891
*17–100.
A force of F = 10 N is applied to the 10-kg ring as shown. If
slipping does not occur, determine the ring’s initial angular
acceleration, and the acceleration of its mass center, G. Neglect
the thickness of the ring.
SOLUTION
Equations of Motion. The mass moment of inertia of the ring about its center of
Kinematics. Since the ring rolls without slipping,
Solving Eqs. (1) and (2)
45
30
0.4 m
G
A
C
F
Ans:
892
17–101.
If the coefficient of static friction at C is μs = 0.3, determine
the largest force F that can be applied to the 5-kg ring,
without causing it to slip. Neglect the thickness of
the ring.
SOLUTION
Equations of Motion: The mass moment of inertia of the ring about its center of
gravity G is I
G=
mr
2=
10
(
0.4
2
)
=
1.60 kg
#
m
2
. Here, it is required that the ring is
Kinematics. Since the ring rolls without slipping,
Solving Eqs. (1) to (4),
45
30
0.4 m
G
A
C
F
893
17–102.
The 25-lb slender rod has a length of 6 ft. Using a collar of
negligible mass, its end A is confined to move along the
smooth circular bar of radius 3
22
ft. End B rests on the
floor, for which the coefficient of kinetic friction is m
B=0.4.
If the bar is released from rest when
u
= 30°, determine the
angular acceleration of the bar at this instant.
SOLUTION
a
B
=a
A
+a
B
>
A
a
A
B
3 2 ft
6 ft
u
894
17–103.
The 15-lb circular plate is suspended from a pin
at A. If the pin is connected to a track which is given an
acceleration aA = 5 ft
>
s2, determine the horizontal and
vertical components of reaction at A and the angular
acceleration of the plate. The plate is originally at rest.
G
A
a
A
2 ft
Ans:
*17–104.
1.5 ft
P
30
If P= 30 lb, determine the angular acceleration of the 50-lb
roller.Assume the roller to be a uniform cylinder and that
no slipping occurs.
SOLUTION
Equations of Motion: The mass moment of inertia of the roller about its mass center
Since the roller rolls without slipping,
Solving Eqs. (1) through (3) yields
Ans:
896
17–105.
1.5 ft
P
30
If the coefficient of static friction between the 50-lb roller
and the ground is determine the maximum force
Pthat can be applied to the handle,so that roller rolls on the
ground without slipping.Also, find the angular acceleration
of the roller.Assume the roller to be a uniform cylinder.
ms=0.25,
SOLUTION
Ans:
17–106.
The uniform bar of mass m and length L is balanced in the
vertical position when the horizontal force P is applied to
the roller at A. Determine the bar’s initial angular
acceleration and the acceleration of its top point B.
SOLUTION
B
L
898
17–107.
Solve Prob. 17–106 if the roller is removed and the
coefficient of kinetic friction at the ground is μk.
SOLUTION
B
L
Ans:
899
*17–108.
SOLUTION
Equations of Motion:The mass moment of inertia of the semicircular disk about its center
Kinematics:Assume that the semicircular disk does not slip at A,then .
Solving Eqs. (1), (2), (3), (4), and (5) yields:
(aA)x=0
The semicircular disk having a mass of 10 kg is rotating at
at the instant . If the coefficient of
static friction at Ais , determine if the disk slips at
this instant.
ms=0.5
u=60°v=4 rad>s
4 (0.4) m
3p
O
G
0.4 m
A
u
v
Ans:
17–109.
SOLUTION
Kinematics: Since the culvert does not slip at A,.Applying the
relative acceleration equation and referring to Fig. b,
Equating the icomponents,
Solving Eqs. (1) and (2) yields
(aA)t=3m>s2
The 500-kg concrete culvert has a mean radius of 0.5 m. If
the truck has an acceleration of , determine the
culvert’s angular acceleration.Assume that the culvert does
not slip on the truck bed, and neglect its thickness.
3m>s24m
0.5m
3m/s2
901
17–110.
The 15-lb disk rests on the 5-lb plate. A cord is wrapped
around the periphery of the disk and attached to the wall at
B. If a torque M = 40 lb
#
ft is applied to the disk, determine
the angular acceleration of the disk and the time needed for
the end C of the plate to travel 3 ft and strike the wall.
Assume the disk does not slip on the plate and the plate
rests on the surface at D having a coefficient of kinetic
friction of μk = 0.2. Neglect the mass of the cord.
SOLUTION
Disk:
Solving,
AB
C
D3 ft
M 40 lb ft
1.25 ft
Ans: