Mechanical Engineering Chapter 19 Problem The Mass The Ship And The Moment Inertia The Vessel About

subject Type Homework Help
subject Pages 14
subject Words 5963
subject Authors Anthony M. Bedford, Wallace Fowler

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page-pf1
Problem 19.68 The mass of the ship is 544,000 kg,
and the moment of inertia of the vessel about its center
of mass is 4 ×108kg-m2. Wind causes the ship to drift
sideways at 0.1 m/s and strike the stationary piling at P.
The coefcient of restitution of the impact is e=0.2.
What is the ship’s angular velocity after the impact?
45 m
16 m
P
Solution: Angular momentum about Pis conserved
(45)(mv) =(45)(mv)+.(1)
The coefcient of restitution is
e=v
P
v,(2)
where v
Pis the vertical component of the velocity of Pafter the
impact. The velocities v
Pand vare related by
v=v
P+(45.(3)
Solving Eqs. (1)(3), we obtain ω=0.00196 rad/s,v=0.0680 m/s,
and v
P=−0.02 m/s.
ω
vP
v
v
ω
= 0
PP
Before impact After impact
45 m 45 m
Problem 19.69 In Problem 19.68, if the duration of
the ship’s impact with the piling is 10 s, what is the
magnitude of the average force exerted on the ship by
the impact?
600
page-pf2
Problem 19.70 In Active Example 19.7, suppose that
the ball Aweighs 2 lb, the bar Bweighs 6 lb, and the
length of the bar is 3 ft. The ball is translating at vA=
10 ft/s before the impact and strikes the bar at h=2 ft.
What is the angular velocity of the bar after the impact
if the ball adheres to the bar?
C
h
A
v
Problem 19.71 The 2-kg sphere Ais moving toward
the right at 4 m/s when it strikes the end of the 5-kg slen-
der bar B. Immediately after the impact, the sphere Ais
moving toward the right at 1 m/s. What is the angular
velocity of the bar after the impact?
O
B
page-pf3
Problem 19.72 The 2-kg sphere Ais moving toward
the right at 4 m/s when it strikes the end of the 5-kg
slender bar B. The coefcient of restitution is e=0.4.
The duration of the impact is 0.002 seconds. Determine
O
mAvA1Ft =mAvA2,Ft+Rt =mBω2
2.
Problem 19.73 The 2-kg sphere Ais moving toward
the right at 10 m/s when it strikes the unconstrained
4-kg slender bar B. What is the angular velocity of the
bar after the impact if the sphere adheres to the bar?
B
0.25 m
The coefcient of restitution is used to relate the relative velocities
before and after the impact. Linear momentum is conserved for the
system.
602
page-pf4
Problem 19.74 The 2-kg sphere Ais moving to the
right at 10 m/s when it strikes the unconstrained 4-kg
slender bar B. The coefcient of restitution of the impact
is e=0.6. What are the velocity of the sphere and the
angular velocity of the bar after the impact?
B
A
0.25 m
Solution: Angular momentum for the system is conserved about
the center of mass of the bar. The coefcient of restitution is used
to relate the relative velocities before and after the impact. Linear
momentum is conserved for the system.
vA2=1.47 m/s to the right.
page-pf5
Problem 19.75 The 5-oz ball is translating with
velocity vA=80 ft/s perpendicular to the bat just before
impact. The player is swinging the 31-oz bat with
angular velocity ω=6πrad/s before the impact. Point
Cis the bat’s instantaneous center both before and after
the impact. The distances b=14 in and y=26 in. The
bat’s moment of inertia about its center of mass is IB=
0.033 slug-ft2. The coefcient of restitution is e=0.6,
and the duration of the impact is 0.008 s. Determine the
magnitude of the velocity of the ball after the impact
and the average force Axexerted on the bat by the
player during the impact if (a) d=0, (b) d=3 in, and
(c) d=8 in.
Ay
Ax
A
d
ω
y
Solution: By denition, the coefcient of restitution is
Pv
A
and (4), and these six equations in six unknowns reduce to four
equations in four unknowns. v
P,v
B,v
A, and ω. Further reductions
may be made by substituting (5) and (6) into (1) and (2); however here
the remaining four unknowns were solved by iteration for values of
Only the values in the rst two columns are required for the problem;
the other values are included for checking purposes. Note: The reaction
reverses between d=3 in. and d=8 in., which means that the point
of zero reaction occurs in this interval.
604
c
2008 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.
page-pf6
Problem 19.76 In Problem 19.75, show that the force
Axis zero if d=IB/(mBy), where mBis the mass of
ymB
Problem 19.77 A 10-lb slender bar of length l=2ft
l
page-pf7
Problem 19.78 A 10-lb slender bar of length l=2ft
is released from rest in the horizontal position at a height
h=1 ft above a peg (Fig. a). A small hook at the end
of the bar engages the peg, and the bar swings from the
peg (Fig. b).
(a) Through what maximum angle does the bar rotate
relative to its position when it engages the peg?
(b) At the instant when the bar has reached the angle
determined in part (a), compare its gravitational
potential energy to the gravitational potential
energy the bar had when it was released from rest.
How much energy has been lost?
l
h
(b)(a)
Problem 19.79 The 1-slug disk rolls at velocity v=
10 ft/s toward a 6-in step. The wheel remains in contact
with the step and does not slip while rolling up onto it.
What is the wheel’s velocity once it is on the step?
18 in
606
page-pf8
Problem 19.80 The 1-slug disk rolls toward a 6-in
step. The wheel remains in contact with the step and
does not slip while rolling up onto it. What is the mini-
mum velocity vthe disk must have in order to climb up
onto the step?
(2) with v3=0, we obtain
Problem 19.81 The length of the bar is 1 m and its
mass is 2 kg. Just before the bar hits the oor, its angu-
lar velocity is ω=0 and its center of mass is moving
downward at 4 m/s. If the end of the bar adheres to the
oor, what is the bar’s angular velocity after the impact?
v
Problem 19.82 The length of the bar is 1 m and its
mass is 2 kg. Just before the bar hits the smooth oor,
its angular velocity is ω=0 and its center of mass is
moving downward at 4 m/s. If the coefcient of restitu-
tion of the impact is e=0.4, what is the bar’s angular
velocity after the impact?
e=0.4, smooth oor.
Angular momentum about the contact point
mvG
L
page-pf9
Problem 19.83 The length of the bar is 1 m and its
mass is 2 kg. Just before the bar hits the smooth oor, it
has angular velocity ωand its center of mass is moving
downward at 4 m/s. The coefcient of restitution of the
impact is e=0.4. What value of ωwould cause the bar
to have no angular velocity after the impact?
Problem 19.84 During her parallel-bars routine, the
velocity of the 90-lb gymnast’s center of mass is 4i10j
(ft/s) and her angular velocity is zero just before she
grasps the bar at A. In the position shown, her moment
of inertia about her center of mass is 1.8 slug-ft2.Ifshe
stiffens her shoulders and legs so that she can be mod-
eled as a rigid body, what is the velocity of her center of
mass and her angular velocity just after she grasps the
bar?
(–8, –22) in
x
y
A
Solution: Let vand ωbe her velocity and angular velocity after
608
page-pfa
Problem 19.85 The 20-kg homogenous rectangular
plate is released from rest (Fig. a) and falls 200 mm
before coming to the end of the string attached at the
corner A(Fig. b). Assuming that the vertical component
of the velocity of Ais zero just after the plate reaches
the end of the string, determine the angular velocity of
the plate and the magnitude of the velocity of the corner
Bat that instant.
300 m
b)
A
200 mm
500 mm
B
A
B
(a)
m
(
Solution: We use work and energy to determine the plate’s down-
ward velocity just before the string becomes taut.
y
page-pfb
Problem 19.86* Two bars Aand Bare each 2 m in
length, and each has a mass of 4 kg. In Fig. (a), bar A
has no angular velocity and is moving to the right at
1 m/s, and bar Bis stationary. If the bars bond together
on impact, (Fig. b), what is their angular velocity ωafter
the impact?
(b)(a)
AA
1 m/s
B
B
ω
'
Solution: Linear momentum is conserved:
610
page-pfc
Problem 19.87* In Problem 19.86, if the bars do not
bond together on impact and the coefcient of restitution
is e=0.8, what are the angular velocities of the bars
after the impact?
v
BP =v
B+l
2ω
B.(6)
Problem 19.88* Two bars Aand Bare each 2 m in
length, and each has a mass of 4 kg. In Fig. (a), bar A
has no angular velocity and is moving to the right at
1 m/s, and Bis stationary. If the bars bond together on
impact (Fig. b), what is their angular velocity ωafter
the impact?
A
B
B
ω
'
page-pfd
Problem 19.89* The horizontal velocity of the landing
airplane is 50 m/s, its vertical velocity (rate of descent)
is 2 m/s, and its angular velocity is zero. The mass
of the airplane is 12 Mg, and the moment of inertia
about its center of mass is 1 ×105kg-m2. When the
rear wheels touch the runway, they remain in contact
with it. Neglecting the horizontal force exerted on the
wheels by the runway, determine the airplane’s angular
velocity just after it touches down. (See Example 19.8.)
0.3 m
1.8 m
Solution: Use a reference frame that moves to the left with the
v
G=v
P+ω×rG/P :
v
Gy j=v
Pxi+
ijk
00ω
0.31.80
.(2)
The jcomponent of this equation is
v
Gy =−0.3ω.(3)
Solving Eqs. (1) and (3), we obtain
ω=0.0712 rad/s.
v'
Gy
Problem 19.90* Determine the angular velocity of the
airplane in Problem 19.89 just after it touches down if
its wheels don’t stay in contact with the runway and the
coefcient of restitution of the impact is e=0.4. (See
Example 19.8.)
612
page-pfe
Problem 19.91* While attempting to drive on an icy
street for the rst time, a student skids his 1260-
of the impact is e=0.5. The moments of inertia of the
after the collision. (See Example 19.9.)
y
1.7 m
0.6 m
B
P
Solution: Car A’s initial velocity is
0=0.6mBv
B+IBω
B(3).
1.70.60
v
AP =(v
A+0.6ω
A)i+1.7ω
Aj.
The x-components of the velocities at Pare related by the coefcient
of restitution:
Problem 19.92* The student in Problem 19.91
claimed he was moving at 5 km/h prior to the collision,
but police estimate that the center of mass of the Rolls-
Royce was moving at 1.7 m/s after the collision. What
was the student’s actual speed? (See Example 19.9.)
page-pff
Problem 19.93 Each slender bar is 48 in. long and
weighs 20 lb. Bar Ais released in the horizontal posi-
tion shown. The bars are smooth, and the coefcient
of restitution of their impact is e=0.8. Determine the
angle through which Bswings afterward.
A
B
28 in
Solution: Choose a coordinate system with the xaxis parallel to
U=h
0Wdh=Wh
2.
3.
The angular velocities at impact:Bydenition, the coefcient of resti-
tution is
e=v
BPx v
AP x
.
Bar Bis at rest initially, from which
vBPx =0,v
BPx =v
BP ,and
v
AP x =v
AP cos β, vAP x =vAP cos β,
from which
e=v
BP v
AP cos β
vAP cos β.
FF
614
c
2008 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.
page-pf10
(1) ω
Bω
A=A
The force reactions at P: From the principle of angular impulse-
momentum,
(2) ω
B+ω
A=ωA.
Solve (1) and (2):
ω
A=(1e)
page-pf11
Problem 19.94* The Apollo CSM (A) approaches the
Soyuz Space Station (B). The mass of the Apollo is
mA=18 Mg, and the moment of inertia about the axis
through the center of mass parallel to the zaxis is IA=
114 Mg-m2. The mass of the Soyuz is mB=6.6 Mg,
and the moment of inertia about the axis through its
center of mass parallel to the zaxis is IB=70 Mg-m2.
The Soyuz is stationary relative to the reference frame
shown and the CSM approaches with velocity vA=
0.2i+0.05j(m/s)and no angular velocity. What is the
angular velocity of the attached spacecraft after docking?
7.3 m
(A)
4.3 m
x
y
(B)
The angular momentum about the origin is conserved:
11.60 0
=v
GAx i+(v
GAy +11.6ω)j,
ijk
Collect terms and substitute into conservation of angular
from which
616
page-pf12
Problem 19.95 The moment of inertia of the pulley
is 0.2 kg-m2. The system is released from rest. Use the
principle of work and energy to determine the velocity
of the 10-kg cylinder when it has fallen 1 m.
5 kg 10 kg
150 mm
UL=hR
0mRgdh=−mRghR.
Since the pulley is one-to-one, hL=−hR, from which
U=UL+UR=(mLmR)ghR.
page-pf13
Problem 19.96 The moment of inertia of the pulley
is 0.2 kg-m2. The system is released from rest. Use
momentum principles to determine the velocity of the
10-kg cylinder 1 s after the system is released.
From the principle of angular impulse-momentum for the pulley:
t2
t1
(TLTR)R dt =IPω2,
Problem 19.97 Arm BC has a mass of 12 kg, and the
moment of inertia about its center of mass is 3 kg-m2.
Point Bis stationary. Arm BC is initially aligned with
the (horizontal) xaxis with zero angular velocity, and a
constant couple Mapplied at Bcauses the arm to rotate
upward. When it is in the position shown, its counter-
clockwise angular velocity is 2 rad/s. Determine M.
x
y
300
mm
40°
B
M
A
C
618
page-pf14
Problem 19.98 The cart is stationary when a constant
force Fis applied to it. What will the velocity of the
cart be when it has rolled a distance b? The mass of the
body of the cart is mC, and each of the four wheels has
mass m, radius R, and moment of inertia I.
F
Solution: From the principle of work and energy, U=T2T1,
where T1=0. The work done is
Problem 19.99 Each pulley has moment of inertia I=
0.003 kg-m2, and the mass of the belt is 0.2 kg. If a
constant couple M=4 N-m is applied to the bottom
pulley, what will its angular velocity be when it has
turned 10 revolutions?
100 mm

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