16–1.
Ans.
Ans.aA=2(4)2+(8.45)2=9.35 m>s2
an=v2r=(3.25)2(0.8) =8.45 m>s2
az=ar=5(0.8) =4m>s2
vA=vr=3.25(0.8) =2.60 m>s
a=5 rad>s2
v=3.25 rad>s
t=0.5 s
a=dv
dt =10 t
The angular velocity of the disk is defined by
where tis in seconds. Determine the
magnitudes of the velocity and acceleration of point Aon
the disk when t=0.5 s.
v=15t2+22rad>s,
A
0.8 m
16–2.
The angular acceleration of the disk is defined by
a=3t2+12
rad
>
s, where t is in seconds. If the disk is
originally rotating at v
0=12
rad
>
s, determine the
magnitude of the velocity and the n and t components of
acceleration of point A on the disk when t
=
2 s.
SOLUTION
0.4 m
0.5 m
B
A
v0
12 rad
/
s
16–3.
The disk is originally rotating at
v
0
=
12 rad
>
s. If it is
subjected to a constant angular acceleration of
a
=
20 rad
>
s2, determine the magnitudes of the velocity
and the n and t components of acceleration of point A at the
instant t
=
2 s.
0.4 m
0.5 m
B
A
v0
12 rad
/
s
16–5.
SOLUTION
Ans.
Angular Velocity: Applying Eq. 16–1. we have
Ans.
Angular Acceleration: Applying Eq. 16–2. we have
Ans.a=dv
dt =8 rad s2
v=du
dt =20 +8t2t=90 s
=740 rad>s
u=20(90) +4
A
902
B
=(34200 rad) *
¢
1 rev
2prad
≤
=5443 rev
The disk is driven by a motor such that the angular position
of the disk is defined by where tis in
seconds. Determine the number of revolutions, the angular
velocity, and angular acceleration of the disk when t=90 s.
u=120t+4t22rad,
0.5 ft
θ
635
16–6.
SOLUTION
Ans.
Ans.t=1.67 s
15=10+3t
v=v0+act
u=20.83rad=20.83
¢
1
2p
≤
=3.32 rev.
(15)2=(10)2+2(3)(u–0)
v2=v0
2+2ac(u–u0)
Awheel has an initial clockwise angular velocity of
a
nd a constant angular acceleration of Determine
t
he number of revolutions it must undergo to acquire a
c
lockwise angular velocity of .What time is
r
equired?
15 rad>s
3rad>s2.
10 rad>s
Ans:
u=3.32
rev
t=1.67
s
636
16–7.
DA
B
C
F
If gear Arotates with a constant angular acceleration of
starting from rest, determine the time
required for gear Dto attain an angular velocity of 600 rpm.
Also, find the number of revolutions of gear Dto attain this
angular velocity. Gears A,B,C, and Dhave radii of 15 mm,
50 mm, 25 mm, and 75 mm, respectively.
aA=90 rad>s2,
SOLUTION
Since gears Cand Bshare the same shaft, .Also, gear Dis in
mesh with gear C.Thus,
The final angular velocity of gear Dis
. Applying the constant acceleration equation,
Ans.
and
Ans. =34.9 rev
uD=(219.32 rad)a1 rev
2p rad b
(20p)2=02+2(9)(uD–0)
vD
2=(vD)0
2+2aD [uD–(uD)0]
t=6.98 s
20p=0+9t
vD
=(vD)0
+aD t
20p rad>s
vD=a600 rev
min ba2p rad
1 rev ba1 min
60 s b =
aD=arC
rDbaC=a25
75 b(27) =9 rad>s2
aD rD=aC rC
aC=aB=27 rad>s2
aB=arA
rBbaA=a15
50 b(90) =27 rad>s2
Ans:
t=6.98
s
u
D=34.9
rev
637
Ans:
a
D=
0.4
rad
>
s
2
*16–8.
DA
B
C
F
If gear Arotates with an angular velocity of
,where is the angular displacement of
g
ear A, measured in radians, determine the angular
acceleration
of gear Dwhen , starting from rest.
Gear
s A,B,C,and Dhave radii of 15 mm, 50 mm, 25 mm,
and 75 mm,
respectively.
uA=3 rad
uA
(uA+1) rad>s
vA
=
SOLUTION
M
otion of Gear A:
At
,
M
otion of Gear D: Gear Ais in mesh with gear B.Thus,
Since
gears Cand Bshare the same shaft .Also, gear Dis in
me
sh with gear C.Thus,
Ans.aD=arC
rD
baC=a25
75
b(1.20) =0.4 rad>s2
aD rD=aC rC
aC=aB=1.20 rad>s2
aB=arA
rBbaA=a15
50 b(4) =1.20 rad>s2
aB rB=aA rA
aA=3+1=4 rad>s2
uA=3 rad
aA=(uA+1)
aA duA=(uA+1) duA
aA duA=(uA+1) d(uA+1)
aA duA=vA dvA
16–9.
At the instant vA
=5 rad>s
, pulley A is given an angular
acceleration a
=
(0.8u) rad
>
s
2
, where
u
is in radians.
Determine the magnitude of acceleration of point B on
pulley C when A rotates 3 revolutions. Pulley C has an inner
hub which is fixed to its outer one and turns with it.
SOLUTION
50 mm
40 mm
B
A
C
vA
aA
639
16–10.
At the instant vA
=5 rad>s
, pulley A is given a constant
angular acceleration aA
=
6 rad
>
s
2
. Determine the
magnitude of acceleration of point B on pulley C when A
rotates 2 revolutions. Pulley C has an inner hub which is
fixed to its outer one and turns with it.
SOLUTION
50 mm
40 mm
B
A
C
vA
aA
Ans:
a
B=
16.5 m
>
s
2
640
16–11.
The cord, which is wrapped around the disk, is given an
acceleration of a
=
(10t) m
>
s
2
, where t is in seconds.
Starting from rest, determine the angular displacement,
angular velocity, and angular acceleration of the disk when
t=3 s
.
SOLUTION
3
a (10t) m/s2
0.5 m
Ans:
a
=
60 rad
>
s
2
v
=90.0 rad>s
u=90.0 rad
641
*16–12.
The power of a bus engine is transmitted using the belt-and-
pulley arrangement shown. If the engine turns pulley A at
vA
=(20t+40) rad>s
, where t is in seconds, determine the
angular velocities of the generator pulley B and the
air-conditioning pulley C when
t=3 s
.
Ans:
v
B=300 rad>s
v
C=600 rad>s
SOLUTION
t=3 s
r
C
0.05 =600 rad
B
D
C
A
25 mm
75 mm
50 mm
100 mm
vA
vB
vC
642
16–13.
The power of a bus engine is transmitted using the belt-and-
pulley arrangement shown. If the engine turns pulley A at
vA
=60 rad>s
, determine the angular velocities of the
generator pulley B and the air-conditioning pulley C. The
hub at D is rigidly connected to B and turns with it.
SOLUTION
r
C
0.05 =360 rad
B
D
A
25 mm
75 mm
50 mm
100 mm
vB
vC
Ans:
v
B=180 rad>s
v
C=360 rad>s
643
16–14.
The disk starts from rest and is given an angular acceleration
where tis in seconds.Determine the
an
gular velocity of the disk and its angular displacement
when
t=4 s.
a
=(2t 2) rad>s2,
SOLUTION
W
hen ,
Ans.
W
hen ,
Ans.u=1
6
(4)4=42.7 rad
t=4 s
u=1
6 t4
Lu
0
du=Lt
0
2
3 t3dt
v=2
3(4)3=42.7 rad>s
t=4 s
v=2
3t3
v=2
3 t320
t
Lv
0
dv=Lt
0
2 t2dt
a=dv
dt =2 t2
0.4 m
P
Ans:
v
=42.7
rad>s
u=42.7
rad
644
16–15.
0.4 m
P
The disk starts from rest and is given an angular acceleration
,where tis in seconds.Determine the
magnitudes of the normal and tangential components of
acceleration of a point Pon the rim of the disk when t=2 s.
a=(5t1>2) rad>s2
SOLUTION
When ,
When ,
Motion of point P: The tangential and normal components of the acceleration of
point P when are
Ans.
Ans.a
n=
v2r
=
9.4282(0.4)
=
35.6 m
>
s2
at=ar=7.071(0.4) =2.83 m>s2
t=2 s
a=5
A
21
2
B
=7.071 rad>s2
t=2 s
v=10
3
A
23
2
B
=9.428 rad>s
t=2 s
v=e10
3 t3
2f rad>s
v20
v
=10
3 t3
2 20
t
Lv
0
dv=Lt
0
5t
1
2dt
dv=adt
Ans:
a
t=
2.83
m
>
s
2
a
n=
35.6
m
>
s
2
645
*16–16.
SOLUTION
0.4 m
P
The disk starts at
v0
= 1 rad>s when
u
= 0, and is given an
angular acceleration a = (0.3u) rad>s2, where u is in radians.
Determine the magnitudes of the normal and tangential
components of acceleration of a point P on the rim of the
disk when u = 1 rev.
Ans.
Ans.
v2
2–0.5 =0.15u2
1
2v221
v
=0.15u220
u
Lv
1
vdv=Lu
0
0.3udu
a=0.3u
At
Ans.
Ans.
ap=2(0.7540)2+(5.137)2=5.19 m>s2
an=v2r=(3.584 rad>s)2(0.4 m) =5.137 m>s2
at=ar=0.3(2p) rad>s (0.4 m) =0.7540 m>s2
v=3.584 rad>s
v=20.3(2p)2+1
u=1 rev =2p rad
v=20.3u2+1
2
Ans:
a
t=
0.7540
m
>
s
2
a
n=
5.137
m
>
s
2
646
16–17.
A motor gives gear A an angular acceleration of
aA
=
(2
+
0.006 u
2
) rad
>
s
2
, where
u
is in radians. If this
gear is initially turning at vA
=15 rad>s
, determine the
angular velocity of gear B after A undergoes an angular
displacement of 10 rev.
B
175 mm
100 mm
A
a
A
v
A
a
B
647
16–18.
A motor gives gear A an angular acceleration of
aA
=
(2t
3
) rad
>
s
2
, where t is in seconds. If this gear is
initially turning at vA
=15 rad>s
, determine the angular
velocity of gear B when
t=3 s
.
SOLUTION
t=0 s
t=3 s
B
175 mm
100 mm
A
a
A
v
A
a
B
Ans:
vB
=31.7 rad>s
d
648
16–19.
SOLUTION
When
Motion of the Beater Brush: Since the brush is connected to the shaft by a non-slip
belt, then
Ans.vB=
¢
rs
rB
≤
vs=a0.25
1b(625) =156 rad>s
vBrB=vsrs
vs=54=625 rad>s
t=4s
vS=
(
t+1 4
)
t2t
0=vS1>42vs
1
Lt
0
dt =Lvs
1
dvS
4vS3>4
Ldt =LdvS
aS
The vacuum cleaner’s armature shaft S rotates with an
angular acceleration of a = 4v
3>4
rad>s
2
, where v is in
rad>s. Determine the brush’s angular velocity when t = 4 s,
starting from v
0
= 1 rad>s, at u = 0. The radii of the shaft
and the brush are 0.25 in. and 1 in., respectively. Neglect the
thickness of the drive belt.
Ans:
v
B=156 rad>s
*16–20.
SOLUTION
A motor gives gear A an angular acceleration of
aA=(4t3) rad>s2, where t is in seconds. If this gear is
initially turning at (vA)0=20 rad>s, determine the angular
velocity of gear B when t=2 s. A
B
0.15 m
0.05 m
( A)0 = 20 rad/s
A
α
ω