Mechanical Engineering Chapter 18 Homework Gof Rolls Without Slipping determine Its Angular Velocity

18–55.

The slender 15-kg bar is initially at rest and standing in the

vertical position when the bottom end A is displaced slightly

to the right. If the track in which it moves is smooth,

determine the speed at which end A strikes the corner D.

The bar is constrained to move in the vertical plane. Neglect

the mass of the cord BC.

SOLUTION

x

2+

y

2=

5

2

Ans:

5 m

B

C

970

*18–56.

B

6 ft

4 ft

0.5 ft

If the chain is released from rest from the position shown,

determine the angular velocity of the pulley after the end B

has risen 2 ft.The pulley has a weight of 50 lb and a radius of

gyration of 0.375 ft about its axis.The chain weighs 6 lb/ft.

SOLUTION

Potential Energy: ,,, and .With

reference to the datum in Fig.a, the gravitational potential energy of the chain at

position 1 and 2 is

Kinetic Energy: Since the system is initially at rest, .The pulley rotates about

a fixed axis, thus,.The mass moment of inertia of

(VG1)2=(VG2)2 =v2 r=v2(0.5)

T1=0

(yG2)2=4 ft(yG1)2=1 ft(yG 2)1=3 ft(yG1)1=2 ft

Conservation of Energy:

T1+V1=T2+V2

Ans:

18–57.

1 ft

O

If the gear is released from rest, determine its angular

velocity after its center of gravity Ohas descended a

distance of 4 ft. The gear has a weight of 100 lb and a radius

of gyration about its center of gravity of .k=0.75 ft

SOLUTION

Potential Energy: With reference to the datum in Fig. a, the gravitational potential

energy of the gear at position 1 and 2 is

Kinetic Energy: Referring to Fig. b,we obtain .The mass momentvO=vrO / IC =v(1)

Conservation of Energy:

Ans:

972

18–58.

The slender 6-kg bar AB is horizontal and at rest and the

spring is unstretched. Determine the stiffness k of the spring

so that the motion of the bar is momentarily stopped when

it has rotated clockwise 90° after being released.

k

AB

C

1.5 m

SOLUTION

Kinetic Energy. The mass moment of inertia of the bar about A is

Potential Energy. With reference to the datum set in Fig. a, the gravitational

potential energies of the bar when it is at positions ➀ and ➁ are

Thus, the initial and final elastic potential energy of the spring are

Conservation of Energy.

973

18–59.

The slender 6-kg bar AB is horizontal and at rest and the

spring is unstretched. Determine the angular velocity of the

bar when it has rotated clockwise 45° after being released.

The spring has a stiffness of

k=12 N>m

.

k

AB

C

1.5 m

SOLUTION

Kinetic Energy. The mass moment of inertia of the bar about A is

Potential Energy. with reference to the datum set in Fig. a, the gravitational potential

energies of the bar when it is at positions ① and ② are

Then, using cosine law,

Thus, the stretch of the spring when the bar is at position ② is

Thus, the initial and final elastic potential energies of the spring are

Conservation of Energy.

Ans:

*18–60.

The pendulum consists of a 6-kg slender rod fixed to a 15-kg

disk. If the spring has an unstretched length of 0.2 m,

determine the angular velocity of the pendulum when it is

released from rest and rotates clockwise 90° from the

position shown. The roller at C allows the spring to always

remain vertical.

0.5 m 0.5 m 0.3 m

k  200 N/m

C

BD

A

0.5 m

SOLUTION

Kinetic Energy. The mass moment of inertia of the pendulum about B is

Potential Energy. with reference to the datum set in Fig. a, the gravitational potential

energies of the pendulum when it is at positions ① and ② are

The stretch of the spring when the pendulum is at positions

Thus, the initial and final elastic potential energies of the

Conservation of Energy.

Ans:

975

18–61.

The 500-g rod AB rests along the smooth inner surface of a

hemispherical bowl. If the rod is released from rest from the

position shown, determine its angular velocity at the instant

it swings downward and becomes horizontal.

SOLUTION

Select datum at the bottom of the bowl.

u=sin–1a0.1

0.2 b=30°

A

B

200 mm

18–62.

The 50-lb wheel has a radius of gyration about its center of

gravity Gof If it rolls without slipping,

determine its angular velocity when it has rotated clockwise

90° from the position shown. The spring AB has a stiffness

and an unstretched length of 0.5 ft.The wheel

is released from rest.

k=1.20 lb/ft

kG=0.7 ft.

SOLUTION

G

B

A

k= 1.20 lb/ft

3ft

1ft

0.5 ft

Ans:

977

18–63.

The system consists of 60-lb and 20-lb blocks Aand B,

respectively,and 5-lb pulleys Cand Dthat can be treated as

thin disks.Determine the speed of block Aafter block Bhas

risen 5 ft, starting from rest. Assume that the cord does not

slip on the pulleys,and neglect the mass of the cord.

SOLUTION

Kinematics: The speed of block Aand Bcan be related using the position

coordinate equation.

Potential Energy: Datum is set at fixed pulley C.When blocks Aand B(pulley D)are

at their initial position, their centers of gravity are located at sAand sB.Their initial

Kinetic Energy: The mass moment inertia of the pulley about its mass center is

Conservation of Energy: Applying Eq. 18–19, we have

0.5 ft

A

C

D

0.5 ft

B

*18–64.

The door is made from one piece, whose ends move along

the horizontal and vertical tracks. If the door is in the open

position, u

=0°,

and then released, determine the speed at

which its end A strikes the stop at C. Assume the door is a

180-lb thin plate having a width of 10 ft.

CA

B

5 ft

3 ft

u

SOLUTION

T1+V1=T2+V2

Ans:

18–65.

The door is made from one piece, whose ends move along

the horizontal and vertical tracks. If the door is in the open

position, u

=0°,

and then released, determine its angular

velocity at the instant

u=30°

. Assume the door is a 180-lb

thin plate having a width of 10 ft.

CA

B

5 ft

3 ft

u

SOLUTION

T1+V1=T2+V2

Ans:

18–66.

SOLUTION

The end Aof the garage door AB travels along the

horizontal track, and the end of member BC is attached to a

spring at C. If the spring is originally unstretched, determine

the stiffness kso that when the door falls downward from

rest in the position shown, it will have zero angular velocity

the moment it closes, i.e., when it and BC become vertical.

Neglect the mass of member BC and assume the door is a

thin plate having a weight of 200 lb and a width and height

of 12 ft. There is a similar connection and spring on the

other side of the door.

15

7 ft

12 ft A

B

C

D

2 ft 6 ft

1 ft

981

18–67.

The system consists of a 30-kg disk, 12-kg slender rod BA,

and a 5-kg smooth collar A. If the disk rolls without slipping,

determine the velocity of the collar at the instant

u=0°

.

The system is released from rest when

u=45°

.

SOLUTION

Kinetic Energy. Since the system is released from rest,

T1=0

. Referring to the

kinematics diagram of the rod at its final position, Fig. a, we found that IC is located

at B. Thus,

(

v

B)2=0

. Also

(

Potential Energy. Datum is set as shown in Fig. a. Here,

SB=2–2 cos 45°=0.5858 m

2 m

A

u

18–67. Continued

Conservation of Energy.

Ans.

Ans:

983

SOLUTION

Kinetic Energy. Since the system is released from rest,

T1=0

. Referring to the

kinematics diagram of the rod at final position with IC so located, Fig. a,

Since the disk rolls without slipping,

Thus, the kinetic energy of the system at final position is

Potential Energy. Datum is set as shown in Fig. a. Here,

SB=2 cos 30°–2 cos 45°=0.3178 m

*18–68.

The system consists of a 30-kg disk A, 12-kg slender rod BA,

and a 5-kg smooth collar A. If the disk rolls without slipping,

determine the velocity of the collar at the instant

u=30°

.

The system is released from rest when

u=45°

.

2 m

A

u

984

Thus, the gravitational potential energies of the disk, rod and collar at initial and

final position are

Conservation of Energy.

T1+V1=T2+V2

*18–68. Continued