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912
18–1.
Q.E.D.=1
2IIC v2
=1
2
A
mr2
G>IC +IG
B
v2However mr2
G>IC +IG=IIC
=1
2m(vrG>IC)2+1
2IGv2
At a given instant the body of mass mhas an angular
velocity and its mass center has a velocity . Show that
its kinetic energy can be represented as , where
is the moment of inertia of the body determined about
the instantaneous axis of zero velocity, located a distance
from the mass center as shown.rG>IC
IIC
T=1
2IICv2
vG
V
IC
G
V
rG/IC
vG
18–2.
M
O
0.5 m
The wheel is made from a 5-kg thin ring and two 2-kg
slender rods. If the torsional spring attached to the wheel’s
center has a stiffness , and the wheel is
rotated until the torque is developed,
determine the maximum angular velocity of the wheel if it
is released from rest.
M=25 N #m
k=2 N #m>rad
18–3.
The wheel is made from a 5-kg thin ring and two 2-kg slender
rods. If the torsional spring attached to the wheel’s center has
a stiffness so that the torque on the center
of the wheel is where is in radians,
determine the maximum angular velocity of the wheel if it is
rotated two revolutions and then released from rest.
uM=12u2N#m,
k=2N#m>rad,
SOLUTION
Ans.v=14.1 rad/s
(4p)2=0.7917v2
0+L4p
0
2udu=1
2(1.583) v2
T1+©U1-2=T2
Io=2c1
12(2)(1)2d+5(0.5)2=1.583
M
O
0.5 m
*18–4.
A force of P
=
60 N is applied to the cable, which causes
the 200-kg reel to turn since it is resting on the two rollers
Aand B of the dispenser. Determine the angular velocity of
the reel after it has made two revolutions starting from rest.
Neglect the mass of the rollers and the mass of the cable.
Assume the radius of gyration of the reel about its center
axis remains constant at kO
=
0.6 m.
SOLUTION
0.75 m
0.6 m
1 m
P
A
O
B
Ans:
18–5.
A force of P=20 N is applied to the cable, which causes
the 175-kg reel to turn since it is resting on the two rollers
A and B of the dispenser. Determine the angular velocity of
the reel after it has made two revolutions starting from rest.
Neglect the mass of the rollers and the mass of the cable.
The radius of gyration of the reel about its center axis is
kG=0.42
m.
500 mm
400 mm
250 mm
30°
P
A
G
B
18–6.
A force of P=20 N is applied to the cable, which causes
the 175-kg reel to turn without slipping on the two rollers A
and B of the dispenser. Determine the angular velocity of
the reel after it has made two revolutions starting from rest.
Neglect the mass of the cable. Each roller can be considered
as an 18-kg cylinder, having a radius of 0.1 m. The radius of
gyration of the reel about its center axis is kG=0.42
m.
SOLUTION
500 mm
400 mm
250 mm
30°
P
A
G
B
18–7.
SOLUTION
Ans.=283 ft #lb
2a50
32.2 (0.6)2b(20)2+1
2a20
32.2 b
2+1
2a30
32.2 b
2
T=1
2IOv2
O+1
2mAv2
A+1
2mBv2
B
1 ft
0.5ft
O
30 lb
B
v 20 rad/s
The double pulley consists of two parts that are attached to
one another. It has a weight of 50 lb and a radius of gyration
about its center of k = 0.6 ft and is turning with an angular
velocity of 20 rad>s clockwise. Determine the kinetic energy
of the system. Assume that neither cable slips on the pulley.
O
919
*18–8.
1 ft
0.5 ft
O
30 lb
B
v
20 rad
/
s
The double pulley consists of two parts that are attached to
one another. It has a weight of 50 lb and a centroidal radius
of gyration of and is turning with an angular
velocity of 20 rad s clockwise. Determine the angular
velocity of the pulley at the instant the 20-lb weight moves
2 ft downward.
>
kO=0.6 ft
SOLUTION
pulley about point Ois slugft2.Thus, the
kinetic energy of the system is
Thus,.Referring to the FBD of the system shown
in Fig.a, we notice that , and do no work while does positive work and
does negative work. When Amoves 2 ft downward, the pulley rotates
Thus, the work of are
Principle of Work and Energy:
Ans.v
=
20.4 rad
>
s
282.61 +[40 +(-30)] =0.7065 v2
T1+U1-2=T2
UWB =-WB SB=-30(1) =-30 ft #lb
UWA =WA SA=20(2) =40 ft #lb
WA and WB
SB=2(0.5) =1 ft c
2
1=SB
0.5
u=SA
rA
=SB
rB
WB
WA
Wp
Ox, Oy
T1=0.7065(202)=282.61 ft #lb
=0.7065v2
=1
2 (0.5590)v2+1
2
¢
20
32.2
≤
[v(1)]2+1
2
¢
30
32.2
≤
[v(0.5)]2
T =1
2 IOv2+1
2 mAvA
2+1
2 mBvB
2
#
=0.5590 IO=mkO
2 =
¢
50
≤
(0.62)
Ans:
v
=20.4 rad>s
18–9.
The disk, which has a mass of 20 kg, is subjected to the
couple moment of M
=(2
u
+4)
N
#
m, where
u
is in
radians. If it starts from rest, determine its angular velocity
when it has made two revolutions.
O
M
300 mm
18–10.
The spool has a mass of 40 kg and a radius of gyration of
kO
=
0.3 m. If the 10-kg block is released from rest,
determine the distance the block must fall in order for the
spool to have an angular velocity
v
=
15 rad
>
s. Also, what
is the tension in the cord while the block is in motion?
Neglect the mass of the cord.
SOLUTION
500 mm
300 mm
O
922
18–11.
The force of T
=
20 N is applied to the cord of negligible
mass. Determine the angular velocity of the 20-kg wheel
when it has rotated 4 revolutions starting from rest. The
wheel has a radius of gyration of kO
=
0.3 m.
SOLUTION
Work. Referring to the FBD of the wheel, Fig. a, only force T does work.
This work is positive since T is required to displace vertically downward,
sT=
u
r=4(2
p
)(0.4) =3.2
p
m
.
UT=TsT=20(3.2
p
)=64
p
J
Principle of Work and Energy.
T1+ΣU1-2=T2
0+64p=0.9 v2
v
=14.94 rad>s=14.9 rad>s
Ans.
O
0.4 m
Ans:
v
=14.9 rad>s
*18–12.
75 mm
A
Determine the velocity of the 50-kg cylinder after it has
de
scended a distance of 2 m. Initially,the system is at rest.
T
he reel has a mass of 25 kg and a radius of gyration about its
center of ma
ss Aof .kA=125 mm
SOLUTION
Ans.v=4.05 m>s
+1
2 (50) v2
0+50(9.81)(2) =1
2 [(25)(0.125)2]
¢
v
0.075
≤
2
T1+©U1-2=T2
924
SOLUTION
18–13.
The 10-kg uniform slender rod is suspended at rest when
the force of F
=
150 N is applied to its end. Determine the
angular velocity of the rod when it has rotated 90° clockwise
from the position shown. The force is always perpendicular
to the rod.
O
3 m
Ans:
v
=6.11 rad>s
SOLUTION
18–14.
The 10-kg uniform slender rod is suspended at rest when
the force of F
=
150 N is applied to its end. Determine the
angular velocity of the rod when it has rotated 180°
clockwise from the position shown. The force is always
perpendicular to the rod.
O
3 m
SOLUTION
18–15.
The pendulum consists of a 10-kg uniform disk and a 3-kg
uniform slender rod. If it is released from rest in the position
shown, determine its angular velocity when it rotates
clockwise 90°.
2 m
M 30 N fi m
A
B
D
0.8 m
927
*18–16.
Amotor supplies a constant torque to the
winding drum that operates the elevator. If the elevator has a
mass of 900 kg,the counterweight Chas a mass of 200 kg,and
the winding drum has a mass of 600 kg and radius of gyration
about its axis of determine the speed of the
elevator after it rises 5 m starting from rest. Neglect the mass
of the pulleys.
k=0.6 m,
M=6kN
#
m
SOLUTION
Ans.
v
=2.10 m s
+1
2[600(0.6)2](
v
0.8)2
0+6000( 5
0.8)-900(9.81)(5) +200(9.81)(5) =1
2(900)(
v
)2+1
2(200)(
v
)2
T1+©U1-2=T2
u=s
r=5
0.8
v
E=
v
C
M
D
C
0.8 m
Ans:
v=2.10 m>s
928
18–17.
s
O
r
v0
The center Oof the thin ring of mass mis given an angular
velocity of . If the ring rolls without slipping, determine
its angular velocity after it has traveled a distance of sdown
the plane.Neglect its thickness.
v0
SOLUTION
1
2 (mr2+mr2)v0
2+mg(s sin u)=1
2 (mr2+mr2)v2
T1+©U1-2=T2
A
r
2
929
18–18.
The wheel has a mass of 100 kg and a radius of gyration
of
kO=0.2 m
. A motor supplies a torque
M=(40
u
+900) N #m
, where
u
is in radians, about the
drive shaft at O. Determine the speed of the loading car,
which has a mass of 300 kg, after it travels
s=4 m
. Initially
the car is at rest when
s=0
and
u=0°
. Neglect the mass of
the attached cable and the mass of the car’s wheels.
SOLUTION
s=0.3u=4
u=13.33 rad
T1+ΣU1-2=T2
[0 +0] +
L13.33
0
(40u+900)du-300(9.81) sin 30° (4) =
1
2(300)vC
2+1
2
c
100(0.20)2
da
vC
0.3
b
2
vC
=7.49 m>s
Ans.
Ans:
vC=7.49 m>s
30
M
s
0.3 m
O
930
18–19.
The rotary screen S is used to wash limestone. When empty
it has a mass of 800 kg and a radius of gyration of
kG=1.75 m
. Rotation is achieved by applying a torque of
M=280 N #m
about the drive wheel at A. If no slipping
occurs at A and the supporting wheel at B is free to roll,
determine the angular velocity of the screen after it has
rotated 5 revolutions. Neglect the mass of A and B.
0.3 m
S
2 m
Ans:
v
=6.92 rad>s
931
*18–20.
A
B
45°
600 mm
P 200 N
u
If and the 15-kg uniform slender rod starts from
rest at , determine the rod’s angular velocity at the
instant just before .u=45°
u=0°
P=200 N
SOLUTION
Then
Thus,
The mass moment of inertia of the rod about its mass center is
.Thus, the final kinetic energy is
Since the rod is initially at rest, . Referring to Fig.b, and do no work,
while does positive work and does negative work. When ,displaces
through a horizontal distance and displaces vertically upwards
through a distance of ,Fig.c.Thus, the work done by and is
Principle of Work and Energy:
Ans.v2=4.97 rad>s
0+[120 -31.22] =3.6v2
2
T1+©U1-2=T2
UW=-Wh =-15(9.81)(0.3 sin 45°) =-31.22 J
UP=PsP=200(0.6) =120 J
WPh=0.3 sin 45°
WsP=0.6 m
Pu=45°WP
NB
NA
T1=0
=3.6v2
2
=1
2 (15)[w2(0.6708)]2+1
2 (0.45) v2
2
T
2=1
2 m(vG)22+1
2 IG v2
2
= 1
12 (15)(0.62)=0.45 kg#m2
IG=1
12 ml2
(vG)2=v2rG>IC =v2(0.6708)
rG>IC =30.32+0.62=0.6708 m
Ans:
v
2=4.97 rad>s
