Mechanical Engineering Chapter 18 Homework Mexerts Constant Force The Ropeif The 100kg

932

18–21.

SOLUTION

vG=(0.02)70 =1.40 ft>s

A yo-yo has a weight of 0.3 lb and a radius of gyration

If it is released from rest, determine how far it

must descend in order to attain an angular velocity

Neglect the mass of the string and assume

that the string is wound around the central peg such that the

mean radius at which it unravels is r=0.02 ft.

v=70 rad>s.

kO=0.06 ft.

O

r

Ans:

933

18–22.

SOLUTION

Kinetic Energy and Work:Since the windlass rotates about a fixed axis,

Thus, the kinetic energy of the system is

Since the system is initially at rest, . Referring to Fig. a,WA,Ax,Ay, and RB

Principle of Work and Energy:

T1+©U1–2=T2

T1=0

vC=vArA

If the 50-lb bucket is released from rest, determine its

velocity after it has fallen a distance of 10 ft.The windlass A

can be considered as a 30-lb cylinder, while the spokes are

slender rods, each having a weight of 2 lb. Neglect the

pulley’s weight. 4ft

0.5 ft

3ft

B

A

Ans:

934

18–23.

The coefficient of kinetic friction between the 100-lb disk

and the surface of the conveyor belt is 0.2. If the

conveyor belt is moving with a speed of when

the disk is placed in contact with it, determine the number

of revolutions the disk makes before it reaches a constant

angular velocity.

vC=6ft>s

mA

SOLUTION

Equation of Motion:In order to obtain the friction developed at point Aof the

Work:elpuoc tnatsnoc a spoleved noitcirf ehT

Principle of Work and Energy:The disk achieves a constant angular velocity

F

f=mkN=0.2(100) =20.0 lb

C

=6ft/sv

A

B

0.5 ft

935

*18–24.

The 30-kg disk is originally at rest, and the spring is

unstretched. A couple moment of

M=80 N #m

is then

applied to the disk as shown. Determine its angular velocity

when its mass center G has moved 0.5 m along the plane.

The disk rolls without slipping.

0.5 m

G

M  80 N  m

k  200 N/mA

SOLUTION

Kinetic Energy. Since the disk is at rest initially,

T1=0

. The disk rolls without

Work. Since the disk rolls without slipping, the friction

F

f does no work. Also when

18–25.

The 30-kg disk is originally at rest, and the spring is

unstretched. A couple moment

M=80 N #m

is then

applied to the disk as shown. Determine how far the center

of mass of the disk travels along the plane before it

momentarily stops. The disk rolls without slipping.

0.5 m

G

M  80 N  m

k  200 N/mA

SOLUTION

Kinetic Energy. Since the disk is at rest initially and required to stop nally,

Principle of Work and Energy.

Ans:

937

18–26.

A

B

1.5 ft

3 ft

u

Two wheels of negligible weight are mounted at corners A

and Bof the rectangular 75-lb plate. If the plate is released

from rest at , determine its angular velocity at the

instant just before .u=0°

u=90°

SOLUTION

Kinetic Energy and Work: Referring Fig. a,

The mass moment of inertia of the plate about its mass center is

Since the plate is initially at rest,.Referring to Fig. b, and do no work,

Principle of Work and Energy:

NB

NA

T1=0

18–27.

The link AB is subjected to a couple moment of

M=40 N #m

. If the ring gear C is fixed, determine the

angular velocity of the 15-kg inner gear when the link has

made two revolutions starting from rest. Neglect the mass

of the link and assume the inner gear is a disk. Motion

occurs in the vertical plane.

150 mm

200 mm

M  40 N  m

AB

C

SOLUTION

Kinetic Energy. The mass moment of inertia of the inner gear about its center

Thus,

Work. Referring to the FBD of the gear system, we notice that M does positive

Principle of Work and Energy.

Ans:

939

*18–28.

SOLUTION

Free Body Diagram:The spring force FsP does negative work since it acts in the

Principle of Work and Energy: The mass moment of inertia of the cylinder about

The 10-kg rod AB is pin-connected at Aand subjected to

acouple moment of M15 Nm.If the rod is released

from rest when the spring is unstretched at 30 ,

determine the rod’s angular velocity at the instant 60 .

As the rod rotates, the spring always remains horizontal,

because of the roller support at C.

u

.

C

B

k=40 N/m

Ans:

940

18–29.

SOLUTION

Kinematics:

vG=0.5vG

The 10-lb sphere starts from rest at 0° and rolls without

slipping down the cylindrical surface which has a radius of

10 ft. Determine the speed of the sphere’s center of mass

at the instant 45°.u

u

10 ft

0.5 ft

θ

Ans:

18–30.

MC

P



750 N

3 m

4 m

A

u

B

Motor Mexerts a constant force of on the rope.

If the 100-kg post is at rest when , determine the

angular velocity of the post at the instant .Neglect

the mass of the pulley and its size, and consider the post as a

slender rod.

u=60°

u=0°

P=750 N

SOLUTION

Kinetic Energy and Work: Since the post rotates about a fixed axis,

The mass moment of inertia of the post about its mass center is

Since the post is initially at rest, .Referring to Fig.a,,,and do no

work, while does positive work and does negative work. When ,

Principle of Work and Energy:

T1+©U1–2=T2

u=60°WP

RC

By

Bx

T

1=0

vG=vrG=v (1.5).

Ans:

18–31.

The linkage consists of two 6-kg rods AB and CD and a

20-kg bar BD. When

u=0°

, rod AB is rotating with an

angular velocity

v=2

rad

>

s. If rod CD is subjected to a

couple moment of

M=30 N #m

, determine

v

AB at the

instant

u=90°

.

SOLUTION

Kinetic Energy. The mass moment of inertia of each link about the axis of rotation

Work. Referring to the FBD of the assembly, Fig. a, the weights

W

b

,

W

c

and couple moment M do positive work when the links

undergo an angular displacement

u

. When u=90°=

p

rad,

Ans:

1.5 m

1 m 1 m

u

vM  30 N  m

B

C

A

D

943

*18–32.

The linkage consists of two 6-kg rods AB and CD and a

20-kg bar BD. When

u=0°

, rod AB is rotating with an

angular velocity

v=2

rad

>

s. If rod CD is subjected to a

couple moment

M=30 N #m

, determine

v

at the instant

u=45°

.

SOLUTION

Kinetic Energy. The mass moment of inertia of each link about the axis of rotation

Initially, v

=2 rad>s

. Then

Work. Referring to the FBD of the assembly, Fig. a, the weights

W

b

,

W

c

and couple moment M do positive work when the links

Principle of Work and Energy.

1.5 m

1 m 1 m

u

vM  30 N  m

B

C

A

D

944

18–33.

The two 2-kg gears Aand Bare attached to the ends of a

3-kg slender bar.The gears roll within the fixed ring gear C,

which lies in the horizontal plane.If a torque is

applied to the center of the bar as shown, determine the

number of revolutions the bar must rotate starting from rest

in order for it to have an angular velocity of

Forthe calculation, assume the gears can be approximated by

thin disks.What is the result if the gears lie in the vertical plane?

vAB =20 rad>s.

10-N #m

SOLUTION

Energy equation (where Grefers to the center of one of the two gears):

,, gnisU

IG=1

mgear =2kg

Mu=T2

400 mm

150 mm

M=10N·m

150 mm

AB

C

Ans:

945

18–34.

SOLUTION

T1+ΣU1–2=T2

The linkage consists of two 8-lb rods AB and CD and

a 10-lb bar AD. When u=0°, rod AB is rotating with an

angular velocity vAB =2 rad>s. If rod CD is subjected to a

couple moment M=15 lb #ft and bar AD is subjected to a

horizontal force P=20 lb as shown, determine vAB at the

instant u=90°.

Ans:

3 ft

2 ft 2 ft

M  15 lb · ft

BC

AD

P  20 lb

u

vAB

18–35.

SOLUTION

T1+ΣU1–2=T2

The linkage consists of two 8-lb rods AB and CD and

a 10-lb bar AD. When u=0°, rod AB is rotating with an

angular velocity vAB =2 rad>s. If rod CD is subjected to a

couple moment M=15 lb #ft and bar AD is subjected to a

horizontal force P=20 lb as shown, determine vAB at the

instant u=45°.

3 ft

2 ft 2 ft

M = 15 lb · ft

AB

BC

AD

P = 20 lb

Ans:

*18–36.

The assembly consists of a 3-kg pulley A and 10-kg pulley B.

If a 2-kg block is suspended from the cord, determine the

block’s speed after it descends 0.5 m starting from rest.

Neglect the mass of the cord and treat the pulleys as thin

disks. No slipping occurs.

A

B

30 mm

100 mm

SOLUTION

T1+V1=T2+V2

Thus,

Ans:

948

18–37.

The assembly consists of a 3-kg pulley A and 10-kg pulley B.

If a 2-kg block is suspended from the cord, determine the

distance the block must descend, starting from rest, in order

to cause B to have an angular velocity of 6 rad

>

s. Neglect

the mass of the cord and treat the pulleys as thin disks. No

slipping occurs.

A

B

30 mm

100 mm

SOLUTION

vC=

v

B (0.1) =0.03

v

A

949

18–38.

SOLUTION

System:

Block:

T1+©U1–2=T2

T1+V1=T2+V2

The spool has a mass of 50 kg and a radius of gyration

If the 20-kg block Ais released from rest,

determine the distance the block must fall in order for the

spool to have an angular velocity Also, what is

the tension in the cord while the block is in motion? Neglect

the mass of the cord.

v=5 rad>s.

kO=0.280 m.

A

0.2 m

O

0.3 m

A

0.2 m

O

0.3 m

Ans:

950

18–39.

0.2 m

O

0.3 m

The spool has a mass of 50 kg and a radius of gyration

. If the 20-kg block Ais released from rest,

determine the velocity of the block when it descends 0.5 m.

kO=0.280 m

SOLUTION

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

potential energy of block Aat position 1 and 2 are

Kinetic Energy: Since the spool rotates about a fixed axis,.

Here, the mass moment of inertia about the fixed axis passes through point Ois

Conservation of Energy:

T1+V1=T2+V2

v=vA

rA

=vA

0.2 =5vA

Ans:

951

*18–40.

An automobile tire has a mass of 7 kg and radius of gyration

If it is released from rest at Aon the incline,

determine its angular velocity when it reaches the horizontal

plane.The tire rolls without slipping.

kG=0.3 m.

SOLUTION

Datum at lowest point.

T1+V1=T2+V2

0.4 m

30°

5m

G

A

B

0.4 m