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

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 coefﬁcient 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′)+Iω′.(1)

The coefﬁcient 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 v′are related by

v′=v′

P+(45)ω′.(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

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

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 coefﬁcient of restitution is e=0.4.

The duration of the impact is 0.002 seconds. Determine

O

mAvA1−Ft =mAvA2,Ft+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 coefﬁcient of restitution is used to relate the relative velocities

before and after the impact. Linear momentum is conserved for the

system.

602

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 coefﬁcient 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 coefﬁcient 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.

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 coefﬁcient 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 deﬁnition, the coefﬁcient of restitution is

P−v′

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 ﬁrst 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

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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

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

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 ﬂoor, 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

ﬂoor, 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 ﬂoor,

its angular velocity is ω=0 and its center of mass is

moving downward at 4 m/s. If the coefﬁcient of restitu-

tion of the impact is e=0.4, what is the bar’s angular

velocity after the impact?

e=0.4, smooth ﬂoor.

Angular momentum about the contact point

mvG

L

Problem 19.83 The length of the bar is 1 m and its

mass is 2 kg. Just before the bar hits the smooth ﬂoor, it

has angular velocity ωand its center of mass is moving

downward at 4 m/s. The coefﬁcient 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 4i−10j

(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 v′and ω′be her velocity and angular velocity after

608

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

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 ω′after

the impact?

(b)(a)

AA

1 m/s

B

ω

‘

Solution: Linear momentum is conserved:

610

Problem 19.87* In Problem 19.86, if the bars do not

bond together on impact and the coefﬁcient of restitution

is e=0.8, what are the angular velocities of the bars

after the impact?

v′

BP =v′

B+l

2ω′

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 ω′after

the impact?

A

B

ω

‘

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

coefﬁcient of restitution of the impact is e=0.4. (See

Example 19.8.)

612

Problem 19.91* While attempting to drive on an icy

street for the ﬁrst 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).

1.7−0.60

v′

AP =(v′

A+0.6ω′

A)i+1.7ω′

Aj.

The x-components of the velocities at Pare related by the coefﬁcient

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.)

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 coefﬁcient

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

0−Wdh=Wh

2.

3.

The angular velocities at impact:Bydeﬁnition, the coefﬁcient 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

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(1) ω′

B−ω′

A=eωA

The force reactions at P: From the principle of angular impulse-

momentum,

(2) ω′

B+ω′

A=ωA.

Solve (1) and (2):

ω′

A=(1−e)

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

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

0−mRgdh=−mRghR.

Since the pulley is one-to-one, hL=−hR, from which

U=UL+UR=(mL−mR)ghR.

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

(TL−TR)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

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=T2−T1,

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