710
16–78.
If the ring gear A rotates clockwise with an angular velocity
of vA
=30 rad>s
, while link BC rotates clockwise with an
angular velocity of vBC
=15 rad>s
, determine the angular
velocity of gear D.
SOLUTION
250 mm
C
vBC 15 rad/s
300 mm
D
A
B
30 rad/
s
vA
16–79.
The mechanism shown is used in a riveting machine.It
consists of a driving piston A, three links, and a riveter which
is attached to the slider block D. Determine the velocity of
Dat the instant shown, when the piston at Ais traveling at
vA=20 m>s.
150 mm 300 mm
v=20m/s
200 mm
A
A
C
D
B
45°
45°
60°30°
45°
SOLUTION
712
*16–80.
The mechanism is used on a machine for the manufacturing
of a wire product. Because of the rotational motion of link
AB and the sliding of block F, the segmental gear lever DE
undergoes general plane motion. If AB is rotating at
v
AB
=
5 rad
>
s, determine the velocity of point E at the
instant shown.
C
A
50 mm
200 mm
20 mm
20 mm
50 mm
45
E
F
D
*16–80. Continued
714
16–81.
In each case show graphically how to locate the
instantaneous center of zero velocity of link AB. Assume
the geometry is known.
SOLUTION
AA
A
B
BB
C
(a)
(b)
v
v
v
16–82.
Determine the angular velocity of link AB at the instant
shown if block Cis moving upward at 12 in.>s.
(5)
Ans.vAB =1.24 rad>s
6.211 =vAB
vB=vAB rAB
=2.1962(2.828) =6.211 in.>s
vB=vBC
(rIC–B)
vBC =2.1962 rad>s
12 =vBC(5.464)
vC=vBC
(rIC –C)
rIC–B=2.828 in.
rIC–C=5.464 in.
A
B
5in. 45
30
4in.
C
V
AB
16–83.
SOLUTION
Bis directed at an angle
with the horizontal. Also, block Cis moving horizontally due to the constraint
of the guide.
Instantaneous Center: The instantaneous center of zero velocity of link BC at the
instant shown is located at the intersection point of extended lines drawn
perpendicular from vBand vC. Using law of sines, we have
The angular velocity of bar BC is given by
Ans.vBC =vB
rB IC
=1.20
0.1768 =6.79 rad s
rC>IC
sin 105° =0.125
sin 30°
rC>IC =0.2415 m
rB>IC
sin 45° =0.125
sin 30°
rB>IC =0.1768 m
30°
The shaper mechanism is designed to give a slow cutting
stroke and a quick return to a blade attached to the slider
at C. Determine the angular velocity of the link CB at the
instant shown, if the link AB is rotating at 4 rad>s.
C
A
45°
125 mm
B
*16–84.
The conveyor belt is moving to the right at v
=
8 ft
>
s, and at
the same instant the cylinder is rolling counterclockwise at
v
=2 rad>s
without slipping. Determine the velocities of
the cylinder’s center C and point B at this instant.
SOLUTION
8
v
v
1 ft
C
B
A
16–85.
The conveyor belt is moving to the right at
v
=
12 ft
>
s, and
at the same instant the cylinder is rolling counterclockwise
at
v
=
6 rad
>
s while its center has a velocity of 4 ft
>
s to the
left. Determine the velocities of points A and B on the disk
at this instant. Does the cylinder slip on the conveyor?
SOLUTION
4
v
v
1 ft
C
B
A
719
16–86.
As the cord unravels from the wheel’s inner hub, the wheel
is rotating at at the instant shown. Determine
the velocities of points Aand B.
v=2 rad>s
Ans.
Ans.
Ans.u=tan–1a2
5b=21.8° R
¬
yA=vrA>IC =2
A
229
B
=10.8 in.>s
yB=vrB>IC =2(7) =14 in.>sT
5in.
2in.
A
=2rad/s
ω
Ans:
720
16–87.
If rod CD is rotating with an angular velocity
v
CD
=
4 rad
>
s, determine the angular velocities of rods AB
and CB at the instant shown.
B
30
C
D
A
vCD
0.4 m
1 m
0.5 m
4 rad/s
*16–88.
SOLUTION
is always directed perpendicular to link AB and its magnitude is
.At the instant shown. vBis directed with an
angle with the horizontal. Also,block Cis moving horizontally due to the
constraint of the guide.
Instantaneous Center: The instantaneous center of zero velocity of bar BC at the
instant shown is located at the intersection point of extended lines drawn
perpendicular from and . Using law of sine, we have
The angular velocity of bar BC is given by
Thus, the velocity of block Cis
Ans.yC=vBC rCIC =1.960(0.6830) =1.34 m s ;
vBC =yB
rB>IC
=1.20
0.6124 =1.960 rad>s
rC>IC
sin 75° =0.5
sin 45°
rC>IC =0.6830 m
rB>IC
sin 60° =0.5
sin 45°
rB>IC =0.6124 m
vC
vB
45°
yB=vAB rAB =6(0.2) =1.20 m>s
If bar AB has an angular velocity , determine
the velocity of the slider block Cat the instant shown.
vAB
=6 rad>s
30°
500 mm
200 mm
AB
=6rad/s
=45°
A
B
C
v
16–89.
Show that if the rim of the wheel and its hub maintain
contact with the three tracks as the wheel rolls, it is necessary
that slipping occurs at the hub A if no slipping occurs at B.
Under these conditions, what is the speed at A if the wheel
has angular velocity
V
?
SOLUTION
IC
is at B.
v
A=
v
(r2–r1) S
Ans.
B
A
v
r2
r1
723
16–90.
Due to slipping, points Aand Bon the rim of the disk have
the velocities shown. Determine the velocities of the center
point Cand point Dat this instant.
SOLUTION
f=10.80°
sin f
0.2667 =sin 135°
1.006
rIC–D=2(0.2667)2+(0.8)2–2(0.2667)(0.8) cos 135° =1.006 ft
v=10
1.06667 =9.375 rad>s
x=1.06667 ft
5x=16 –10x
1.6 –x
5=x
10
C
A
B
F
D
E
v
B
10 ft
/
s
v
A
5ft/s
0.8 ft
30
45
Ans:
vC=2.50
ft>sd
vD=9.43
ft>s
u=55.8° h
16–91.
SOLUTION
=9.375(1.06667 –0.8)
vC=v(rIC –C)
v=10
1.06667 =9.375 rad>s
x=1.06667 ft
5x=16 –10x
1.6 –x
5=x
10
Due to slipping, points Aand Bon the rim of the disk have
the velocities shown. Determine the velocities of the center
point Cand point Eat this instant.
C
A
B
F
D
E
v
B
10 ft
/
s
v
A
5ft/s
0.8 ft
30
45
725
*16–92.
Member AB is rotating at vAB
=6 rad>s.
Determine the
velocity of point D and the angular velocity of members
BPD and CD.
SOLUTION
200 mm
B
A
D
C
P
250 mm
200 mm
200 mm200 mm
vAB 6 rad/s
6060
16–93.
Member AB is rotating at vAB
=6 rad>s.
Determine the
velocity of point P, and the angular velocity of member BPD.
200 mm
B
A
D
C
P
250 mm
200 mm
200 mm200 mm
vAB 6 rad/s
6060
16–94.
The cylinder B rolls on the fixed cylinder A without slipping.
If connected bar CD is rotating with an angular velocity
v
CD
=
5 rad
>
s, determine the angular velocity of cylinder
B. Point C is a fixed point.
SOLUTION
0.3
B
CA
D
0.1 m
0.3 m
vCD 5 rad/s
728
16–95.
As the car travels forward at 80 ft
/
s on a wet road, due to
slipping, the rear wheels have an angular velocity
Determine the speeds of points A, B, and C
caused by the motion.
v=100 rad>s.
SOLUTION
Ans.
Ans.
Ans.vB=1.612(100) =161 ft s60.3°
b
vC=2.2(100) =220 ft s ;
vA=0.6(100) =60.0 ft s:
r=80
100 =0.8 ft
80 ft/s
100 rad/s
1.4 ft A
C
B
*16–96.
The pinion gear A rolls on the fixed gear rack B with an
angular velocity v
=8 rad>s.
Determine the velocity of the
gear rack C.
SOLUTION
150 mm
A
B
C
v