PROBLEM 16.71 (Continued)
(b) Eq. (2):
10 1
15 0.1(32.2)(1.5972)
k
v v gt
µ
= −
= −
1
15 5.1429
v= −
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PROBLEM 16.72
Solve Problem 16.71, assuming that the bowler projects the ball with the same forward
velocity but with a backspin of 18 rad/s.
PROBLEM 16.71 A bowler projects an 8–in.–diameter ball weighing 12 lb along an
alley with a forward velocity v0 of 15 ft/s and a backspin
ω
0 of 9 rad/s. Knowing that
the coefficient of kinetic friction between the ball and the alley is 0.10, determine
(a) the time t1 at which the ball will start rolling without sliding, (b) the speed of the
ball at time t1, (c) the distance the ball will have traveled at time t1.
PROBLEM 16.72 (Continued)
( )
1
3
1
15 (18)
21.8634 s
7 0.1(32.2)
11.863 st=
(b) Eq. (2):
10
15 0.1(32.2)(1.8634)
t
k
vv g
µ
= −
= −
1
15 6.000
9 ft/s
v= −
=
19 ft/sv=
(c)
22
0.1(32.2 ft/s ) 3.22 ft/s
k
ag
µ
= = =
2
1 01 1
22
1
21
(15 ft/s)(1.8634 s) (3.22 ft/s )(1.8634 s)
2
s v t at= −
= −
27.95 5.59
22.36 ft
= −
=
1
22.4 fts=
PROBLEM 16.73
A uniform sphere of radius r and mass m is placed with no initial velocity on
a belt that moves to the right with a constant velocity v1. Denoting by
µ
k the
coefficient of kinetic friction between the sphere and the belt, determine
(a) the time t1 at which the sphere will start rolling without sliding, (b) the
linear and angular velocities of the sphere at time t1.
PROBLEM 16.74
A sphere of radius r and m has a linear velocity v0 directed to the left and no
angular velocity as it is placed on a belt moving to the right with a constant
velocity v1. If after first sliding on the belt the sphere is to have no linear
velocity relative to the ground as it starts rolling on the belt without sliding,
determine in terms of v1 and the coefficient of kinetic friction
k
µ
between
the sphere and the belt (a) the required value of v0, (b) time t1 at which the
sphere will start rolling on the belt, (c) the distance the sphere will have
moved relative to the ground at time t1.
PROBLEM 16.74 (Continued)
Distance when
1
:tt=
2
01 1
1
2
s v t at= −
2
11
1
2
1
22
21
()
55 2 5
42
25 25
k
kk
k
vv
sv g
gg
v
µ
µµ
µ
= −
2
1
2
25
k
v
g
µ
PROBLEM 16.75
Show that the couple
Iα
of Figure 16.15 can be eliminated by attaching the vectors
t
ma
and
n
ma
at a Point P called the center of percussion, located on line OG at a distance
2/GP k r=
from the mass center of the body.
PROBLEM 16.76
A uniform slender rod of length L = 900 mm and mass m = 4 kg is suspended from a
hinge at C. A horizontal force P of magnitude 75 N is applied at end B. Knowing that
225 mm,r=
determine (a) the angular acceleration of the rod, (b) the components
of the reaction at C.
PROBLEM 16.77
In Problem 16.76, determine (a) the distance
r
for which the horizontal component of the reaction at C is
zero, (b) the corresponding angular acceleration of the rod.
PROBLEM 16.78
A uniform slender rod of length L = 36 in. and weight W = 4 lb hangs freely from a hinge
at A. If a force P of magnitude 1.5 lb is applied at B horizontally to the left (h = L),
determine (a) the angular acceleration of the rod, (b) the components of the reaction at A.