91962_06_s16_p0513-0640 6/8/09 2:09 PM Page 513
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they curr
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
•16–1. A disk having a radius of 0.5 ft rotates with an
initial angular velocity of 2 rad>s and has a constant angular
acceleration of 1 rad>s2. Determine the magnitudes of the
velocity and acceleration of a point on the rim of the disk
when t = 2 s.
v = v0 + ac t;
v = 2 + 1(2) = 4 rad>s
v = rv; v = 0.5(4) = 2 ft>s Ans.
at = ra ; at = 0.5(1) = 0.5 ft>s2
an = v2 r; an = (4)2(0.5) = 8 ft>s2
a = 282 + (0.5)2 = 8.02 ft>s2 Ans.
16–2. Just after the fan is turned on, the motor gives the
blade an angular acceleration a = (20e - 0.6t) rad>s2, where t
is in seconds. Determine the speed of the tip P of one of the
blades when t = 3 s. How many revolutions has the blade 1.75 ft
turned in 3 s? When t = 0 the blade is at rest. P
dv = a dt
v t
L0 L0
dv = 20e - 0.6t dt
v = - e 2 = 33.3 A 1 - e - 0.6t B
20 - 0.6t t
0.6 0
v = 27.82 rad>s when t = 3s
vp = vr = 27.82(1.75) = 48.7 ft>s Ans.
du = v dt
33.3 A 1 - e - 0.6t B dt
u t
L0 L0
du =
b e - 0.6t b 2 = 33.3c3 + a b A e - 0.6(3) - 1 B d
3
u = 33.3 at + a
1 1
0.6 0 0.6
u = 53.63 rad = 8.54 rev Ans.
513
,91962_06_s16_p0513-0640 6/8/09 2:09 PM Page 514
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they current
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
16–3. The hook is attached to a cord which is wound
around the drum. If it moves from rest with an
acceleration of 20 ft>s2, determine the angular acceleration 2 ft
of the drum and its angular velocity after the drum has
completed 10 rev. How many more revolutions will the
drum turn after it has first completed 10 rev and the hook
continues to move downward for 4 s?
a ! 20 ft
Angular Motion: The angular acceleration of the drum can be determine by
applying Eq. 16–11.
at = ar; 20 = a(2) a = 10.0 rad>s2 Ans.
Applying Eq. 16–7 with ac = a = 10.0 rad>s2 and u = (10 rev) * a b
2p rad
1 rev
= 20p rad, we have
v2 = v20 + 2ac (u - u0)
v2 = 0 + 2(10.0)(20p - 0)
v = 35.45 rad>s = 35.4 rad>s Ans.
The angular displacement of the drum 4 s after it has completed 10 revolutions can
be determined by applying Eq. 16–6 with v0 = 35.45 rad>s.
1
u = u0 + v0 t + a t2
2 c
(10.0) A 42 B
1
= 0 + 35.45(4) +
2
= (221.79 rad) * a b = 35.3rev
1 rev
Ans.
2p rad
514
,91962_06_s16_p0513-0640 6/8/09 2:10 PM Page 515
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they curr
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*16–4. The torsional pendulum (wheel) undergoes
oscillations in the horizontal plane, such that the angle of
rotation, measured from the equilibrium position, is given
by u = (0.5 sin 3t) rad, where t is in seconds. Determine the
maximum velocity of point A located at the periphery of
the wheel while the pendulum is oscillating. What is the
acceleration of point A in terms of t?
Angular Velocity: Here, u = (0.5 sin 3t) rad>s. Applying Eq. 16–1, we have
du
v = = (1.5 cos 3t) rad>s
dt
By observing the above equation, the angular velocity is maximum if cos 3t = 1.
Thus, the maximum angular velocity is vmax = 1.50 rad>s. The maximum speed of
2 ft
point A can be obtained by applying Eq. 16–8. u
(yA)max = vmax r = 1.50(2) = 3.00 ft>s Ans.
Angular Acceleration: Applying Eq. 16–2, we have
dv
a = = (-4.5 sin 3t) rad>s2
dt
The tangential and normal components of the acceleration of point A can be
determined using Eqs. 16–11 and 16–12, respectively.
at = ar = (-4.5 sin 3t)(2) = (-9 sin 3t) ft>s2
an = v2 r = (1.5 cos 3t)2 (2) = A 4.5 cos2 3t B ft>s2
Thus,
aA = A -9 sin 3tut + 4.5 cos2 3tun B ft>s2 Ans.
•16–5. The operation of reverse gear in an automotive G
transmission is shown. If the engine turns shaft A at
vA = 40 rad>s, determine the angular velocity of the drive H
vA ! 40 r
shaft, vB. The radius of each gear is listed in the figure. vB
D A
B
C
rA vA = rC vC : 80(40) = 40vC vC = vD = 80 rad>s
rG ! 80 mm
vE rE = vD rD : vE(50) = 80(40) vE = vF = 64 rad>s rC ! rD ! 4
F E rE ! rH ! 5
vF rF = vB rB : 64(70) = vB (50) vB = 89.6 rad>s rF ! 70 mm
vB = 89.6 rad>s Ans.
515
, 91962_06_s16_p0513-0640 6/8/09 2:10 PM Page 516
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they current
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
16–6. The mechanism for a car window winder is shown in
the figure. Here the handle turns the small cog C, which vw
rotates the spur gear S, thereby rotating the fixed-connected
D
lever AB which raises track D in which the window rests.
The window is free to slide on the track. If the handle is 20 mm
wound at 0.5 rad>s, determine the speed of points A and E 0.5 rad/s
C
and the speed vw of the window at the instant u = 30°. A
50 mm
SB
200 mm E
F u
vC = vC rC = 0.5(0.02) = 0.01 m>s 200 mm
vC 0.01
vS = = = 0.2 rad>s
rS 0.05
vA = vE = vS rA = 0.2(0.2) = 0.04 m>s = 40 mm>s Ans.
Points A and E move along circular paths. The vertical component closes the
window.
vw = 40 cos 30° = 34.6 mm>s Ans.
16–7. The gear A on the drive shaft of the outboard motor
has a radius rA = 0.5 in. and the meshed pinion gear B on the
propeller shaft has a radius rB = 1.2 in. Determine the
angular velocity of the propeller in t = 1.5 s, if the drive shaft
rotates with an angular acceleration a = (400t3) rad>s2,
where t is in seconds. The propeller is originally at rest and
the motor frame does not move. A
B
2.20 in.
P
Angular Motion: The angular velocity of gear A at t = 1.5 s must be determined
first. Applying Eq. 16–2, we have
dv = adt
vA 1.5 s
L0 L0
dv = 400t3 dt
vA = 100t4 |1.5
0
s
= 506.25 rad>s
However, vA rA = vB rB where vB is the angular velocity of propeller. Then,
vA = a b(506.25) = 211 rad>s
rA 0.5
vB = Ans.
rB 1.2
516
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they curr
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
•16–1. A disk having a radius of 0.5 ft rotates with an
initial angular velocity of 2 rad>s and has a constant angular
acceleration of 1 rad>s2. Determine the magnitudes of the
velocity and acceleration of a point on the rim of the disk
when t = 2 s.
v = v0 + ac t;
v = 2 + 1(2) = 4 rad>s
v = rv; v = 0.5(4) = 2 ft>s Ans.
at = ra ; at = 0.5(1) = 0.5 ft>s2
an = v2 r; an = (4)2(0.5) = 8 ft>s2
a = 282 + (0.5)2 = 8.02 ft>s2 Ans.
16–2. Just after the fan is turned on, the motor gives the
blade an angular acceleration a = (20e - 0.6t) rad>s2, where t
is in seconds. Determine the speed of the tip P of one of the
blades when t = 3 s. How many revolutions has the blade 1.75 ft
turned in 3 s? When t = 0 the blade is at rest. P
dv = a dt
v t
L0 L0
dv = 20e - 0.6t dt
v = - e 2 = 33.3 A 1 - e - 0.6t B
20 - 0.6t t
0.6 0
v = 27.82 rad>s when t = 3s
vp = vr = 27.82(1.75) = 48.7 ft>s Ans.
du = v dt
33.3 A 1 - e - 0.6t B dt
u t
L0 L0
du =
b e - 0.6t b 2 = 33.3c3 + a b A e - 0.6(3) - 1 B d
3
u = 33.3 at + a
1 1
0.6 0 0.6
u = 53.63 rad = 8.54 rev Ans.
513
,91962_06_s16_p0513-0640 6/8/09 2:09 PM Page 514
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they current
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
16–3. The hook is attached to a cord which is wound
around the drum. If it moves from rest with an
acceleration of 20 ft>s2, determine the angular acceleration 2 ft
of the drum and its angular velocity after the drum has
completed 10 rev. How many more revolutions will the
drum turn after it has first completed 10 rev and the hook
continues to move downward for 4 s?
a ! 20 ft
Angular Motion: The angular acceleration of the drum can be determine by
applying Eq. 16–11.
at = ar; 20 = a(2) a = 10.0 rad>s2 Ans.
Applying Eq. 16–7 with ac = a = 10.0 rad>s2 and u = (10 rev) * a b
2p rad
1 rev
= 20p rad, we have
v2 = v20 + 2ac (u - u0)
v2 = 0 + 2(10.0)(20p - 0)
v = 35.45 rad>s = 35.4 rad>s Ans.
The angular displacement of the drum 4 s after it has completed 10 revolutions can
be determined by applying Eq. 16–6 with v0 = 35.45 rad>s.
1
u = u0 + v0 t + a t2
2 c
(10.0) A 42 B
1
= 0 + 35.45(4) +
2
= (221.79 rad) * a b = 35.3rev
1 rev
Ans.
2p rad
514
,91962_06_s16_p0513-0640 6/8/09 2:10 PM Page 515
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they curr
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*16–4. The torsional pendulum (wheel) undergoes
oscillations in the horizontal plane, such that the angle of
rotation, measured from the equilibrium position, is given
by u = (0.5 sin 3t) rad, where t is in seconds. Determine the
maximum velocity of point A located at the periphery of
the wheel while the pendulum is oscillating. What is the
acceleration of point A in terms of t?
Angular Velocity: Here, u = (0.5 sin 3t) rad>s. Applying Eq. 16–1, we have
du
v = = (1.5 cos 3t) rad>s
dt
By observing the above equation, the angular velocity is maximum if cos 3t = 1.
Thus, the maximum angular velocity is vmax = 1.50 rad>s. The maximum speed of
2 ft
point A can be obtained by applying Eq. 16–8. u
(yA)max = vmax r = 1.50(2) = 3.00 ft>s Ans.
Angular Acceleration: Applying Eq. 16–2, we have
dv
a = = (-4.5 sin 3t) rad>s2
dt
The tangential and normal components of the acceleration of point A can be
determined using Eqs. 16–11 and 16–12, respectively.
at = ar = (-4.5 sin 3t)(2) = (-9 sin 3t) ft>s2
an = v2 r = (1.5 cos 3t)2 (2) = A 4.5 cos2 3t B ft>s2
Thus,
aA = A -9 sin 3tut + 4.5 cos2 3tun B ft>s2 Ans.
•16–5. The operation of reverse gear in an automotive G
transmission is shown. If the engine turns shaft A at
vA = 40 rad>s, determine the angular velocity of the drive H
vA ! 40 r
shaft, vB. The radius of each gear is listed in the figure. vB
D A
B
C
rA vA = rC vC : 80(40) = 40vC vC = vD = 80 rad>s
rG ! 80 mm
vE rE = vD rD : vE(50) = 80(40) vE = vF = 64 rad>s rC ! rD ! 4
F E rE ! rH ! 5
vF rF = vB rB : 64(70) = vB (50) vB = 89.6 rad>s rF ! 70 mm
vB = 89.6 rad>s Ans.
515
, 91962_06_s16_p0513-0640 6/8/09 2:10 PM Page 516
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they current
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
16–6. The mechanism for a car window winder is shown in
the figure. Here the handle turns the small cog C, which vw
rotates the spur gear S, thereby rotating the fixed-connected
D
lever AB which raises track D in which the window rests.
The window is free to slide on the track. If the handle is 20 mm
wound at 0.5 rad>s, determine the speed of points A and E 0.5 rad/s
C
and the speed vw of the window at the instant u = 30°. A
50 mm
SB
200 mm E
F u
vC = vC rC = 0.5(0.02) = 0.01 m>s 200 mm
vC 0.01
vS = = = 0.2 rad>s
rS 0.05
vA = vE = vS rA = 0.2(0.2) = 0.04 m>s = 40 mm>s Ans.
Points A and E move along circular paths. The vertical component closes the
window.
vw = 40 cos 30° = 34.6 mm>s Ans.
16–7. The gear A on the drive shaft of the outboard motor
has a radius rA = 0.5 in. and the meshed pinion gear B on the
propeller shaft has a radius rB = 1.2 in. Determine the
angular velocity of the propeller in t = 1.5 s, if the drive shaft
rotates with an angular acceleration a = (400t3) rad>s2,
where t is in seconds. The propeller is originally at rest and
the motor frame does not move. A
B
2.20 in.
P
Angular Motion: The angular velocity of gear A at t = 1.5 s must be determined
first. Applying Eq. 16–2, we have
dv = adt
vA 1.5 s
L0 L0
dv = 400t3 dt
vA = 100t4 |1.5
0
s
= 506.25 rad>s
However, vA rA = vB rB where vB is the angular velocity of propeller. Then,
vA = a b(506.25) = 211 rad>s
rA 0.5
vB = Ans.
rB 1.2
516
