17.71 is in the position shown here, use instantaneous
centers to determine the horizontal velocity of B.
A
B
1 rad/s
O
Solution: The strategy is to determine the intersection of lines
perpendicular to the motions at Aand B. The velocity of Ais parallel
the intersecting lines and the base are similar, and thus the interior
angles are known for the larger triangle. From the law of sines
C
6)=30 in. from which 30ωAB =6 in/s, from which ωAB =0.2 rad/s.
check.]. The vector from the instantaneous center to point Bis
Problem 17.73 The angle θ=45◦, and the bar OQ is
rotating in the counterclockwise direction at 0.2 rad/s.
Use instantaneous centers to determine the velocity of
the sleeve P.2 ft
O
Q
P
θ
2 ft
Solution: The velocity of Qis
P
O
Instantaneous
p
Problem 17.74 Bar AB is rotating in the counterclock-
wise direction at 5 rad/s. The disk rolls on the horizontal
surface. Determine the angular velocity of bar BC.C
0.4 m 0.2 m 0.2 m
D
A
B
5 rad/s
374
Problem 17.75 Bar AB rotates at 6 rad/s in the clock-
wise direction. Use instantaneous centers to determine
the angular velocity of bar BC.
6 rad/s
A
B
4 in
3 in
The vector location of a point on a line perpendicular to the velocity
of 14 in. and interior angles of 45◦, from which the coordinates of C′
are (14, 14) in. check.]. The angular velocity of bar BC is determined
from the known velocity at B. The vector from the instantaneous center
to point Bis
Problem 17.76 The crank AB is rotating in the clock-
wise direction at 2000 rpm (revolutions per minute).
(a) At the instant shown, what are the coordinates of
the instantaneous center of the connecting rod BC ?
(b) Use instantaneous centers to determine the angular
velocity of the connecting rod BC at the
instant shown. B
A
50 mm
50 mm
175 mm
y
x
C
Solution:
Q
376
Problem 17.77 The disks roll on the plane surface.
The left disk rotates at 2 rad/s in the clockwise direction.
Use the instantaneous centers to determine the angular
velocities of the bar and the right disk.
2 rad/s
1 ft1 ft
3 ft
Problem 17.78 Bar AB rotates at 12 rad/s in the clock-
wise direction. Use instantaneous centers to determine
the angular velocities of bars BC and CD.
C
B
350
mm
200
mm
12 rad/s
378
Problem 17.79 The horizontal member ADE support-
ing the scoop is stationary. The link BD is rotating in the
clockwise direction at 1 rad/s. Use instantaneous centers
to determine the angular velocity of the scoop. 1 ft 6 in
2 ft 6 in
1 ft
5 ft
2 ft
C
B
DE
A
Scoop
ωCE =vC
rCE =2.2
1.5=1.47 rad/s
Problem 17.80 The disk is in planar motion. The
directions of the velocities of points Aand Bare shown.
The velocity of point Ais vA=2 m/s.
(a) What are the coordinates of the disk’s instanta-
neous center?
(b) Determine the velocity vBand the disk’s angular
velocity. B
Ax
y
(0.5, 0.4)m
30
70°
B
A
Solution:
0=vAx−ωyc(1)
A
vB
y
30°
380
Problem 17.81 The rigid body rotates about the zaxis
with counterclockwise angular velocity ω=4 rad/s and
counterclockwise angular acceleration α=2 rad/s2. The
distance rA/B =0.6m.
(a) What are the rigid body’s angular velocity and
angular acceleration vectors?
(b) Determine the acceleration of point Arelative to
point B,first by using Eq. (17.9) and then by using
Eq. (17.10).
x
y
BA
ω
rA/B
α
Solution:
y
Problem 17.82 The bar rotates with a counterclock-
wise angular velocity of 5 rad/s and a counterclockwise
angular acceleration of 30 rad/s2. Determine the accel-
eration of A(a) by using Eq. (17.9) and (b) by using
Eq. (17.10).
x
y
A
2 m
30 rad/s2
5 rad/s
30°
Solution:
382
Problem 17.83 The bar rotates with a counterclock-
wise angular velocity of 20 rad/s and a counterclockwise
angular acceleration of 6 rad/s2.
(a) By applying Eq. (17.10) to point Aand the fixed
point O, determine the acceleration of A.
(b) By using the result of part (a) and Eq. (17.10),
to points Aand B, determine the acceleration
point B.x
y
AB
O
1 m 1 m
20 rad/s 6 rad/s2
Solution:
Problem 17.84 The helicopter is in planar motion in
the x–yplane. At the instant shown, the position of its
center of mass Gis x=2m,y=2.5 m, its velocity is
vG=12i+4j(m/s), and its acceleration is aG=2i+
3j(m/s2). The position of point Twhere the tail rotor
is mounted is x=−3.5m,y=4.5 m. The helicopter’s
angular velocity is 0.2 rad/s clockwise, and its angular
acceleration is 0.1 rad/s2counterclockwise. What is the
acceleration of point T?
y
x
T
G
Problem 17.85 Point Aof the rolling disk is moving
toward the right and accelerating toward the right. The
magnitude of the velocity of point Cis 2 m/s, and the
magnitude of the acceleration of point Cis 14 m/s2.
Determine the acceleration of points Band D. (See
Active Example 17.5.)
x
y
A
45⬚
C
D
B
300 mm
Solution: First the velocity analysis
384
Problem 17.86 The disk rolls on the circular surface
with a constant clockwise angular velocity of 1 rad/s.
What are the accelerations of points Aand B?
Strategy: Begin by determining the acceleration of the
center of the disk. Notice that the center moves in a cir-
cular path and the magnitude of its velocity is constant. 0.4 m
A
B
y
x
Solution:
Problem 17.87 The length of the bar is L=4ftand
1.8 rad/s and its angular acceleration is α=6 rad/s2.
The endpoints of the bar slide on the plane surfaces.
points of the bar to determine their accelerations.
Solution: Call the top point Dand the bottom point B.
Put in the known constraints
Equating components we have
Now we can use either point as a base point to find the acceleration
of point G. We will use Bas the base point.
Problem 17.88 The angular velocity and angular
acceleration of bar AB are ωAB =2 rad/s and αAB =
10 rad/s2. The dimensions of the rectangular plate are
12 in ×24 in. What are the angular velocity and angular
acceleration of the rectangular plate? 12 in
B
D
α
ω
AB
20 in
1.2 rad/s. Use Eq. (17.10) to determine the accelerations. The accel-
eration of point Ais
ijk
386
Problem 17.89 The ring gear is stationary, and the
sun gear has an angular acceleration of 10 rad/s2in the
counterclockwise direction. Determine the angular accel-
eration of the planet gears.
20 in
Sun gear
34 in
Planet gears (3)
Ring gear
7 in
Solution: The strategy is to use the tangential acceleration at the
aST =α×rS=
ijk
0010
020 0
=−200i(in/s2).
This is also the tangential acceleration of the planet gear at the point
of contact. At the contact with the ring gear, the planet gears are
stationary, hence the angular acceleration of the planet gear satisfies
αP×(−2rP)=
ij k
00αP
0−14 0
=−200i
from which
αP=−200
14 =−14.29 (rad/s2)(clockwise).
Problem 17.90 In Example 17.6, what is the acceler-
ation of the midpoint of bar BC?
Solution: From Example 17.6 we know that
Problem 17.91 The 1-m-diameter disk rolls, and point
Bof the 1-m-long bar slides, on the plane surface. Deter-
mine the angular acceleration of the bar and the accel-
eration of point B.
A
O
B
4 rad/s
10 rad/s2
0.50 0
The motion at point B is constrained to be parallel to the xaxis. The
line perpendicular to the velocity of B is parallel to the yaxis. The line
rA/C =rA−rC=−0.866i−0.866j(m). The angular velocity of the
0.866 =−2.31 (rad/s).
The acceleration of the center of the rolling disk is aO=−αRi=
to the xaxis. Separate components:
aB=−13 +0.5αAB −ω2
AB (0.866),
388
Problem 17.92 If θ=45◦and sleeve Pis moving to
the right with a constant velocity of 2 m/s, what are the
angular accelerations of the bars OQ and PQ?
O
Q
P
θ
1.2 m
Solution:
vQy =0.848ωoQ (2)
Also,
Problem 17.93 Consider the system shown in Prob-
lem 17.92. If θ=50◦and bar OQ has a constant clock-
wise angular velocity of 1 rad/s, what is the acceleration
of sleeve P?1.2 m
O
Q
P
θ
1.2 m
Solution:
We need another eqn. To get it, we use the velocity relationships and
and (2), we solve to get
390
Problem 17.94 The angle θ=60◦, and bar OQ has
a constant counterclockwise angular velocity of 2 rad/s.
What is the angular acceleration of the bar PQ?
Q
200 mm 400 mm
0.2 cos 60◦0.2 sin 60◦0
0.4 cos β−0.4 sin β0
−(−0.555)2(0.4 cos βi−0.4 sin βj).
Equating jcomponents
0=−0.8 sin 60◦+0.4αPQ cos β+(0.555)20.4 sin β.
Solving, we obtain αPQ =1.77 rad/s2.
x
O60°
P
β
vp
ap
Problem 17.95 At the instant shown, the piston’s velo-
city and acceleration are vC=−14i(m/s) and aC=
−2200i(m/s2). What is the angular acceleration of the
crank AB?
B
A
50 mm
50 mm
175 mm
y
x
C
Solution: The velocity analysis:
392