Problem 19.100 The ring gear is fixed. The mass
and moment of inertia of the sun gear are mS=
22 slugs and IS=4400 slug-ft2. The mass and moment
of inertia of each planet gear are mP=2.7 slugs and
IP=65 slug-ft2. A couple M=600 ft-lb is applied to
the sun gear. Use work and energy to determine the
angular velocity of the sun gear after it has turned 100
revolutions.
20 in
M
Sun gear
Planet gears (3)
Ring gear
34 in
7 in
Solution: Denote the radius of planetary gear, RP=7 in. =
0.5833 ft, the radius of sun gear RS=20 in. =1.667 ft, and angular
620
Problem 19.101 The moments of inertia of gears A
and Bare IA=0.014 slug-ft2, and IB=0.100 slug-ft2.
k=0.2 ft-lb/rad. If the spring is unstretched and the
surface supporting the 5-lb weight is removed, what is
the velocity of the weight when it has fallen 3 in?
10 in
3 in
0.25 ft, are the radii of the gears and pulley. Choose a coordinate
system with ypositive upward. From the conservation of energy T+
V=const. Choose the datum at the initial position, such that V1=0,
T1=0, from which T2+V2=0 at any position 2. The gear Brotates
energy in the spring is
Vspring =−θA
0
Mdθ =θA
0
kθdθ =1
2kθ2
A=0.2778 ft-lb.
The force due to the weight is negative, from which the potential
energy of the weight is
Vweight =−−s
0
(−W)dy =−Ws =−1.25 ft-lb.
Problem 19.102 Consider the system in Prob-
lem 19.101.
(a) What maximum distance does the 5-lb weight fall
when the supporting surface is removed?
(b) What maximum velocity does the weight achieve?
0
0
2kθ2
Vweight =−−s
(−W)dy =−Ws,
sv−max =0.5625 ft.
Problem 19.103 Each of the go-cart’s front wheels
weighs 5 lb and has a moment of inertia of 0.01 slug-ft2.
The two rear wheels and rear axle form a single rigid
body weighing 40 lb and having a moment of inertia of
0.1 slug-ft2. The total weight of the rider and go-cart,
including its wheels, is 240 lb. The go-cart starts from
rest, its engine exerts a constant torque of 15 ft-lb on the
rear axle, and its wheels do not slip. Neglecting friction
and aerodynamic drag, how fast is the go-cart moving
when it has traveled 50 ft?
6 in 4 in
0.5=100 rad, from which the work done is
U=θA
Mdθ =MθA=15(100)ft-lb.
4.020v2ft-lb. Substitute into U=T2and solve: v=19.32 ft/s.
15 in.
16 in. 60 in.
AB
Problem 19.104 Determine the maximum power and
the average power transmitted to the go-cart in
Problem 19.103 by its engine.
Solution: The maximum power is Pmax =MωAmax, where
ωAmax =vmax
R. From which Pmax =Mvmax
R. Under constant torque,
622
Problem 19.105 The system starts from rest with the
4-kg slender bar horizontal. The mass of the suspended
cylinder is 10 kg. What is the angular velocity of the bar
when it is in the position shown?
2 m
3 m
45°
Solution: From the principle of work and energy: U=T2−T1,
Problem 19.106 The 0.1-kg slender bar and 0.2-kg
cylindrical disk are released from rest with the bar
horizontal. The disk rolls on the curved surface. What is
the angular velocity of the bar when it is vertical?
120 mm
40 mm
624
Problem 19.107 A slender bar of mass mis released
from rest in the vertical position and allowed to fall.
Neglecting friction and assuming that it remains in
contact with the floor and wall, determine the bar’s
angular velocity as a function of θ.
l
θ
Problem 19.108 The 4-kg slender bar is pinned to
2-kg sliders at Aand B. If friction is negligible and
the system starts from rest in the position shown, what
is the bar’s angular velocity when the slider at Ahas
fallen 0.5 m?
45°
A
1.2 m
Solution: Choose a coordinate system with the origin at the initial
(c) use the principle of work and energy to determine the angular
vA
626
The kinematic relations: From kinematics, vB=vA+ω×rB/A. The
10.23 =1.77 rad/s
Problem 19.109 A homogeneous hemisphere of mass
mis released from rest in the position shown. If it rolls
on the horizontal surface, what is its angular velocity
when its flat surface is horizontal?
R
The work done is
mg R−5
8R=3
8mgR.
Work and energy is
3
8mgR =1
2mv2+1
2Iω2
=1
2m5R
8ω2
+1
283
320 mR2ω2.
Solving for ω, we obtain
ω=15g
13R.
3R
8
Problem 19.110 The homogeneous hemisphere of
mass mis released from rest in the position shown.
It rolls on the horizontal surface. What normal force
is exerted on the hemisphere by the horizontal surface
at the instant the flat surface of the hemisphere is
horizontal?
628
Problem 19.111 The slender bar rotates freely in the
horizontal plane about a vertical shaft at O. The bar
weighs 20 lb and its length is 6 ft. The slider Aweighs
2 lb. If the bar’s angular velocity is ω=10 rad/s and
the radial component of the velocity of Ais zero when
r=1 ft, what is the angular velocity of the bar when
r=4 ft? (The moment of inertia of Aabout its center
A
r
ω
Problem 19.112 A satellite is deployed with angular
velocity ω=1 rad/s (Fig. a). Two internally stored
antennas that span the diameter of the satellite are then
of 1.2-m radius and each antenna as a 10-kg slender bar,
determine ω′.
ω
(a)
(b)
ω
‘
2.4 m 2.4 m
Solution: Assume (I) in configuration (a) the antennas are folded
The angular momentum is conserved,
Problem 19.113 An engineer decides to control the
angular velocity of a satellite by deploying small masses
attached to cables. If the angular velocity of the satellite
in configuration (a) is 4 rpm, determine the distance d
in configuration (b) that will cause the angular velocity
to be 1 rpm. The moment of inertia of the satellite is
I=500 kg-m2and each mass is 2 kg. (Assume that the
cables and masses rotate with the same angular veloc-
ity as the satellite. Neglect the masses of the cables and
the mass moments of inertia of the masses about their
centers of mass.)
d d
1 rpm
2 m 2 m
4 rpm
(a) (b)
Solution: From the conservation of angular momentum,
4 rpm
Problem 19.114 The homogenous cylindrical disk of
mass mrolls on the horizontal surface with angular
velocity ω. If the disk does not slip or leave the slanted
surface when it comes into contact with it, what is the
angular velocity ω′of the disk immediately afterward?
ω
β
R
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Problem 19.115 The 10-lb slender bar falls from rest
0.6, the duration of the impact is 0.1 s, and b=1 ft.
Determine the average force exerted on the bar at Bas
a result of the impact.
b
3 ft
AB
Solution: Choose a coordinate system with the origin at Aand
From the principle of work and energy, U=T2−T1, where T1=0.
The center of mass of the bar falls a distance h=L
2.The work done
by the weight of the bar is U=mg L
2.The kinetic energy is
T2=1
2Iω2, where I=mL2
3. Substitute into U=T2and solve:
ω=−
3g
L, where the negative sign on the square root is chosen to
be consistent with the choice of coordinates. By definition, the coef-
ficient of restitution is e=v′
B−v′
A
vA−vB
, where vA,v
′
Aare the velocities
of the bar at a distance bfrom Abefore and after impact. Since the
projection Bis stationary before and after the impact, vB=v′
B=0,
from which v′
A=−evA. From kinematics, vA=bω, and v′
A=bω′,
from which ω′=−eω. The principle of angular impulse-momentum
about the point Ais
t2−t1
0
bFBdt =(I ω′−Iω) =mL2
3(ω′−ω),
Problem 19.116 The 10-lb bar falls from rest in the
vertical position and hits the smooth projection at B.
The coefficient of restitution of the impact is e=0.6
and the duration of the impact is 0.1 s. Determine the
distance bfor which the average force exerted on the
Problem 19.117 The 1-kg sphere Ais moving at 2 m/s
when it strikes the end of the 2-kg stationary slender
bar B. If the velocity of the sphere after the impact is
0.8 m/s to the right, what is the coefficient of restitution?
B
(1) mAvA=mAv′
A+mBv′
CM, where v′
CM is the velocity of the cen-
ter of mass of the bar after impact, and vA,v
′
Aare the velocities of
ω′
632
Problem 19.118 The slender bar is released from rest
in the position shown in Fig. (a) and falls a distance
h=1 ft. When the bar hits the floor, its tip is supported
by a depression and remains on the floor (Fig. b). The
length of the bar is 1 ft and its weight is 4 oz. What is
angular velocity ωof the bar just after it hits the floor? h
ω
45°
Problem 19.119 The slender bar is released from rest
The center of mass of the bar also falls a distance hbefore impact. The
Gy =−evGy −L
2ω′cos θ.
Problem 19.120 The slender bar is released from rest
and falls a distance h=1 m onto the smooth floor.
The length of the bar is 1 m and its mass is 2 kg.
The coefficient of restitution of the impact is e=0.4.
Determine the angle θfor which the angular velocity of
the bar after it hits the floor is a maximum. What is the
maximum angular velocity?
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Problem 19.121 A nonrotating slender bar Amoving
with velocity v0strikes a stationary slender bar B. Each
bar has mass mand length l. If the bars adhere when they
collide, what is their angular velocity after the impact?
A
B
l
–
v′
B=
2.
Substitute into the expression for the conservation of angular momen-
tum to obtain:
2v0
Problem 19.122 An astronaut translates toward a non-
rotating satellite at 1.0i(m/s) relative to the satellite. Her
mass is 136 kg, and the moment of inertia about the
axis through her center of mass parallel to the zaxis
is 45 kg-m2. The mass of the satellite is 450 kg and its
moment of inertia about the zaxis is 675 kg-m2. At the
instant the astronaut attaches to the satellite and begins
moving with it, the position of her center of mass is
(−1.8, −0.9, 0) m. The axis of rotation of the satellite
after she attaches is parallel to the zaxis. What is their
angular velocity?
y
x
1 m /s
1.0i(m/s).
ijk
ijk
(3) 0.9mAvAx =−1.8mAv′
Ay +0.9mAv′
Ax +(IA+IS)ω′.
636
Problem 19.123 In Problem 19.122, suppose that the
design parameters of the satellite’s control system
require that the angular velocity of the satellite not
exceed 0.02 rad/s. If the astronaut is moving parallel
to the xaxis and the position of her center of mass
when she attaches is (−1.8, −0.9, 0) m, what is the
maximum relative velocity at which she should approach
the satellite?
(2)0=mAv′
Ay +mSv′
Sy
(3)0.9mAvAx =−1.8mAv′
Ay +0.9mAv′
Ax +(IA+IS)ω′
(4)v
′
Ax =v′
Sx +0.9ω′,
(5)v
′
Ay =v′
Sy −1.8ω′.
Problem 19.124 A 2800-lb car skidding on ice strikes
a concrete abutment at 3 mi/h. The car’s moment of
inertia about its center of mass is 1800 slug-ft2. Assume
that the impacting surfaces are smooth and parallel to the
yaxis and that the coefficient of restitution of the impact
is e=0.8. What are the angular velocity of the car and
the velocity of its center of mass after the impact?
y
x
2 ft
3 mi/h
Problem 19.125 A 170-lb receiver jumps vertically to
receive a pass and is stationary at the instant he catches
the ball. At the same instant, he is hit at Pby a 180-lb
linebacker moving horizontally at 15 ft/s. The wide re-
ceiver’s moment of inertia about his center of mass is
7 slug-ft2. If you model the players as rigid bodies and
assume that the coefficient of restitution is e=0, what
is the wide receiver’s angular velocity immediately after
the impact?
14 in
638