Download SOLUTION

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19–1.
mvG
The rigid body (slab) has a mass m and rotates with an
angular velocity V about an axis passing through the fixed
point O. Show that the momenta of all the particles
composing the body can be represented by a single vector
having a magnitude mvG and acting through point P, called
the center of percussion, which lies at a distance
rP>G = k2G>rG>O from the mass center G. Here kG is the
radius of gyration of the body, computed about an axis
perpendicular to the plane of motion and passing through G.
V
vG
rG/O
O
SOLUTION
HO = (rG>O + rP>G) myG = rG>O (myG) + IG v,
where IG = mk2G
rG>O (myG) + rP>G (myG) = rG>O (myG) + (mk2G) v
rP>G =
k2G
yG>v
However, yG = vrG>O
rP>G =
or rG>O =
yG
v
k2G
rG>O
Q.E.D.
985
rP/G
G
P
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
19–2.
At a given instant, the body has a linear momentum
L = mvG and an angular momentum H G = IG V computed
about its mass center. Show that the angular momentum of
the body computed about the instantaneous center of zero
velocity IC can be expressed as H IC = IIC V , where IIC
represents the body’s moment of inertia computed about
the instantaneous axis of zero velocity. As shown, the IC is
located at a distance rG>IC away from the mass center G.
mvG
G
rG/IC
SOLUTION
HIC = rG>IC (myG) + IG v,
IC
where yG = vrG>IC
= rG>IC (mvrG>IC) + IG v
= (IG + mr2G>IC) v
Q.E.D.
= IIC v
986
IGV
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
19–3.
Show that if a slab is rotating about a fixed axis perpendicular
to the slab and passing through its mass center G, the angular
momentum is the same when computed about any other
point P.
V
P
G
SOLUTION
Since yG = 0, the linear momentum L = myG = 0. Hence the angular momentum
about any point P is
HP = IG v
Since v is a free vector, so is H P.
Q.E.D.
987
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*19–4.
The 40-kg disk is rotating at V = 100 rad>s. When the force
P is applied to the brake as indicated by the graph. If the
coefficient of kinetic friction at B is mk = 0.3, determine
the time t needed to stay the disk from rotating. Neglect the
thickness of the brake.
P (N)
500
P
t (s)
2
300 mm
300 mm
B
Solution
Equilibrium. Since slipping occurs at brake pad, Ff = mkN = 0.3 N.
Referring to the FBD the brake’s lever, Fig. a,
150 mm
200 mm
V
O
A
a+ ΣMA = 0; N(0.6) - 0.3 N(0.2) - P(0.3) = 0
N = 0.5556 P
Thus,
Ff = 0.3(0.5556 P) = 0.1667 P
Principle of Impulse and Momentum. The mass moment of inertia of the disk about
1
1
its center O is IO = mr 2 = (40) ( 0.152 ) = 0.45 kg # m2.
2
2
IOv1 + Σ
Lt1
t2
MOdt = IOv2
It is required that v2 = 0. Assuming that t 7 2 s,
0.45(100) +
0.025
L0
L0
L0
t
[ - 0.1667 P(0.15)]dt = 0.45(0)
t
P dt = 45
t
P dt = 1800
1
(500)(2) + 500(t - 2) = 1800
2
Ans.
t = 4.60 s
Since t 7 2 s, the assumption was correct.
Ans:
t = 4.60 s
988
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
19–5.
The impact wrench consists of a slender 1-kg rod AB which
is 580 mm long, and cylindrical end weights at A and B that
each have a diameter of 20 mm and a mass of 1 kg. This
assembly is free to turn about the handle and socket,
which are attached to the lug nut on the wheel of a car. If
the rod AB is given an angular velocity of 4 rad>s and it
strikes the bracket C on the handle without rebounding,
determine the angular impulse imparted to the lug nut.
C
B
Iaxle =
L
300 mm
A
SOLUTION
300 mm
1
1
(1)(0.6 - 0.02)2 + 2 c (1)(0.01)2 + 1(0.3)2 d = 0.2081 kg # m2
12
2
Mdt = Iaxle v = 0.2081(4) = 0.833 kg # m2>s
Ans.
Ans:
L
989
M dt = 0.833 kg # m2 >s
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
19–6.
The airplane is traveling in a straight line with a speed of
300 km> h, when the engines A and B produce a thrust of
TA = 40 kN and TB = 20 kN, respectively. Determine the
angular velocity of the airplane in t = 5 s. The plane has a
mass of 200 Mg, its center of mass is located at G, and its
radius of gyration about G is kG = 15 m.
TA 40 kN 8 m
A
G
B
TB 20 kN
SOLUTION
8m
Principle of Angular Impulse and Momentum: The mass moment of inertia of the
airplane about its mass center is IG = mkG2 = 200 A 103 B A 152 B = 45 A 106 B kg # m2.
Applying the angular impulse and momentum equation about point G,
t2
Izv1 + ©
Lt1
MGdt = IGv2
0 + 40 A 103 B (5)(8) - 20 A 103 B (5)(8) = 45 A 106 B v
Ans.
v = 0.0178 rad>s
Ans:
v = 0.0178 rad>s
990
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
19–7.
The double pulley consists of two wheels which are attached
to one another and turn at the same rate. The pulley has a
mass of 15 kg and a radius of gyration of kO = 110 mm. If
the block at A has a mass of 40 kg, determine the speed
of the block in 3 s after a constant force of 2 kN is applied
to the rope wrapped around the inner hub of the pulley. The
block is originally at rest.
2 kN
200 mm
75 mm
O
SOLUTION
Principle of Impulse and Momentum: The mass moment inertia of the pulley about
point O is IO = 15 A 0.112 B = 0.1815 kg # m2. The angular velocity of the pulley and
yB
= 5yB. Applying Eq. 19–15,
the velocity of the block can be related by v =
0.2
we have
a a syst. angular momentumb
O1
+ a a syst. angular impulseb
O1-2
= a a syst. angular momentumb
(a + )
A
O2
0 + [40(9.81)(3)](0.2) - [2000(3)](0.075)
= - 40yB(0.2) - 0.1815(5yB)
Ans.
yB = 24.1 m>s
Ans:
vB = 24.1 m>s
991
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*19–8.
The assembly weighs 10 lb and has a radius of gyration
kG = 0.6 ft about its center of mass G. The kinetic energy
of the assembly is 31 ft # lb when it is in the position shown.
If it is rolling counterclockwise on the surface without
slipping, determine its linear momentum at this instant.
0.8 ft
G
1 ft
1 ft
SOLUTION
IG = (0.6)2 a
T =
10
b = 0.1118 slug # ft2
32.2
1
1 10
a
b v 2 + (0.1118) v2 = 31
2 32.2 G
2
(1)
vG = 1.2 v
Substitute into Eq. (1),
v = 10.53 rad>s
vG = 10.53(1.2) = 12.64 ft>s
L = mvG =
10
(12.64) = 3.92 slug # ft>s
32.2
Ans.
Ans:
L = 3.92 slug # ft>s
992
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
19–9.
The disk has a weight of 10 lb and is pinned at its center O.
If a vertical force of P = 2 lb is applied to the cord wrapped
around its outer rim, determine the angular velocity of the
disk in four seconds starting from rest. Neglect the mass of
the cord.
0.5 ft
O
SOLUTION
t2
(c + )
IO v1 + ©
Lt1
P
MO dt = IOv2
0 + 2(0.5)(4) = c
1 10
a
b (0.5)2 dv2
2 32.2
Ans.
v2 = 103 rad>s
Ans:
v2 = 103 rad>s
993
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
19–10.
The 30-kg gear A has a radius of gyration about its center of
mass O of kO = 125 mm. If the 20-kg gear rack B is
subjected to a force of P = 200 N, determine the time
required for the gear to obtain an angular velocity of
20 rad>s, starting from rest. The contact surface between the
gear rack and the horizontal plane is smooth.
0.15 m
O
B
A
P 200 N
SOLUTION
Kinematics: Since the gear rotates about the fixed axis, the final velocity of the gear
rack is required to be
(vB)2 = v2rB = 20(0.15) = 3 m>s :
Principle of Impulse and Momentum: Applying the linear impulse and momentum
equation along the x axis using the free-body diagram of the gear rack shown in Fig. a,
+ B
A:
t2
m(vB)1 + ©
Fxdt = m(vB)2
Lt1
0 + 200(t) - F(t) = 20(3)
(1)
F(t) = 200t - 60
The mass moment of inertia of the gear about its mass center is IO =
mkO2 = 30(0.1252) = 0.46875 kg # m2. Writing the angular impulse and momentum
equation about point O using the free-body diagram of the gear shown in Fig. b,
t2
IOv1 + ©
MOdt = IOv2
Lt1
0 + F(t)(0.15) = 0.46875(20)
(2)
F(t) = 62.5
Substituting Eq. (2) into Eq. (1) yields
Ans.
t = 0.6125 s
Ans:
t = 0.6125 s
994
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
19–11.
The pulley has a weight of 8 lb and may be treated as a thin
disk. A cord wrapped over its surface is subjected to forces
TA = 4 lb and TB = 5 lb. Determine the angular velocity
of the pulley when t = 4 s if it starts from rest when t = 0.
Neglect the mass of the cord.
0.6 ft
SOLUTION
Principle of Impulse and Momentum: The mass moment inertia of the pulley about its
8
1
mass center is Io = a
b (0.62) = 0.04472 slug # ft2. Applying Eq. 19–14, we have
2 32.2
t
IOv1 + © 1t12Modt = Iov2
(a +)
TB = 5 lb
TA = 4 lb
0 + [5(4)](0.6) - [4(4)](0.6) = 0.04472v2
Ans.
v2 = 53.7 rad>s
Ans:
v2 = 53.7 rad>s
995
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*19–12.
The 40-kg roll of paper rests along the wall where the
coefficient of kinetic friction is mk = 0.2. If a vertical force
of P = 40 N is applied to the paper, determine the angular
velocity of the roll when t = 6 s starting from rest. Neglect
the mass of the unraveled paper and take the radius of
gyration of the spool about the axle O to be kO = 80 mm.
B
P 40 N
13
12
5
A
Solution
O
120 mm
Principle of Impulse and Momentum. The mass moment of inertia of the paper roll
about its center is IO = mk 2O = 40 ( 0.082 ) = 0.256 kg # m2. Since the paper roll is
required to slip at point of contact, Ff = mkN = 0.2 N. Referring to the FBD of the
paper roll, Fig. a,
Lt1
+ ) m[(vO)x]1 + Σ
(S
(+ c)
t2
Fx dt = m[(vO)x]2
0 + N(6) - FABa
FAB =
m[(vO)y]1 + Σ
Lt1
5
b(6) = 0
13
13
N(1)
5
t2
Fy dt = m[(vO)y]2
0 + 0.2 N(6) + FABa
12
b(6) + 40(6) - 40(9.81)(6) = 0
13
0.2 N +
12
F = 352.4(2)
13 AB
Solving Eqs. (1) and (2)
N = 135.54 N FAB = 352.4 N
Subsequently
a+ IO v1 + Σ MO dt = IO v2
L
0.256(0) + 40(0.12)(6) - 0.2(135.54)(0.12)(6) = 0.256v
Ans.
v = 36.26 rad>s = 36.3 rad>s
Ans:
v = 36.3 rad>s
996
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
19–13.
The slender rod has a mass m and is suspended at its end A
by a cord. If the rod receives a horizontal blow giving it an
impulse I at its bottom B, determine the location y of the
point P about which the rod appears to rotate during the
impact.
A
P
SOLUTION
l
y
Principle of Impulse and Momentum:
t2
IG v1 + ©
(a + )
MG dt = IG v2
Lt1
l
1
0 + I a b = c ml2 d v
2
12
+ b
a:
I
1
I = mlv
6
B
t2
m(yAx)1 + ©
0 +
Lt1
Fx dt = m(yAx)2
1
mlv = mvG
6
yG =
l
v
6
Kinematics: Point P is the IC.
yB = v y
Using similar triangles,
vy
=
y
l
v
6
y -
l
2
y =
2
l
3
Ans.
Ans:
y =
997
2
l
3
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
19–14.
The rod of length L and mass m lies on a smooth horizontal
surface and is subjected to a force P at its end A as shown.
Determine the location d of the point about which the rod
begins to turn, i.e, the point that has zero velocity.
P
L
A
d
Solution
L
+ ) m(vGx)1 + Σ Fx dt = m(vGx)2
(S
0 + P(t) = m(vG)x
(+ c)
m(vGy)1 + Σ Fy dt = m(vGy)2
L
0 + 0 = m(vG)y
(a+) (HG)1 + Σ MG dt = (HG)2
L
L
1
0 + P(t)a b =
mL2v
2
12
vG = yv
L
1
m(vG)x a b =
mL2v
2
12
(vG)x =
L
v
6
y =
L
6
d =
L
L
2
+
= L
2
6
3
Ans.
Ans:
d =
998
2
l
3
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
19–15.
z
A 4-kg disk A is mounted on arm BC, which has a negligible
mass. If a torque of M = 15e0.5t2 N # m, where t is in seconds,
is applied to the arm at C, determine the angular velocity of
BC in 2 s starting from rest. Solve the problem assuming
that (a) the disk is set in a smooth bearing at B so that it
rotates with curvilinear translation, (b) the disk is fixed to
the shaft BC, and (c) the disk is given an initial freely
spinning angular velocity of V D = 5 -80k6 rad>s prior to
application of the torque.
250 mm
A
60 mm
B
M
C
#
(5e0.5 t) N m
SOLUTION
a)
(Hz)1 + ©
L
Mz dt = (Hz)2
2
0+
L0
5e0.5 tdt = 4(vB)(0.25)
5 0.5 t 2
e 2 = vB
0.5
0
vB = 17.18 m>s
Thus,
17.18
= 68.7 rad>s
0.25
vBC =
Ans.
b)
(Hz)1 + ©
2
0 +
L0
L
Mz dt = (Hz)2
1
5e0.5 tdt = 4(vB)(0.25) + c (4)(0.06)2 d vBC
2
Since vB = 0.25 vBC, then
Ans.
vBC = 66.8 rad>s
c)
(Hz)1 + ©
L
Mz dt = (Hz)2
2
1
1
- c (4)(0.06)2 d(80) +
5e0.5 tdt = 4(vB)(0.25) - c (4)(0.06)2 d(80)
2
2
L0
Since vB = 0.25 vBC,
Ans.
vBC = 68.7 rad>s
Ans:
(a) vBC = 68.7 rad>s
(b) vBC = 66.8 rad>s
(c) vBC = 68.7 rad>s
999
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*19–16.
The frame of a tandem drum roller has a weight of 4000 lb
excluding the two rollers. Each roller has a weight of 1500 lb
and a radius of gyration about its axle of 1.25 ft. If a torque
of M = 300 lb # ft is supplied to the rear roller A, determine
the speed of the drum roller 10 s later, starting from rest.
1.5 ft
SOLUTION
1.5 ft
A
Principle of Impulse and Momentum: The mass moments of inertia of the rollers
1500
about their mass centers are IC = ID =
A 1.252 B = 72.787 slug # ft2. Since the
32.2
v
v
= 0.6667v. Using the free-body
=
rollers roll without slipping, v =
r
1.5
diagrams of the rear roller and front roller, Figs. a and b, and the momentum
diagram of the rollers, Fig. c,
M
B
300 lb ft
t2
(HA)1 + ©
MAdt = (HA)2
Lt1
0 + 300(10) - Cx(10)(1.5) =
1500
v(1.5) + 72.787(0.6667v)
32.2
(1)
Cx = 200 - 7.893v
and
t2
(HB)1 + ©
Lt1
MBdt = (HB)2
1500
v(1.5) + 72.787(0.6667v)
32.2
0 + Dx(10)(1.5) =
(2)
Dx = 7.893v
Referring to the free-body diagram of the frame shown in Fig. d,
+
:
m C (vG)x D 1 + ©
t2
Lt1
Fxdt = m C (vG)x D 2
0 + Cx(10) - Dx(10) =
4000
v
32.2
(3)
Substituting Eqs. (1) and (2) into Eq. (3),
(200 - 7.893v)(10) - 7.893v(10) =
4000
v
32.2
Ans.
v = 7.09 ft>s
Ans:
v = 7.09 ft>s
1000
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
19–17.
The 100-lb wheel has a radius of gyration of kG = 0.75 ft.
If the upper wire is subjected to a tension of T = 50 lb,
determine the velocity of the center of the wheel in 3 s,
starting from rest.The coefficient of kinetic friction between
the wheel and the surface is mk = 0.1.
T
0.5 ft
G
1 ft
SOLUTION
Principle of Impulse and Momentum: We can eliminate the force F from the
analysis if we apply the principle of impulse and momentum about point A. The
100
100
mass moment inertia of the wheel about point A is IA =
A 0.752 B +
A 0.52 B
32.2
32.2
= 2.523 slug # ft2. Applying Eq. 19–14, we have
m A yGy B 1 + ©
(+ c)
t2
Lt1
Fy dt = m A yGy B 2
0 + N(t) - 100(t) = 0
N = 100 lb
t2
IA v1 + ©
(a + )
Lt1
MA dt = IA v2
[1]
0 + [50(3)](1) - [0.1(100)(3)](0.5) = 2.523 v2
Kinematics: Since the wheel rolls without slipping at point A, the instantaneous
center of zero velocity is located at point A. Thus,
yG = v2 rG/IC
v2 =
yG
yG
= 2yG
=
rG/IC
0.5
[2]
Solving Eqs. [1] and [2] yields
Ans.
yG = 26.8 ft>s
v2 = 53.50 rad s
Ans:
vG = 26.8 ft>s
1001
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
19–18.
The 4-kg slender rod rests on a smooth floor. If it is kicked
so as to receive a horizontal impulse I = 8 N # s at point A
as shown, determine its angular velocity and the speed of its
mass center.
2m
A
1.75 m
SOLUTION
+ 2
1;
60
m(vGx)1 + ©
L
Fx dt = m(vGx)2
I
8N s
0 + 8 cos 60° = 4(vG)x
(vG)x = 1 m>s
(+ c)
m(vGy)1 + ©
L
Fy dt = m(vGy)2
0 + 8 sin 60° = 4(vG)y
(vG)y = 1.732 m>s
vG = 2(1.732)2 + (1)2 = 2 m>s
(a + )
(HG)1 + ©
L
Ans.
MG dt = (HG)2
0 + 8 sin 60°(0.75) =
1
(4)(2)2 v
12
Ans.
v = 3.90 rad s
Ans:
vG = 2 m>s
v = 3.90 rad>s
1002
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
19–19.
The double pulley consists of two wheels which are
attached to one another and turn at the same rate. The
pulley has a mass of 15 kg and a radius of gyration
kO = 110 mm. If the block at A has a mass of 40 kg,
determine the speed of the block in 3 s after a constant
force F = 2 kN is applied to the rope wrapped around the
inner hub of the pulley. The block is originally at rest.
Neglect the mass of the rope.
200 mm
(HO)1 + ©
75 mm
F
A
SOLUTION
(c + )
O
L
MO dt = (HO)2
0 + 2000(0.075)(3) - 40(9.81)(0.2)(3) = 15(0.110)2v + 40(0.2v) (0.2)
v = 120.4 rad>s
Ans.
vA = 0.2(120.4) = 24.1 m>s
Ans:
vA = 24.1 m>s
1003
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*19–20.
The 100-kg spool is resting on the inclined surface for which
the coefficient of kinetic friction is mk = 0.1. Determine the
angular velocity of the spool when t = 4 s after it is released
from rest. The radius of gyration about the mass center is
kG = 0.25 m.
0.4 m
0.2 m
G
Solution
A
Kinematics. The IC of the spool is located as shown in Fig. a. Thus
30
vG = vrG>IC = v(0.2)
Principle of Impulse and Momentum. The mass moment of inertia of the spool
about its mass center is IG = mk 2G = 100 ( 0.252 ) = 6.25 kg # m2. Since the spool is
required to slip, Ff = mkN = 0.1 N. Referring to the FBD of the spool, Fig. b,
t2
a+ m[(vG)y]1 + Σ Fy dt = m[(vG)y]2
Lt1
0 + N(4) - 100(9.81) cos 30°(4) = 0
N = 849.57 N
t2
Q m[(vG)x]1 + Σ Fx dt = m[(vG)x]2
Lt1
+
0 + T(4) + 0.1(849.57)(4) - 100(9.81) sin 30°(4) = 100[ - v(0.2)]
(1)
T + 5v = 405.54
t2
a+ IG v1 + Σ MG dt = IG v2
Lt1
0 + 0.1(849.57)(0.4)(4) - T(0.2)(4) = - 6.25 v2
(2)
0.8T - 6.25v2 = 135.93
Solving Eqs. (1) and (2),
Ans.
v = 18.39 rad>s = 18.4 rad>s b
T = 313.59 N
Ans:
v = 18.4 rad>s b
1004
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
19–21.
The spool has a weight of 30 lb and a radius of gyration
kO = 0.45 ft. A cord is wrapped around its inner hub and the
end subjected to a horizontal force P = 5 lb. Determine the
spool’s angular velocity in 4 s starting from rest. Assume
the spool rolls without slipping.
0.9 ft
O
P
5 lb
0.3 ft
A
SOLUTION
(+ c)
m(vy) + ©
L
Fy dt = m(vy)2
0 + NA(4) - 30(4) = 0
NA = 30 lb
+ )
(:
m(vx)1 + ©
L
Fxdt = m(vx)2
0 + 5(4) - FA(4) =
(c + )
(HG)1 + ©
L
30
vG
32.2
MG dt = (HG)2
0 + FA(4)(0.9) - 5(4)(0.3) =
30
(0.45)2 v
32.2
Since no slipping occurs
Set vG = 0.9 v
FA = 2.33 lb
Ans.
v = 12.7 rad>s
Also,
(c + )
(HA)1 + ©MA dt = (HA)2
0 + 5(4)(0.6) = c
30
30
(0.45)2 +
(0.9)2 d v
32.2
32.2
Ans.
v = 12.7 rad>s
Ans:
v = 12.7 rad>s
1005
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
19–22.
The two gears A and B have weights and radii of gyration of
WA = 15 lb, kA = 0.5 ft and WB = 10 lb, kB = 0.35 ft,
respectively. If a motor transmits a couple moment to gear
B of M = 2(1 - e - 0.5t ) lb # ft, where t is in seconds,
determine the angular velocity of gear A in t = 5 s, starting
from rest.
0.8 ft
A
M
0.5 ft
B
Solution
vA(0.8) = vB(0.5)
vB = 1.6vA
Gear B:
L
( c + ) (HB)1 + Σ MB dt = ( HB ) 2
0 +
L0
5
2 ( 1 - e -0.5t ) dt -
L
0.5F dt = c a
6.328 = 0.5
Gear A:
L
F dt + 0.06087vA
0 = 0.8
L
(2), and solving for vA.
10
b(0.35)2 d ( 1.6vA )
32.2
F dt - 0.1165vA
L
(2)
L
F dt between Eqs. (1) and
( a+ ) (HA)1 + Σ MA dt = ( HA ) 2 Eliminate
0 +
L
0.8F dt = c a
(1)
15
b(0.5)2 d vA vA = 47.3 rad>s
32.2
Ans.
Ans:
vA = 47.3 rad>s
1006
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
19–23.
The hoop (thin ring) has a mass of 5 kg and is released down
the inclined plane such that it has a backspin v = 8 rad>s
and its center has a velocity vG = 3 m>s as shown. If the
coefficient of kinetic friction between the hoop and the
plane is mk = 0.6, determine how long the hoop rolls before
it stops slipping.
ω = 8 rad/s
G
vG = 3 m /s
SOLUTION
b+
mvx1 + a
L
0.5 m
30°
Fxdt = mvx2
5(3) + 49.05 sin 30° (t) - 25.487t = 5vG
a+
(HG)1 + a
L
MG dt = (HG)2
- 5(0.5)2 (8) + 25.487(0.5)(t) = 5(0.5)2 a
vG
b
0.5
Solving,
vG = 2.75 m>s
Ans.
t = 1.32 s
Ans:
t = 1.32 s
1007
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*19–24.
The 30-kg gear is subjected to a force of P = (20t) N, where
t is in seconds. Determine the angular velocity of the gear at
t = 4 s, starting from rest. Gear rack B is fixed to the
horizontal plane, and the gear’s radius of gyration about its
mass center O is kO = 125 mm.
150 mm
P (20t) N
O
B
A
SOLUTION
Kinematics: Referring to Fig. a,
vO = vrO>IC = v(0.15)
Principle of Angular Impulse and Momentum: The mass moment of inertia of
the gear about its mass center is IO = mkO2 = 30(0.1252) = 0.46875 kg # m2 .
Writing the angular impulse and momentum equation about point A shown in
Fig. b,
t2
(HA)1 + ©
Lt1
MA dt = (HA)2
4s
0 +
L0
1.5t2 2
4s
0
20t(0.15)dt = 0.46875v + 30 [v(0.15)] (0.15)
= 1.14375v
Ans.
v = 21.0 rad>s
Ans:
v = 21.0 rad>s
1008
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
19–25.
The 30-lb flywheel A has a radius of gyration about its center
of 4 in. Disk B weighs 50 lb and is coupled to the flywheel by
means of a belt which does not slip at its contacting surfaces.
If a motor supplies a counterclockwise torque to the
flywheel of M = (50t) lb # f t, where t is in seconds,
determine the time required for the disk to attain an angular
velocity of 60 rad>s starting from rest.
M (50t) lbft
9 in.
6 in.
A
B
SOLUTION
Principle of Impulse and Momentum: The mass moment inertia of the flywheel
4 2
30
about point C is IC =
a b = 0.1035 slug # ft2. The angular velocity of the
32.2 12
rB
0.75
v =
(60) = 90.0 rad>s. Applying Eq. 19–14 to the
flywheel is vA =
rA B
0.5
flywheel [FBD(a)], we have
t2
IC v1 + ©
t
(a + )
0 +
L0
50t dt +
C
L
Lt1
MC dt = IC v2
T2 (dt) D (0.5) -
25t2 + 0.5
L
C
L
T1(dt) D (0.5) = 0.1035(90)
(1)
(T2 - T1)dt = 9.317
1 50
a
b(0.752)
2 32.2
= 0.4367 slug # ft2. Applying Eq. 19–14 to the disk [FBD(b)], we have
The mass moment inertia of the disk about point D is ID =
t2
ID v1 + ©
(a +)
0 +
C
L
T1 (dt) D (0.75) L
MD dt = ID v2
Lt1
C
L
T2 (dt) D (0.75) = 0.4367(60)
(2)
(T2 - T1)dt = - 34.94
Substitute Eq. (2) into Eq. (1) and solving yields
Ans.
t = 1.04 s
Ans:
t = 1.04 s
1009
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
19–26.
If the shaft is subjected to a torque of M = (15t2) N # m,
where t is in seconds, determine the angular velocity of the
assembly when t = 3 s, starting from rest. Rods AB and BC
each have a mass of 9 kg.
M
1m
1m
(15t2) N m
A
C
B
SOLUTION
Principle of Impulse and Momentum: The mass moment of inertia of the rods
1
1
ml2 =
(9) A 12 B = 0.75 kg # m2. Since the
about their mass center is IG =
12
12
assembly rotates about the fixed axis, (vG)AB = v(rG)AB = v(0.5) and
(vG)BC = v(rG)BC = va 212 + (0.5)2 b = v(1.118). Referring to Fig. a,
t2
c+
(Hz)1 + ©
3s
0 +
5t3 2
L0
3s
0
Lt1
Mz dt = (Hz)2
15t2dt = 9 C v(0.5) D (0.5) + 0.75v + 9 C v(1.118) D (1.118) + 0.75v
= 15v
Ans.
v = 9 rad>s
Ans:
v = 9 rad>s
1010
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
19–27.
The double pulley consists of two wheels which are attached
to one another and turn at the same rate. The pulley has a
mass of 15 kg and a radius of gyration of
kO = 110 mm. If the block at A has a mass of 40 kg and the
container at B has a mass of 85 kg, including its contents,
determine the speed of the container when t = 3 s after it is
released from rest.
200 mm
O
75 mm
C
A
Solution
The angular velocity of the pulley can be related to the speed of
vA
container B by v =
= 13.333 vB. Also the speed of block A
0.075
vA = v(0.2) = 13.33 vB(0.2) - 2.667 vB,
B
(a+ )( ΣSyst. Ang. Mom. ) O1 + ( Σ Syst. Ang. Imp. ) O(1 - 2) = ( Σ Syst. Ang. Mom. ) O2
0 + 40(9.81)(0.2)(3) - 85(9.81)(0.075)(3)
=
3 15(0.110)2 4 (13.333 vB)
+ 85 vB(0.075) + 40(2.667 vB)(0.2)
Ans.
vB = 1.59 m>s
Ans:
vB = 1.59 m>s
1011
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*19–28.
The crate has a mass mc. Determine the constant speed v0 it
acquires as it moves down the conveyor. The rollers each
have a radius of r, mass m, and are spaced d apart. Note that
friction causes each roller to rotate when the crate comes in
contact with it.
d
A
SOLUTION
30°
The number of rollers per unit length is 1/d.
Thus in one second,
v0
rollers are contacted.
d
If a roller is brought to full angular speed of v =
v0
in t0 seconds, then the moment
r
of inertia that is effected is
v0
v0
1
I¿ = I a b(t0) = a m r2ba bt0
d
2
d
Since the frictional impluse is
F = mc sin u then
a + (HG)1 + ©
L
MG dt = (HG)2
v0
v0
1
0 + (mc sin u) r t0 = c a m r2 b a b t0 d a b
r
2
d
v0 =
A
mc
(2 g sin u d) a b
m
Ans.
Ans:
v0 =
1012
A
(2 g sin u d) a
mc
b
m
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
19–29.
The turntable T of a record player has a mass of 0.75 kg and
a radius of gyration kz = 125 mm. It is turning freely at
vT = 2 rad>s when a 50-g record (thin disk) falls on it.
Determine the final angular velocity of the turntable just
after the record stops slipping on the turntable.
z
150 mm
T
vT 2 rad/s
Solution
(Hz)1 = (Hz)2
0.75(0.125)2(2) = c 0.75(0.125)2 +
v = 1.91 rad>s
1
(0.050)(0.150)2 d v
2
Ans.
Ans:
v = 1.91 rad>s
1013
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
19–30.
The 10-g bullet having a velocity of 800 m>s is fired into the
edge of the 5-kg disk as shown. Determine the angular
velocity of the disk just after the bullet becomes embedded
into its edge. Also, calculate the angle u the disk will swing
when it stops. The disk is originally at rest. Neglect the mass
of the rod AB.
A
2m
B
v 800 m/s
0.4 m
Solution
Conservation of Angular Momentum. The mass moment of inertia of the disk about
1
its mass center is (IG)d = (5) ( 0.42 ) = 0.4 kg # m2. Also, (vb)2 = v2 1 25.92 2 and
2
(vd)2 = v2(2.4). Referring to the momentum diagram with the embedded bullet,
Fig. a,
Σ(HA)1 = Σ(HA)2
Mb(vb)1(rb)1 = (IG)d v2 + Md(vd)2(r2) + Mb(vb)2(rb)2
0.01(800)(2.4) = 0.4v2 + 5 3 v2(2.4) 4 (2.4) + 0.01 3 v2 1 25.92 2 4 1 25.92 2
Ans.
v2 = 0.6562 rad>s = 0.656 rad>s
Kinetic Energy. Since the system is required to stop finally, T3 = 0. Here
T2 =
=
1
1
1
(I ) v22 + Md(vd)22 + Mb(vb)22
2 Gd
2
2
1
1
1
(0.4) ( 0.65622 ) + (5)[0.6562(2.4)]2 + (0.01) 3 0.6562 1 25.92 2 4 2
2
2
2
= 6.2996 J
Potential Energy. Datum is set as indicated on Fig. b.
Here f = tan-1 a
0.4
b = 9.4623. Hence
2.4
y1 = 2.4 cos u yb = 25.92 cos (u - 9.4623°)
Thus, the gravitational potential energy of the disk and bullet with reference to the
datum is
(Vg)d = Md g (yd) = 5(9.81)( - 2.4 cos u) = -117.72 cos u
(Vg)b = Mbg(yb) = 0.01(9.81) 3 1 - 25.92 cos (u - 9.4623°) 4
= - 0.098125.92 cos (u - 9.4623°)
At u = 0°,
[(Vg)d]2 = - 117.72 cos 0° = -117.72 J
[(Vg)b]2 = - 0.098125.92 cos (0 - 9.4623°) = - 0.23544 J
Conservation of Energy.
T2 + V2 = T3 + V3
6.2996 + ( -117.72) + ( -0.23544) = 0 + ( - 117.72 cos u)
+
3 - 0.098125.92 cos (u
117.72 cos u + 0.098125.92 cos (u - 9.4623°) = 111.66
- 9.4623°) 4
Solved by numerically,
Ans.
u = 18.83° = 18.8°
1014
Ans:
v2 = 0.656 rad>s
u = 18.8°
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
19–31.
The 10-g bullet having a velocity of 800 m>s is fired into the
edge of the 5-kg disk as shown. Determine the angular
velocity of the disk just after the bullet becomes embedded
into its edge. Also, calculate the angle u the disk will swing
when it stops. The disk is originally at rest. The rod AB has a
mass of 3 kg.
A
2m
B
v 800 m/s
0.4 m
Solution
Conservation of Angular Momentum. The mass moments of inertia of the disk and
1
rod about their respective mass centers are (IG)d = (5) ( 0.42 ) = 0.4 kg # m2 and
2
1
(IG)r =
(3) ( 22 ) = 1.00 kg # m2. Also, (vb)2 = v2 ( 25.92 ) , (vd)2 = v2(2.4) and
12
( vr ) 2 = v2 (1). Referring to the momentum diagram with the embedded bullet, Fig. a,
Σ(HA)1 = Σ(HA)2
Mb(vb)1(rb)1 = (IG)d v2 + Md(vd)2(rd) + (IG)r v2 + mr(vr)2 (r) + Mb(vb)2(rb)2
0.01(800)(2.4) = 0.4v2 + 5[v2(2.4)](2.4) + 1.00v2 + 3[v2 (1)] (1)
+ 0.01 3 v2 ( 25.92 ) 4 ( 25.92 ) 4
Ans.
v2 = 0.5773 rad>s = 0.577 rad>s
Kinetic Energy. Since the system is required to stop finally, T3 = 0. Here
1
1
1
1
1
(IG)d v22 + Md (vd)22 + ( IG ) r v22 + Mr ( vr ) 22 + Mb ( vb ) 22
2
2
2
2
2
1
1
1
1
1
2
2
= (0.4) ( 0.5773 ) + (5)[0.5773(2.4)] + (1.00)( 0.57732 ) + (3)[0.5773(1)]2 + (0.01) 3 0.5773( 25.92) 4 2
2
2
2
2
2
= 5.5419 J
T2 =
1015
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
19–31. Continued
Potential Energy. Datum is set as indicated on Fig. b.
0.4
b = 9.4623°. Hence
2.4
yd = 2.4 cos u, yr = cos u, yb = 25.92 cos (u - 9.4623°)
Here f = tan-1a
Thus, the gravitational potential energy of the disk, rod and bullet with reference to
the datum is
(Vg)d = Md g yd = 5(9.81)( - 2.4 cos u) = - 117.72 cos u
(Vg)r = Mr g yr = 3(9.81)( - cos u) = - 29.43 cos u
(Vg)b = mb g yb = 0.01(9.81) 3 - 25.92 cos (u - 94623°) 4
= - 0.098125.92 cos(u - 9.4623°)
At u = 0°,
[(Vg)d]2 = - 117.72 cos 0° = - 117.72 J
[(Vg)r]2 = - 29.43 cos 0° = -29.43 J
[(Vg)b]2 = - 0.098125.92 cos (0° - 9.4623°) = - 0.23544 J
Conservation of Energy.
T2 + V2 = T3 + V3
5.5419 + ( -117.72) + ( - 29.43) + ( - 0.23544) = 0 + ( -117.72 cos u)
+ ( - 29.43 cos u) +
3 - 0.098125.92 cos (u
147.15 cos u + 0.098125.92 cos (u - 9.4623°) = 141.84
- 9.4623°) 4
Solved numerically,
Ans.
u = 15.78° = 15.8°
Ans:
v2 = 0.577 rad>s
u = 15.8°
1016
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*19–32.
The circular disk has a mass m and is suspended at A by the
wire. If it receives a horizontal impulse I at its edge B,
determine the location y of the point P about which the disk
appears to rotate during the impact.
A
y
P
a
Solution
I
B
Principle of Impulse and Momentum. The mass moment of inertia of the disk about
1
1
its mass center is IG = mr 2 = ma2
2
2
a+ IGv1 + Σ
MG dt = IG v2
1
0 + Ia = a ma2 bv
2
1
I = ma v(1)
2
+ )
(S
Lt1
t2
m 3 ( vG ) x 4 , + Σ
Lt1
t2
Fx dt = m 3 ( vG ) x 4 2
0 + I = mvG(2)
Equating eqs. (1) and (2),
1
ma v = mvG
2
a
vG = v
2
Kinematics. Here, IC is located at P, Fig. b. Thus, vB = v rB>IC = v(2a - y). Using
similar triangles,
a - y
2a - y
a - y
2a - y
=
; =
vG
vB
a
v(2a - y)
v
2
1
y = a
2
Ans.
Ans:
y =
1017
1
a
2
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
19–33.
0.20 m
0.65 m
The 80-kg man is holding two dumbbells while standing on a
turntable of negligible mass, which turns freely about a
vertical axis. When his arms are fully extended, the turntable is rotating with an angular velocity of 0.5 rev>s.
Determine the angular velocity of the man when he retracts
his arms to the position shown. When his arms are fully
extended, approximate each arm as a uniform 6-kg rod
having a length of 650 mm, and his body as a 68-kg solid
cylinder of 400-mm diameter. With his arms in the retracted
position, assume the man as an 80-kg solid cylinder of 450-mm
diameter. Each dumbbell consists of two 5-kg spheres of
negligible size.
0.3 m
0.3 m
SOLUTION
Conservation of Angular Momentum: Since no external angular impulse acts on the
system during the motion, angular momentum about the axis of rotation (z axis) is
conserved. The mass moment of inertia of the system when the arms are in the fully
extended position is
(Iz)1 = 2 c 10(0.852) d + 2 c
1
1
(6)(0.652) + 6(0.5252) d + (68)(0.2 2)
12
2
= 19.54 kg # m2
And the mass moment of inertia of the system when the arms are in the retracted
position is
(Iz)2 = 2 c10(0.32) d +
1
(80)(0.2252)
2
= 3.825 kg # m2
Thus,
(Hz)1 = (Hz)2
(Iz)1v1 = (Iz)2v2
19.54(0.5) = 3.825v2
Ans.
v2 = 2.55 rev>s
Ans:
v2 = 2.55 rev>s
1018
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
19–34.
The platform swing consists of a 200-lb flat plate suspended
by four rods of negligible weight. When the swing is at rest,
the 150-lb man jumps off the platform when his center of
gravity G is 10 ft from the pin at A. This is done with a
horizontal velocity of 5 ft>s, measured relative to the swing
at the level of G. Determine the angular velocity he imparts
to the swing just after jumping off.
A
10 ft
11 ft
SOLUTION
(a + )
(HA)1 = (HA)2
G
1 200
200
150
0 + 0 = c a
(11)2 d v - c a
b (4)2 +
b (5 - 10v) d (10)
12 32.2
32.2
32.2
4 ft
Ans.
v = 0.190 rad>s
Ans:
v = 0.190 rad>s
1019
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
19–35.
The 2-kg rod ACB supports the two 4-kg disks at its ends. If
both disks are given a clockwise angular velocity
1vA21 = 1vB21 = 5 rad>s while the rod is held stationary
and then released, determine the angular velocity of the rod
after both disks have stopped spinning relative to the rod
due to frictional resistance at the pins A and B. Motion is in
the horizontal plane. Neglect friction at pin C.
0.75m
B
(VB)1
0.75m
C
0.15 m
A
0.15 m
(VA)1
SOLUTION
c+
H1 = H2
1
1
1
2c (4) (0.15)2d(5) = 2c (4)(0.15)2d v + 2[4(0.75 v)(0.75)] + c (2)(1.50)2d v
2
2
12
Ans.
v = 0.0906 rad>s
Ans:
v = 0.0906 rad>s
1020
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*19–36.
The satellite has a mass of 200 kg and a radius of gyration
about z axis of kz = 0.1 m, excluding the two solar panels
A and B. Each solar panel has a mass of 15 kg and can be
approximated as a thin plate. If the satellite is originally
spinning about the z axis at a constant rate
vz = 0.5 rad>s when u = 90°, determine the rate of spin if
both panels are raised and reach the upward position,
u = 0°, at the same instant.
z
vz
B
y
u 90
A
1.5 m
0.2 m
x
0.3 m
Solution
Conservation of Angular Momentum. When u = 90°, the mass moment of inertia of
the entire satellite is
Iz = 200 ( 0.12 ) + 2c
1
(15) ( 0.32 + 1.52 ) + 15 ( 0.952 ) d = 34.925 kg # m2
12
when u = 0°, I′z = 200 ( 0.12 ) + 2c
Thus
1
(15) ( 0.32 ) + 15 ( 0.22 ) d = 3.425 kg # m2
12
(Hz)1 = (Hz)2
Iz(vz)1 = I′z(vz)2
34.925(0.5) = 3.425(vz)2
Ans.
(vz)2 = 5.0985 rad>s = 5.10 rad>s
Ans:
(vz)2 = 5.10 rad>s
1021
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
19–37.
Disk A has a weight of 20 lb. An inextensible cable is
attached to the 10-lb weight and wrapped around the disk.
The weight is dropped 2 ft before the slack is taken up. If
the impact is perfectly elastic, i.e., e = 1, determine the
angular velocity of the disk just after impact.
0.5 ft
A
SOLUTION
For the weight
T1 + V1 = T2 + V2
0 + 10(2) =
10 lb
1 10
a
b v 22
2 32.2
v2 = 11.35 ft>s
(HA)2 = (HA)3
mv2 (0.5) + 0 = m y3 (0.5) + IA v
10
10
20
a 32.2
b (11.35)(0.5) + 0 = a 32.2
b v3 (0.5) + c 12 a 32.2
b (0.5)2 d v
( + T)
e =
0.5 v - v3
v2 - 0
1 =
0.5 v - v3
11.35 - 0
[1]
11.35 = 0.5 v - v3
[2]
Solving Eqs.[1] and [2] yields:
Ans.
v = 22.7 rad>s
v3 = 0
Ans:
v = 22.7 rad>s
1022
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
19–38.
The plank has a weight of 30 lb, center of gravity at G, and it
rests on the two sawhorses at A and B. If the end D is raised
2 ft above the top of the sawhorses and is released from
rest, determine how high end C will rise from the top of the
sawhorses after the plank falls so that it rotates clockwise
about A, strikes and pivots on the sawhorses at B, and
rotates clockwise off the sawhorse at A.
2 ft
B
A
G
C
3 ft
1.5 ft
1.5 ft
D
3 ft
SOLUTION
Establishing a datum through AB, the angular velocity of the plank just before striking
B is
T1 + V1 = T2 + V2
1 1 30
2
30
b(9)2 +
(1.5)2 d (vCD)22 + 0
0 + 30c (1.5) d = c a
6
2 12 32.2
32.2
(vCD)2 = 1.8915 rad>s
(vG)2 = 1.8915(1.5) = 2.837 m>s
(c +)
c
(HB)2 = (HB)3
1 30
30
30
1 30
a
b(9)2 d(1.8915) (2.837)(1.5) = c a
b(9)2 d(vAB)3 +
(v ) (1.5)
12 32.2
32.2
2 32.2
32.2 G 3
Since (vG)3 = 1.5(vAB)3
(vAB)3 = 0.9458 rad>s
(vG)3 = 1.4186 m>s
T3 + V3 = T4 + V4
1 1 30
1 30
c a
b(9)2 d (0.9458)2 + a
b (1.4186)2 + 0 = 0 + 30hG
2 12 32.2
2 32.2
hG = 0.125
Thus,
hC =
6
(0.125) = 0.500 ft
1.5
Ans.
Ans:
hC = 0.500 ft
1023
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
19–39.
The 12-kg rod AB is pinned to the 40-kg disk. If the disk is
given an angular velocity vD = 100 rad>s while the rod is
held stationary, and the assembly is then released, determine
the angular velocity of the rod after the disk has stopped
spinning relative to the rod due to frictional resistance at
the bearing B. Motion is in the horizontal plane. Neglect
friction at the pin A.
2m
vD
A
B
0.3 m
Solution
Conservation of Angular Momentum
Initial: Since the rod is stationary and the disk is not translating, the total angular
momentum about A equals the angular momentum of the disk about B, where for
1
1
the disk IB = mDr 2 = (40)(0.3)2 = 1.80 kg # m2.
2
2
(HA)1 = IBv1 = 1.80(100) = 180 kg # m2 >s
1
1
Final: The rod and disk move as a single unit for which IA = mRl 2 + mDr 2 + mDl 2
3
2
1
1
2
2
2
2
#
= (12)(2) + (40)(0.3) + (40)(2) = 177.8 kg m .
3
2
(HA)2 = IA v2 = 177.8 v2
Setting (HA)1 = (HA)2 and solving,
Ans.
v2 = 1.012 rad>s = 1.01 rad>s
Ans:
v2 = 1.01 rad>s
1024
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*19–40.
A thin rod of mass m has an angular velocity V 0 while
rotating on a smooth surface. Determine its new angular
velocity just after its end strikes and hooks onto the peg and
the rod starts to rotate about P without rebounding. Solve
the problem (a) using the parameters given, (b) setting
m = 2 kg, v0 = 4 rad>s, l = 1.5 m.
P
ω0
l
SOLUTION
(a)
© (HP)0 = ©(HP)1
c
1
1
ml2 d v0 = c ml2 d v
12
3
v =
1
v
4 0
(b) From part (a)
Ans.
v =
1
1
v = (4) = 1 rad s
4 0
4
Ans.
Ans:
1
v0
4
v = 1 rad>s
v =
1025
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
19–41.
Tests of impact on the fixed crash dummy are conducted
using the 300-lb ram that is released from rest at u = 30°,
and allowed to fall and strike the dummy at u = 90°. If the
coefficient of restitution between the dummy and the ram is
e = 0.4, determine the angle u to which the ram will
rebound before momentarily coming to rest.
u
10 ft
10 ft
SOLUTION
Datum through pin support at ceiling.
T1 + V1 = T2 + V2
0 - 300(10 sin 30°) =
1 300
a
b (v)2 - 300(10)
2 32.2
v = 17.944 ft>s
+ 2
1:
e = 0.4 =
v¿ - 0
0 - (- 17.944)
v¿ = 7.178 ft>s
T2 + V2 = T3 + V3
1 300
a
b(7.178)2 - 300(10) = 0 - 300(10 sin u)
2 32.2
Ans.
u = 66.9°
Ans:
u = 66.9°
1026
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
19–42.
z
The vertical shaft is rotating with an angular velocity of
3 rad>s when u = 0°. If a force F is applied to the collar so
that u = 90°, determine the angular velocity of the shaft.
Also, find the work done by force F. Neglect the mass of
rods GH and EF and the collars I and J. The rods AB and
CD each have a mass of 10 kg.
D
0.3 m
0.3 m
0.3 m
E
I
u
Conservation of Angular Momentum: Referring to the free-body diagram of the
assembly shown in Fig. a, the sum of the angular impulses about the z axis is zero.
Thus, the angular momentum of the system is conserved about the axis. The mass
moments of inertia of the rods about the z axis when u = 0° and 90° are
G
u
A
C
SOLUTION
(Iz)1 = 2 c
0.3 m
0.1 m
v
F
0.1 m
J
H
F
1
(10)(0.62) + 10(0.3 + 0.1)2 d = 3.8 kg # m2
12
(Iz)2 = 2 c 10(0.12) d = 0.2 kg # m2
Thus,
(Hz)1 = (Hz)2
3.8(3) = 0.2v2
Ans.
v2 = 57 rad>s
Principle of Work and Energy: As shown on the free-body diagram of the assembly,
Fig. b, W does negative work, while F does positive work. The work of W is
UW = -Wh = -10(9.81)(0.3) = - 29.43 J. The initial and final kinetic energy of the
1
1
1
assembly is T1 = (Iz)1v1 2 = (3.8)(32) = 17.1 J and T2 = (Iz)2v2 2 =
2
2
2
1
2
(0.2)(57 ) = 324.9 J. Thus,
2
T1 + ©U1 - 2 = T2
17.1 + 2(- 29.43) + UF = 324.9
Ans.
UF = 367 J
Ans:
v2 = 57 rad>s
UF = 367 J
1027
B
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
19–43.
The mass center of the 3-lb ball has a velocity of
(vG)1 = 6 ft>s when it strikes the end of the smooth 5-lb
slender bar which is at rest. Determine the angular velocity
of the bar about the z axis just after impact if e = 0.8.
z
2 ft
A
2 ft
B
SOLUTION
Conservation of Angular Momentum: Since force F due to the impact is internal to
the system consisting of the slender bar and the ball, it will cancel out. Thus, angular
momentum is conserved about the z axis. The mass moment of inertia of the slender
(yB)2
5
1
a
b A 42 B = 0.2070 slug # ft2. Here, v2 =
bar about the z axis is Iz =
.
12 32.2
2
Applying Eq. 19–17, we have
0.5 ft
O
(vG)1
6 ft/s
G
r
0.5 ft
(Hz)1 = (Hz)2
C mb (yG)1 D (rb) = Iz v2 + C mb (yG)2 D (rb)
a
(yB)2
3
3
b (6)(2) = 0.2070 c
d + a
b(yG)2(2)
32.2
2
32.2
(1)
Coefficient of Restitution:Applying Eq. 19–20, we have
e =
0.8 =
(yB)2 - (yG)2
(yG)1 - (yB)1
(yB)2 - (yG)2
6 - 0
(2)
Solving Eqs. (1) and (2) yields
(yG)2 = 2.143 ft>s
(yB)2 = 6.943 ft>s
Thus, the angular velocity of the slender rod is given by
v2 =
(yB)2
6.943
=
= 3.47 rad>s
2
2
Ans.
Ans:
v2 = 3.47 rad>s
1028
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*19–44.
The pendulum consists of a slender 2-kg rod AB and 5-kg
disk. It is released from rest without rotating. When it falls
0.3 m, the end A strikes the hook S, which provides a
permanent connection. Determine the angular velocity of
the pendulum after it has rotated 90°. Treat the pendulum’s
weight during impact as a nonimpulsive force.
0.5 m
A
S
B
0.2 m
0.3 m
SOLUTION
T0 + V0 = T1 + V1
0 + 2(9.81)(0.3) + 5(9.81)(0.3) =
1
1
(2)(vG)21 + (5)(vG)21
2
2
(vG)1 = 2.4261 m>s
©(Hs)1 = ©(Hs)2
2(2.4261)(0.25) + 5(2.4261)(0.7) = c
1
1
(2)(0.5)2 + 2(0.25)2 + (5)(0.2)2 + 5(0.7)2 dv
12
2
v = 3.572 rad>s
T2 + V2 = T3 + V3
1 1
1
c (2)(0.5)2 + 2(0.25) + (5)(0.2)2 + 5(0.7)2 d (3.572)2 + 0
2 12
2
=
1 1
1
c (2)(0.5)2 + 2(0.25)2 + (5)(0.2)2 + 5(0.7)2 dv2
2 12
2
+ 2(9.81)(- 0.25) + 5(9.81)(- 0.7)
Ans.
v = 6.45 rad>s
Ans:
v = 6.45 rad>s
1029
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
19–45.
The 10-lb block slides on the smooth surface when the
corner D hits a stop block S. Determine the minimum
velocity v the block should have which would allow it to tip
over on its side and land in the position shown. Neglect the
size of S. Hint: During impact consider the weight of the
block to be nonimpulsive.
B
v
1 ft
C
A
B
D
S D
C
1 ft
A
SOLUTION
Conservation of Energy: If the block tips over about point D, it must at least achieve
the dash position shown. Datum is set at point D. When the block is at its initial and
final position, its center of gravity is located 0.5 ft and 0.7071 ft above the datum. Its
initial
and
final
potential
energy
are
and
10(0.5) = 5.00 ft # lb
#
10(0.7071) = 7.071 ft lb. The mass moment of inertia of the block about point D is
ID =
1 10
10
a
b A 12 + 12 B + a
b A 20.52 + 0.52 B 2 = 0.2070 slug # ft2
12 32.2
32.2
1
1
The initial kinetic energy of the block (after the impact) is ID v22 = (0.2070) v22.
2
2
Applying Eq. 18–18, we have
T2 + V2 = T3 + V3
1
(0.2070) v22 + 5.00 = 0 + 7.071
2
v2 = 4.472 rad>s
Conservation of Angular Momentum:Since the weight of the block and the normal
reaction N are nonimpulsive forces, the angular momentum is conserves about
point D. Applying Eq. 19–17, we have
(HD)1 = (HD)2
(myG)(r¿) = ID v2
ca
10
b y d (0.5) = 0.2070(4.472)
32.2
Ans.
y = 5.96 ft>s
Ans:
v = 5.96 ft>s
1030
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
19–46.
Determine the height h at which a billiard ball of mass m
must be struck so that no frictional force develops between
it and the table at A. Assume that the cue C only exerts a
horizontal force P on the ball.
P
C
r
h
A
SOLUTION
For the ball
+ ) mn + © F dt = mn
(;
1
2
1
(1)
0 + P(¢ t) = mn2
a+
(HA)1 + © 1 MA dt = (HA)2
2
0 + (P)¢ t(h) = c m r2 + m r2 d v2
5
(2)
(3)
Require n2 = v2 r
Solving Eqs. (1)–(3) for h yields
h =
7
r
5
Ans.
Ans:
h =
1031
7
r
5
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
19–47.
The pendulum consists of a 15-kg solid ball and 6-kg rod. If it
is released from rest when u1 = 90°, determine the angle u2
after the ball strikes the wall, rebounds, and the pendulum
swings up to the point of momentary rest. Take e = 0.6.
A
100 mm
u
2m
Solution
300 mm
Kinetic Energy. The mass moment of inertia of the pendulum about A is
1
2
IA = (6) ( 22 ) + c (15) ( 0.32 ) + 15 ( 2.32 ) d = 87.89 kg # m2.
3
5
Thus,
T =
1
1
I v2 = (87.89) v2 = 43.945v2
2 A
2
Potential Energy. With reference to datum set in Fig. a, the gravitational potential
energy of the pendulum is
Vg = mrgyr + msgys
= 6(9.81)( - cos u) + 15(9.81)( -2.3 cos u)
= - 397.305 cos u
Coefficient of Restitution. The velocity of the mass center of the ball is
vb = vrG = v(2.3). Thus
+ )e =
(d
( vw ) 2 - ( vb ) 2′
0 - [ - v2′ (2.3)]
; 0.6 =
v2(2.3) - 0
( vb ) 2 - ( vw ) 1
(1)
v2′ = 0.6v2
Conservation of Energy. Consider the pendulum swing from the position u = 90° to
u = 0° just before the impact,
(Vg)1 = - 397.305 cos 90° = 0
(Vg)2 = - 397.305 cos 0° = - 397.305 J
T1 = 0 T2 = 43.945v22
Then
T1 + V1 = T2 + V2
0 + 0 = 43.945v22 + ( - 397.305)
v2 = 3.0068 rad>s
Thus, just after the impact, from Eq. (1)
v2′ = 0.6(3.0068) = 1.8041 rad>s
Consider the pendulum swing from position u = 90° just after the impact to u,
(Vg)2 ′ = (Vg)2 = - 397.305 J
(Vg)3 = - 397.305 cos u
T2 ′ = 43.945 ( 1.80412 ) = 143.03 J
T3 = 0 (required)
Then
T2 ′ + V2 ′ = T3 + V3
143.03 + ( - 397.305) = 0 + ( - 397.305 cos u)
Ans.
u = 50.21° = 50.2°
1032
Ans:
u = 50.2°
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*19–48.
The 4-lb rod AB is hanging in the vertical position. A 2-lb
block, sliding on a smooth horizontal surface with a velocity
of 12 ft/s, strikes the rod at its end B. Determine the velocity of
the block immediately after the collision. The coefficient
of restitution between the block and the rod at B is e = 0.8.
A
3 ft
12 ft/s
B
SOLUTION
Conservation of Angular Momentum: Since force F due to the impact is internal to
the system consisting of the slender rod and the block, it will cancel out. Thus,
angular momentum is conserved about point A. The mass moment of inertia of the
4
1
4
slender rod about point A is IA =
a
b (32) +
(1.52) = 0.3727 slug # ft2.
12 32.2
32.2
(vB)2
Here, v2 =
. Applying Eq. 19–17, we have
3
(HA)1 = (HA)2
[mb (vb)1](rb) = IAv2 + [mb (vb)2](rb)
a
(vB)2
2
2
b (12)(3) = 0.3727 c
d + a
b (vb)2 (3)
32.2
3
32.2
[1]
Coefficient of Restitution: Applying Eq. 19–20, we have
+ )
(:
e =
(vB)2 - (vb)2
(vb)1 - (vB)1
0.8 =
(vB)2 - (vb)2
12 - 0
[2]
Solving Eqs. [1] and [2] yields
(vb)2 = 3.36 ft s :
Ans.
(vB)2 = 12.96 ft s :
Ans:
(vb)2 = 3.36 ft>s S
1033
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
19–49.
The hammer consists of a 10-kg solid cylinder C and 6-kg
uniform slender rod AB. If the hammer is released from rest
when u = 90° and strikes the 30-kg block D when u = 0°,
determine the velocity of block D and the angular velocity of
the hammer immediately after the impact. The coefficient of
restitution between the hammer and the block is e = 0.6.
u
100 mm
B
SOLUTION
150 mm
Conservation of Energy: With reference to the datum in Fig. a, V1 = (Vg)1 =
WAB(yGAB)1 + WC(yGC)1 = 0 and V2 = (Vg)2 = - WAB(yGAB)2 - WC(yGC)2 =
- 6(9.81)(0.25) - 10(9.81)(0.55) = - 68.67 J. Initially, T1 = 0. Since the hammer
rotates
about
the
fixed
axis,
and
(vGAB)2 = v2rGAB = v2(0.25)
(vGC)2 = v2rGC = v2(0.55). The mass moment of inertia of rod AB and cylinder C
1
1
about their mass centers is IGAB =
ml2 =
(6)(0.52) = 0.125 kg # m2 and
12
12
1
1
m(3r2 + h2) =
(10) C 3(0.052) + 0.152 D = 0.025 kg # m2. Thus,
IC =
12
12
T2 =
=
A
500 mm
1
1
1
1
I
v 2 + mAB(vGAB)2 2 + IGC v22 + mC(vGC)22
2 GAB 2
2
2
2
1
1
1
1
(0.125)v22 + (6) C v2(0.25) D 2 + (0.025)v22 + (10) C v2(0.55) D 2
2
2
2
2
= 1.775 v22
Then,
T1 + V1 = T2 + V2
0 + 0 = 1.775v2 2 + ( -68.67)
v2 = 6.220 rad>s
Conservation of Angular Momentum: The angular momentum of the system is
conserved point A. Then,
(HA)1 = (HA)2
0.125(6.220) + 6[6.220(0.25)](0.25) + 0.025(6.220) + 10[6.220(0.55)](0.55)
= 30vD(0.55) - 0.125v3 - 6[v3(0.25)](0.25) - 0.025v3 - 10[v3(0.55)](0.55)
(1)
16.5vD - 3.55v3 = 22.08
1034
C 50 mm
D
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
19–49. Continued
Coefficient of Restitution: Referring to Fig. c, the components of the velocity of the
impact point P just before and just after impact along the line of impact are
C (vP)x D 2 = (vGC)2 = v2rGC = 6.220(0.55) = 3.421 m>s :
v3rGC = v3 (0.55) ; . Thus,
+
:
e =
and
C (vP)x D 3 = (vGC)3 =
(vD)3 - C (vP)x D 3
C (vP)x D 2 - (vD)2
0.6 =
(vD)3 -
C - v3(0.55) D
3.421 - 0
(2)
(vD)3 + 0.55v3 = 2.053
Solving Eqs. (1) and (2),
(vD)3 = 1.54 m>s
Ans.
v3 = 0.934 rad>s
Ans:
(vD)3 = 1.54 m>s
v3 = 0.934 rad>s
1035
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
19–50.
The 20-kg disk strikes the step without rebounding.
Determine the largest angular velocity v1 the disk can have
and not lose contact with the step, A.
◊1
200 mm
A
30 mm
Solution
Conservation of Angular Momentum. The mass moment of inertia of the disk about
1
1
its mass center is IG = mr 2 = (20) ( 0.22 ) = 0.4 kg # m2. Since no slipping occurs,
2
2
vG = vr = v(0.2). Referring to the impulse and momentum diagram, Fig. a, we
notice that angular moment is conserved about point A since W is nonimpulsive.
Thus,
(HA)1 = (HA)2
20[v1(0.2)](0.17) + 0.4 v1 = 0.4 v2 + 20[v2(0.2)](0.2)
(1)
v1 = 1.1111 v2
Equations of Motion. Since the requirement is the disk is about to lose contact with the
0.17
step when it rotates about A, NA ≃ 0. Here u = cos-1 a
b = 31.79°. Consider the
0.2
motion along n direction,
+ R ΣFn = M(aG)n; 20(9.81) cos 31.79° = 203v22(0.2) 4
v2 = 6.4570 rad>s
Substitute this result into Eq. (1)
Ans.
v1 = 1.1111(6.4570) = 7.1744 rad>s = 7.17 rad>s
Ans:
v1 = 7.17 rad>s
1036
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
19–51.
The solid ball of mass m is dropped with a velocity v1 onto
the edge of the rough step. If it rebounds horizontally off
the step with a velocity v2, determine the angle u at which
contact occurs. Assume no slipping when the ball strikes the
step. The coefficient of restitution is e.
u
r
v1
v2
SOLUTION
Conservation of Angular Momentum: Since the weight of the solid ball is a
nonimpulsive force, then angular momentum is conserved about point A. The mass
2
moment of inertia of the solid ball about its mass center is IG = mr2. Here,
5
y2 cos u
v2 =
. Applying Eq. 19–17, we have
r
(HA)1 = (HA)2
C mb (yb)1 D (r¿) = IG v2 + C mb (yb)2 D (r–)
y2 cos u
2
b + (my2)(r cos u)
(my1)(r sin u) = a mr2 b a
r
5
y2
5
= tan u
y1
7
(1)
Coefficient of Restitution:Applying Eq. 19–20, we have
e =
e =
0 - (yb)2
(yb)1 - 0
- (y2 sin u)
- y1 cos u
y2
e cos u
=
y1
sin u
(2)
Equating Eqs. (1) and (2) yields
5
e cos u
tan u =
7
sin u
7
tan2 u = e
5
u = tan - 1 ¢
7
e≤
A5
Ans.
Ans:
u = tan - 1a
1037
7
eb
A5
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*19–52.
V1
150 mm
The wheel has a mass of 50 kg and a radius of gyration of
125 mm about its center of mass G. Determine the
minimum value of the angular velocity V1 of the wheel, so
that it strikes the step at A without rebounding and then
rolls over it without slipping.
G
A
SOLUTION
Conservation of Angular Momentum: Referring to Fig. a, the sum of the angular
impulses about point A is zero. Thus, angular momentum of the wheel is conserved
about this point. Since the wheel rolls without slipping, (vG)1 = v1r = v1(0.15) and
(vG)2 = v2r = v2(0.15). The mass moment of inertia of the wheel about its mass
center is IG = mkG2 = 50(0.1252) = 0.78125 kg # m2. Thus,
(HA)1 = (HA)2
50[v1(0.15)](0.125) + 0.78125v1 = 50[v2(0.15)](0.15) + 0.78125v2
(1)
v1 = 1.109v2
Conservation of Energy: With reference to the datum in Fig. a, V2 = (Vg)2 =
W(yG)2 = 0 and V3 = (Vg)3 = W(yG)3 = 50(9.81)(0.025) = 12.2625 J. Since the
wheel is required to be at rest in the final position, T3 = 0. The initial kinetic energy
1
1
1
1
of the wheel is T2 = m(vG)2 2 + IGv2 2 = (50)[v2(0.15)]2 + (0.78125)(v2 2) =
2
2
2
2
0.953125v2 2. Then
T2 + V2 = T3 + V3
0.953125v22 + 0 = 0 + 12.2625
v2 = 3.587 rad>s
Substituting this result into Eq. (1), we obtain
Ans.
v1 = 3.98 rad>s
1038
25 mm
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
19–53.
The wheel has a mass of 50 kg and a radius of gyration of
125 mm about its center of mass G. If it rolls without
slipping with an angular velocity of V1 = 5 rad>s before it
strikes the step at A, determine its angular velocity after it
rolls over the step. The wheel does not loose contact with
the step when it strikes it.
150 mm
V1
G
A
25 mm
SOLUTION
Conservation of Angular Momentum: Referring to Fig. a, the sum of the angular
impulses about point A is zero. Thus, angular momentum of the wheel is conserved
about this point. Since the wheel rolls without slipping, (vG)1 = v1r = (5)(0.15) =
0.75 m>s and v2 = v2r = v2(0.15). The mass moment of inertia of the wheel about its
mass center is IG = mkG2 = 50(0.1252) = 0.78125 kg # m2. Thus,
(HA)1 = (HA)2
50(0.75)(0.125) + 0.78125(5) = 50[v2(0.15)](0.15) + 0.78125v2
(1)
v2 = 4.508 rad>s
Conservation of Energy: With reference to the datum in Fig. a, V2 = (Vg)2 =
W(yG)2 = 0 and V3 = (Vg)3 = W(yG)3 = 50(9.81)(0.025) = 12.2625 J. The initial
1
1
1
kinetic energy of the wheel is T = mvG 2 + IGv2 = (50)[v(0.15)]2 +
2
2
2
1
2
2
2
(0.78125)v = 0.953125v . Thus, T2 = 0.953125v2 = 0.953125(4.5082) = 19.37 J
2
and T3 = 0.953125v32.
T2 + V2 = T3 + V3
19.37 + 0 = 0.953125v32 + 12.2625
Ans.
v3 = 2.73 rad>s
Ans:
v3 = 2.73 rad>s
1039
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
19–54.
The rod of mass m and length L is released from rest
without rotating. When it falls a distance L, the end A strikes
the hook S, which provides a permanent connection.
Determine the angular velocity v of the rod after it has
rotated 90°. Treat the rod’s weight during impact as a
nonimpulsive force.
A
L
L
S
Solution
T1 + V1 = T2 + V2
0 + mgL =
1 2
mv + 0
2 G
vG = 22gL
H1 = H2
L
1
m22gL a b = mL2(v2)
2
3
v2 =
3 22gL
2 L
T2 + V2 = T3 + V3
1 1 2 9(2gL)
1 1
L
a mL b
+ 0 = a mL2 bv2 - mga b
2
2 3
2 3
2
4L
3
1
L
gL = L2v2 - ga b
4
6
2
v =
A
g
7.5 L
Ans.
Ans:
v =
1040
A
7.5
g
L
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
19–55.
B
The 15-lb rod AB is released from rest in the vertical
position. If the coefficient of restitution between the floor
and the cushion at B is e = 0.7, determine how high the end
of the rod rebounds after impact with the floor.
2 ft
SOLUTION
A
T1 = V1 = T2 + V2
0 + 15(1) =
1 1 15
c a
b (2)2 d v22
2 3 32.2
v2 = 6.950 rad>s
A+TB
e =
Hence (vB)2 = 6.950(2) = 13.90 rad>s
0 - (vB)3
;
(vB)2 - 0
0.7 =
0 - (vB)3
13.90
(vB)3 = -9.730 ft>s = 9.730 ft>s
v3 =
c
(vB)3
9.730
=
= 4.865 rad>s
2
2
T3 + V3 = T4 + V4
1 1 15
c a
b (2)2 d(4.865)2 = 0 + 15(hG)
2 3 32.2
hG = 0.490 ft
Ans.
hB = 2hG = 0.980 ft
Ans:
hB = 0.980 ft
1041
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*19–56.
A ball having a mass of 8 kg and initial speed of
v1 = 0.2 m>s rolls over a 30-mm-long depression. Assuming
that the ball rolls off the edges of contact first A, then B,
without slipping, determine its final velocity v2 when it
reaches the other side.
v2
v1
SOLUTION
B
0.2
= 1.6 rad>s
0.125
15
b = 6.8921°
u = sin - 1 a
125
v1 =
y2
v2 =
= 8y2
0.125
0.2 m/s
A
125 mm
30 mm
h = 125 - 125 cos 6.8921° = 0.90326 mm
T1 + V1 = T2 + V2
1 2
1
(8)(0.2)2 + c (8)(0.125)2 d (1.6)2 + 0
2
2 5
= - (0.90326)(10 - 3)8(9.81) +
1
1 2
(8)v2(0.125)2 + c (8)(0.125)2 d(v)2
2
2 5
v = 1.836 rad>s
(HB)2 = (HB)3
2
c (8)(0.125)2 d (1.836) + 8(1.836)(0.125) cos 6.892°(0.125 cos 6.892°)
5
- 8(0.22948 sin 6.892°)(0.125 sin 6.892°)
2
= c (8)(0.125)2 d v3 + 8(0.125)v3 (0.125)
5
v3 = 1.7980 rad>s
T3 + V3 = T4 + V4
1 2
1
c (8)(0.125)2 d (1.7980)2 + (8)(1.7980)2(0.125)2 + 0
2 5
2
= 8(9.81)(0.90326(10 - 3)) +
+
1 2
c (8)(0.125)2 d(v4)2
2 5
1
(8)(v4)2(0.125)2
2
v4 = 1.56 rad>s
So that
Ans.
y2 = 1.56(0.125) = 0.195 m>s
Ans:
v2 = 0.195 m>s
1042
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
19–57.
A solid ball with a mass m is thrown on the ground such that
at the instant of contact it has an angular velocity V 1 and
velocity components 1vG2x1 and 1vG2y1 as shown. If the
ground is rough so no slipping occurs, determine the
components of the velocity of its mass center just after
impact. The coefficient of restitution is e.
(vG)y1
V1
(vG)x1
G
r
SOLUTION
Coefficient of Restitution (y direction):
A+TB
e =
0 - (yG)y 2
(yG)y2 = - e(yG)y1 = e(yG)y1
(yG)y1 - 0
c
Ans.
Conservation of angular momentum about point on the ground:
(c + )
(HA)1 = (HA)2
2
2
- mr 2v1 + m(vG)x 1r = mr2v2 + m(vG)x 2 r
5
5
Since no slipping, (vG)x2 = v2 r then,
5 a (vG)x 1 v2 =
2
v rb
5 1
7r
Therefore
(yG)x 2 =
5
2
a (yG)x 1 - v1 rb
7
5
Ans.
Ans:
(vG)y2 = e(vG)y1 c
(vG)x2 =
1043
5
2
a(v ) - v1r b d
7 G x1
5
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
19–58.
The pendulum consists of a 10-lb solid ball and 4-lb rod. If
it is released from rest when u0 = 0°, determine the angle
u1 of rebound after the ball strikes the wall and the
pendulum swings up to the point of momentary rest. Take
e = 0.6.
0.3 ft
A
θ
2 ft
SOLUTION
IA =
0.3 ft
1
2 10
10
4
a
b (2)2 + a
b (0.3)2 + a
b (2.3)2 = 1.8197 slug # ft2
3 32.2
5 32.2
32.2
Just before impact:
T1 + V1 = T2 + V2
0 + 0 +
1
(1.8197)v2 - 4(1) - 10(2.3)
2
v = 5.4475 rad>s
vP = 2.3(5.4475) = 12.529 ft>s
Since the wall does not move,
+ b
a:
e = 0.6 =
(vP) - 0
0 - ( - 12.529)
(vP) = 7.518 ft>s
v¿ =
7.518
= 3.2685 rad>s
2.3
T3 + V3 = T4 + V4
1
(1.8197)(3.2685)2 - 4(1) - 10(2.3) = 0 - 4(1) sin u1 - 10(2.3 sin u1)
2
Ans.
u1 = 39.8°
Ans:
u1 = 39.8°
1044