902
17–111.
4 (0.4) m
3p
SOLUTION
For roll A.
For roll B
Kinematics:
Solving Eqs. (1)–(5) yields:
The semicircular disk having a mass of 10 kg is rotating
at v = 4 rad>s at the instant u = 60º. If the coefficient of
static friction at A is m = 0.5, determine if the disk slips at
this instant.
VO
G
0.4 m
A
u
v
Ans:
*17–112.
4 ft
0.5 ft
G
O
v
The circular concrete culvert rolls with an angular velocity
of when the man is at the position shown. At
this instant the center of gravity of the culvert and the man is
located at point G,and the radius of gyration about Gis
Determine the angular acceleration of the
culvert. The combined weight of the culvert and the man is
500 lb.Assume that the culvert rolls without slipping, and the
man does not move within the culvert.
kG=3.5 ft.
v=0.5 rad>s
SOLUTIONS
Equations of Motion: The mass moment of inertia of the system about its mass
Kinematics: Since the culvert rolls without slipping,
Applying the relative acceleration equation and referrring to Fig. b,
Equation the iand jcomponents,
Subtituting Eqs. (2) and (3) into Eq. (1),
Ans:
904
17–113.
v0
r
The uniform disk of mass mis rotating with an angular
velocity of when it is placed on the floor. Determine the
initial angular acceleration of the disk and the acceleration
of its mass center.The coefficient of kinetic friction between
the disk and the floor is .mk
v0
SOLUTION
Equations of Motion. Since the disk slips, the frictional force is .The mass
F
f=mkN
Ans:
905
17–114.
v
0
r
The uniform disk of mass mis rotating with an angular
velocity of when it is placed on the floor. Determine the
time before it starts to roll without slipping.What is the
angular velocity of the disk at this instant? The coefficient
of kinetic friction between the disk and the floor is .mk
v0
SOLUTION
Equations of Motion: Since the disk slips, the frictional force is .The mass
moment of inertia of the disk about its mass center is .
Kinematics: At the instant when the disk rolls without slipping,.Thus,
vG=v
r
IG=1
r2
F
f=mkN
Ans:
17–115.
SOLUTION
For A:
For B:
Solving,
90 mm
B
A
D
C
A cord is wrapped around each of the two 10-kg disks.
If they are released from rest, determine the angular
acceleration of each disk and the tension in the cord C.
Neglect the mass of the cord.
Ans:
907
*17–116.
The disk of mass mand radius rrolls without slipping on the
circular path.Determine the normal force which the path
exerts on the disk and the disk’s angular acceleration if at
the instant shown the disk has an angular velocity of V.
SOLUTION
Equation of Motion: The mass moment of inertia of the disk about its center of
Kinematics: Since the semicircular disk does not slip at A, then and
. Substitute into Eq. [1] yields
(aG)t=ar(aG)t=ar
yG=vr
R
r
Ans:
908
17–117.
The uniform beam has a weight W. If it is originally at rest
while being supported at Aand Bby cables, determine the
tension in cable Aif cable Bsuddenly fails. Assume the
beam is a slender rod.
SOLUTION
Also,
+c©F
y=m(aG)y;T
A–W=-
W
gaG
BA
L
––
4
L
––
2
L
––
4
7
909
17–118.
SOLUTION
The 500-lb beam is supported at Aand Bwhen it is
subjected to a force of 1000 lb as shown. If the pin support
at Asuddenly fails, determine the beam’s initial angular
acceleration and the force of the roller support on the beam.
For the calculation, assume that the beam is a slender rod
so that its thickness can be neglected.
BA
8ft2ft
1000 lb
3
4
5
Ans:
910
17–119.
SOLUTION
d=rsin 30° =r
2
The solid ball of radius rand mass mrolls without slipping
down the trough. Determine its angular acceleration.60°
30°
45°
30°
Ans:
911
*17–120.
By pressing down with the finger at B, a thin ring having a
mass m is given an initial velocity and a backspin when
the finger is released. If the coefficient of kinetic friction
between the table and the ring is determine the distance
the ring travels forward before backspinning stops.
mk,
V0
v0
SOLUTION
(c
+)v=v0+act
+c©F
y=0; N
A–mg =0
B
A
0
0
v
ω
r
Ans: