Mechanical Engineering Chapter 22 Homework The Surface Smooth And The Springs Are

1210

22–21.

If the wire AB is subjected to a tension of 20 lb, determine

the equation which describes the motion when the 5-lb

weight is displaced 2 in. horizontally and released from rest.

Ans:

SOLUTION

L′KL

32.2

x=A sin pt +B cos pt

6 ft

A

1211

22–22.

The bar has a length land mass m.It is supported at its ends

by rollers of negligible mass.If it is given a small displacement

and released, determine the natural frequency of vibration.

SOLUTION

Moment of inertia about point O:

AB

R

l

Ans:

22–23.

The 20-kg disk, is pinned at its mass center O and supports

the 4-kg block A. If the belt which passes over the disk is

not allowed to slip at its contacting surface, determine the

natural period of vibration of the system.

SOLUTION

Equation of Motion. The mass moment of inertia of the disk about its mass

When the system is in equilibrium,

u=0°

. Then

Substitute this result into Eq. (1), we obtain

Since

a=u

$

, the above equation becomes

k = 50 N/m

A

300 mm

k  200 N/

m

O

Ans:

1213

Ans:

*22–24.

The 10-kg disk is pin connected at its mass center. Determine

the natural period of vibration of the disk if the springs have

sufficient tension in them to prevent the cord from slipping

on the disk as it oscillates. Hint: Assume that the initial

stretch in each spring is

d

O.

SOLUTION

Equation of Motion. The mass moment of inertia of the disk about its mass centerO

a

+

ΣM0=I0

a;

–80(

d

0+0.15

u

)(0.15) +80(

d

0–0.15

u

)(0.15) =0.1125

a

Since

a=u

$

, this equation becomes

Comparing to that of standard form, vn

=2

32 rad

>

s. Then

k  80 N/m

O

150 mm

22–25.

If the disk in Prob. 22–24 has a mass of 10 kg, determine the

natural frequency of vibration. Hint: Assume that the initial

stretch in each spring is

d

O.

Ans:

SOLUTION

Equation of Motion. The mass moment of inertia of the disk about its mass centerO

Since

a=u

$

, this equation becomes

k  80 N/m

O

150 mm

1215

22–26.

A flywheel of mass m, which has a radius of gyration about

its center of mass of , is suspended from a circular shaft

that has a torsional resistance of . If the flywheel

is given a small angular displacement of and released,

determine the natural period of oscillation.

u

M=Cu

kO

SOLUTION

Equation of Motion: The mass moment of inertia of the wheel about point Ois

. Referring to Fig. a,

Comparing this equation to the standard equation, the natural circular frequency of

the wheel is

IO=mkO2

L

O

Ans:

1216

22–27.

SOLUTION

TOis the equilibrium force.

Thus, for small ,

u

The 6-lb weight is attached to the rods of negligible mass.

Determine the natural frequency of vibration of the weight

when it is displaced slightly from the equilibrium position

and released.

3ft

2ft

k=5lb/ft

O

Ans:

*22–28.

AB

2.50 m

1.83 m

O

G2

G1

The platform AB when empty has a mass of 400 kg, center

of mass at , and natural period of oscillation .

If a car,having a mass of 1.2 Mg and center of mass at ,is

placed on the platform, the natural period of oscillation

becomes . Determine the moment of inertia of

the car about an axis passing through .G2

t2=3.16 s

G2

t1=2.38 sG1

SOLUTION

Free-body Diagram: When an object arbitrary shape having a mass mis pinned at O

Equation of Motion: Sum moment about point Oto eliminate and .

Kinematics: Since and if is small, then substituting these

values into Eq. (1), we have

When the platform is empty,,and .

Substituting these values into Eq. (3), we have

.Substituting these values into Eq. (3), we have

Thus, the mass moment inertia of the car about its mass center is

1407.55

l=2.50 mm=400 kgt=t1=2.38 s

u

sin u=ua =d2u

dt2=u

$

Oy

Ox

22–29.

The plate of mass m is supported by three symmetrically

placed cords of length l as shown. If the plate is given a

slight rotation about a vertical axis through its center and

released, determine the natural period of oscillation.

SOLUTION

ΣMz=Iz

a –3(T sin f)R=

1

2

mR2u

$

ΣFz=0

120

R

l

120

22–30.

SOLUTION

T+V=const.

k500 N/mk500 N/m

3kg

Determine the differential equation of motion of the 3-kg block

when it is displaced slightly and released. The surface is smooth

and the springs are originally unstretched.

1220

22–31.

SOLUTION

For small ,,then

sin u=uu

y=1(8)(2) +2(8)(2)

8(2) +8(2) =1.5 ft

Determine the natural period of vibration of the pendulum.

Consider the two rods to be slender, each having a weight

of 8 .lb>ft

O

2ft

Ans:

1221

*22–32.

Determine

the natural period of vibration of the 10-lb

semicircular disk.

SOLUTION

Datum at initial level of center of gravity of disk.

F

or small

u, sin u=u

0.5ft

Ans:

1222

22–33.

If the 20-kg wheel is displaced a small amount and released,

determine the natural period of vibration. The radius of

gyration of the wheel is

kG=0.36 m

. The wheel rolls

without slipping.

k



500 N

/

m

G

0.5 m

SOLUTION

Energy Equation. The mass moment of inertia of the wheel about its mass center

When the disk undergoes a small angular displacement

u

, the spring stretches

s=

u

(1) =

u, Fig. a. Thus, the elastic potential energy is

Thus, the total energy is

Since

u

#

≠0

, then

u

n

22–34.

Determine the differential equation of motion of the 3-kg

spool.

Assume that it does not slip at the surface

of

contact as it oscillates.The radius of gyration of the spool

about its center of mass is

kG=125 mm.

SOLUTION

k400N/m

G

200mm

100 mm

22–35.

Determine the natural period of vibration of the 3-kg

sphere. Neglect the mass of the rod and the size of

the sphere.

SOLUTION

By statics,

T(0.3) =3(9.81)(0.3)

E=T+V

300 mm300 mm

k500 N/m

O

Ans:

*22–36.

If the lower end of the 6-kg slender rod is displaced a small

amount and released from rest, determine the natural

frequency of vibration. Each spring has a stiffness of

k=200 N>m

and is unstretched when the rod is hanging

vertically.

SOLUTION

Energy Equation. The mass moment of inertia of the rod about O is

with reference to the datum set in Fig. a, the gravitational potential energy is

u

Thus, the total energy is

Time Derivative. Taking the first time derivative of the above equation

Since

u

#

≠0

,

u

O

k

2 m

1226

22–37.

The disk has a weight of 30 lb and rolls without slipping on

the horizontal surface as it oscillates about its equilibrium

position. If the disk is displaced, by rolling it counterclockwise

0.2 rad, determine the equation which describes its

oscillatory motion and the natural period when it is released.

SOLUTION

Energy Equation. The mass moment of inertia of the disk about its center of gravity

When the disk undergoes a small angular displacement

u

the spring stretches

s=

u

r=

u

(0.5)

, Fig. a. Thus, the elastic potential energy is

Thus, the total energy is

Time Derivative. Taking the time derivative of the above equation,

Since

u

#

≠0

, then

u

Comparing to that of standard form,

0.5 ft

k  80 lb/ft

Ans:

1227

22–38.

The machine has a mass mand is uniformly supported by

four springs,each having a stiffness k.Determine the natural

period of vertical vibration.

SOLUTION

T+V=const.

G

kk

22–39.

SOLUTION

f=1.5umax

0.75

The slender rod has a weight of . If it is supported in

the horizontal plane by a ball-and-socket joint at Aand a

cable at B, determine the natural frequency of vibration

when the end Bis given a small horizontal displacement

and then released.

4lb>ft

0.75 ft

1.5 ft

A

B

Ans:

1229

*22–40.

If the slender rod has a weight of 5 lb, determine the natural

frequency of vibration. The springs are originally

unstretched.

SOLUTION

Energy Equation:When the rod is being displaced a small angular displacement of ,

Time Derivative:Taking the time derivative of Eq.[1], we have

Since , then

u

Z0

u

k=4lb/ft

O

2ft

1ft

Ans: