 .571 ≈ 55.2
From eqn. 7.6.14, tan φ 1 =
m
From eqn. 7.6.18, γ = 1
m2
sin θ
γ + cosθ
 Q  m1  
1 − 1 +

 T  m2  
−
−
1
2
1
1  1  1  2
γ = 1 − 1 +   = 0.3015
4  4  4 
sin θ
tan 45 =
.3015 + cos θ
.3015 + cosθ = sin θ
(since sin θ > cos θ , θ > 45 )
2
2
2
.3015 = sin θ − 2sin θ cosθ + cos θ
Using the identity 2sin θ cos θ = sin 2θ
sin 2θ = 1 − .30152 = 0.9091
Since θ > 45 , 2θ > 90 : 2θ = sin −1 .9091 = 114.62
θ = 57.3
92
P1’
7.18
Conservation of momentum:
P1 = P1′ cos φ + P2′ cos (ψ − φ
φ
ψ
P1
)
0 = P1′ sin φ − P2′ sin (ψ − φ )
From Appendix B for sin (α + β ) and cos (α + β ) :
P2’
P1 = P1′ cos φ + P2′ ( cosψ cos φ + sinψ sin φ )
0 = P1′ sin φ − P2′ ( sin ψ cos φ − cosψ sin φ )
P12 = P1′2 cos 2 φ + P2′2 ( cos 2 ψ cos 2 φ + 2 cosψ cos φ sin φ sinψ + sin 2 ψ sin 2 φ )
+2 P1′ P2′ ( cos 2 φ cosψ + cos φ sinψ sin φ )
0 = P1′2 sin 2 φ + P2′2 ( sin 2 ψ cos 2 φ − 2sinψ cos φ cosψ sin φ + cos 2 ψ sin 2 φ )
−2 P1′ P2′ ( sin φ sinψ cos φ − cosψ sin 2 φ )
Adding: P12 = P1′2 + P2′2 + 2 P1′ P2′ cosψ
Conservation of energy:
P12 P1′2 P2′2
=
+
+Q
2m 2m 2m
1
1
2 P1′ P2′ cosψ
Q=
P12 − P1′2 − P2′2 ) =
(
2m
2m
P′ P ′ cosψ
Q= 1 2
m
(
)
1
1
m1v12
T1′ = m1v1′ 2
2
2
2
T ′ v′
let r = 1 = 12
… ratio of scattered particle to incident particle energy
T1 v1
Looking at Figure 7.6.2 …
v1′ ⋅ v1′ = ( v1′ − vcm ) ⋅ ( v1′ − vcm )
7.19
T1 =
2
v1′ 2 = v1′ 2 + vcm
− 2v1′ vcm cos φ1
2
+ 2v1′vcmγ
hence v1′ 2 = v1′ 2 − vcm
2v′v γ
v1′ 2 v
−
+ 1 2cm
2
v1
v
v1
v1′ = v1
but
scattered particle are the same.
∴r =
where γ = cos φ1
2
cm
2
1
v1′
m2
α
=
=
v1 m2 + m1 1 + α
…the center of mass speeds of the incident and
…from equation 7.6.12 where α =
m2
m1
93
vcm
1
m1
=
=
Equation 7.6.11
v1 m2 + m1 1 + α
Thus
1
α2
1
2γ
v1′
α2
1
2γ
r=
−
+
r2
=
−
+
2
2
2
2
α
+
α
1
1
v
+
(
)
(
)
α
α
+
α
+
α
1
1
1
1
+
+
(
) (
)
(
) (
)
1
Simplifying
2γ 12  1 − α 
r−
r +
=0
1+α
 1+α 
Let x 2 = r and solving the resulting quadratic for x
1
γ
1
2
2


x=
+
γ − (1 − α )  2
1+α 1+α 
Squaring
1

1  2
2
2
2
2
2γ + α − 1 + 2γ (γ + α − 1) 2 
r=x =
2 
(1 + α ) 

1
 2

2
2
2
2
2
1
2
1
γ
α
γ
γ
α
+
−
+
+
−
(
)




And, after a little algebra, we get the desired solution
∆T1
2
2γ 
γ + γ 2 + α 2 − 1
=
−
2


T1 1 + α (1 + α )
Now
7.20
1
∆T1
= 1− r = 1−
2
T1
(1 + α )
From Equation 7.6.15 … γ =
m1v1
m1
v1
=
v1′ ( m1 + m2 ) m2 (1 + m1 m2 ) v1′
v1
…
v1′
Now we solve for
1
 2T  2
v1 = 
 and now solving for v1′ starting with Equation 7.6.9 …
 m1 
m2
1
1
v1 we get …
µ v1′2 = µ v12 − Q and using v1′ =
m1 + m2
2
2
1
1
T
−Q
= m1v12
−Q =
2
(1 + m1 m2 )
(1 + m1 m2 )


2
T
− Q .

m1 (1 + m1 m2 )  (1 + m1 m2 )

Thus, solving for γ …
v1′2 =
94
1
γ=
m1
m2
( 2T m1 ) 2
1
2
( 2 m1 ) (1 + m1


T
− Q
m2 ) 
 (1 + m1 m2 )

1
2
1
2
Finally…
γ=
7.21
m1
m2
1
1
 Q (1 + m1 m2 )  2
1 −

T


The time of flight, τ = constant—so τ =
v1 τ 
γ + γ 2 + α 2 − 1


1+α
As an example, let v1 τ = 1 and we have
r
but from problem 7.19 above
v1′
r = v1′ τ =
r1 = γ
α =1
1
α =2
r2 = γ + γ 2 + 3 

3
1
α =4
r3 = γ + γ 2 + 15 

5
1
α = 12
r4 = γ + γ 2 + 143 

13 
Below is a polar plot of these four curves.
7.22
pp scattering
p–D
p – He
p–C
From eqn. 7.7.6, Fu − Fg = mv + vm
since v = constant, v = 0
m = λ z = λv ,
λ = mass per unit length
Fg = ( λ z ) g

v2 
Fu = λ zg + ( λ v ) v = g λ  z + 
g

Fu is equal to the weight of a length z +
v2
of chain.
g
95
7.23
4
m = π r3ρ
3
m = 4π r 2 ρ r ∝ π r 2 z
where v = z
r = kz
k a constant of proportionality
r = r + kz
From eqn. 7.7.6, mg = mv + vm
4 3
4
π r ρ g = π r 3 ρ z + 4π r 2 ρ ( kz ) z
3
3
2
3kz
g = z+
r
3z 2
z=g−
r
z+
k
3z 2
For r = 0 , z = g −
z
A series solution is used for this differential equation:
∞
z 2 = ∑ an z n
n=0
z=
dz dz dz
dz 1 d ( z 2 )
= ⋅ =z =
dt dz dt
dz 2 dz
2
d (z )
= ∑ an nz n −1
dz
n
z2
= ∑ an z n −1
z
n
1
∴ z = ∑ an nz n −1 = g − 3∑ an z n −1
2 n
n
1
For n = 1 : a1 = g − 3a1
2
2
a1 = g
7
1
For n ≠ 1 : nan = −3an
2
Since n is an integer, an = 0 for n ≠ 1
2
z2 = g z
7
32
 g
z = g −  g z =
z7
 7
96
7.24
From eqn. 7.7.6, mg = mv + vm , where m and v refer to the portion of the chain
hanging over the edge of the table.
m = λ z and v = z
where λ is the mass per unit length of chain
m = λ z and v = z
dz dz dz
dz 1 d ( z 2 )
= ⋅ =z =
dt dz dt
dz 2 dz
2
 1 d (z ) 
λ z g = λ z
 + z (λ z )
 2 dz 
z=
1 d (z2 )
z2
=g−
z
2 dz
Because of the initial condition z = b ≠ 0 , a normal power series solution to this
differential equation (…as in Prob. 7.22) does not work. Instead, we use the
Method of Frobenius …
z=
∞
z 2 = ∑ an z n + s
n=0
2
d (z )
= ∑ an ( n + s ) z n + s −1
dz
n
z2
= ∑ an z n + s −1
z
n
1
z = ∑ an ( n + s ) z n + s −1 = g − ∑ an z n + s −1
2 n
n
Equality can be attained for an ≠ 0 at n = 0 and n = 3
n ≠ 0,3
…otherwise an = 0
1
a s = −a
2
s = −2
1
z = ∑ an ( n − 2 ) z n −3 = g − ∑ an z n −3
2 n
n
1
For n = 3 , a3 = g − a3
2
2
a3 = g
3
1
For all n ≠ 0 , 3: an ( n − 2 ) = − an
2
an = 0 ,
n ≠ 0,3 .
2
z 2 = a z −2 + gz
3
For n = 0 ,
97
At t = 0 , z = 0 , and z = b
a 2 gb
0= 2 +
b
3
2
a = − gb3
3
2 b3 2
z 2 = − g 2 + gz
3 z
3
2 
b3  2 g
At z = a , z 2 = g  a − 2  = 2 ( a 3 − b3 )
a  3a
3 
1
 2g
2
z =  2 ( a 3 − b3 ) 
 3a

7.25
Initially, the upward buoyancy force balances the weight of the balloon and sand.
FB − ( M + m ) g = 0
(1)
Let m = m ( t ) − the mass of sand at time t where 0 ≤ t ≤ t

t
(2)
m = m 1 − 
 t 
The velocity of sand relative to the balloon is zero upon release so
V = 0 in equation 7.7.5 … there is no upward “rocket-thrust.”
As sand is released, the net upward force is the difference between the initial
buoyancy force, FB, and the weight of the balloon and remaining sand. Let y be
the subsequent displacement of the balloon, so equation 7.7.5 reduces to F = ma
dv
FB − ( M + m ) g = ( M + m )
dt
and using (1) and (2) above we get
( M + m ) gt
dv
m gt
=
= −g +
dt ( M + m ) t − m t
(M + m )t − m t
whose solution is:
( M + m ) gt ln 1 − m t 
dy
v=
= − gt −


dt
m
 (M + m )t 
g
m


y = C − ∫  gt + ln (1 − kt )  dt ,
k=
k
t (M + m


1
gt
tdt
= C − gt 2 − ln (1 − kt ) − g ∫
k
2
1 − kt
Integrating by parts
)
98
1
gt
g 
1 
= C − gt 2 − ln (1 − kt ) − ∫  −1 +
 dt
k
k 
2
1 − kt 
1
gt
gt g
= C − gt 2 − ln (1 − kt ) + + 2 ln (1 − kt )
k
k k
2
1
g
gt
= C − gt 2 + 2 (1 − kt ) ln (1 − kt ) +
k
k
2
but y = 0 at t = 0 so C = 0
gt 1
g
y = − gt 2 + 2 (1 − kt ) ln (1 − kt )
k 2
k
and at t = t
(a)
(b)
(c)
 M
gt 2 
2M + m ) m + 2M ( M + m ) ln 
2 (
2m 
M +m
(M + m ) − m 
gt 
v=
( M + m ) ln

m 
M

H=
letting ε =



m
<< 1 we have
M
gt 2
( 2 + ε ) ε − 2 (1 + ε ) ln (1 + ε ) 
2ε 2 


gt 2 
ε2 ε3
= 2  2ε + ε 2 − 2 (1 + ε )  ε − + − …  
2ε 
2 3


H=
H≅
gt 2
ε
6
Similarly:
gt
(1 + ε ) ln (1 + ε ) − ε 
ε 

 
gt 
ε2 ε3
1
ε
ε
=
+
−
+ − … − ε 
(
)


ε 
2 3

 
v=
1
gt ε
2
H = 327 m ;
≈
(d)
7.26
v = 9.8ms −1
m = − k or m = m − k t
Burn-out occurs at time T =
εm
k
So – the rocket equation (7.7.7) becomes
dv
m = − Vm
(-) since V is oppositely directed to v
dt
99
dv
Vk
=+
dt
m −kt
Thus
dt
= −V ln ( m − k t ) + C1
m −kt
Now, v = 0@ t = 0 so C1 = V ln m
v = Vk ∫
where C1 is a constant.
 ( m − kt ) 
εm
Hence v = −V ln 
0≤t ≤
,
k
 m

Let y be the displacement at the time t so
 ( m − kt ) 
y = −V ∫ ln 
 dt + C2
 m

Integrating the above expression by parts
(m − k t ) 
t dt
+ C2
y = −Vt ln 
 − Vk ∫
m −kt
 m

(m − k t ) 

m 
= −Vt ln 
 dt + C2
 + V ∫ 1 −
 m −kt 
 m

(m − k t ) 
Vm
= −Vt ln 
ln ( m − k t ) + C2
 + Vt +
k
 m

−Vm ln m
since y = 0 at t = 0 , C2 =
and we have
k
(m − k t ) 
V
y = Vt + ( m − k t ) ln 

k
 m

At burn-out t = T =
εm
so
k
mV
(a)
y (ε ) = D =
ε + (1 − ε ) ln (1 − ε ) 
k 
ε cannot exceed 1.0 although it can approach 1.0 for small payloads
(b)
lim
mV
y (ε ) =
Thus ymax =
ε →1
k
7.27
From eqn. 7.7.5, − kv = mv − Vm
Since V is opposite in direction from v
− kv = mv + Vm
m
m
v =− v− V
V
k
k
m
m
α=
and since m < 0 , α = −
k
k
v
100
v −α V = α
m
m dv
m dv dm
dv
v =α
=α
= αm
m
m dt
m dm dt
dm
dm
dv
=
α m v − Vα
v
1 m dm
dv
=
∫
∫
m
0
α
m
v − Vα
1 m
 v − Vα 
ln   = ln 

α m 
 −V α 
1
 m α
v
+1
  =−
Vα
m 
1


α


m 

v = Vα 1 −  
 m  


7.28
From eqn. 7.7.5, − mg = mv − Vm
since V is opposite in direction from v ,
− mg = mv + Vm
− mgdt = mdv + Vdm
dm
dm
m=
so dt =
dt
m
dm
− mg
= mdv + Vdm
m
g V
dv = − dm  + 
m m
ve
mp
g V
dv
=
−
∫0
∫m dm  m + m 
m
g
ve = ( m − m p ) + V ln
m
mp
m f = m − mp ≈ m
ve =
 mf
g
m f + V ln 1 +
 m
m
p

v
V
m p = payload mass
m f = fuel mass



 m  v gm
ln  1 + f  = e −
 m  V V m
p 

mf
v g m 
= exp  e −
 −1
mp
V V m 
101
1 g m 
= exp  −
 −1
mp
 k kve m 
mf
For V = kve ,
ve ≈ 11
From chap. 2, Section 2.3 …
For m = 0.01 m s −1
and k =
km
s
1
:
4




9.8
= exp  4 −
 −1
1
mp

(11, 000 )( −.01) 


4
mf
= 77
mp
mf
7.29 We can use Equation 7.7.9 to calculate the final velocity attained by the ion rocket
during the 100 hour burn. Assuming the rocket starts from rest (even if the ion rocket is
turned on while in Earth orbit, the initial rocket speed v0 ≈ 10−4 c ≈ 0 . Thus …
m
v ≈ V ln 0 and m0 = mF + m p = 2m p + m p = 3m p
mp
v ≈ V ln 3 = 0.1099 c . The final rocket velocity is a little more than 10% c.
4 LY
L
T=
=
= 36.4 yr
0.1099 c 0.1099 LY yr
We again use Equation 7.7.9 …
mF + m p
m
= Vion ln 2 for the ion rocket. For the chemical
v ≈ Vion ln 0 = Vion ln
mp
mp
rocket …
mF + m p
. Setting these two equations equal …
v ≈ Vchem ln
mp
7.30
Vchem ln
mF + m p
mF + m p
mp
= 2(
= Vion ln 2 . Solving for mF …
Vion Vchem )
=2
( 0.1c 10 c ) = 210 ≈ 10500 which demonstrates the virtue of
−5
4
mp
ejecting mass at high velocity!
----------------------------------------------------------------------------------------------------------
102
CHAPTER 8
MECHANICS OF RIGID BODIES:
PLANAR MOTION
m
and centered at
3
 b b
b b
 − ,  , ( 0, 0 ) , and  ,  …
2 2
 2 2
1   b  m 
 b   m 
xcm =  −     + 0 +      = 0
m   2  3 
 2   3 
8.1 (a) For each portion of the wire having a mass
y
b
ycm =
x
1  b   m 
 b   m  b
    + 0 +     =

m  2   3 
 2   3  3
b
1
(b)
ds = xdy = ( b 2 − y 2 ) 2 dy
ycm =
ycm =
1
1 b
2
2 2
ρ
y
b
−
y
(
) dy
m ∫0
−
ρ
∫
2
y =b
y =0
1
(b2 − y 2 ) 2 d (b2 − y 2 )
1 2
πb ρ
4
4b
3π
4b
From symmetry, xcm =
3π
ycm =
(c)
The center of mass is on the y-axis.
1
ds = 2 xdy = 2 ( by ) 2 dy
b
ycm =
∫
∫
0
b
0
ycm =
(d)
1
2 ρ y ( by ) 2 dy
1
2
2 ρ ( by ) dy
=
∫
b
∫
b
0
0
3
y 2 dy
1
2
y dy
3b
5
The center of mass is on the z-axis.
dv = π r 2 dz = π ( x 2 + y 2 ) dz = π bzdz
103
b
zcm
∫
=
∫
0
ρ zπ bzdz
b
0
ρπ bzdz
b
∫ z dz
=
∫ zdz
2
0
b
0
2
zcm = b
3
(e)
The center of mass is on the z-axis.
α is the half-angle of the apex of the cone. rD is the radius of
the base at z = 0 and r is radius of a circle at some arbitrary z in
a plane parallel to the base.
r
r
tan α = D =
, a constant
b b−z
2
dv = π r 2 dz = π ( b − z ) tan 2 α dz
1
1
m = ρ π rD2b = πρ b3 tan 2 α
3
3
zcm
∫
=
b
0
zcm =
ρ zπ ( b − z ) tan 2 α dz
2
1
πρ b3 tan 2 α
3
xcm
3
b3
∫ ( b z − 2bz
b
2
0
2
+ z 3 ) dz
b
4
b
8.2
=
∫ ρ xdx = ∫ cx dx
=
∫ ρ dx ∫ cxdx
2
0
b
0
2b
xcm =
3
8.3 The center of mass is on the z-axis. Consider the sphere with the cavity to be made
of a (i) solid sphere of radius a and mass M s , with its center of
mass at z = 0 , and (ii) a solid sphere the size of the cavity, with
a
mass − M c and center of mass at z = − . The actual sphere with
2
the cavity has a mass m = M s − M c and center of mass zcm .
0=
1
Ms
  a

 M c  − 2  + m zcm 

 

3
4 a
4
M s = π a3 ρ , M c = π   ρ
3
3 2
3
3
1  a   a  
 a   
0 = 3    −  +  a 3 −    zcm 
a  2   2  
 2   
104
zcm =
a
14
2
2
m  b 
b 
8.4 (a) I z = ∑ mi R =   + 0 +   
3  2 
 2  
i
2
i
Iz =
mb 2
6
(b)
ds = rdθ dr , R = r sin θ
I z = ∫ R 2 ρ ds
Iz = ρ ∫
r =b
r =0
Iz =
ρb
4
4
r rdr ∫
2
θ=
π
4
θ =−
π
∫ π sin
4
−
2
π
sin 2 θ dθ
4
θ dθ
4
θ sin 2θ 

2
 ∫ sin θ dθ = 2 − 4 
ρb4  π 1 
Iz =
 − 
4  4 2
1
m = ρπ b 2
4
mb 2
Iz =
(π − 2 )
4π

x2 
ds = hdx =  b −  dx
b 

Where the parabola intersects the line y = b ,
(c)
1
x = ( by ) 2 = ±b
b
b 

x2 
x4 
I y = ∫ x 2 ρ  b −  dx = ρ ∫  bx 2 −  dx
−b
−b
b 
b 


4
I y = ρb4
15
b

x2 
4
m = ∫ ρ  b −  dx = ρ b 2
−b
b 
3

1
I y = mb 2
5
(d)
dv = 2π RhdR
h =b−z
3
105
R = (x + y
2
1
2 2
) = ( bz )
1
2
1
1  b 2
dR =   dz
2 z 
b
I z = ∫ R ρ dv = ∫ bz ρ 2π ( bz )
2
0
1
2
1
1 b 2
( b − z )   dz
2 z 
b
1
I z = πρ b 2 ∫ ( bz − z 2 ) dz = πρ b5
0
6
b
m = ∫ ρ dv = ∫ ρ 2π ( bz )
1
2
0
1
1 b 2
( b − z )   dz
2 z 
b
1
m = πρ b ∫ ( b − z ) dz = πρ b3
0
2
1
I z = mb 2
3
(e)
α is the half-angle of the apex of the cone. rD is the
radius of the base at z = 0 and r is radius of a circle at
some arbitrary z in a plane parallel to the base.
tan α =
dv = 2π RhdR = 2π
( b − z ) RD zdR
RD
R
, a constant
=
b b−z
b
b
z
R
−
( ) D , dR = − RD dz , and the limits of integration for R = 0 → R
Since R =
D
b
b
correspond to z = b → 0
0
(b − z )
b
2
I z = ∫ R ρ dv = ∫
2
I z = +2πρ
RD4
b4
b
2
RD2
∫ ( b z − 3b z
b
0
3
2 2
ρ 2π
( b − z ) RD z  − RD  dz
b
+ 3bz 3 − z 4 ) dz =



b 
1
πρ RD4b
10
1
m = ρ π RD2b
3
3
I z = mRD2
10
106
8.5
Consider the sphere with the cavity to be made of a (i) solid sphere of radius a
and and mass M s , with its center of mass at z = 0 , and (ii) a solid sphere the size of the
a
cavity, with mass − M c and center of mass at z = − . The actual sphere with the cavity
2
has a mass m = M s − M c and center of mass zcm .
3
4
4 a
74 3
m = M s − M c = π a3 ρ − π   ρ =
πa ρ
3
3 2
83
8
1
M s = m and M c = m
7
7
From eqn. 8.3.2,
I s = Ic + I
2
2
2
a
M s a2 = M c   + I
5
5
2
28 
2  1  a 2 31
I =  m  a 2 −  m  = ma 2
57 
5  7  4 70
8.6
The moment of inertia about one of the straight edges is
I z = ∫ R 2 ρ dv where R 2 = x 2 + y 2 . From Appendix F …
dv = r 2 sin θ dr dθ dφ
R 2 = x 2 + y 2 = r 2 sin 2 θ
Let a = radius of sphere
z
R
θ
Iz = ∫
r
φ
r =a
r =0
y
Iz = ρ
π
θ=
π
π
φ=
2
2
θ =0 φ =0
∫ ∫
r =a
θ=
2 ∫ ∫θ
r =0
π
2
=0
r 2 sin 2 θρ r 2 sin θ dr dθ dφ
r 4 sin 3 θ drdθ
π
1
ρπ a 5 ∫ 2 sin 3 θ dθ
0
10


cos3 θ
3
=
− cos θ 
sin
d
θ
θ
∫
3


Iz =
x
2
ρπ a 5
30
14 3
1
m=
π a ρ = π a3 ρ
83
6
2
I z = ma 2
5
Iz =
107
8.7
For a rectangular parallelepiped:
dv = hdxdy
y
R = ( x2 + y
h is the length of the box in the z-direction
1
2 2
)
I z = ∫ R 2 ρ dv = ∫
x=a
∫ (x
y =b
x =− a
y =− b
2
+ y 2 ) ρ hdxdy

2b3 
4
2
2
2
I z = ρ h ∫  2bx +
 dx = ρ hab ( a + b )
−a
3 
3

m = ρ ( 2a )( 2b ) h = 4 ρ abh
a
x
2b
Iz =
z
m 2
a + b2 )
(
3
2a
For an elliptic cylinder:
y
1
Again dv = hdxdy , and R = ( x 2 + y 2 ) 2
On the surface,
x2 y 2
+
=1
a 2 b2
1
2b
x
 y2 2
x = ± a 1 − 2 
 b 
1
z
I z = ∫ R 2 dv = ∫
2a
y =b
y =− b
∫
 y2  2
x = a 1− 2 
 b 


1
 y2  2
x =− a  1− 2 
 b 


(x
2
+ y 2 ) ρ hdxdy
3
1


2 2
2 2




2
y
y
3
2

I z = ρ h∫
a 1 − 2  + 2a 1 − 2  y dy
−b  3

 b 
 b 


3
1
2
b

aa b 2
2 2
2
2 2
= 2 ρ h  2 ∫ ( b − y ) dy + ∫ ( b − y ) y 2 dy 
−
−
b
b
b  3b

From a table of integrals:
3
3
1
3 2
3
y 2
y
2
2 2
2 2
2
2 2
−
=
−
+
−
+ b 4 sin −1
b
y
dy
b
y
b
y
b
y
)
(
)
(
)
∫(
4
8
8
b
1
3
1
2
4
y 2
b y 2
b
−1 y
2
2
2
2
2
∫ ( b − y ) 2 y dy = − 4 ( b − y ) 2 + 8 ( b − y ) 2 + 8 sin b
4
a  a2  3
 b  1
I z = 2 ρ h  2  b 4π  + π  = ρ hπ ab ( a 2 + b 2 )
b  3b  8
 8  4
b
108
m = ρ h (π ab )
Iz =
m 2
a + b2 )
(
4
For an ellipsoid:
dv = hdxdy ,
R2 = x2 + y 2
and on the surface,
x2 y 2 z 2
+
+ =1
a 2 b2 c2
1
 x2 y2  2
z = ±c 1 − 2 − 2 
b 
 a
y
1
2b
x
 x2 y2  2
h = 2 z = 2c  1 − 2 − 2 
b 
 a
x2 y 2
In the xy plane 2 + 2 = 1
a
b
1
 y2 2
x = ± a 1 − 2 
 b 
2a
z
1
1
 y2  2
x = a 1− 2 
 b 


 x2 y2  2
2
2
+
I z = R 2 ρ dv = ∫ ∫
x
y
ρ
2
c
1 (
) 1 − a 2 − b2  dxdy
y =− b
 y2  2


x =− a  1− 2 
y =b

b 
1
1


2 2
2




2 ρ c y =b   2 a 2 y 2  2  2 2
a
y

2
2
2
Iz =
a − 2  − x  x dx + y ∫  a − 2  − x  dx 



∫
∫
a y =− b  
b 
b 






From a table of integrals:
1
2 2
3
1
2
x 2
k2x 2
k4
x
2
2
2
2 2
−
=
−
−
+
−
+
sin −1
k
x
x
dx
k
x
k
x
)
(
)
(
)
∫(
4
8
8
k
1
1
2
x 2
k
−1 x
2
2
2
∫ ( k − x ) 2 dx = 2 ( k − x ) 2 + 2 sin k
2
2
2ρc b  a4  y2 
y2  
2 a 
−
+
−
π
1
1
Iz =
y




 π  dy
2  b2  
a ∫− b  8  b 2 


2
2
4
4
b a 
2y
y 
y 
I z = ρπ ac ∫  1 − 2 + 4  + y 2 − 2  dy
−b
b
b 
b 
4
4
I z = ρπ abc ( a 2 + b 2 )
15
4
m
For an ellipsoid, m = ρ π abc , so I z = ( a 2 + b 2 )
3
5
109
8.8
(See Figure 8.4.1)
Note that l ′ + l is the distance from 0′ to 0 , defined as d
2
From eqn. 8.4.13, kcm = ll ′
2
kcm
+ l 2 = ll ′ + l 2 = l ( l ′ + l )
2
kcm
+ l 2 = ld
2
From eqn. 8.4.9b, k 2 = kcm
+ l2
k 2 = ld
From eqn. 8.4.6, TD = 2π
TD = 2π
k2
gl
d
g
Period of a simple pendulum: T = 2π
8.9
Period of real pendulum: TD = 2π
a
M
I
Mgl
(eqn. 8.4.5)
Where I = moment of inertia
l = distance to CM of physical pendulum
a = distance to CM of bob
b = radius of bob
location of CM of physical pendulum:
(a − b) + M − m a
m
(
)
m a b
m
2
l=
= a−  + = a−
(a + b)
m + ( M − m)
M 2 2
2M
m
m b

−
l = a 1 −
 2M 2M a 
Moment of inertia:
2
2 

I bob = ( M − m ) a 2 + ( M − m ) b 2 = ( M − m )  a 2 + b 2 
5
5 

m

= Ma 2  1 −
 M
I rod
2
 2 b 
1
+

2 
 5 a 
m
1
1
2
= m ( a − b ) = Ma 2 
3
3
 M
2
 b 
1 −  
 a  

m   2 b2  1 m
∴ I = I bob + I rod = Ma  1 −  1 +
+
2 
 M   5 a  3 M
m
b
and β =
letting α =
M
a
2
2
 b 
1 −  
 a  
110
1
2
 2
2
 2 2 1
 Ma (1 − α )  1 + 5 β  + 3 α (1 − β )  





TD = 2π 

 α αβ 


Mga 1 − −


 2 2 
(a)
(b)
8.10
TD
=
T
2
1
2
(1 − α ) 1 + β 2  + α (1 − β )

5
 3
≈ 1−
 α αβ 
1 − −

 2 2 
M = 1kg
a = 1.27m
β = 0.0394
m = 10 g
α = 0.01
TD
1
≈ 1 − α = 0.9992
12
T
1
α to 1st order in α
12
b = 5cm
(actually 0.9994 using complete expression)
The period of the “seconds” pendulum is
I
T2 = 2π
= 2s
Mgl
The period of the modified pendulum is
I′
 n 
T ′ = 2π
= 2

M ′gl ′
 n − 20 
where I ′ , M ′ , l ′ refer to parameters of pendulum with m attached and
n ( = 24 × 60 × 60 ) is the number of seconds in a day.
Ml + mlm
M
where lm is the distance of the attached mass m from the pivot point.
I ′ = I + mlm2
l′ =
−2
π 2 I ′ π 2 I + π 2 mlm2
 20 
So 1 −  =
=
n 
M ′gl ′ ( Ml + mlm ) g

=
Mgl + π 2 mlm2
( Ml + mlm ) g
( Ml + mlm ) g
 20 
Thus 1 −  =
n  ( Mgl + π 2 mlm2 )

2
 α lm 
1 +

40
m
l 
Or 1 −
α=
≈ 
2
2
M
n  π α lm 
1 +

gl 

Solving for α gives the approximate result
111
40 l
n lm
m
α=
≅
2
M  π lm 
− 1

 g

Letting lm = 1.3m ; l = 1.0m ; we obtain α ≅ 1.15 ⋅10−3
8.11
(a) I ⊥cm = ma 2 (all mass in rim)
I ⊥rim = ma 2 + ma 2 = 2ma 2
∴T = 2π
(b)
I z = I x + I y = 2 I &cm = ma 2
∴ I &cm =
I ⊥ rim
2a
= 2π
mga
g
( = I ⊥cm )
ma 2
2
ma 2
3
hence I &rim =
+ ma 2 = ma 2
2
2
I
3a
∴T = 2π &rim = 2π
mga
2g
8.12
G
T
a
G
mg
x
8.13
mxcm = mg − T
I cmω = aT
xcm = aω
2
I cm = ma 2
5
I ω
12
x 
2
mxcm = mg − cm = mg −  ma 2 cm  = mg − mxcm
5
a
a5
a 
5
xcm = g
7
When two men hold the plank, each supports
mg
.
2
When one man lets go:
mg − R = mxcm and R
From table 8.3.1,
I cm =
ω =
l
= I cmω
2
ml 2
12
Rl 12 6 R
⋅
=
2 ml 2 ml
112
l
3R
xcm = ω =
2
m
 3R 
mg − R = m 
 = 3R
 m
mg
R=
4
6 R 6  mg 
= 
xend = lω = l

ml m  4 
3
xend = g
2
8.14
For a solid sphere:
4
2
M s = π a 3 ρ and I s = M s a 2
3
5
2
( kcm2 )s = 5 a 2
For subscript c representing a solid sphere the size of the cavity, from eqn. 8.3.2:
I s = I + Ic
3
2
2  4 3  2 2 4  a   a 
2 4 31
I =  π a ρ  a −  π   ρ    = ⋅ ⋅ π a5 ρ
53
5  3  2    2 
5 3 32

3
4
4 a
4 7
m = π a3 ρ − π   ρ = ⋅ π a3 ρ
3
3 2
3 8
I 2 31 8
31
2
kcm
= = ⋅ ⋅ a2 = a2
70
m 5 32 7
From eqn. 8.6.11, for a sphere rolling down a rough inclined plane:
g sin θ
xcm =
2
 kcm

1+  2 
a 
x
=
xs
(k )
1+
2
cm s
2
a
k2
1 + cm2
a
2
7
1+
5 = 5
=
31 101
1+
70 70
x 98
=
xs 101
113
Energy is conserved:
a
8.15
m2
m1
8.16
2
1
1
1  x 
m1 x 2 + m2 x 2 + I   + m2 gx − m1 gx = E
2
2
2 2
I
+ m2 xx
+ 2 xx
+ m2 gx − m1 gx = 0
m1 xx
a
( m1 − m2 ) g
x=
I
m1 + m2 + 2
a
x
While the cylinders are in contact:
mv 2
b
f r = cm = mg cos θ − R
r
G
2
R
mvcm
r = a + b , so
= mg cos θ − R
vcm
a+b
From conservation of energy:
1 2 1 2
a
mg ( a + b ) = mvcm
+ Iω + mg ( a + b ) cos θ
2
2
G
1
mg
From table 8.3.1, I = ma 2
2
vcm
ω=
a
2

1 2 1  1 2   vcm
mvcm +  ma   2  = mg ( a + b )(1 − cos θ )
2
22
 a 
2
mvcm
4
= mg (1 − cos θ )
a+b 3
When the rolling cylinder leaves, R = 0 :
4
mg cos θ = mg (1 − cos θ )
3
7
4
cos θ =
3
3
4
θ = cos −1
7
8.17
mx = N 2
my = N1 − mg
114
1 2 1 2
mvcm + Iω + mgy = mgyD
2
2
l
l
x = cos θ ,
x = − θ sin θ ,
2
2
l
l
y = sin θ ,
y = θ cos θ ,
2
2
l 2
θ cos θ + θsin θ
2
l
y = −θ 2 sin θ + θ cos θ
2
2
2
2 2
 l
 l
 lθ
2
= x 2 + y 2 =  − θ sin θ  +  θ cosθ  =
vcm
4
 2
 2

x=
(
)
(
)
ml 2
, and ω = θ
12
1 l 2θ 2 1 ml 2 2
l
l
+
θ + mg sin θ = mg sin θ D
m
2
4
2 12
2
2
l 2
θ = g ( sin θ D − sin θ )
3
I=
1
 3g
2
θ =  ( sin θ D − sin θ ) 
 l

−
1
1  3g
3g
 2  3g 
θ =  ( sin θ D − sin θ )    ( − cos θ )θ = − cosθ
2 l
2l
  l 
ml 

 3g 
 3g 
N 2 = mx = −  cos θ   ( sin θ D − sin θ ) + sin θ  −  cos θ 
2 
 l 
 2l 

Separation occurs when N 2 = 0 :
1
2

θ = sin −1  sin θ D 
sin θ D − sin θ − sin θ = 0 ,
2
3

8.18
Rx = mx
Ry − mg = my
l
l
l
x = sin θ , x = θ cos θ , x = −θ 2 sin θ + θ cosθ
2
2
2
l
l
l
y = cos θ , y = − θ sin θ ,
y = − θ 2 cos θ + θsin θ
2
2
2
1 2 1 2
l
l
mvcm + Iθ + mg cos θ = mg
2
2
2
2
2
ml
l
I=
vcm = θ ,
2
12
(
)
(
)
115
m l 2θ 2 1 ml 2 2
l
+
θ = mg (1 − cos θ )
2 4
2 12
2
2
lθ
= g (1 − cos θ )
3
1
 3g
2
θ =  (1 − cosθ ) 
 l

−
1
1  3g
3g
 2  3g 
θ =  (1 − cosθ )    sin θθ = sin θ
2 l
2l
  l 
ml 
3g
3g

Rx =
( − sin θ )   (1 − cosθ ) + cosθ  sin θ  

2 
 l 
 2l

3mg
Rx =
sin θ ( 3cos θ − 2 )
4
ml 
 3g 
 3g

Ry = mg −
cos θ   (1 − cos θ ) + sin θ  sin θ  

2 
 l 
 2l

Ry = mg −
3mg 
sin 2 θ 
2
 cos θ − cos θ +

2 
2 
mg
2
( 3cosθ − 1)
4
The reaction force constrains the tail of the rocket from sliding backward for Rx > 0 :
3cos θ − 2 > 0
2
θ < cos −1
3
The rocket is constrained from sliding forward for Rx < 0 :
2
θ > cos −1
3
Ry =
8.19
mx = −mg sin θ − µ mg cos θ
x = − g ( sin θ + µ cos θ )
x
G
f
G
mg
ω=
5 µ g cos θ
t
2
a
θ
1 2
xt :
Since acceleration is constant, x = xDt + 2
gt 2
x = vDt −
( sin θ + µ cosθ )
2
2
Meanwhile ( µ mg cos θ ) a = Iω = ma 2ω
5
5 µ g cos θ
ω =
2
a
116
The ball begins pure rolling when v = aω …
5 µ g cos θ
v = vD + xt = vD − g ( sin θ + µ cos θ ) t = a
t
2
a
2vD
t=
g ( 2sin θ + 7 µ cos θ )
At that time:
2
2vD2
g 4vD ( sin θ + µ cos θ )
−
x=
g ( 2sin θ + 7 µ cosθ ) 2 g 2 ( 2sin θ + 7 µ cos θ )2
x=
8.20
2vD2 ( sin θ + 6 µ cos θ )
g ( 2sin θ + 7 µ cos θ )2
mx = µ mg
x = µg
x = µ gt , and x =
1
µ gt 2
2
2 2
ma ω = − µ mga
5
5 µg
ω = −
2 a
5 µg
ω = ωD −
t
2 a
Slipping ceases to occur when v = aω …
5
µ gt = aω D − µ gt
2
2 aω D
t=
7 g
Iω =
 2 aω D 
1
x = µg 

2
 7 µg 
x=
2
2 a 2ω D2
49 µ g
 1

 1

Let the moments of inertia of A and B be I a  = M a a 2  and I b  = M bb 2  .
 2

 2

The angular velocity of A is α while that of B is β − α + φ (remember that in two
dimensions, angular velocity is the rate of change of an angle between an line or direction
fixed to the body and one fixed in space). For rolling contact, lengths traveled along the
perimeters of the disks A and B must be equal to the arc length traveled along the track
C.
aφ = b β = ( a + 2b )(α − φ )
8.21
117
so that φ =
( a + 2b )α
2 (a + b)
and β =
a ( a + 2b ) α
2b ( a + b )
After some algebra … the angular velocity of B is found to be …
aα
ω B = β − α + φ =
2b
For A , we take moments about O and for B we take moments about its center.
Call TA and TB the components of the reaction forces tangent to A and B (the
“upward-going” TA acts on disk B . The “downward-going” TA acts on disk A)
Thus
K − TA a = I Aα
(Torque on Disk A)
α
−TAb − TB b = − I B β − α + φ = − I B a
2b
1
TA − TB = M B ( a + b ) α − φ = M B aα
2
Eliminate TA and TB
α
K = 2 ( 4b 2 I A + M B a 2b 2 + a 2 I B )
4b
Integrating this equation gives :
α
Kt = 2 ( 4b 2 I A + M B a 2b 2 + a 2 I B )
4b
Putting ω A = α at t = tD gives
(
(
)
(Torque on Disk B)
)
(Force on Disk B)
4b 2 KtD
ωA =
( 4b2 I A + M B a 2b2 + a 2 I B )
Putting in values for I A and I B gives
4 KtD
ωA =
3


a 2  2M A + M B 
2


Since the angular velocity of B is ω B = β − α + φ =
ωB =
a
ωA =
2b
aα
, we have
2b
2 KtD
3


ab  2 M A + M B 
2


118
From section 8.7 (see Figure 8.7.1), the instantaneous center of rotation is the
l
point O . If x is the distance from the center of mass to O and is the distance from
2
the center of mass to the center of percussion O′ , then from eqn. 8.7.10 …
Ml 2  1  l 2
l I
x   = cm =
 =
12  M  12
2 M
l
x=
6
8.22
8.23 In order that no reaction occurs between the table surface and the ball, the ball
must approach and recede from its collision with the cushion by rolling without slipping.
Using a prime to denote velocity and rotational velocity after the collision:
Pˆ
′ = vcm −
vcm
m
Pˆ
′ = ω cm − ( h − a )
ω cm
I
′ = aω cm
′ .
The conditions for no reaction are vcm = aω cm and vcm
aω cm −


Pˆ
Pˆ
= a ω cm − ( h − a ) 
m
I


1 1
= a (h − a)
m I
2
I = ma 2
5
2ma 2
7
h=
+a = a
5ma
5
d
a=
2
7
h= d
10
8.24
During the collision, angular momentum about the point 0 is conserved:
m′v l ′ = Iθ
D
m′vDl ′
I
After the collision, energy is conserved:
1 2
l
l
Iθ − mg − m′gl ′ = − mg cos θ D − m′gl ′ cos θ D
2
2
2
2 2 2
1 m′ vD l ′
 ml

= g (1 − cos θ D ) 
+ m′l ′ 
I
2
 2

θ =
119
I=
ml 2
+ m′l ′2
3
1
2
2
1 
 ml
  ml
2
′
′
′
′
θ
2
1
cos
−
+
+
vD =
g
m
l
m
l
(
)


D 

m′l ′ 
 2
 3

8.25
The effect of rod BC acting on rod AB is impulse + P̂′ .
The effect of AB on BC is − P̂′ .
mv1 = Pˆ + Pˆ ′
mv2 = − Pˆ ′
l
l
Iω1 = Pˆ − Pˆ ′
Iω 2 = − Pˆ ′
2
2
2
ml
I=
12
6 ˆ ˆ
6
ω1 =
ω 2 = − P̂′
P − P′
ml
ml
l
l
vB = v1 − ω1
vB = v2 + ω 2
2
2
Pˆ + Pˆ ′ l  6  ˆ ˆ
−   P − P′
vB =
m
2  ml 
Pˆ ′ l  6 ˆ 
vB = − +  −
P′ 
m 2  ml 
Pˆ + Pˆ ′ − 3 Pˆ − Pˆ ′ = − Pˆ ′ − 3Pˆ ′
(
)
(
)
(
(
)
)
8 Pˆ ′ = 2 Pˆ
Pˆ
Pˆ ′ =
4
5Pˆ
4m
9 Pˆ
ω1 =
2ml
Pˆ
4m
3Pˆ
ω2 = −
2ml
v1 =
v2 = −
Pˆ
m
----------------------------------------------------------------------------------------------------------vB = −
120
CHAPTER 9
MOTION OF RIGID BODIES
IN THREE DIMENSIONS
9.1 (a) I xx = ∫ ( y 2 + z 2 ) dm
dm = ρ dxdy and m = 2a 2 ρ
z
I xx = ∫
y=a
y =0
∫
x=2 a
x=0
(y
2
+ 0 ) ρ dxdy
a
= ∫ 2a ρ y 2 dy
0
I xx =
2 4 ma 2
ρa =
3
3
y
I yy = ∫ ( x 2 + z 2 ) dm = ∫
2a
x
a
=∫
a
0
y=a
y =0
∫
x=2 a
x=0
x 2 ρ dxdy
8a 3 ρ
dy
3
8a ρ 4ma
=
3
3
From the perpendicular axis theorem:
5ma 2
I zz = I xx + I yy +
3
4
I yy =
2
I xy = I yx = − ∫ xydm = − ∫
= −∫
a
0
y =a
y =0
∫
x=2a
x =0
xy ρ dxdy
4a 2 ρ
ydy
2
ma 2
2
I xz = I zx = − ∫ xzdm = 0 = I yz = I zy
I xy = I yx = − ρ a 4 = −
2
1
, cos β =
, cos γ = 0
5
5
From equation 9.1.10 …
ma 2  4  4ma 2  1   ma 2   2   1 
I=

 +
  + 2 −
3 5
3  5   2   5   5 
2
= ma 2
15
ω ˆ
G
(c) ω = ω iˆ cos α + ˆj cos β =
2i + ĵ
5
From equation 9.1.29 …
(b) cos α =
(
)
(
)
121
G
 2ω ma 2 ω  ma 2  
 2ω  ma 2  ω 4ma 2 
ˆ
L = iˆ 
j
⋅
+
−
+
⋅




−
+
3
2
2
3 
5
5
5
5







G ω ma 2
L=
iˆ + 2 ˆj
6 5
(d) From equation 9.1.32:
1 G G
ω ω ma 2
1
T = ω ⋅L =
⋅
( 2 + 2 ) = ma 2ω 2
2
15
2 5 6 5
(
)
(a)
9.2
G
ω=
ω ˆ ˆ ˆ
i + j+k
3
(
)
m ( 2a )
2
I xx = 2 I rod = 2
= ma 2 = I yy = I zz
12
3
I xy = − ∫ xydm = 0 since, for each rod, either x or
2
a
-a
a
-a
a
-a
y or both are 0. The same is true for the other
products of inertia.
From equation 9.1.29:
G 2
ω ˆ ˆ ˆ
L = ma 2 ⋅
i + j+k
3
3
From equation 9.1.32,
1 ω 2ma 2ω 2 2 2
ma 2ω 2
T=
(1 + 1 + 1 ) = 3
2 3 3 3
(b)
From equation 9.1.10, with the moments of
(
)
2ma 2
and the products of inertia equal to 0:
3
2ma 2
2
I=
cos 2 α + cos 2 β + cos 2 γ ) = ma 2
(
3
3
(c)
For the x-axis being any axis through the center of the
lamina and in the plane of the lamina, and the y-axis also in
the plane of the lamina …
I xx = I yy due to the similar geometry of the mass
inertia equal to
distributions with respect to the x- and y-axes.
From the perpendicular axis theorem:
I zz = I xx + I yy
I zz = 2 I xx
From Table 8.3.1, I zz =
I xx =
ma 2
6
ma 2
12
122
9.3 (a) From equation 9.2.13, tan 2θ =
2 I xy
I xx − I yy
ma 2
4ma 2
ma 2
, I yy =
, I xy = −
3
3
2
tan 2θ = 1
2θ = 45D θ = 22.5D
The 1-axis makes an angle of 22.5D with the x-axis.
(b) From symmetry, the principal axes are parallel to the sides of the lamina and
perpendicular to the lamina, respectively.
From Prob. 9.1, I xx =
9.4 (a) From symmetry, the coordinate axes are principal axes.
From Table 8.3.1:
z
G
ω
13
m
2
2
I1 = ( 2a ) + ( 3a )  = ma 2
 12
12 
10
m
2
I 2 =  a 2 + ( 3a )  = ma 2


12
12
5
m
2
y
I 3 =  a 2 + ( 2a )  = ma 2


12
12
ω
G
ω=
( eˆ1 + 2eˆ2 + 3eˆ3 )
14
x
From equation 9.2.5,
2
2
1 13 2 ω 10 2 4ω 2 5
2 9ω 
T =  ma ⋅
+ ma ⋅
+ ma ⋅

2 12
14 12
14 12
14 
=
7
ma 2ω 2
24
(b) From equation 9.2.4,
G
ω 
2ω 
3ω 
 13
 10
 5
+ eˆ2  ma 2 ⋅
+ eˆ3  ma 2 ⋅
L = eˆ1  ma 2 ⋅



14 
14 
14 
 12
 12
 12
G ma 2ω
L=
(13eˆ1 + 20eˆ2 + 15eˆ3 )
12 14
G G
(1)(13) + ( 2 )( 20 ) + ( 3)(15 ) 
ω ⋅L
cos θ =
= 
1
1
ω L
(12 + 22 + 32 ) 2 ⋅ (132 + 202 + 152 ) 2
=
98
(11,116 )
1
2
= 0.9295
θ = 21.6D
9.5 (a) Select coordinate axes such that the axis of the rod is the 3-axis, its center is at the
123
G
origin, and ω lies in the 1, 3 plane.
ml 2
From Table 8.3.1, I1 = I 2 =
, I3 = 0
12
G
ω = ω ( eˆ1 sin α + eˆ3 cos α )
3
G
ω
α
2
1
(b)
From equation 9.2.4,
G ml 2
ml 2ω
ω ( eˆ1 sin α + 0 + 0 ) = eˆ1
sin α
L=
12
12
G
G ml 2ω
sin α
L is perpendicular to the rod, and L =
12
G
Since ω is constant, from equations. 9.3.5 …
N1 = 0 + 0
 ml 2 
N 2 = 0 + (ω cos α )(ω sin α ) 

 12 
N3 = 0 + 0
G
ml 2ω 2
sin α cos α
N = eˆ2
12
G
G
N is perpendicular to the rod ( ê3 direction) and to L ( eˆ1 direction), and
G ml 2ω 2
N =
sin α cos α
12
ω
G
9.6 From Problem 9.4 … ω =
( eˆ1 + 2eˆ2 + 3eˆ3 )
14
13
10
5
I1 = ma 2 , I 2 = ma 2 , and I 3 = ma 2
12
12
12
From eqns. 9.3.5:
ω2
ma 2
ma 2ω 2
N1 = 0 +
( 2 )( 3)
( 5 − 10 ) = −
( 5)
14
12
28
ω2
ma 2
ma 2ω 2
3
1
13
−
5
=
N2 = 0 +
( )( )
(
)
( 4)
14
12
28
ω2
ma 2
ma 2ω 2
N3 = 0 +
(1)( 2 )
(10 − 13) = −
14
12
28
1
2 2
G ma 2ω 2 2
ma ω
5 + 42 + 12 ) 2 =
42
N =
(
28
28
9.7 Multiplying equations 9.3.5 by ω1 , ω 2 , and ω3 ,respectively …
0 = I1ω1ω1 + ω1ω 2ω 3 ( I 3 − I 2 )
0 = I 2ω 2ω 2 + ω1ω 2ω3 ( I1 − I 3 )
124
0 = I 3ω 3ω3 + ω1ω 2ω 3 ( I 2 − I1 )
Adding, 0 = I1ω1ω1 + I 2ω 2ω 2 + I 3ω 3ω3
d 1

I1ω12 + I 2ω 22 + I 3ω 32 ) 
(

dt  2

d
From equation 9.2.5, 0 = [Trot ]
dt
Trot = constant
Multiplying equations 9.3.5 by I1ω1 , I 2ω 2 , and I 3ω 3 , respectively:
0=
0 = I12ω1ω1 + I1ω1ω 2ω 3 ( I 3 − I 2 )
0 = I 22ω 2ω 2 + I 2ω1ω 2ω 3 ( I1 − I 3 )
0 = I 32ω 3ω 3 + I 3ω1ω 2ω 3 ( I 2 − I1 )
Adding, 0 = I12ω1ω1 + I 22ω 2ω 2 + I 32ω 3ω3
1 d 2 2
0=
I1 ω1 + I 22ω 22 + I 32ω32 )
(
2 dt
d
From equation 9.2.4, 0 = ( L2 )
dt
2
L = constant
9.8 From equations 9.3.5 for zero torque …
0 = I1ω1 + ω 2ω 3 ( I 3 − I 2 )
0 = I 2ω 2 + ω1ω3 ( I1 − I 3 )
From the perpendicular axis theorem, I 3 = I1 + I 2
0 = I1ω1 + ω 2ω3 I1
0 = I 2ω 2 − ω1ω 3 I 2
Multiplying by
ω1
I1
and
ω2
I2
, respectively:
0 = ω1ω1 + ω1ω 2ω 3
0 = ω 2ω 2 − ω1ω 2ω 3
Adding, 0 = ω1ω1 + ω 2ω 2 =
1 d
ω12 + ω 22 )
(
dt
2
ω12 + ω 22 = constant
If I1 = I 2 , from the third of Euler’s equations … ( I 3ω 3 = 0 )
ω3 = constant
9.9 (a) From symmetry … I s = I 3 and I1 = I 2 = I
From the perpendicular axis theorem, I 3 = I1 + I 2 or I s = 2 I
From eqn 9.5.8, Ω = ( 2 − 1) ω cos α
125
For α = 45D , Ω =
ω
2
2π
= 2 s = 1.414 s
Ω
ω
G
T1 is the period of precession of ω about ê3 .
From equation 9.6.12,
3
For
G
ω
45o
2π
= 1s , T1 =
 2

I

 1  
− 1 cos 2 α  = ω 1 +  2 − 1 
 
1
2

 


I





φ = ω 1 + 

φ = ω
2
1
1
2
2
s
2
2
2
1
2
5
2
2π
T2 = = 0.4 s = 0.632 s
φ
G
T2 is the period of wobble of ê3 about L .
I1 = I 2 = I =
(b)
m  2 a 2  17 ma 2
a +  = ⋅
12 
16  16 12
m 2
ma 2
2
2
a
+
a
=
⋅
(
) 12
12


 2
 1 15 ω
From equation 9.5.8, Ω = 
− 1 ω
=
2 17 2
 17 
 16 
2π 17
=
T1 =
2 s = 1.603s
Ω 15
I3 = I s =
1
  22 ⋅162   1  2  2
− 1 
From equation 9.6.12, φ = ω 1 + 
 
2
17
  2  
 
φ = 1.5072 ω
2π
1
T2 = =
s = 0.663s
φ 1.5072
9.10
From equation 9.6.10, tan θ =
I
tan α .
Is
G
Since I s > I , θ < α and the angle between the axis of rotation (ω ) and the
G
invariable line L is α − θ .
( )
From your favorite table of trigonometric identities …
tan α − tan θ
tan (α − θ ) =
1 + tan α tan θ
126

I 
1 −  tan α
Is 
tan (α − θ ) = 
I
1 + tan 2 α
Is
 ( I s − I ) tan α 

2
 I s + I tan α 
α − θ = tan −1 
Is
a maximum.
I
I
tan α
For s = 2 , tan (α − θ ) =
2 + tan 2 α
I
d tan (α − θ )
1
2 tan 2 α
2 − tan 2 α
=
−
=
2 + tan 2 α ( 2 + tan 2 α )2 ( 2 + tan 2 α )2
d tan α
9.11
α − θ is a maximum for
d tan (α − θ )
= 0 = 2 − tan 2 α
d tan α
−1
α = tan 2 = 54.7D
 1
1 −  2
2
2
=
tan (α − θ ) ≤ 
2
1
4
1+
2
2
 2
D
α − θ ≤ tan −1 
 = 19.5
4


At the maximum,
( )
G
G
9.12 (a) From Problem 9.10, for I s > I , the angle between ω and L is …
 ( I s − I ) tan α 

2
 I s + I tan α 
α − θ = tan −1 
From Problem 9.9(a), I s = 2 I and tan α = tan 45D = 1
1
1
α − θ = tan −1
= tan −1 = 18.4D
3
( 2 + 1)
ma 2
17 ma 2
and I = ⋅
12
16 12
  17  
  2 − 16  
  = tan −1 15 = 17.0D
−1 
α − θ = tan 
49
 2 + 17 

16 
(b) From Prob. 9.9(b), I s = 2 ⋅
127
Is
= 1.00327
I
G
G
It follows from equation 9.6.10 that, for I s > I , the angle between ω and L is:
From Example 9.6.2, α = 0.2′′ and
9.13
I

tan α 
 Is

α − θ = α − tan −1 
For α so small,
I
 I
Iα
tan α ≈ α , and tan −1  tan α  ≈ tan α ≈
Is
 Is
 Is
Iα I s − I
α −θ ≈ α −
=
α
Is
Is
.00327
0.2 = 0.00065 arcsec
α −θ ≈
1.00327
From Table 8.3.1, I s =
9.14
3
G
ω0
G
ω
1
ma 2
ma 2
and I =
2
4
G
maω
G
ωD = ω ê3 and Pˆ =
eˆ3
4
Selecting the 2-axis through the point of
impact, during the collision …
G
maω
ma 2ω
G G
∫ Ndt = r × pˆ = aeˆ2 × 4 eˆ3 = 4 eˆ1
G
The only non-zero component of N is N1 .
2
During the collision ω 2 = 0 , so integrating
G
P
the first of Euler’s equations 9.3.5 …
∫ N1dt = I1ω1
Immediately after the collision,
ma ω  4 

 =ω
4  ma 2 
( ω3 still is equal to ω , and ω 2 = 0 )
G
G
After the collision, N = 0 . From Prob. 9.8, with I1 = I 2 , for N = 0 ,
ω1 =
2
ω3 = constant and ω12 + ω 22 = constant .
(ω
tan α =
2
1
1
+ ω 22 ) 2
ω3
= constant
Using the values of ωi immediately after the collision …
(ω
tan α =
2
1
+ 0) 2
ω
=1
α = 45D
128
G G
From equation 9.5.8, with ω = ω D + ω1eˆ1 = ω ( eˆ1 + eˆ3 ) :
Ω = ( 2 − 1)
(
)
2ω ( cos 45D ) = ω
From equation 9.6.12 …
1
φ = 2ω 1 + ( 22 − 1)( cos 2 45D )  2 = ω 5
9.15
G
G
N = −cω = −c (ω1eˆ1 + ω 2 eˆ2 + ω 3eˆ3 )
Say the 3 axis is the symmetry axis, I 3 = I s , I1 = I 2 = I
For the third component of equations 9.3.5 …
−cω3 = I sω 3 + 0
3
G
ω
ω 3
c
=−
α
ω3
Is
ω
c
ln 3 = − t
Is
(ω 3 )D
2
ct
Is
For the first two components of equations
9.3.5 …
−cω1 = Iω1 + ω 2ω 3 ( I s − I ) , and
G
N
1
ω 3 = (ω 3 )D e
−
−cω 2 = Iω 2 + ω1ω 3 ( I − I s )
Rearranging terms and multiplying by ω1 and ω 2 respectively …
c
ω1ω1 + ω12 + ω1ω 2ω 3 ( I s − I ) = 0
I
c
ω 2ω 2 + ω 22 − ω1ω 2ω 3 ( I s − I ) = 0
I
c
Adding, ω1ω1 + ω 2ω 2 + (ω12 + ω 22 ) = 0
I
1
d
2c
ω12 + ω 22 ) = −
(
2
2
I
(ω1 + ω 2 ) dt
ln
ω12 + ω 22
2c
=− t
2
2
(ω1 + ω 2 )D I
(ω
2
1
+ ω 22 ) = (ω12 + ω 22 ) e
(ω12 + ω
−
2 ct
I
D
1
2 2
2
)
(ω
tan α =
2
1
1
ct
= (ω12 + ω 22 ) 2 e I
D
+ω
ω3
1
2 2
2
)
= ( tan α D ) e
1 1 
− ct  − 
 I Is 
129
1 1
For I s > I ,  −
 I Is

 > 0 and α decreases with time.

ma 2
9.16 (a) From table 8.3.1… I s =
about the symmetry (spin) axis.
2
a2
G
I
=
m
about an axis through
cm disk
S
4
z
the center of mass of the disk and ⊥
a
to the spin axis. From the parallel
axis theorem, equation 8.3.21, …
2
ma 2
a
I disk =
+ m   about the pivot
4
2
point.
2
ma
moment of inertia of
I rod =  
2 3
the rod about the pivot point. Thus …
o
a/2
45
y
x
I = I disc + I rod
2
2
 ma 2
2
a  ma
=
+ m    +   = ma 2
 2    2  3 3
 4
From equation 9.7.10, φ =
I s S + ( I S − 4 MglI cos θ )
2
s
2
1
2
2 I cos θ
1


2
2


I
I
cos
Mgl
1
4
θ




2
s
s
φ =
 
 S ±   S −
I
2 cos θ  I
 I 
 


2
ma
Is
1
3
= 2 2 = , cosθ = cos 45D =
2ma
I
4
2
3
m  a 

m +  

9
Ml 
2  2  9
=
=
= cm −1
2
2ma
8a 80
I
3
1

2
2 2


1
3
3
9
cm
1
60
rpm
2


 


 
−1  
φ =
 ( 900 rpm ) ±   ( 900 ) − 4  cm   980 2  


−1   
s   2   2π s   
2 4
 80

 4 


φ = 15.1rpm or 939 rpm
130
(b) From equation 9.7.12, I s2 S 2 ≥ 4MglI
 3m   a 
I 4 Ml  2   2  3
3
= ,
=
=
=
cm −1
2
ma
Is 3 Is
2a 20
2
4Mgl I
 3

 4   60 rpm 
S ≥
⋅ = 4  cm −1  ( 980 cm ⋅ s −2 )   
−1 
Is
Is
 20

 3   2π s 
S ≥ 267 rpm
2
2
9.17
(Neglecting the pencil head) From Table 8.3.1,
2
b
m 
mb 2
2

=
Is =
2
8
The moment of inertia of the pencil about an axis thru its
center of mass and ⊥ to its symmetry axis from equation
 b 

 
2
 b2 a 2 
a
2
8.3.26 ... I c = m    +  = m  + 
 4
12 
 16 12 




2
b
a
2
 b2 a 2 
a
From the parallel axis theorem, I = I c + m   = m  + 
2
 16 3 
4MglI
From equation 9.7.12, S 2 ≥
I s2
2
2
a b a 
4mg   m  + 
 2   16 3 
S2 ≥
m 2b 4
64
1
1
16  ga  b 2 a 2   2 16  ( 980 )( 20 )  12 202   2
−1
S ≥ 2   +  = 2 
 +
  rad ⋅ s
b  2  16 3  
3 
(1)  2
 16
S ≥ 18, 294 rad ⋅ s −1 = 2910 rps
9.18
I s = I rim + I spokes + I hub
ma 2
2
3
a
ma 2
I spokes = mspokes
=
, assuming the spokes to be thin rods
3
12
I hub = 0 , assuming its radius is small
I rim = mrim a 2 =
131
From the perpendicular axis theorem, equation 8.3.14, I =
From equation 9.10.14, S 2 >
Is
2
Imga
I s ( I s + ma 2 )
1

2
1


1
6
mga
g
2
 =
S>


2
 2  7 ma 

 19 a 
2
 
 + ma 
  12 

For rolling without slipping, v = aS
1

 30   2
1
6 × 32 ×   

 6 ga  2
 2   ft ⋅ s −1 = 3.55 ft ⋅ s −1
v>
 =
 19 
 19 ×12 


ma 2
If the spokes and hub are neglected, I s =
2
1

2
1


2
1
mga
g


 =
S>


2
 2  ma 

 3a 
2
 
 + ma 
  2 

1

 30   2
1
32
×
 
 ga  2 
 2   ft ⋅ s −1 = 3.65 ft ⋅ s −1

v>  =
 3 
 3 × 12 


G
9.19 From equations 9.3.5 with N = 0 …
I1ω1 + ( I 3 − I 2 ) ω 2ω 3 = 0
I 2ω 2 + ( I1 − I 3 ) ω1ω 3 = 0
I 3ω 3 + ( I 2 − I1 ) ω1ω 2 = 0
Differentiating the first equation with respect to t:
I1ω1 + ( I 3 − I 2 )(ω 2ω 3 + ω 2ω3 ) = 0
From the second and third equations:
(I − I )
(I − I )
ω 2 = 3 1 ω1ω 3 , and ω 3 = 1 2 ω1ω 2
I2
I3
132
(I − I )

(I − I )
I1ω1 + ( I 3 − I 2 )  1 2 ω1ω 22 + 3 1 ω1ω 32  = 0
I2
 I3

 ( I − I )( I − I ) 
 ( I − I )( I − I ) 
ω1 + K1ω1 = 0 , K1 = −ω 22  3 2 2 1  + ω 32  3 2 3 1 
I1 I 3
I1 I 2




(a) For ω3 large and ω 2 = 0 , K1 > 0 so ω1 + K1ω1 = 0 is the harmonic oscillator
equation. ω1 oscillates, but remains small. Motion is stable for initial rotation
about the 3 axis if the 3 axis is the principal axis having the largest or smallest
moment of inertia.
(b) For ω3 = 0 and ω 2 large, K1 < 0 so ω1 + K1ω1 = 0 is the differential equation for
kt
− kt
exponential growth of ω1 with time: ω1 = Ae 1 + Be 1 . Motion is unstable
for the initial rotation mostly about the principal axis having the median moment
of inertia.
9.20 I xy = ∑ mi xi yi = 0 since either xi or yi is zero for all six particles. Similarly, all the
i
other products of inertia are zero. Therefore the
coordinate axes are principle axes.
2
2
I xx = ∑ mi ( yi2 + zi2 ) = m 0 + 0 + b 2 + ( −b ) + c 2 + ( −c ) 


i
(0,0,c)
I xx = 2m ( b 2 + c 2 )
(0,b,0)
I yy = 2m ( a 2 + c 2 )
I zz = 2m ( a 2 + b 2 )
(a,0,0)
b 2 + c 2
I

I = 2m  0
 0

ω
G
ω=
(a + b + c
2
2
1
2 2
)
( aeˆ1 + beˆ2 + ceˆ3 ) =
G IG
From equation 9.1.28, L = I ω
b 2 + c 2
G
2mω

0
L=
1 
2
2
2 2
( a + b + c )  0
0
a +c
0
2
0
a2 + c2
0
ω
1
( a 2 + b2 + c2 ) 2


0 
a 2 + b 2 
0
a 
b 
 
 c 
 a 
 
0  b 
2
a + b 2   c 
0
2
133
 a ( b2 + c 2 )


G
2mω
2
2


a
c
L=
b
+
(
)
1
2
2
2 2 
( a + b + c ) c ( a 2 + b2 )


G
1 G
From equation 9.1.32, T = ω ⋅ L
2
a ( b2 + c 2 )


1
2mω 2
b ( a 2 + c 2 ) 
T=
a
b
c
[
]
2 ( a 2 + b2 + c2 )


c ( a 2 + b 2 ) 


2
mω
 a 2 ( b2 + c 2 ) + b2 ( a 2 + c 2 ) + c 2 ( a 2 + b2 )
T= 2
2
2 

a +b +c
2mω 2
T= 2
a 2b 2 + a 2 c 2 + b 2 c 2 )
2
2 (
a +b +c
9.21 For Problem 9.1:
 1
 3

I
1
2
I = ma −
 2

 0


1
2
4
3
−
0

0

0,


5

3
 2
ω  
1
ω=
5 
G
0
 
1
 1

 3 − 2 0

 2
G I G ma 2ω  1 4
 
L = Iω =
0 1
−


2
3
5

  0 
5
 0

0


3

2 1
1
 3−2 
6
1


 
2
2
ma ω 
4  ma ω  1  ma 2ω  
2
=
−1 +
=
=
3
5 
5  3  6 5  


 
0
 0 
 0 


 
1
1 G G 1 ω ma 2ω
T = ω ⋅L =
( 2 1 0 )  2 
2
2 5 6 5
0
 
=
ma 2ω 2
ma 2ω 2
2
+
2
+
0
=
(
)
60
15
134
For Problem 9.4:
 13 0 0 
1
I ma 2 
ω  
G

I =
0 10 0  ,
2
ω=
12 
14  

 0 0 5
 3
13 0 0   1 
 13 
2
G I G ma 2ω 
   ma ω  
L = Iω =
0 10 0   2  =
20
12 14 
12 14  



 0 0 5 3
 15 
 13 
1 G G 1 ω ma 2ω
T = ω ⋅L =
(1 2 3)  20 
2
2 14 12 14
 15 
 
=
ma 2ω 2
7
(13 + 40 + 45) = ma 2ω 2
24 (14 )
24
9.22 Since the coordinate axes are axes of symmetry, they are principal axes and all
products of inertia are zero.
G
From Table 8.3.1,
z
ω
m
m
2
2
I xx = ( 2b ) + ( 2c )  = ( b 2 + c 2 )
 3
12 
m
m
I yy = ( a 2 + c 2 ) ,
I zz = ( a 2 + b 2 )
3
3
2
2
y
b + c
0
0 
2c
I m

0 
I =  0
a2 + c2
3
2a
 0
0
a 2 + b 2 

x
2b
a 
ω
ω
G
b 
ω=
aeˆ1 + beˆ2 + ceˆ3 ) = 2
1 (
2
2  
(a + b + c ) c 
( a 2 + b2 + c 2 ) 2
 
G IG
From equation 9.1.28, L = I ω
b 2 + c 2
0
0  a 
G
mω


2
2
0
0   b 
L=
a +c
1 
3 ( a 2 + b 2 + c 2 ) 2  0
0
a 2 + b 2   c 
a ( b2 + c 2 )


G
mω
2
2

L=
b ( a + c )
1


3 ( a 2 + b2 + c2 ) 2 
2
2 
c ( a + b ) 
1 G G
From equation 9.1.32, T = ω T ⋅ L
2
135
a ( b2 + c 2 )


1
mω 2
b ( a 2 + c 2 ) 
T=
a
b
c
[
]
2 3 ( a 2 + b2 + c2 )


c ( a 2 + b 2 ) 


T=
mω 2
 a 2 ( b 2 + c 2 ) + b 2 ( a 2 + c 2 ) + c 2 ( a 2 + b 2 )
2
2
2 

6(a + b + c )
T=
mω 2
a 2b 2 + a 2 c 2 + b 2 c 2 )
2
2
2 (
3( a + b + c )
With the origin at one corner, from the parallel axis theorem:
m ( b2 + c 2 )
4m 2 2
I xx =
+ m (b2 + c2 ) =
(b + c )
3
3
4m 2 2
4m 2
I yy =
a +c ),
I zz =
(
( a + b2 )
3
3
I xy = − ∫ xydm = − ∫ xy ρ dV
I xy = − ρ ∫
x=2a
x =0
∫
y =2b
y =0
∫
z =2c
z =0
xydxdydz = −8 ρ a 2b 2 c
m = ρ ( 2a )( 2b )( 2c ) = 8 ρ abc , so I xy = − mab
I xy = − mac ,
I yz = − mbc
4 2 2
 3 (b + c )

I
I = m  −ab


 − ac

−ab
4 2 2
(a + c )
3
−bc
9.23 (See Figure 9.7.1)
Lz = ( Lx′ ) z + ( Ly′ ) z + ( Lz ′ ) z
(



−bc 


4 2
2 
+
a
b
(
)
3
− ac
)
Lz = Ly′ sin θ + Lz′ cos θ = Iφ sin θ sin θ + ( I s S ) cos θ
9.24 (See Figure 9.7.1) If the top precesses without nutation, it must do so at θ = θ D
where V (θ D ) is a minimum of V (θ ) …
dV (θ )
=0
dθ θ =θ
D
( L − Lz′ cosθ )
V (θ ) = z
2
2 I sin θ
2
+ mgl cos θ
(See equation 9.8.7)
136
dV
dθ
− cosθ D ( Lz − Lz′ cos θ D ) + Lz′ sin 2 θ D ( Lz − Lz ′ cosθ D )
=
− mgl sin θ D = 0
I sin 3 θ D
2
θ =θ D
let γ = Lz − Lz′ cos θ D
Then γ 2 cos θ D − γ Lz ′ sin 2 θ D + mglI sin 4 θ D = 0 and solving for γ
1




Lz ′ sin θ D 
4mglI cos θ D 2 
γ=
1 ± 1 −
 
2 cos θ D  
L2z′
 


now Lz ′ is large since ψ is large and the precession rate is small, so we can expand
the term in square root above and use the (-) solution since γ must be positive …
2

2mglI cos θ D 
1 − 1 +

L2z′


2
mglI sin θ D
γ = Lz − Lz′ cosθ D ≈
Lz′
From equations 9.7.2, 9.7.5 and 9.7.7 …
Lz = Iφ 2 sin 2 θ 0 + I s φ cos θ 0 + ψ cosθ 0
L cos θ = I (φ cosθ + ψ ) cosθ
γ≈
Lz ′ sin 2 θ D
2 cos θ D
(
z′
0
s
)
0
and …
so …
0
γ = ( I sin 2 θ D + I s cos 2 θ D ) φ + I s ψ cosθ D − I s φ cos 2 θ D − I s ψ cosθ D
= Iφ sin 2 θ D ≈
mglI sin 2 θ D
mglI sin 2 θ D
=
Lz′
I s ψ + φ cos θ D
(
)
and since ψ >> 0 , we can ignore
the φ term in the denominator and we have …
mgl
φ ≈
I sψ
mgl
= 0 and φ
=
the top will precess without nutation
Hence, if ψ large, θ
θ =θ1
θ =θ1
I sψ
at θ1 = θ D the place where V (θ ) = min .
9.25
a
b1 + b2 = 3  
2
1
b1
= sin 30D =
b2
2
a
a
b1 =
b2 =
2 3
3
a 2 = H 2 + b22
a 2 2a 2
H = a −b = a − =
3
3
2
2
2
2
2
137
2
a
3
Thus, the coordinates of the 4 atoms are:
H=
( 0, 0, H )
Oxygen:
a  
a 

 −b1 , , 0  ;  −b1 , − , 0 
2  
2 

( b2 , 0, 0 )
Hydrogen:
Carbon:
(a) The axes 1, 2, 3 are principal axes if the products of inertia are zero.
a 

 −a  
− I xy = ∑ mi xi yi = 0 +  −b1  +  −b1 
 + 0 ≡ 0
2 

 2 
− I yz = 0 + 0 + 0 + 0 ≡ 0
− I xz = 0 + 0 + 0 + 0 ≡ 0
The 1,2,3 axes are principal axes.
(b) Find principal moments
2
2
a
 a  67 2
I1 = I xx = ∑ mi ( y + z ) = 16 H + 1  + 1  =
a
6
2
2
89
I 2 = I yy = ∑ mi ( xi2 + zi2 ) = 16 H 2 + 1( b12 ) + 1( b12 ) + 12b22 = a 2
6
2
2

14
a  
a 
I 3 = I zz = ∑ mi ( xi2 + yi2 ) = 0 +  b12 +    +  b12 +    + b22 = a 2




3
2  
2 

2
i
 67
 6

I = 0


 0


2
i
2

0

89
0  a2

6

14 
0

3
0
(c) I 3 < I1 < I 2 therefore rotation about the1-axis is unstable (see discussion Sec. 9.4)
----------------------------------------------------------------------------------------------------------
138
CHAPTER 10
LAGRANGIAN MECHANICS
10.1
x ( t ) = x ( 0, t ) + αη ( t )
Solution …
x ( t ) = x ( 0, t ) + αη ( t )
where x ( 0, t ) = sin ω t and x ( 0, t ) = ω cos ω t
1 2
1
1
mx
V = kx 2 = mω 2 x 2
2
2
2
t2
m
J (α ) = ∫ ( x 2 − ω 2 x 2 ) dt
2 t1
T=
so:
t
=
m 2
2
2
(ω cos ω t + αη ) − ω 2 ( sin ω t + αη )  dt
∫

2 t1
t
t
t
2
m2
α 2m 2 2
J (α ) = ∫ ω 2 ( cos 2 ω t − sin 2 ω t ) dt + mαω ∫ (η cos ω t − ωη sin ω t ) dt +
(η − ω 2η 2 ) dt
∫
2 t1
2
t1
t1
Examine the term linear in α :
t2
∫ (η cos ω t − ωη sin ω t ) dt = η ( t ) cos ω t
t1
t2
t1
t2
t2
t1
t1
+ ∫ ωη sin ω tdt − ∫ ωη sin ω tdt ≡ 0
(1 term vanishes at both endpoints: η ( t2 ) = η ( t1 ) = 0 )
st
t
so
t
2
2
1
1
J (α ) = mω 2 ∫ cos 2ω tdt + mα 2 ∫ (η 2 − ω 2η 2 ) dt
2
2
t1
t1
t
2
1
1
= mω [sin 2ω t2 − sin 2ω t1 ] + mα 2 ∫ (η 2 − ω 2η 2 ) dt
4
2
t1
which is a minimum at α = 0
10.2
V = mgz
1
T = m ( x 2 + y 2 + z 2 )
2
1
L = T − V = m ( x 2 + y 2 + z 2 ) − mgz
2
∂L
∂L
∂L
= mx ,
= my ,
= mz
∂x
∂y
∂z
d  ∂L 
  = mx ,
dt  ∂x 
d  ∂L 
  = my ,
dt  ∂y 
d  ∂L 
  = mz
dt  ∂z 
139
∂L ∂L
=
= 0,
∂x ∂y
∂L
= −mg
∂z
∂L d  ∂L 
− 
=0
∂qi dt  ∂qi 
mx = const
my = const
From equations 10.4.5 …
mx = 0 ,
my = 0 ,
mz = − mg
10.3 Choosing generalized coordinate x as linear displacement down the inclined
plane (See Figure 8.6.1), for rolling without slipping …
x
ω=
a
2
1
1
1
12
7
  x 
T = mx 2 + Iω 2 = mx 2 +  ma 2    = mx 2
2
2
2
25
  a  10
For V = 0 at the initial position of the sphere,
V = −mgx sin θ
7
L = T − V = mx 2 + mgx sin θ
10
d  ∂L  7
∂L 7
= mx ,
  = mx
dt  ∂x  5
∂x 5
∂L
= mg sin θ
∂x
7
mx = mg sin θ
5
5
x = g sin θ
7
5
xcm = g sin θ
From equations 8.6.11 - 8.6.13 … 7
10.4 (a) For x the distance of the hanging block below the edge of the table:
1
1
T = mx 2 + mx 2 = mx 2 and V = −mgx
2
2
L = T − V = mx 2 + mgx
d  ∂L 
∂L
= 2mx ,
  = 2mx
∂
x
dt
 ∂x 
x
∂L
= mg
∂x
2mx = mg
g
x=
2
140
1
m′ 
m′g 2

x  x
(b) T = mx 2 + m′x 2 =  m +  x 2 and V = −mgx −  m′  g = −mgx −
x
2
2 
2l

l  2
m′ 
m′g 2

L = T − V =  m +  x 2 + mgx +
x
2 
2l

d  ∂L 
∂L
x
= ( 2m + m′ ) x ,
  = ( 2m + m′ ) dt  ∂x 
∂x
m′g
∂L
x
= mg +
l
∂x
m′g
x
( 2m + m′) x = mg +
l
g  ml + m′x 
x= 

l  2m + m′ 
The four masses have positions:
m1 :
x1 + x2
m2 :
l1 − x1 + x2
m3 :
l2 − x2 + x3
m4 :
l2 − x2 + l3 − x3
10.5
V = − g  m1 ( x1 + x2 ) + m2 ( l1 − x1 + x2 )
+ m3 ( l2 − x2 + x3 ) + m4 ( l2 − x2 + l3 − x3 ) 
1
2
2
m1 ( x1 + x2 ) + m2 ( − x1 + x2 )
2
2
2
+ m3 ( − x2 + x3 ) + m4 ( − x2 − x3 ) 

T=
L = T −V =
1
1
1
2
2
2
m1 ( x1 + x2 ) + m2 ( − x1 + x2 ) + m3 ( − x2 + x3 )
2
2
2
1
2
+ m4 ( − x2 − x3 ) + gx1 ( m1 − m2 ) + gx2 ( m1 + m2 − m3 − m4 ) + gx3 ( m3 − m4 ) + const.
2
∂L
= m1 ( x1 + x2 ) − m2 ( − x1 + x2 )
∂x1
= ( m1 + m2 ) x1 + ( m1 − m2 ) x2
d ∂L
= ( m1 + m2 ) x1 + ( m1 − m2 ) x2
dt ∂x1
∂L
= g ( m1 − m2 )
∂x1
( m1 + m2 ) x1 + ( m1 − m2 ) x2 = g ( m1 − m2 )
141
∂L
= m1 ( x1 + x2 ) + m2 ( − x1 + x2 ) − m3 ( − x2 + x3 ) − m4 ( − x2 − x3 )
∂x2
∂L
= g ( m1 + m2 − m3 − m4 )
∂x2
( m1 − m2 ) x1 + ( m1 + m2 + m3 + m4 ) x2 + ( m4 − m3 ) x3 = g ( m1 + m2 − m3 − m4 )
∂L
= m3 ( − x2 + x3 ) − m4 ( − x2 − x3 )
∂x3
∂L
= g ( m3 − m4 )
∂x3
( m4 − m3 ) x2 + ( m3 + m4 ) x3 = g ( m3 − m4 )
m1 = m , m2 = 4m , m3 = 2m , and m4 = m :
3
5mx1 − 3mx2 = −3mg ,
x1 = ( x2 − g )
5
−3mx1 + 8mx2 − mx3 = 2mg
1
− mx2 + 3mx3 = mg ,
x3 = ( x2 + g )
3
Substituting into the second equation:
9
9
1
1
− x2 + g + 8
x2 − x2 − g = 2 g
5
5
3
3
88
8
g
x2 =
x2 = g ,
15
15
11
3  10 
6
x1 =  − g  = − g
5  11 
11
1  12  4
x3 =  g  = g
3  11  11
For
Accelerations:
m1 :
m2 :
m3 :
m4 :
5
g
11
7
− x1 + x2 = g
11
3
− x2 + x3 = g
11
5
− x2 − x3 = − g
11
x1 + x2 = −
10.6 See figure 10.5.3, replacing the block with a ball. The square of the speed of the
ball is calculated in the same way as for the block in Example 10.5.6.
′ cos θ
v 2 = x 2 + x ′2 + 2 xx
The ball also rotates with angular velocity ω so …
142
T=
1 2 1 2 1
mv + Iω + Mx 2
2
2
2
x ′
2
. I = ma 2
5
a
1
1
1
′ cosθ ) + mx ′2 + Mx 2
T = m ( x 2 + x ′2 + 2 xx
2
5
2
′
V = −mgx sin θ , for V = 0 at the initial position of the ball.
1 
7
 1
′ cos θ  + Mx 2 + mgx′ sin θ
L = T − V = m  x 2 + x ′2 + 2 xx
2 
5
 2
d  ∂L  7
∂L 7
= mx ′ + mx cos θ ,

 = mx′ + mx cos θ
∂x′ 5
dt  ∂x ′  5
∂L
= mg sin θ
∂x′
7
mx′ + mx cos θ = mg sin θ
5
5
x′ = ( g sin θ − x cos θ )
7
d  ∂L 
∂L
= mx + mx ′ cos θ + Mx ,
x + mx′ cos θ
  = ( m + M ) ∂x
dt  ∂x 
∂L
=0
∂x
( m + M ) x + mx′ cosθ = 0
For rolling without slipping, ω =
5
5
mg sin θ cos θ − mx cos 2 θ = 0
7
7
5mg sin θ cos θ
x=
5m cos 2 θ − 7 ( m + M )
( m + M ) x +
10.7 Let x be the slant height of the particle …
v 2 = x 2 + x 2ω 2
1
1
T = mv 2 = m ( x 2 + x 2ω 2 )
2
2
x
V = mgx sin θ = mgx sin ω t
1
L = T − V = m ( x + x 2ω 2 ) − mgx sin ω t
θ = ωt
2
d  ∂L 
∂L
= mx ,
  = mx
dt  ∂x 
∂x
∂L
= mxω 2 − mg sin ω t
∂x
mx = mxω 2 − mg sin ω t
x − ω 2 x = − g sin ω t
143
The solution to the homogeneous equation
x − ω 2 x = 0 , is x = Aeω t + Be −ω t
Assuming a particular solution to have the form x p = C sin ω t ,
−ω 2C sin ω t − ω 2C sin ω t = − g sin ω t
g
C=
2ω 2
g
x = Aeω t + Be −ω t +
sin ω t
2ω 2
At time t = 0 , x = xD and x = 0
xD = A + B
g
0 = ω A −ωB +
2ω
1
g 
A =  xD − 2 
2
2ω 
1
g 
B =  xD +

2
2ω 2 
g 1
g
1


x = xD  ( eω t + e −ω t )  − 2  ( eω t − e −ω t )  +
2
2
 2ω  2
 2ω sin ω t
From Appendix B, we use the identities for hyperbolic sine and cosine to obtain
g
g
x = xD cosh ω t − 2 sinh ω t +
sin ω t
2ω
2ω 2
10.8 In order that a particle continues to move in a plane in a rotating coordinate system,
it is necessary that the axis of rotation be perpendicular to the plane of motion.
G
For motion in the xy plane, ω = k̂ω .
G G G G
ˆ + ˆjy + kˆω × ix
ˆ + ˆjy
v = v′ + ω × r ′ = ix
G
v = iˆ ( x − ω y ) + ˆj ( y + ω x )
(
)
(
)
1 G G m 2
mv ⋅ v = ( x − 2 xω y + ω 2 y 2 + y 2 + 2 yω x + ω 2 x 2 )
2
2
L = T −V
d  ∂L 
∂L
= m ( x − ω y ) ,
x − ω y ) ,
  = m ( ∂x
dt  ∂x 
∂L
∂V
= m ( yω + ω 2 x ) −
∂x
∂x
∂V
m ( x − ω y ) = m ( yω + ω 2 x ) −
∂x
2
Fx = m ( x − 2ω y − ω x )
T=
∂L
= m ( y + ω x ) ,
∂y
d  ∂L 
y + ω x ) ,
  = m ( dt  ∂y 
144
∂L
∂V
= m ( − xω + ω 2 y ) −
∂y
∂y
m ( y + ω x ) = m ( − xω + ω 2 y ) −
Fy = m ( y + 2ω x − ω 2 y )
∂V
∂y
G
G
For comparison, from equation 5.3.2 ( A0 = 0 and ω = 0 )
G
G
G G
G G G
F = ma′ + 2mω × v′ + mω × (ω × r ′ )
G
ˆ + ˆjy + 2m ω kˆ × ix
ˆ + ˆjy + m ω kˆ × ω kˆ × ix
ˆ + ˆjy 
F = m ix


2
Fx = m ( x − 2ω y − ω x )
(
)
(
)
(
)
Fy = m ( y + 2ω x − ω 2 y )
10.9 Choosing the axis of rotation as the z axis …
G G G G
ˆ + ω kˆ × ix
ˆ
ˆ + ˆjy + kz
ˆ + ˆjy + kz
v = v′ + ω × r ′ = ix
G
ˆ
v = iˆ ( x − ω y ) + ˆj ( y + ω x ) + kz
(
)
(
)
1 G G m 2
mv ⋅ v = ( x − 2 xω y + ω 2 y 2 + y 2 + 2 yω x + ω 2 x 2 + z 2 )
2
2
L = T −V
d  ∂L 
∂L
= m ( x − ω y )
x − ω y ) ,
  = m ( ∂x
dt  ∂x 
∂L
∂V
= m ( yω + ω 2 x ) −
∂x
∂x
∂V
m ( x − ω y ) = m ( yω + ω 2 x ) −
∂x
2
Fx = m ( x − 2ω y − ω x )
T=
∂L
d  ∂L 
= m ( y + ω x ) ,
y + ω x )
  = m ( ∂y
dt  ∂y 
∂L
∂V
= m ( − xω + ω 2 y ) −
∂y
∂y
∂V
m ( y + ω x ) = m ( − xω + ω 2 y ) −
∂y
Fy = m ( y + 2ω x − ω 2 y )
d  ∂L 
∂L
∂L
∂V
= mz ,
= mz ,
=−


dt  ∂z 
∂z
∂z
∂z
∂V
mz = −
= Fz
∂z
G
G
From equation 5.3.2 ( A0 = 0 and ω = 0 ) …
145
G
G
G G
G G G
F = ma′ + 2mω × v′ + mω × (ω × r ′ )
G
ˆ + 2mω kˆ × ix
ˆ + mω kˆ × ω kˆ × ix
ˆ 
ˆ + ˆjy + kz
ˆ + ˆjy + kz
ˆ + ˆjy + kz
F = m ix


2
Fx = m ( x − 2ω y − ω x )
(
)
(
)
(
)
Fy = m ( y − 2ω x − ω 2 y )
Fz = m
z
10.10
1
m r 2 + r 2θ 2
2
1
2
V = k ( r − lD ) − mgr cos θ
2
m 2 2 2 k
2
L = T −V =
r + r θ − ( r − lD ) + mgr cos θ
2
2
d  ∂L 
∂L
= mr ,
  = mr
dt  ∂r 
∂r
∂L
= mrθ 2 − k ( r − lD ) + mg cos θ
∂r
(
T=
r
)
(
θ
mr = mr 2 − k ( r − lD ) + mg cos θ
)
∂L
∂L
= mr 2θ
= −mgr sin θ
∂θ
∂θ
d
mr 2θ = − mgr sin θ
dt
(
)
a
a
( 2θ + sin 2θ ) y = (1 − cos 2θ )
4
4
at θ = 0
x& y = 0
1
T = m ( x 2 + y 2 )
V = mgy
2
a
aθ
x = θ + θ cos 2θ =
(1 + cos 2θ )
2
2
a
y = θ sin 2θ
2
2 2
ma θ 
mga
2
L = T −V =
(1 + cos 2θ ) + sin 2 2θ  −
(1 − cos 2θ )

8
4
ma 2θ 2
mga
=
(1 − cos 2θ )
[1 + 2 cos 2θ + 1] −
8
4
ma 2θ 2
mga
cos 2 θ  −
sin 2 θ 
=
2
2 
where we used the trigonometric identies … 2 cos 2 θ = 1 + cos 2θ and 2sin 2 θ = 1 − cos 2θ
10.11 (See Example 4.6.2)
x=
(
)
146
s = a sin θ
so
s = aθ cos θ
ms2 mg 2 1 2 1 2
mg
L=
−
s = ms − ks where k =
a
2
2a
2
2
The equation of motion is thus
k
g
s + s = 0 or s+ s=0
-a simple harmonic oscillator
m
a
Let
Coordinates:
10.12
x = a cos ω t + b sin θ
y = a sin ω t − b cos θ
x = − aω sin ω t + bθ cos θ
y = aω cos ω t + bθ sin θ
L = T −V =
1
m ( x 2 + y 2 ) − mgy
2
m 2 2
 a ω + b 2θ 2 + 2bθaω sin (θ − ω t )  − mg ( a sin ω t − b cos θ )
2
d ∂L
= mb 2θ + mbaω θ − ω cos (θ − ω t )
dt ∂θ
∂L
= mbθaω cos (θ − ω t ) − mgb sin θ
∂θ
∂L d ∂L
−
= 0 is
The equation of motion
∂θ dt ∂θ
a
g
θ − ω cos (θ − ω t ) + sin θ = 0
b
b
Note – the equation reduces to equation of simple pendulum if ω → 0 .
=
(
)
Coordinates:
x = l cos ω t + l cos (θ + ω t )
10.13
y = l sin ω t + l sin (θ + ω t )
x = −ω l sin ω t − θ + ω l sin (θ + ω t )
(
(
)
)
y = ω l cos ω t + θ + ω l cos (θ + ω t )
2
1
1
m ( x 2 + y 2 ) = ml 2 ω 2 + θ + ω  + ...


2
2
... ml 2ω θ + ω sin ω t ( sin θ cos ω t + sin ω t cos θ ) + cos ω t ( cos θ cos ω t − sin ω t sin θ ) 
(
L =T =
(
)
)
147
2
1 2 2
ml ω + θ + ω  + ml 2ω θ + ω cos θ


2
∂L d ∂L
−
=0
∂θ dt ∂θ
ml 2 θ − ωθ sin θ + ml 2ω θ + ω sin θ = 0
θ + ω 2 sin θ = 0
(
=
(
(a)
(b)
)
)
(
(
)
)
The bead executes simple harmonic motion (θ > 0 ) about a point diametrically
opposite the point of attachment.
(c)
The effective length is " l " =
g
ω2
G
10.14 v = ˆjat + lθ iˆ cos θ + ˆj sin θ
(
)
= ilˆ θ cos θ + ˆj ( at + lθ sin θ )
1 G G m
T = mv ⋅ v = ( l θ cos θ + a t
2
2
2
2
2
2 2
+ 2atlθ sin θ + l 2θ 2 sin 2 θ
)
1

V = mg  at 2 − l cos θ 
2

 at 2

m 2 2
− l cosθ 
L = T −V =
l θ + a 2t 2 + 2atlθ sin θ − mg 
2
 2

(
)
d  ∂L 
∂L
2 = ml 2θ + matl sin θ ,
  = ml θ + mal sin θ + matlθ cos θ
∂θ
dt  ∂θ 
∂L
= matlθ cos θ − mgl sin θ
∂θ
ml 2θ + mal sin θ + matlθ cos θ = matlθ cosθ − mgl sin θ
a+g
θ +
sin θ = 0
l
For small oscillations, sin θ ≈ θ
a+g
θ +
θ =0
l
2π
l
TD =
= 2π
ω
a+g
1 2 1 2
mx + Iθ
2
2
1
12

T = mx 2 +  ma 2 θ 2
2
25

V = −mgx
10.15 (a)
T=
and
I=
2 2
ma
5
148
1 2 1 2 2
mx + ma θ + mgx
2
5
T
x
The equation of constraint is …
f ( x,θ ) = x − aθ = 0
The 2 Lagrange equations with multipliers are …
∂L d ∂L
∂f
−
+λ
=0
x
dt
x
x
∂
∂
∂
θ = θt
∂L d ∂L
∂f
−
+λ
=0
∂θ dt ∂θ
∂θ
∂L
d ∂L
∂L
∂f
= mx
= mx
= mg
λ =λ
dt ∂x
∂x
∂x
∂x
mg − mx + λ = 0
and from the θ-equation …
mg
2
− ma 2θ − λ a = 0
5
x
and substituting into the above …
Differentiating the equation of constraint … θ =
a
5
2
x = g and λ = − mg
7
7
(b)
The generalized force that is equivalent to the tension T is …
∂f
2
Qx = λ
= λ = − mg
∂x
7
L = T −V =
G
10.16 For v′ the velocity of a differential mass element, dm, of the spring at a distance
x′ below the support …
m′
G x′ G
dm = dx′
v′ = v ,
x
x
2
m′ 1
x 1  x′  m′
1 2
1 2
2
T = mv + ∫
( v′) dm = mx + ∫0  x  dx′
0 2
2
2
2 x  x
1
1 m′ 2
T = mx 2 +
x
2
2 3
m′
1
2
V = k ( x − l ) − mgx − ∫ gx′dm′
0
2
x
1
m′
2
= k ( x − l ) − mgx − ∫ gx′ dx′
0
x
2
′
1
m
2
V = k ( x − l ) − mgx − gx
2
2
′
1
m 
1
m′ 
2

L = T − V =  m +  x 2 − k ( x − l ) +  m +  gx
2
3 
2
2 

∂L 
m′ 
d  ∂L  
m′ 
=  m +  x ,
=  m +  x,


∂x 
dt  ∂x  
3 
3
149
For
∂L
m′ 

= −k ( x − l ) +  m +  g
∂x
2 

m′ 
m′ 


x = −k ( x − l ) +  m +  g
 m +  3 
2


g
m′ 
y = x −l − m+ ,
k
2 
m′ 

y + ky = 0
 m +  3 

The block oscillates about the point x = l +
g
m′ 
m+  …
k
2 
with a period …
m′ 

m+ 

2π
3 
TD =
= 2π 
ω
k
10.17 Note: 4 objects move – their coordinates are labeled xi : The coordinate of the
movable, massless pulley is labeled x p .
Two equations of constraint:
f1 ( x1 , x p ) = x p − ( l − x1 ) = 0
f 2 ( x2 , x3 , x p ) = ( x2 + x3 ) − ( 2 x p + l ′ ) = 0
1
1
1
m1 x12 − m1 gx1 + mx22 − m2 gx2 + m3 x32 − m3 gx3
2
2
2
∂f
∂L d ∂L
−
+ ∑λj j = 0
∂qi dt ∂qi
∂qi
j
L = T −V =
Thus: (1)
− m1 g − m1 x1 + λ1 = 0
− m2 g − m2 x2 + λ2 = 0
(2)
− m3 g − m3 x3 + λ2 = 0
(3)
Now – apply Lagrange’s equations to the movable pulley – note m p → 0
So:
(4)
λ1
∂f1
∂f
+ λ2 2 = 0 or λ1 − 2λ2 = 0
∂x p
∂x p
Now - x p can be eliminated between f1 and f 2
f 2 = 2 f1 = x2 + x3 + 2 x1 − ( 2l + l ′ ) = 0
(5)
Thus x2 + x3 + 2 x1 = 0
With a little algebra – we can solve (1) –(5) for the 5 unknowns x1 , x2 , x3 , λ1 , and λ2
150
2g
λ1 =
λ2 =
 1 1 1
1 
+
 + 

 m1 4  m2 m3  
As the check, let m2 = m3 = m and m1 = 2m .
g
 1 1 1
1 
+
 + 

 m1 4  m2 m3  
Thus, there is no acceleration and λ1 = 2mg
10.18 (See Example 5.3.3)
(a)
r = reˆr
ˆr + rθeˆθ
r = re
Constraint:
f (θ ) = θ − ω t = 0
so … θ = ω and θ = 0
1
T = m r 2 + r 2θ 2 = L
2
∂L d ∂L
∂L d ∂L
∂f
−
= 0 and
−
+λ
=0
∂r dt ∂r
∂θ dt ∂θ
∂θ
r = ω 2r
−2mrrθ − mr 2θ + λ = 0
r = Aeωt + Be −ω t
r ( 0) = 0
(
)
ωt
r = ω Ae − ω Be
−ω t
A=
r ( 0 ) = ω l
λ = 2mrrω
so … A + B = 0
ω A − ω B = ωl
2A = l
apply constraint
l
2
l ω t −ω t
(e − e )
2
r = l sinh ω t 

r = ω l cosh ω t 
B=−
l
2
thus .. r =
now r = l at t = T so
(b)
λ = 2mω 2l 2 sinh ω t cosh ω t
T=
1
ω
sinh −1 1 =
0.88
ω
There are 2 ways to calculate F …
(i) See Example 5.3.3 …
F = 2mω x ′
l
ωl
and
x ′ = ( e ω t + e −ω t )
x ′ = ( e ω t − e −ω t )
2
2
2
F = 2mω l cosh ω t
∂f
(ii) λ
is the generalized force, in this case – a torque T, acting on the bead
∂θ
∂f
T = rF = λ
∂θ
151
λ ∂f 2mω 2l 2 sinh ω t cosh ω t
=
r ∂θ
l sinh ω t
2
= 2mω l cosh ω t
F=
10.19 (See Example 4.6.1) The equation of constraint is
m 2 2 2
V = mgr cos θ
r + r θ
2
m 2 2 2
L=
r + r θ − mgr cos θ
2
∂L d ∂L
∂f
−
+λ
=0
∂r dt ∂r
∂r
∂f
∂f
=1
=0
∂r
∂θ
T=
θ
r
a
Thus
∂L d ∂L
∂f
=0
−
+λ
∂θ dt ∂θ
∂θ
mrθ 2 − mg cosθ − mr + λ = 0
mgr sin θ − mr 2θ − 2mrrθ = 0
f ( r ,θ ) = r − a = 0
(
)
(
)
Now r = a , r = r = 0 so
2
maθ − mg cos θ + λ = 0
mga sin θ − ma 2θ = 0
g
sin θ
and
a
g
so
∫ θdθ = a ∫ sin θ dθ or
hence,
λ = mg ( 3cosθ − 2 )
θ =
and when λ → 0
dθ
dθ
2
θ
g
g
= − cosθ +
2
a
a
θ = θ
particle falls off hemisphere at
2
θ D = cos −1  
3
10.20 Let x2 mark the location of the center of curvature of the movable surface relative
to a fixed origin. This point defines the position of m2. r1 marks the position of the
particle of mass m1. r is the position of the particle relative to the movable center of
curvature of m2.
1
1
Kinetic energy T = m2 x22 + m1r1 2
2
2
r1 ⋅ r1 = ( x2 + r ) ⋅ ( x2 + r ) = x22 + r 2 + 2 x2 ⋅ r
but r = reˆr + rθeˆθ
1
1
ˆr + rθeˆθ 
m2 x22 + m1  x22 + r 2 + r 2θ 2 + 2 x2 ⋅ re

2
2
1
1
= m2 x22 + m1  x22 + r 2 + r 2θ 2 + 2 x2 rθ sin θ − 2 x2 r cos θ 
2
2
∴T =
(
)
152
V = − mgy = −mgr sin θ
Potential energy
L = T −V
Equation of constraint f ( r ,θ ) = r − a = 0
∴
∂f
∂f
∂f
=1
=0
=1
∂θ
∂x2
∂r
∂L d  ∂L 
∂f
− 
=0
+λ
∂xi dt  ∂xi 
∂xi
d
x2 − m1 x1 − m1  rθ sin θ − r cosθ  = 0
− m2 x2 )
dt
d
m2 x2 − m1 x1 − m1  rθ sin θ + rθsin θ + rθ 2 cosθ − r cosθ + rθ sin θ  = 0
dt
using constraint
r = a = constant ; r = r =0
m1
x2 = −
a θsin θ + θ 2 cos θ
(1)
m1 + m2
θ − sin θ + x rθ cosθ + gr cosθ = 0
θ)
− r 2θ − 2rr
x r sin θ − x r sin θ − x rθ cosθ + xr
Lagrange’s equations
(
2
)
2
2
2
using constraint
r = a = constant ; r = r =0)
x
g
−θ − 2 sin θ + cos θ = 0
a
a
r)
−
r + x2 cos θ − x2θ sin θ + rθ 2 + x2θ sin θ + g sin θ +
Apply constraint…
x2 cos θ + aθ 2 + g sin θ +
λ
(2)
λ
m1
=0
m1
Now, plug solution for x2 (1) into equation (2):
m1
g
θ −
θsin θ + θ 2 cosθ sin θ − cosθ = 0
m1 + m2
a
g
θ(1 − f m1 sin 2 θ ) − θ 2 f m1 sin θ cosθ − cosθ = 0
a
m1
where f m1 =
m1 + m2
dθ dθ dθ dθ d  1 2 
=
=θ
=
Now θ =
 θ 
dt dθ dt
dθ dθ  2 
So – let x = θ 2
dx
2g
1 − f m1 sin −2 θ  − 2 xf m1 sin θ cosθ −
cosθ = 0
dθ
a
Let
y = 1 − f m1 sin 2 θ
(
y
=0
(3)
)
dx
dy 2 g d ( sin θ )
+x
=
dθ
dθ
a
dθ
153
2g
Hence: d ( yx ) =
d ( sin θ ) and
a
yx (θ =θ )
(
∫
D
yx θ = 0
)
d ( yx ) =
2g
a
sin θ
∫ d ( sin θ )
0
x = θ 2 = 0 at θ = 0D so
2g
yx (θ ) =
sin θ
a
2g
sin θ
x (θ ) = θ 2 =
a 1 − f m1 sin 2 θ 
Now we can solve for θ and plug θ 2 , θ into (3) to obtain λ (θ )
but

d  1 2  g d 
sin θ


 θ =
dθ  2  a dθ 1 − f m1 sin 2 θ 
After some algebra – yields
(1 + fm1 sin 2 θ )
g
θ = cosθ
2
a
(1 − fm1 sin 2 θ )
θ =
The solution for λ (θ ) is thus determined from equation (3)
g
λ
θ 2 − f m θsin θ cos θ + θ 2 cosθ  + sin θ = −
a
m1a
1
collecting terms:
θ 2 (1 − f m cos 2 θ ) − θ f m sin θ cosθ +
1
1
g
λ
sin θ = −
a
m1a
Plug θ , θ into the above --- plus --- a lot of algebra yields …
2
2
λ (θ ) 2 f m2 sin θ − f m1 sin θ cos θ 1 − f m1 sin θ 
−
=
+ sin θ
2
m1 g
1 − f m1 sin 2 θ 
m2
where f m2 =
m1 + m2
As a check, let m2 → ∞ so it is immoveable … then f m2 → 1 and f m1 → 0
and we have
−
λ (θ )
m1 g
→ 3sin θ
… which checks
1 2 1
mv = m R 2 + R 2φ 2 + z 2
2
2
m 2
L = T −V =
R + R 2φ 2 + z 2 − V
2
d  ∂L 
∂L
∂V
∂L
,
= mRφ 2 −
= mR
= mR ,


∂R
∂R
dt  ∂R 
∂R
10.21 (a)
(
T=
(
)
)
154
= mRφ 2 − ∂V
mR
∂R
2
mR − mRφ = QR
d  ∂L 
∂L
∂V
2 =−
  = mR φ + 2mRRφ ,
∂φ
∂φ
dt  ∂φ 
∂V
mR 2φ + 2mRRφ = −
= Qφ
∂φ
d  ∂L 
∂L
∂V
∂L
=−
= mz ,
  = mz ,
∂z
∂z
dt  ∂z 
∂z
∂V
mz = −
= Qz
∂z
G
G
G
For F = ma , using the components of a from equation 1.12.3:
− Rφ 2 ,
FR = m R
Fφ = m 2 Rφ + Rφ , Fz = m
z
∂L
= mR 2φ ,
∂φ
(
)
(
)
From Section 10.2, since R and z are distances, QR and Qz are forces. However, since
φ is an angle, Qφ is a torque. Since Fφ is coplanar with and perpendicular to R ,
Qφ = RFφ
(b)
… and all equations agree.
v 2 = r 2 + r 2θ 2 + r 2φ 2 sin 2 θ
m 2 2 2 2 2 2
L = T −V =
r + r θ + r φ sin θ − V
2
d  ∂L 
∂L
∂L
∂V
= mr ,
= mrθ 2 + mrφ 2 sin 2 θ −
= mr ,


dt  ∂r 
∂r
∂r
∂r
∂V
= Qr
mr − mrθ 2 − mrφ 2 sin 2 θ = −
∂r
d  ∂L 
∂L
∂L
∂V
2 = mr 2θ ,
= mr 2φ 2 sin θ cosθ −
  = mr θ + 2mrrθ ,
∂θ
∂θ
dt  ∂θ 
∂θ
∂V
= Qθ
mr 2θ + 2mrrθ − mr 2φ 2 sin θ cosθ = −
∂θ
∂L
d  ∂L 
2 2
2 2
2 2
φ
sin
θ
,
=
mr
  = mr φ sin θ + 2mrrφ sin θ + 2mr θφ sin θ cosθ ,
∂φ
dt  ∂φ 
∂L
∂V
=−
∂φ
∂φ
sin θ cosθ = − ∂V = Q
mr 2φsin 2 θ + 2mrrφ sin 2 θ + 2mr 2θφ
φ
∂φ
Qr is a force Fr . Qθ and Qφ are torques.
(
)
Since φ is in the xy plane, the moment arm for φ is r sin θ ; i.e. Qφ = r sin θ Fφ .
Qθ = rFθ .
From equation 1.12.14 …
155
(
)
m ( rθ + 2rθ − rφ sin θ cosθ ) = F
cosθ ) = F
m ( rφsin θ + 2rφ sin θ + 2rθφ
m r − rφ 2 sin 2 θ − rθ 2 = Fr
2
θ
φ
The equations agree.
∂V ∂V
∂V
=
= 0 and
= −F .
∂θ ∂φ
∂r
For spherical coordinates, using equation 1.12.12:
v 2 = r 2 + r 2φ 2 sin 2 θ + r 2θ 2
m 2 2 2 2
L = T −V =
r + r φ sin θ + r 2θ 2 − V ( r )
2
d  ∂L 
∂L
∂L
∂V
= mr ,
= mrφ 2 sin 2 θ + mrθ 2 −
= mr ,


dt  ∂r 
∂r
∂r
∂r
2
2
2
mr − mrφ sin θ − mrθ = Fr
d  ∂L 
∂L
2 ∂L = mr 2φ 2 sin θ cosθ
= mr 2θ ,
  = mr θ + 2mrrθ ,
dt  ∂θ 
∂θ
∂θ
2 2
2
mr θ + 2mrrθ − mr φ sin θ cos θ = 0
∂L
= mr 2φ sin 2 θ ,
∂φ
d  ∂L 
2 2
2 2
  = mr φ sin θ + 2mrrφ sin θ + 2mr φθ sin θ cosθ
dt  ∂φ 
∂L
=0
∂φ
sin θ cos θ = 0
mr 2φsin 2 θ + 2mrrφ sin 2 θ + 2mr 2φθ
10.22 For a central field, V = V ( r ) , so
(
)
10.23 Since θ = α = constant, there are two degrees of freedom, r and θ .
vr = r , vφ = rφ sin α
1 2 1
mv = m r 2 + r 2φ 2 sin α
2
2
V = mgr cos α
m 2 2 2 2
L = T −V =
r + r φ sin α − mgr cos α
2
d  ∂L 
∂L
∂L
= mrφ 2 sin 2 α − mg cos α
= mr ,
= mr ,


dt  ∂r 
∂r
∂r
2
2
mr = mrφ sin α − mg cos α
∂L
∂L
2 2
φ
sin
α
,
=0
=
mr
∂φ
∂φ
(
T=
(
)
)
156
d
mr 2φ sin α = 0
dt
mr 2φ sin α = A = constant
(
Say
)
A2
− mg cos α
mr 3
d
dr
1 d 2
r
r = r = r =
2 dr
dt
dr
m
A 2 dr
− mg cos α dr
d ( r 2 ) =
2
m r3
mr 2
A2
=−
− mgr cos α + C
2
2mr 2
The constant of integration C is the total energy of the particle: kinetic
energy due to the component of motion in the radial direction, kinetic energy due to the
component of motion in the angular direction, and the potential energy.
A2
U (r ) =
+ mgr cos α
2mr 2
For φ ≠ 0 , A ≠ 0 , and turning points occur at r = 0
mr =
A2
Then 0 = −
− mgr cos α + C
2mr 2
A2
( mg cos α ) r 3 − Cr 2 + = 0
2m
The above equation is quadratic in r ( A 2 ∝ r 4 ) and has two roots.
l2
=0,
2m
defines the turning points. For the particle to remain on a single horizontal circle, there
2
must be two roots with r = rD . Thus ( r − rD ) divided into the above expression leaves a
10.24 Note that the relation obtained in Problem 10.23, ( mg cos α ) r 3 − Cr 2 +
term that is linear in r.
r + ( 2rD − c )
r 2 − 2rrD + rD2 r 3 − Cr 2
+
A 2 2m
r 3 − 2r 2 rD + rrD2
( 2rD − C ) r 2 − rD2 r
( 2rD − C ) r 2 − 2rD ( 2rD − C ) r + rD2 ( 2rD − C )
 2rD ( 2rD − C ) − rD2  + A 2 2m − rD2 ( 2rD − C )
For the remainder to vanish, both terms must equal zero.
4rD2 − 2rDC − rD2 = 0
3
C = rD
2
157
A2
− 2rD3 + CrD2 = 0
2m
r
A2
= D
2
2mrD
2
And with A = mr 2φ sin α
D
D
1

2
1
φD = 

2
 mrD sin α 
A2
− mg cos α
mr 3
For small oscillations about rD , r = rD + ε .
r = ε
From Prob. 10.23, mr =
−3
1 1 ε
= 1 + 
r 3 rD3  rD 
A 2  3ε
mε = 3 1 −
mrD 
rD
≈
1  3ε 
1 − 
rD3 
rD 

 − mg cos α

2
3A
A2
− mg cos α
mε + 4 ε =
mrD
mrD3
2π
m 2 rD4
2π
T=
= 2π
=
2
ω
φD sin α
3A
1
3
1 2
G G m
x + yA
y + zA
z)
mv + qv ⋅ A = ( x 2 + y 2 + z 2 ) + q ( xA
2
2
d  ∂L 
dA
∂L
= mx + qAx ,
= mx + q x


dt  ∂x 
dt
∂x
dAx
∂A
∂A
∂A
Using the hint,
= x x + y x + z x
dt
∂x
∂y
∂z
∂Ay
 ∂A
∂L
∂A 
= q  x x + y
+ z z 
∂x
∂x
∂x 
 ∂x
∂A
 ∂A
 ∂A
∂A
∂A 
∂A 
mx + q  x x + y x + z x  = q  x x + y y + z z 
∂y
∂z 
∂x
∂x 
 ∂x
 ∂x
10.25 L =
  ∂Ay ∂Ax   ∂Ax ∂Az
mx = q  y 
−
−
 − z 
∂y   ∂z
∂x
  ∂x
G
G
G
G
= q  y ∇ × A − z ∇ × A 

z
y

G G
G
= q v × ∇ × A 

x
(
)
(
(



)
)
158
G G
mx = q v × B
(
)
x
Due to the cyclic nature of the Cartesian coordinates, i.e., iˆ × ˆj = kˆ …
G G
my = q v × B
y
G G
mz = q v × B
z
G
G G
Altogether, mr = q v × B
(
(
(
)
)
10.26 V = mgz
)
1
m ( x 2 + y 2 + z 2 )
2
p
p
p
x = x
similarly, y = y and z = z
m
m
m
T=
∂T
= mx ,
∂x
1
H = T +V =
( px2 + p y2 + pz2 ) + mgz
2m
∂H px
=
= x
∂px m
∂H
px = constant
= 0 = − p x ,
∂x
d
p x = ( mx ) = mx = 0
similarly, p y = constant , or my = 0
dt
∂H p y
=
= y
∂p y m
px =
∂H pz
=
= z
∂pz m
d
∂H
p z = ( mz ) = mz = − mg
= mg = − p z ,
dt
∂z
These agree with the differential equations for projectile motion in Section 4.3.
10.27 (a) Simple pendulum …
V = −mgl cos θ
1 2 2
ml θ
2
p
θ = θ2
ml
T=
∂T
= ml 2θ ,
∂θ
pθ2
− mgl cos θ
H = T +V =
2ml 2
p
∂H
= θ2 = θ
∂pθ ml
∂H
= mgl sin θ = − pθ
∂θ
pθ =
159
(b) Atwood’s machine …
1
1
1 x 2
m1 x 2 + m2 x 2 + I 2 [Includes the pulley]
2
2
2 a
p
x =
I 

 m1 + m2 + 2 
a 

V = −m1 gx − m2 g ( l − x )
p=
T=
I 
∂T 
=  m1 + m2 + 2  x ,
a 
∂x 
p2
− ( m1 − m2 ) gx − m2 gl
I 

2  m1 + m2 + 2 
a 

p
∂H
=
= x
I 
∂p 
 m1 + m2 + 2 
a 

∂H
p = ( m1 − m2 ) g
= − ( m1 − m2 ) g = − p ,
∂x
H = T +V =
(c) Particle sliding down a smooth inclined plane …
1
V = −mgx sin θ
T = mx 2
2
p
∂T
p=
x =
= mx ,
m
∂x
2
p
− mgx sin θ
H = T +V =
2m
∂H p
= = x
∂p m
∂H
p = mg sin θ
= − mg sin θ = − p ,
∂x
10.28
(a)
F
r
L = T (qi , qi ) − T (qi , t )
…note, potential energy is time dependent.
k
∂V
so
F =−
V = ∫ Fdr = − e − β t
r
∂r
1 G G 1
T = mv ⋅ v = m r 2 + r 2θ 2
2
2
(
)
160
∂L
= mr
∂r
∂L
pθ =
= mr 2θ
∂θ
pr =
pr
m
p
θ = θ 2
mr
r =
p2
p2
k
H = ∑ p i qi − L = pr r + pθθ − r − θ 2 − e − β t
2m 2mr
r
Substituting for r and θ …
pθ 2
pr 2
k
+
− e −βt
H=
2
2m 2mr
r
(b) H = T + V … which is time-dependent
(c) E is not conserved
10.29 Locate center of coordinate system at C.M. The potential is independent of the
center of mass coordinates. Therefore, they are ignorable.
(r1,θ1)
(r2,θ2)
H = ∑ p i qi − L
L = T −V
1
1
1
2
= m1 r12 + r12θ12 + m2 r2 2 + r2 2θ2 2 − k ( r1 + r2 − l )
2
2
2
where l is the length of the relaxed spring and k is the
spring constant.
p
∂L
p1r =
= mr1
r1 = 1r
∂r1
m1
p
∂L
p1θ = = mr12θ1
θ1 = 1θ 2
m1r1
∂θ1
… and similarly for m2
(
)
(
)
p 2
p 2
p1r 2
p 2
1
+ 1θ 2 + 2 r + 2θ 2 − k (r1 + r2 − l ) 2
2m1 2m1r1 2m2 2m2 r2 2
Equations of motion …
First, θ1 , θ2 are ignorable coordinates, so p1θ , p2θ are each conserved.
∂H
r1
= − p1r = k ( r1 + r2 − l ) = − m1
∂r1
∂H
r2
= − p 2 r = k ( r1 + r2 − l ) = − m2 ∂r2
p1r = p 2 r
∆p r = 0
∆pr = constant.
The radial momenta are equal and opposite.
H=
161
10.30 L =
1 2 1 2
mx − kx
2
2
t2
t2
t2
1
1

0 = δ ∫ Ldt = ∫ δ Ldt = ∫ δ  mx 2 − kx 2  dt
t1
t1
t1
2
2

0=∫
t2
t1
d
δx
dt
δ x =
∫
t2
t1
( mxδ x − kxδ x ) dt
t2
= ∫ mx
mxδ xdt
t1
t2
d
(δ x )
(δ x ) dt = ∫t1 mxd
dt
Integrating by parts:
∫
t2
t1
= mxδ x
mxδ xdt
t2
t1
t2
− ∫ δ x d ( mx )
t1
δ x = 0 at t1 and t2
d ( mx ) =
∫
t2
t1
d
( mx ) dt = mx dt
dt
t2
mxδ x dx = − ∫ δ x mx dt
t1
0=∫
t2
t1
( −mxδ x − kxδ x ) dt
mx + kx = 0
10.31 (a) L = − m0 c 2 1 −
v2
−V
c2
1
Let γ =
1−
v2
c2
∂L
= γ m0 x ≡ px
This is the generalized momentum for part (b).
∂x
Thus, Lagrange’s equations for the x-component are …
d ∂L ∂L d
∂V
−
= px −
=0
dt ∂x ∂x dt
∂x
… and so on for the y and z components.
(b) H = ∑ vi pi − L
i
p
but … vi = i
γ m0
So … H = ∑
i
H=
pi 2 c 2 m0 c 2
+
+V
γ m0c 2
γ
p 2 c 2 m0 c 2
+
+V
γ m0 c 2
γ
162
H=
(c)
1
p 2 c 2 + m0 2 c 4 ) + V
2 (
γ m0c
Now, if T = γ m0 c 2 then we have …

p 2c 2 
p c + m0 c = m0 c 1 + 2 4 
 m0 c 
 γ 2 m0 2 v 2 c 2 
2 4
2 2
2
= m0 2 c 4  1 +
 = m0 c (1 + γ v c )
2 4
m0 c 

2 2
2 4
2 4



v2 c2 
1
= m0 c  1 +
= m0 2c 4 
= γ 2 m0 2 c 4
2
2 
2
2 
 1− v c 
 1− v c 
Thus …
H = γ m0 c 2 + V = T + V
2 4

m0 c 2
1 v2
2
≈
m
c
1
+
+ ... 

0
2
2
2
1− v c
 2c

1
T ≈ m0 v 2 + m0 c 2
2
------------------------------------------------------------------------------------------------------------
(d)
T=
163
Chapter 11
11.1
(a)
(b)
(c)
k 2 k2
x +
2
x
2
k
At equilibrium,
V ′ = kx − 2
x
x = k1 3
2k 2
V ′′ = k + 3
x
′′
V x = k1 3 = k + 2k = 3k > 0
Stable
V ( x) = kxe −bx
V ′ = ke− bx − bkxe− bx
ke − bx − bkxe− bx = 0
At equilibrium
1
x=
b
V ′′ = −bke −bx − bkxe − bx + b 2 kxe −bx
V ′′ x =1 b = −2bke−1 + bke−1 = −bke−1 < 0
Unstable
V ( x) =
V′ = 0
V ( x) = k ( x 4 − b 2 x 2 )
V ′′ = k ( 4 x 3 − 2b 2 x )
At equilibrium
x = 0, ± b
k ( 4 x 3 − 2b 2 x ) = 0
2
V ′′ = k (12 x − 2b 2 )
2
(d)
V ′′ x =0 = −2kb 2 < 0
Unstable
V ′′ x =±b
Stable
2
= k ( 6b 2 − 2b 2 ) = 4kb 2 > 0
for case (a) ω 2 =
3k
m
for case (c) at x = ± b
11.2
T=
2
2π
ω
ω2 =
= 2π
4kb 2
m
m 2π
s
=
3k
3
T=
2π
ω
= 2π
m
=π s
4kb 2
V ( x, y ) = k ( x 2 + y 2 − 2bx − 4by )
164
∂V
= 2k ( x − b)
∂x
at equilibrium
x=b
and
∂V
= 2k ( y − 2b)
∂y
y = 2b
∂V
∂V
∂ 2V
= 2k
=
2
k
=
0
∂x 2
∂x∂y
∂y 2
k11 = 2k > 0
k12 = k21 = 0
k22 = 2k
2
k11
k21
2
k12
= 4k 2 − 0 > 0
k22
The equilibrium is stabe.
11.3
1
V ( x) = − kx 2
2
dV ( x)
dx
F ( x) = −
= kx = mx = mx
dx
dx
kx dx = mx dx
x
v
x0
0
k 2
v2
x − x02 ) = m
(
2
2
∫ kxdx = ∫ mx dx
12
dx
= k m ( x 2 − x02 )
dt
x
t
dx
=
∫x x 2 − x 2 1 2 ∫0 α dt ; where α = k m
0 (
0)
v=
)
(
ln x + x 2 − x02 − ln x0 = α t
x + x 2 − x02 = x0 eα t
x 2 − x02 = x02 e 2α t − 2 xx0 eα t + x 2
x = x0
eα t + e −α t
= x0 cosh α t
2
11.4
Let the length of the unstretched, elastic cord
be d. Then
2l
y
d = 2 l 2 + y2
1
2
V = k ( d − 2l ) − mgy
2
m
165
V=
(
)
k
4l 2 + 4 y 2 − 8l l 2 + y 2 + 4l 2 − mgy
2
)
(
V = 2k 2l 2 + y 2 − 2l l 2 + y 2 − mgy
The first term, 4kl 2 , is an additive constant to the potential energy, so with appropriate
adjustment of the zero reference point …
)
(
V ( y ) = 2k y 2 − 2l l 2 + y 2 − mgy
−1 2
dV ( y )
= 2k  2 y − 2 yl ( l 2 + y 2 )  − mg


dy
At equilibrium, the above expression is zero, so …
4kly
4ky −
− mg = 0
l 2 + y2
4kly
4ky − mg =
l 2 + y2
16k 2 y 2 − 8kmgy + m 2 g 2 =
16k 2l 2 y 2
l 2 + y2
16k 2 y 4 − 8kmgy 3 + (16k 2l 2 + m2 g 2 − 16k 2l 2 ) y 2 − 8kl 2 mgy + l 2 m2 g 2 = 0
y 4 mg 3 m 2 g 2 2 mg
m2 g 2
=0
−
y
+
y
−
y
+
16k 2l 4
2kl 2
16k 2l 2
l 4 2kl 4
y
mg
and
letting u =
a=
4kl
l
4
3
2 2
u − 2au + a u − 2au + a 2 = 0
11.5
V = mg ( h1 + h2 )
a
h1 = b cos θ
h2 = d sin (θ + ϕ )
h2 = d ( sin θ cos ϕ + cos θ sin ϕ )
φ
h2
d
θ
cos ϕ =
θ
h1
b
bθ
d
sin ϕ =
a
d
h2 = bθ sin θ + a cos ϕ
V = mg ( a + b ) cos θ + bθ sin θ 
V ′ = mg ( a + b )( − sin θ ) + b sin θ + bθ cos θ 
166
V ′ = mg [ − a sin θ + bθ cos θ ]
V ′′ = mg [b cos θ − bθ sin θ − a cos θ ]
V0′′ = mg ( b − a )
Equilibrium …
Stable Unstable
a <b
a >b
11.6
cos θ = 1 −
sin θ = θ −
θ2
2!
+
θ4
4!
− ...
θ3 θ5
− ...
5!

 θ2 θ4



θ3 θ5
V = mg ( a + b )  1 − + − ...  + bθ  θ − + − ...  
2! 4!
3! 5!





a − b 2 a − 3b 4


θ +
θ − ...
V = mg  a + b −
2
24


For a = b,
3!
+
a


V = mg  2a − θ 4 +...
12


mga 3
θ + ... terms in higher order of θ
3
V ′′ = − mga θ 2 + ...
V ′′′ = − 2mga θ + ...
V ′′′′ = − 2mga < 0 ∴ Equilibrium is unstable
V′=−
11.7
a
h2
h1
φθ
θ b
The center of mass (CM) of the hemisphere is
3
a from the flat side (see Equation 8.1.8).
8
The height of CM) above the point of contact
between the two hemispheres is designated
by h2 in the figure. h1 is the height of the
point of contact above the ground.
3


V = mg ( h1 + h2 ) = mg b cos θ + a cos θ − a cos (θ + ϕ )  and aϕ = bθ
8


167

3
bθ  

V = mg ( h1 + h2 ) = mg ( a + b ) cos θ − a cos  θ +  
8
a 



3a  a + b    a + b   
o
V ′ = mg  − ( a + b ) sin θ + 
 sin  
 θ   Equilibrium occurs at θ = 0
a
a
8

 
 

2

3a  a + b 
  a + b  
V ′′ = mg  − ( a + b ) cos θ + 
 cos  
θ 
8  a 
  a   

2

3a  a + b   mg
V0′′= mg  − ( a + b ) + 
( a + b )( 3b − 5a )
 =
8  a   8a

V0′′ > 0 for 3b > 5a
Therefore, the equilibrium is stable for a <
3b
5
11.8
From Problem 11.4, we have
(
V ( y ) = 2k y − 2l l + y
2
2
2
)
1

2 2 


y
2
2
− mgy = 2k  y − 2l 1 + 2   − mgy

l  



1

y
y2 2
1 y2 1 y4
… 1 + 2  =1 +
− 4 + ...
Expanding the square root for small
2
l
l
2
l
8l



y4 
V ( y ) ≈ 2k  y 2 − 2l 2 − y 2 + 2  − mgy
4l 

V′ ≈
2k 3
y − mg
l2
1
at equilibrium, V ′ = 0
V ′′ =
 mgl 2  3
⇒ y =

 2k 
6k 2
y
l2
2
V ′′ y = mgl 2
(
2
1
6k  mgl 2  3
mg  3
1
3 
=
=
6
k



2 
2 k )3
l  2k 
 2l 
1
2
1
1
V ′′
 k 3  g  3
 g 3  k 6
= 6    = 6    
ω=
m
 m   2l 
 2l   m 
168
11.9
From Problem 11.5, V0′′ = mg ( b − a )
V0′′
= g (b − a )
m
2π
2π
T0 =
=
ω
g (b − a )
ω=
11.10
mg
( a + b )( 3b − 5a )
8a
m
2a
T0 = 2π
= 4π
g ( a + b )( 3b − 5a )
V0′′
From Problem 11.7, V0′′ =
11.11
b
θ
1 2 1 2
T = mvcm
+ Iϕ
2
2
vcm = ( b − a ) θ
a
φ
2
I = ma 2
5
The relationship between the angles, θ and ϕ (see Figure), can be determined from the
condition that there is no slipping as the ball rolls in the hemisphere, so the length of roll
measured along the ball, a (θ + ϕ ) , must equal the length of roll measured along the
hemisphere, bθ … so we have …
b θ = a (θ + ϕ ) and ∴ b θ = a θ + ϕ
(
)
b−a and … ϕ = 
θ
 a 
2 2
1
1 2 2 ( b − a ) θ 1
2
2
2
T = m ( b − a ) θ 2 +
ma
= m ( b − a )  1 +  θ 2
2
a
2
25
2
 5
V = − mg ( b − a ) cos θ
V ′ = mg ( b − a ) sin θ
equilibrium @ θ = 0o
169
V ′′ = mg ( b − a ) cos θ
mg ( b − a )
V0′′
5g
=
=
7
2
M
7 (b − a )
m (b − a )
5
ω=
T0 =
∴V0′′ = mg ( b − a )
2π
ω
7 (b − a )
5g
= 2π
11.12
The potential energy of the satellite shaped like a “thin rod” is (See Example 11.2.2,
Figure 11.2.1) …
M e dm
m
but dm = dx where 2a is the length of the rod.
2a
r
a
GM e m
GM e m a dx
V =∫ −
dx = −
−a
2a ∫− a r
r 2a
V = ∫ −G
1
r = ( r0 2 + x 2 + 2r0 x cos φ ) 2
1
1
r = ( r0 2 + x 2 ) 2 (1 + ε cos φ ) 2 where ε =
2xr0
r0 2 + x 2
1
GM e m a (1 + ε cos φ ) 2 dx
V =−
1
2a ∫− a
2
2 2
r
+
x
(0 )
−
For r0 >> x, r0 2 + x 2 ≈ r0 2 ,
ε≈
1
2x
and r ≈ r0 (1 + ε cos φ ) 2
r0
Thus, for small ε , the expression for the potential energy, V, can be approximated …
2

GM e m a  1  2 x 
3  2x 
2

 dx
−
+
1
cos
cos
V ≈−
φ
φ





2ar0 ∫− a  2  r0 
8  r0 


2
3
GM e 
3cos φ 2a 
V ≈−
 2a + 0 +

3 
2ar0 
2r0 2
GM e  a 2 cos 2 φ 
V ≈−
1 +

r0 
2r0 2 
GM e  a 2 
GM e a 2
=
V′≈
2
cos
φ
sin
φ
sin 2φ


r0  2r0 2 
2r03
170
Equilibrium @ φ = 0
GM e a 2
cos 2φ
r03
1
M = I = ma 2
3
V ′′ ≈
V0′ ≈
GM e a 2
r03
V0′′
3GM e
=
M
r03
ω=
T0 =
and
2π
ω
= 2π
r03
3GM e
11.13
The amplitude of the symmetric component is A1 and the amplitude of the anti-symmetric
component is A2 (See Equations 11.3.19a through 11.3.20b).
1
2
A12 = [ x1 (0) + x2 (0)]
4
A0
1
A1 = [ x1 (0) + x2 (0) ] =
2
2
1
2
A2 2 = [ x2 (0) − x1 (0) ]
4
A0
1
∴ A1 = A2
A2 = [ x2 (0) − x1 (0) ] =
2
2
From Equation 11.3.18 the solution for x1 is …
A0
x1 (t ) =
( cos ω1t + cos ω2t ) (The phase δ 2 is 180o, which insures that x1 (0) = A0 )
2
A0 
(ω1 + ω2 ) t cos (ω1 − ω2 ) t 
x1 (t ) =
 2 cos

2 
2
2

ω + ω2
ω − ω2
and ∆ = 1
ω= 1
Letting …
2
2
x1 (t ) = A0 ( cos ω t cos ∆t )
From Equation 11.3.18, the solution for x2 is …
A0
x2 (t ) =
( cos ω1t − cos ω2t )
2
A0 
(ω1 + ω2 ) t sin (ω1 − ω2 ) t 
x2 (t ) =
 2sin

2 
2
2

x2 (t ) = A0 ( sin ω t sin ∆t )
171
11.14
At time t = 0 and short times thereafter … cos ∆t ≈ 1 and sin ∆t ≈ 0 . Thus, …
x1 ≈ A0 cos ω t and x2 ≈ 0 . This situation occurs again when ∆t = 2π .
1
1


1  k + 2 k ′  2  k  2 
∆=
= 
 − 
2
2  m   m  


1
1


1  k  2  2 k ′  2 
∆ =   1 +
−
1


2  m  
k 


ω2 − ω1
1
k′
1  2k ′ 
 2k ′  2
for k ′ << k , 1 +
 −1 ≈ 1+ 
 + ... − 1 =
k 
2 k 
k

1
1  k 2  k′ 
∆=    
2m  k 
1
2π
 m  2 2k 2π  2k 
= 2π  
=
T=
 
∆
 k  k ′ ω1  k ′ 
 2k 
T = T1  
 k′ 
11.15
1
T = ml 2 θ12 + θ2 2
2
(
)
V = mgl (1 − cos θ1 ) + (1 − cos θ 2 )  −
l
θ1
θ2
θ1 =θ 2 = 0
r = r0 + l sin θ 2 − l sin θ1
kl cos θ1
∂V
= mgl sin θ1 −
2
∂θ1
( r0 + l sin θ 2 − l sin θ1 )
2kl 2 cos 2 θ1
kl sin θ1
∂ 2V
= mgl cos θ1 −
−
2
3
∂θ12
( r0 + l sin θ 2 − l sin θ1 ) ( r0 + l sin θ 2 − l sin θ1 )
r
∂ 2V
k11 = 2
∂θ1
k
r
2kl 2
= mgl − 3
r0
2kl 2 cos θ1 cos θ 2
∂ 2V
=
∂θ 2 ∂θ1 ( r0 + l sin θ 2 − l sin θ1 )3
172
∂ 2V
k12 =
∂θ1∂θ 2
θ1 =θ 2 = 0
∂ 2V
=
∂θ 2 ∂θ1
θ1 =θ 2 = 0
2kl 2
= 3
r0
kl cos θ 2
∂V
= mgl sin θ 2 +
2
∂θ 2
( r0 + l sin θ 2 − l sin θ1 )
2kl 2 cos 2 θ 2
kl sin θ 2
∂ 2V
cos
=
−
−
θ
mgl
2
2
3
∂θ 2 2
( r0 + l sin θ 2 − l sin θ1 ) ( r0 + l sin θ 2 − l sin θ1 )
k22 =
∂ 2V
∂θ 2 2
= mgl −
θ1 =θ 2 = 0
2kl 2
r03
Thus, from Equation 11.3.37a or b, we have
1
V =  k11θ12 + 2k12θ1θ 2 + k22θ 2 2 
2
But from Equation 11.3.9, for the coupled oscillator, we have
1
1
1
V = kx12 + k ′ ( x12 − 2 x1 x2 + x2 2 ) + kx12
2
2
2
The forms of the potential energy function are similar with …
k11 = k + k ′ = k22 and k12 = − k ′
In the case here …
2kl 2
k = mgl and k ′ = − 3
r0
-----------------------------------------------------------------------------------------
173
Chapter 11 (continued)
11.16
1
1
T = m1 x12 + m2 x2 2
2
2
1
1
1
2
V = k1 x12 + k ′ ( x2 − x1 ) + k2 x2 2
2
2
2
L = T −V
d ∂L
∂L
x1
= m1 = − k1 x1 + k ′ ( x2 − x1 )
dt ∂x1
∂x1
m1 x1 + k1 x1 − k ′ ( x2 − x1 ) = 0
d ∂L
∂L
x2
= m2 = − k2 x2 − k ′ ( x2 − x1 )
dt ∂x2
∂x2
m2 x2 + k2 x2 + k ′ ( x2 − x1 )
−m1ω 2 + k1 + k ′
−k ′
=0
−k ′
− m2ω 2 + k2 + k ′
m1m2ω 4 −  m2 ( k1 + k ′ ) + m1 ( k2 + k ′ )  ω 2 + ( k1 + k ′ )( k2 + k ′ ) − k ′2 = 0
m1m2ω 4 −  m2 ( k1 + k ′ ) + m1 ( k2 + k ′ )  ω 2 + k1k2 ( k1 + k2 ) k ′ = 0
m2 ( k1 + k ′ ) + m1 ( k2 + k ′ ) ±  m2 ( k1 + k ′ ) + m1 ( k2 + k ′ )  − 4m1m2  k1k2 + ( k1 + k2 ) k ′
=0
ω2 =
2m1m2
2
For
ω =
2
ω2 =
m1 = m, m2 = 2m, k1 = k , k2 = 2k , k ′ = 2k
2m ( 3k ) + m ( 4k ) ±
( 6mk + 4mk )
2 ( 2m 2 )
2
− 4 ( 2m 2 )( 2k 2 + 6k 2 )
10 k 6 k
±
4 m 4m
ω = 2ω0 and ω0 where ω0 =
k
m
174
11.17
k
m
x1
k
2m
x2
Note: As discussed in Section 3.2, the effect of any constant external
force on a harmonic oscillator is to shift the equilibrium position. x1
and x2 are the positions of the harmonic oscillator masses away from
their respective “shifted” equilibrium positions.
1
1
T = mx12 + ( 2m ) x2 2
2
2
1 2 1
2
V = kx1 + k ( x2 − x1 )
2
2
L = T −V
d ∂L
∂L
= mx1 ,
= − kx1 + k ( x2 − x1 )
dt ∂x1
∂x1
mx1 + 2kx1 − kx2 = 0
d ∂L
∂L
= 2mx2 ,
= − k ( x2 − x1 )
dt ∂x2
∂x2
2mx2 + kx2 − kx1 = 0
The secular equation (11.4.12) is thus
−mω 2 + 2k
−k
=0
−k
−2mω 2 + k
2m 2ω 4 − 5mkω 2 + 2k 2 + k 2 = 0
The eigenfrequencies are thus …
5 ± 17  k 
ω2 =
 
4 m
The homogeneous equations (Equations 11.4.10) for the two components of the jth
eigenvector are …
 − mω 2 + 2k
  a1 j 
−k

  = 0
−k
−2mω 2 + k   a2 j 

For the first eigenvector (the anti-symmetric mode, j = 1) …
5 + 17  k 
Inserting ω12 =
  into the first of the two homogeneous equations yields
4 m
 5 + 17

k + 2k  a11 = ka21
−
4


3 − 17
a11
4
Letting a11 = 1, then a21 = -0.281 (Thus, in the anti-symmetric normal mode, the
amplitude of the vibration of the second mass is 0.281 that of the first mass and 180o out
of phase with it.)
a21 =
175
For the second eigenvector (the symmetric mode, j = 2) …
5 − 17  k 
Inserting ω2 2 =
  into the first of the two homogeneous equations yields
4 m
 5 − 17

k + 2k  a12 = ka22
−
4


3 + 17
a12
4
Letting a12 = 1, then a22 = 1.781 (Thus, in the symmetric normal mode, the amplitude of
the vibration of the second mass is 1.781 that of the first mass and in phase with it.)
a22 =
The two eigenvectors (Equation 11.4.13 and see accompanying Table) are …
a 
 1 
Q1 =  11  cos (ω 1 t − δ1 ) = 
 cos (ω 1 t − δ1 )
 −0.281
 a21 
a 
 1 
Q2 =  12  cos (ω2 t − δ 2 ) = 
 cos (ω2 t − δ 2 )
 1.781
 a22 
11.18
2
1
1
T = ml12θ 2 + m l1θ + l2φ
2
2
V = − mgl1 cos θ − mg ( l1 cos θ + l2 cos φ )
(
)
For small angular displacements …
2
 θ2 
 φ2 
m
m L = T − V ≈ l12θ 2 +
l1θ + l2φ + 2mgl1 1 −  + mgl2 1 − 
2 
2
2
2 


(
)
d ∂L
∂L
= ml12θ + ml1 l1θ + l2φ ,
= −2mgl1θ
dt ∂θ
∂θ
2l1θ + l2φ + 2 gθ = 0
d ∂L
∂L
= ml2 l1θ + l2φ ,
= − mgl2φ
dt ∂φ
∂φ
l1θ + l2φ + gφ = 0
The secular equation (Equation 11.4.12) is …
−2l1ω 2 + 2 g
−l2ω 2
=0
−l1ω 2
−l2ω 2 + g
(
(
)
)
2l1l2ω 4 − 2 g ( l1 + l2 ) ω 2 + 2 g 2 − l1l2ω 4 = 0
Solving for the eigenfrequencies ω 2 …
2 g ( l1 + l2 ) ± 4 g 2 ( l1 + l2 ) − 8l1 l2 g 2
2
ω =
2
2l1 l2
=
(
g
l1 + l2 + l12 + l2 2
l1 l2
)
176
The homogeneous equations (Equations 11.4.10) for the two components of the jth
eigenvector are …
 −2l1ω 2 + 2 g
−l2ω 2   a1 j 

  = 0
2
−l2ω 2 + g   a2 j 
 −l1ω
Inserting the larger eigenfrequency (the (+) solution for ω 2 above) into the upper
homogeneous equation yields the solution for the components of the 1st eigenvector …
( −2l1ω1 2 + 2 g ) a11 − l2ω1 2 a21 = 0
 l +l + l 2 +l 2
1
2
2 ga11 1 − 1 2
l

2

l + l + l12 + l2 2
 = ga21 1 2
l1

l2 − l1 − l12 + l2 2
a21 = a11
l2
for the higher frequency, anti-symmetric mode.
Inserting the smaller eigenfrequency (the (-) solution for ω 2 ) into the upper
homogeneous equation yields the solution for the components of the 2nd eigenvector …
( −2l1ω2 2 + 2 g ) a12 − l2ω2 2 a22 = 0
l2 − l1 + l12 + l2 2
a22 = a12
for the lower frequency, symmetric mode.
l2
Again, we let a11 = 1 and a21 = 1, since only ratios of the components of a given eigen
vector can be determined. The two eigenvectors are thus (Equation 11.4.12 and
accompanying table)
1




Q1 =  l2 − l1 − l12 + l2 2  cos (ω 1 t − δ1 ) … anti-symmetric


l2


1




Q2 =  l2 − l1 + l12 + l2 2  cos (ω 2 t − δ 2 ) … symmetric


l2


As a check, set l1 = l2 = l and compare with the solution for Example 11.3.1.
11.19
1
m ( x12 + x2 2 + x32 )
2
1
2
2
V = k  x12 + ( x2 − x1 ) + ( x3 − x2 ) + x32 


2
L = T −V
d ∂L ∂L
−
=0
dt ∂xi ∂xi
T=
x1
x2
x3
177
mx1 + kx1 − k ( x2 − x1 ) = 0
mx1 + 2kx1 − kx2 = 0
mx2 + k ( x2 − x1 ) − k ( x3 − x2 ) = 0
− kx1 + mx2 + 2kx2 − kx3 = 0
mx3 + k ( x3 − x2 ) + kx3 = 0
− kx2 + mx3 + 2kx3 = 0
The secular equation (Equation 11.4.12) is …
− mω 2 + 2k
−k
0
−mω 2 + 2k
−k
−k
0
( −mω
2
−k
=0
2
− mω + 2k
+ 2k ) − 2k 2 ( −mω 2 + 2k ) = 0
3
− mω 2 + 2k = 0, or
( −mω
2
+ 2k ) − 2k 2 = 0
2
2k
= 2 ω0 2
m
2
− mω + 2k = ± 2 k
k
ω2 = 2± 2
= 2 ± 2 ω0 2
m
ω2 =
(
)
(
)
From Equation 11.5.17 (N = 1 and n = 3)
ω1 = 2ω0 sin
Because sin
ω1 = 2ω0
π
8
θ
2
=
1 − cos θ
2
1 − cos π 4
= 2ω0 1 − 2 2
2
ω1 = ω0 2 − 2
2π
2
= 2ω0
= 2ω0
8
2
3π
1 − cos 3π 4
ω3 = 2ω0 sin
= 2ω0
8
2
ω2 = 2ω0 sin
ω3 = 2ω0 1 + 2 2 = ω0 2 + 2
178
11.20
X
Generalized coordinates: X, s ( = aθ )
M
and
x = X + a sin θ
x = X + aθ cos θ
θ
a
y = a (1 − cos θ )
y = aθ sin θ
1 2 1
MX + m ( x 2 + y 2 )
2
2
1 2 1  = MX + m X + aθ cos θ
2
2 
V = mgy = mga (1 − cos θ )
T=
y
m
(x,y)
x
(
) + ( aθ sin θ ) 
2
2
For small oscillations, in terms of the generalized coordinates X and s …
2
1
1
1
T ≈ MX 2 + m ( X + s )
V ≈ mgs 2
2
2
2a
2
1 2 1
1
L = T − V ≈ MX + m ( X + s ) − mgs 2
2
2
2a
Lagrange’s equations of motion yield …
g
X + s+ s=0
( M + m ) X + ms = 0
a
Assuming …
X = Aeiωt
s = Beiωt
we obtain the matrix equation …

ω2
ω2 − g a  A

   = 0
2
2
M
m
m
ω
ω
+
(
)

 B
Setting the determinant of the above matrix equal to zero yields…
g (m + M )
ω12 = 0 ,
ω2 2 =
M
The mode corresponding to ω1 is …
 0 − g a  A
implies that A = B = 0

  = 0
0  B
0
Thus, mode 1 exhibits no oscillation! It is a pure translation with …
θ = 0 and
X = A1 t + A2
The mode corresponding to ω2 is …
ω2 2 A + ( ω2 2 − g a ) B = 0
179
ω2
B
m+M
= 2 2
=−
A ω2 − g a
m
Setting A = 1 , we have for the 2nd mode …
m + M iω2 t
iω t
s = aθ = −
e
X =e 2
and
m
This mode corresponds to an oscillation about the CM where …
( m + M ) X = −ms ( = −maθ )
or…
The normal mode vectors are …
 A t + A2 
Q1 =  1

 0 
1


Q2 = m + M
 −
m

and

 eiω2 t


11.21
(a) We can solve for the normal modes using
Equation 11.4.9 …
( K − ω2M ) a = 0
X
M
θ
r
m
K and M are the potential energy and kinetic
energy matrices respectively. a is a twocomponent vector whose elements are the
amplitudes of the coordinates q. The kinetic
energy in Example 11.3.2, assuming small
displacements from equilibrium, are …
2
1
1
θ
T ≈ mX 2 + m X 2 + rθ + 2 Xr
2
2
1
1
θ + 1 m rθ 2
T ≈ 2mX 2 + m 2 Xr
2
2
2
(
)
( )
(
)
( )
or, in matrix form
X
where q =  
 rθ 
 2 1
Thus, M = m 

 1 1
The potential energy is …
1
g
2
V ≈ kX 2 + m ( rθ )
2
2r
1 k 
2mg
1
2
2
V≈
2mX 2 +
m ( rθ )  = mω0 2  2 X 2 + ( rθ ) 



2 2m 
kr
 2
k
k
=
where ω0 2 =
M + m 2m
1
T = qMq
2
180
 2 0
Thus K = mω0 2 
 The above matrix equation is thus …
0 1
a

0
2 2
2  2 1   1 
 mω0 
 − mω 
  a  = 0
0 1
 1 1   2 

 2 (ω0 2 − ω 2 )
−ω 2   a1 
=0


2
2
2  a 

ω
ω
ω
−
−
2


0


From the top equation in the matrix equation above, we get …
2 (ω0 2 − ω 2 ) a1 − ω 2 a2 = 0
2
2
a2 2 (ω0 − ω )
=
a1
ω2
The amplitudes for the two normal modes can be found by setting ω 2 = ω12 or ω2 2 .
In each case we set a1 = 1 , which we are free to do since only ratios of the amplitudes can
be determined.
For ω 2 = ω12 = 2 − 2 ω0 2 we obtain …
(
a2 =
)
2 (ω0 2 − ω12 )
ω
2
1
(
)
2 1 − 2 − 2 
= 2
= 
2− 2
(
)
Thus, for mode 1, the lower frequency, symmetric mode …
 1  iω t
q1 =   e 1
 2
(
)
For ω 2 = ω2 2 = 2 + 2 ω0 2 we obtain …
a2 =
2 ( ω0 2 − ω 2 2 )
ω
2
1
(
)
2 1 − 2 + 2 
 =− 2
= 
2+ 2
(
)
Thus, for mode 2, the higher frequency, anti-symmetric mode …
 1  iω 2 t
q2 = 
e
− 2
(b) In this case,
m
<< 1 or M >> m . We also assume that the spring is “slack”,
m+M
k
g
<< , an assumption not stated in the problem (which needs to be rectified in
M +m
r
the next edition, I suppose)
k
k
g
Also, 2mg ≠ kr , so we let ω0 2 =
and Ω0 2 =
≈
M +m M
r
i.e.,
The kinetic energy is
181
(
1 2 1
MX + m X 2 +
2
2
1
1
T ≈ ( M + m ) X 2 + m
2
2
and the M-matrix is
 M m
M ≈

 m m
The potential energy is
1
1 g
2
V ≈ kX 2 + m ( rθ )
2
2 r
The K-matrix is thus
 ω0 2 M
0 
K ≈

Ω02 m 
 0
T≈
( rθ )
2
θ
+ 2 Xr
)
(( rθ) + 2 Xr θ) ≈ 12 MX
2
=
2
1
θ + 1 m rθ
+ m 2 Xr
2
2
(
)
( )
2
1
1
M ω0 2 + m Ω 0 2
2
2
To solve for ω 2 , we set K − ω 2 M = 0
(ω
2
0
−ω2 ) M
−ω 2 m
−ω 2 m
=0
( Ω0 2 − ω 2 ) m
which yields …
ω0 2 Ω 0 2 + ω 4 − ω 2Ω 0 2 − ω 2ω0 2  mM − ω 4 m 2 = 0
ω 4 ( M − m ) − ω 2 (ω0 2 + Ω0 2 ) M + ω0 2 Ω0 2 M = 0
Neglecting m with respect to M and simplifying yields …
ω 4 − ω 2 (ω0 2 + Ω 0 2 ) + ω0 2 Ω0 2 = 0
Solving for ω 2 …
ω
2
(ω
=
ω2 =
0
2
+ Ω0 2 )
2
2
(ω0 + Ω02 )
±
±
(ω
2
0
(ω
2
0
+ Ω0 2 ) − 4ω0 2Ω 0 2
− Ω0
2
2
)
2
Thus, we get …
2
2
ω12 = ω0 2 and ω2 2 = Ω 0 2
Now, we solve for the amplitudes a of the normal mode vectors …
( K − ω2M ) a = 0
 (ω0 2 − ω 2 ) M
−ω 2 m   a1 

  = 0
2
2
2

−ω m
( Ω0 − ω ) m   a2 

Using the first of the above matrix equations for ω 2 = ω12 = ω0 2 gives …
a2 = 0 and a1 = 1
Using the second equation for ω 2 = ω2 2 = Ω0 2 gives …
182
a1 = 0 and a2 = 1
Thus, the normal modes are approximately …
1
0
Q1 =   ei ω0 t and Q2 =   ei Ω0 t
0
1
Note that we could have guessed this almost immediately. The above assumption is
tantamount to omitting the cross elements −ω 2 m . This completely eliminates the small
coupling between the two oscillators, which reduces the matrix K − ω 2 M to the purely
diagonal terms …
 (ω0 2 − ω 2 ) M

0
 , which leads directly to the above solution.


0
( Ω02 − ω 2 ) m 

22.
2m
We “scale” the force constants and masses to 1 unit,
namely,
m = 1 and k = 1 .
θ1
k
k
m
θ3
K
θ2
m
Let k ′ = K and l = 1
such that 3l = circumference
So: T =
1 2 2 2
2θ1 + θ 2 + θ3
2
(
)
1
1
1
1
1
1
2
2
2
2
2
2
(θ1 − θ 2 ) + (θ 2 − θ1 ) + (θ1 − θ3 ) + (θ3 − θ1 ) + K (θ 2 − θ3 ) + K (θ3 − θ 2 )
2
2
2
2
2
2
Collecting terms …
1
V =  4θ12 + 2 (1 + K ) θ 2 2 + 2 (1 + K ) θ 32 − 4θ1θ 2 − 4θ1θ 3 − 4 Kθ 2θ 3 
2
K - Matrix
M - Matrix
-2
-2 
4
 2 0 0




K =  -2 2+ 2K
-2K 
M =  0 1 0
 -2 -2K
0 0 1
2+ 2K 



V=
The ttansformation matrix A that diagonalizes these matrices is made up of the three
eigenvectors Qi whose amplitudes are ai … we guess that …
1. Uniform rotation
 1
θ1 = θ 2 = θ 3 or a1 = 1
 1
 
183
2. Anti-symmetric oscillation of 2&3, while 1 remains fixed
0
θ1 = 0; θ3 = −θ 2 or a1 =  1 
 −1 
 
3. Anti-symmetric oscillation of 1&2 together with respect to 3
1
θ1 = 1; θ3 = θ 2 or a1 =  −1
 −1 
 
1 0 1


Thus, A =  1 1 -1
 1 -1 -1


 1 1 1


and A =  0 1 -1
 1 -1 -1


We can now diagonalize Kand M …
-2
-2  1 0 1 
 1 1 1  4




AKA
=  0 1 -1 -2 2+ 2K
-2K  1 1 -1 = K diag
 1 -1 -1 -2 -2K

2+ 2K 


 1 -1 -1
0
0
0
 4 0 0




K diag =  0 4 + 8 K
 Likewise M diag =  0 2 0 
0
 0 0 4
16 



The eigenfrequencies are ωi 2 =  K diag M diag 
i
2
2
2
ω2 = 0; ω2 = 2 + 4 K ; ω3 = 4
The general solution can be generated from the following table …
Q1 Q2 Q3
θ1
θ2
θ3
a1
a1
a1
0
a2
− a2
a3
− a3
− a3
1
0
1
 
 
where Q1 = a1 1 cos (ω1 t − δ1 ) , Q2 = a2  1  cos (ω2 t − δ 2 ) , Q3 = a3  −1 cos (ω3 t − δ 3 )
1
 −1
 −1 
 
 
 
Thus, θ1 = a1 cos (ω2 t − δ 2 )
+ a3 cos (ω3 t − δ 3 )
θ1 = a1 cos (ω2 t − δ 2 ) + a2 cos (ω2 t − δ 2 ) − a3 cos (ω3 t − δ 3 )
θ3 = a1 cos (ω2 t − δ 2 ) − a2 cos (ω2 t − δ 2 ) − a3 cos (ω3 t − δ 3 )
184
Initial conditions are: θ1 = θ 0 = 10o ; θ 2 = θ 3 = 0; θ1 = θ2 = θ3 = 0
The conditions generate 6 equations with 6 unknowns and solving gives …
θ1 =
θ2 =
θ2 =
θ0
2
cos ω1 t +
θ0
2
θ0
2
cos ω1 t −
cos ω1 t −
θ0
2
θ0
2
θ0
2
cos ω3 t
cos ω3 t
cos ω3 t
11.23 See Ex. 11.4.1, page 497-498
The amplitudes of the eigenvectors are …
1
 1
1


 
 


a1 =  1 a2 =  0  a3 =  − 2m M 
 1
 −1


1
 
 


K - Matrix
 K -K

K =  -K 2K
 0 -K

K diag
K diag
M diag
M diag
M - Matrix
m
0 0



M =0 M 0
 0 0 m


0 

-K 
K 
1
1   K −K
1


= AKA
= 1
−1   − K 2 K
0
 1 − 2m M 1   0 − K


0
0
0



= 0
2K
0

 0 2 K + 8K m M 8K m2 M 2 


1

= AMA
= 1
1

 2m + M

= 0
 0

1
0
− 2m M
1  m 0

−1 0 M
1 
 0 0
0  1 1
1



− K   1 0 − 2m M 

K   1 −1
1

0  1 1
1



0   1 0 − 2m M 

1
m   1 −1

0
0


2m
0

0 2m + 4m 2 M 
So, we have …
ω12 = 0 , ω2 2 =
ω32 =
2K K
= , and …
2m m
2
2
2 K + 8K m M + 8 K m 2 M 2 K (1 + 4m M + 4m M )
=
2m + 4 m 2 M
m (1 + 2m M )
185
K (1 + 2 m M )
K
ω3 =
= (1 + 2 m M )
m (1 + 2 m M ) m
2
2
11.24
y2
M
y3
m
α
x2
α
y1
a
a
x3
m
x1
Select coordinates (xi , yi) as shown, then …
1
1
T = m ( x12 + y12 + x32 + y32 ) + M ( x2 2 + y 2 2 )
2
2
The potential energy depends only on the compression (or stretching) of the two springs
connecting each m to M (hydrogen to sulfur). Let δ a1 and δ a2 be incremental changes
in the distances a1 or HS (1 → 2) and a2 or HS (3 → 2) . We have …
1
δ a1 = ( x1 − x2 ) sin α + ( y2 − y1 ) cos α =
( x1 − x2 + y2 − y1 )
2
1
δ a2 = ( x2 − x3 ) sin α + ( y2 − y3 ) cos α =
( x2 − x3 + y2 − y3 )
2
1
2
2
V = k (δ a1 ) + (δ a2 ) 

2 
We can reduce the degrees of freedom from 6 to 3 by ignoring the two translational
modes and the rotational mode. Thus we consider only vibrational modes. The
coordinates must obey the following constraints …
No center of mass motion:
m ( y1 + y3 ) + My2 = 0 and m ( x1 + x3 ) + Mx2 = 0
No angular momentum about any point. We choose that point to be the sulfur atom (M)…
my3 a sin α − my1a sin α − mx3a sin α − mx1a sin α = 0
186
y3 − y1 − x3 − x1 = 0
We introduce three generalized coordinates Q, q1 and q2 that should be close to what we
guess would be normal modes …
Q = x1 + x3 ; q1 = x1 − x3 ; q2 = y1 + y3
Solve for xi , yi in terms of Q, q1 and q2 using the above 3 equations of constraint …
1
m
1
x1 = ( Q + q1 ) ; x2 = − Q ; = x1 − x3 ; x3 = ( Q − q1 )
2
2
M
1
m
1
y1 = ( q2 − Q ) ; y2 = − q2 ;
y3 = ( Q + q2 )
2
M
2
Thus, the kinetic energy, in terms of these generalized coordinates, is …
1 
m
1
1 µ
T = m 1 +  Q 2 + mq12 + m q2 2
2  M
4
4 M
where µ = M + 2m is the mass of the H2S molecule
The potential energy is …
2
V=
1 
m
1
1 µ2
1 µ
k 1 +  Q 2 + kq12 + k 2 q2 2 − k
q1q2
2  M
8
8 M
4 M
Note, that in the Lagrangian L = T − V , the only cross term is one involving q1q2 .
therefore, Q is a normal mode with eigenfrequency given by the ratio …
k
m
ωQ 2 = KQ M Q = 1 + 
m M 
Constructing the residual 2x2 K and M matrices involving only q1 and q2 terms, which we
will call Kq and Mq gives …
−µ M 
0 
1 1
1 1
Kq = 
and Mq = 

2
2
4  −µ M µ M 
2 0 µ M 
We have omitted the factors k and m, which we’ll replace in the final solution,
remembering that the eigenfrequencies that we find as a solution to Kq - ω 2 Mq = 0 will
be multiples of k / m.
2
1 1 − 2ω
2
Kq - ω Mq =
4 −µ M
−µ M
µ M ( µ M − 2ω 2 )
=0
which reduces to …
2ω 2  2ω 2 − (1 + µ M )  = 0 which has the non-trivial solution …
1
k
(1 + µ M ) in which, we have put back the factor k / m. These two modes are
2
m
“degenerate.”
ω1,2 2 =
187
Plugging ω 2 back into the matrix equation ( Kq - ω 2 Mq ) q = 0 gives …
q1 = −q2 or − x1 + x3 = y1 + y3
which can be satisfied in a variety of ways, for example, with x1 = x3 and
y1 = − y3 , etc
Pictorially, the three normal modes are …
Q-mode: ant-symmetric about y-axis:
x1 = x3 and y1 = − y3
ωQ 2 =
k
m
1 + 
m M 
Breathing mode: symmetric about y-axis:
x1 = − x3 and y1 = y3
k
ω12 =
(1 + µ M )
2m
Stretching mode: symmetric about y-axis:
x1 = − x3 and y1 = y3
k
ω2 2 =
(1 + µ M )
2m
-------------------------------------------------------------------------------------------------11.25
(a)
Plug each of these functions into the wave equation and it is satisfied!
(b)
Q = q1 + q2 = e iωt e −ikx + e (
i ω +∆ω )t
e
− i ( k +∆k ) x
− i ( ∆ω )t − ( ∆k ) x  2
+ i ( ∆ω )t − ( ∆k ) x  2

= e i(ω +∆ω 2)t e −i( k +∆k 2) x e 
+e 


The real part of the above is …
 ( ∆ω ) t − ( ∆k ) x 

∆ω  
∆k  
Q = 2 cos 
 cos  ω +
t − k +
x
2
2  
2  



(c)
 ( ∆ω ) t − ( ∆k ) x 
≈ 2 cos 
 cos (ωt − kx )
2


The group speed is …
dx
ug =
(the phase of the amplitude remains the same)
dt
188
ug =
∆ω
∆k
11.26
From Equation 11.5.17 ...
Nπ
2π
3π
4π
sin
sin
sin
sin
ωN
10 , ω2 =
10 = 1.28, ω3 =
10 = 2.62, ω4 =
10 = 3.08
=
π
π
π
ω1
ω1 sin
ω1 sin
ω1 sin π
sin
10
10
10
10
189
11.27
F = k ∆l is the tension in the cord
l + ∆l
d=
is its stretched length
n +1
From the equation following Equation 11.5.7…
F
K=
d
From Equation 11.6.4c …
k
v 2 = ( l + ∆l ) ∆ l
m
1
vtrans
1
 k 2
=   ( l + ∆l ) ∆l  2
m
for transverse waves
For longitudinal vibrations, we use Y, the tension in the cord per unit stretched length
k ∆l
Y=
= k ( l + ∆l )
∆l ( l + ∆l )
Y
(Equation 11.6.8)
d
kd 2 k ( l + ∆l ) nd
v2 =
=
(Equation 11.6.9a)
mn
m
K=
nd ≈ ( n + 1) d = l + ∆l
1
vlong
 k 2
=   ( l + ∆l )
m
11.28
From Equation 11.6.7b …
1
 F 2
v= 
µ
As in problem 11.27, F = k ∆l
1
 k ∆l  2
vtrans = 

 µ 
From Equation 11.6.9b …
1
2
Y 
v =   and from problem 11.27 … Y = k ( l + ∆l ) so … vlong
µ
1
 k ( l + ∆l )  2
=

µ


190
11.29
The general solution to the wave equation (Equation 11.6.10) that yields a standing wave
of any arbitrary shape can be obtained as a linear combination of standing sine waves of a
form given by Equation 11.6.14, i.e.,
∞
 2π x 
y ( x, t ) = ∑ ( An sin ωn t + Bn cos ωn t ) sin 

n =1
 λn 
2π
where ωn =
Tn
since the speed of a wave is …
1
λn
ω λ  F 2
v=
= n n = 0 
2π
Tn
µ
we have …
1
 F  2 2π
ωn =  0 
 µ  λn
But the wavelength of the standing wave is constrained by the fixed endpoints of the
string, i.e. …
2l
λn = , so …
n
1
 F  2 nπ
ωn =  0 
µ l
Now, at t = 0, the wave starts from rest in the configuration specified, so…
nπ x
y ( x, 0) = ∑ Bn sin
l
From the discussion of Fourier analysis in Appendix G or in Section 3.9, the Fourier
coefficients are given by …
l
2
nπ x
Bn = ∫ y ( x, 0) sin
dx
l 0
l
Since the string starts from rest, we have …
∂y ( x, t )
nπ x
= ∑ ωn An sin
≡0
∂t t =0
l
Therefore, An = 0
The initial configuration is shown in the Figure P11.29. Thus …
2a
l
y ( x, 0) =
x
0< x<
l
2
191
2a
l
< x<l
(l − x )
l
2
We can now determine Bn …
l 2
l
2  2a
nπ x
2a
nπ x 
sin
sin
Bn = 
x
dx
l
x
dx 
+
−
(
)
l  l ∫0
l
l l∫2
l

Using integration tables, we obtain the result …
n −1
nπ

 8a
Bn = ( −1) 2  2 2 sin
n = 1,3,5,...
2

n π
Bn = 0
n = 2, 4, 6,...
y ( x, 0) =
Thus, the general solution is …
5π x
8a 
π vt
πx 1
3π vt
3π x 1
5π vt

− cos
+ cos
− ... 
y ( x, t ) = 2  cos
sin
sin
sin
l
l 9
l
l
25
l
l
π 

None of the harmonics which have a node at the midpoint have been stimulated … only
the odd harmonics have been excited.
11.30
By analogy with the generation of the traveling sine wave of Equation 11.6.14 from
Equation 11.6.13, we get …
π ( x − vt )  1  3π ( x + vt )
3π ( x − vt ) 
4a  π ( x + vt )
y ( x, t ) = 2  sin
+ sin
+ sin

 −  sin
l
l
l
l
π 

 9
+
5π ( x − vt )  
1  5π ( x + vt )
+ sin
 sin
 ...
25 
l
l
 
----------------------------------------------------------------------------------------------------
192