Mechanical Engineering Chapter 21 Problem The Stepped Disk Weighs And Itsf Moment Inertia Slugft Rolls The

Problem 21.63 The stepped disk weighs 20 lb and its

moment of inertia is I=0.6 slug-ft2. It rolls on the

horizontal surface. The disk is initially stationary with

the spring unstretched, and at t=0 a constant force

F=10 lb is applied as shown. Determine the position

of the center of the disk as a function of time. 8 in.

16 in.

16 lb/ft

8 lb-s/ft

F

Fx=−kS −cdx

dt +F+f.

x=8F

9k1−edt d

ωd

sin ωdt+cos ωdt

Problem 21.64* An electric motor is bolted to a metal

table. When the motor is on, it causes the tabletop to

vibrate horizontally. Assume that the legs of the table

behave like linear springs, and neglect damping. The

total weight of the motor and the tabletop is 150 lb.

When the motor is not turned on, the frequency of

horizontal vibration of the tabletop and motor is 5 Hz.

When the motor is running at 600 rpm, the amplitude of

the horizontal vibration is 0.01 in. What is the magnitude

of the oscillatory force exerted on the table by the motor

at its 600-rpm running speed?

768

Problem 21.65 The moments of inertia of gears A

and Bare IA=0.014 slug-ft2and IB=0.100 slug-ft2.

Gear Ais connected to a torsional spring with constant

k=2 ft-lb/rad. The system is in equilibrium at t=0

when it is subjected to an oscillatory force F(t) =

4 sin 3tlb. What is the downward displacement of the

5-lb weight as a function of time?

F(t)

5 lb

6 in

AB

10 in

3 in

Solve:

Fd=IB

RWRA=6.667 ft−1,

and M=W

g+IB

(RW)2+RB

RWRA2

IA=2.378 slug,

which is the equilibrium point. Make the transformation ˜xp=xp−

xpw. The equation of motion about the equilibrium point is

0=ωAh−12

770

Problem 21.66* A 1.5-kg cylinder is mounted on a

sting in a wind tunnel with the cylinder axis transverse

to the direction of ﬂow. When there is no ﬂow, a 10-N

vertical force applied to the cylinder causes it to deﬂect

0.15 mm. When air ﬂows in the wind tunnel, vortices

subject the cylinder to alternating lateral forces. The

velocity of the air is 5 m/s, the distance between vortices

is 80 mm, and the magnitude of the lateral forces is

1 N. If you model the lateral forces by the oscillatory

function F(t) =(1.0)sin ω0tN, what is the amplitude

of the steady-state lateral motion of the sphere? 80 mm

5=0.016 s, from which the

period of a sinusoidal-like disturbance is τ=2(δt) =0.032 s, from

0=0.6667

[Note: This is a small deﬂection (113 microns) but the associated aero-

Problem 21.67 Show that the amplitude of the partic-

ular solution given by Eq. (21.31) is a maximum when

The maximum (or minimum) is found from

4[3ω2

0−ω2+2d2]ω0=˜ω0

Problem 21.68* A sonobuoy (sound-measuring

device) ﬂoats in a standing-wave tank. The device is

a cylinder of mass mand cross-sectional area A. The

water density is ρ, and the buoyancy force support-

ing the buoy equals the weight of the water that would

occupy the volume of the part of the cylinder below the

surface. When the water in the tank is stationary, the

buoy is in equilibrium in the vertical position shown at

the left. Waves are then generated in the tank, causing

the depth of the water at the sonobuoy’s position rela-

tive to its original depth to be d=d0sin ω0t. Let ybe

the sonobuoy’s vertical position relative to its original

position. Show that the sonobuoy’s vertical position is

governed by the equation

d2y

dt2+Aρg

my=Aρg

md0sin ω0t.

yd

Solution: The volume of the water displaced at equilibrium is

V=Ah where Ais the cross-sectional area, and his the equilibrium

immersion depth. The weight of water displaced is ρVg =ρgAh,so

Substitute:

d2y

772

Problem 21.69 Suppose that the mass of the sonobuoy

in Problem 21.68 is m=10 kg, its diameter is 125 mm,

and the water density is ρ=1025 kg/m3.Ifd=

0.1 sin 2tm, what is the magnitude of the steady-state

vertical vibrations of the sonobuoy?

yd

Problem 21.70 The mass weighs 50 lb. The spring con-

stant is k=200 lb/ft, and c=10 lb-s/ft. If the base is

subjected to an oscillatory displacement xiof amplitude

10 in. and frequency ωi=15 rad/s, what is the resulting

steady-state amplitude of the displacement of the mass

relative to the base?

c

x

k

xi

m

W=128.7(rad/s)2,

and (see Eq. (21.38),

Problem 21.71 The mass in Fig. 21.21 is 100 kg. The

spring constant is k=4 N/m, and c=24 N-s/m. The

base is subjected to an oscillatory displacement of fre-

quency ωi=0.2 rad/s. The steady-state amplitude of the

displacement of the mass relative to the base is measured

and determined to be 200 mm. What is the amplitude of

the displacement of the base? (See Example 21.7.)

c

x

k

xi

m

d2x

dt2+2ddx

dt +ω2x=a(t)

2=k

a(t) =−d2xi

dt2=xiω2

isin(ωit−φ).

Ep=0.2=ω2

ixi

(ω2−ω2

i)2+4d2ω2

i

0.8333 =0.24 m

Problem 21.72 A team of engineering students builds

the simple seismograph shown. The coordinate ximea-

sures the local horizontal ground motion. The coordi-

nate xmeasures the position of the mass relative to

the frame of the seismograph. The spring is unstretched

when x=0. The mass m=1 kg, k=10 N/m, and c=

2 N-s/m. Suppose that the seismograph is initially sta-

tionary and that at t=0 it is subjected to an oscil-

latory ground motion xi=10 sin 2tmm. What is the

amplitude of the steady-state response of the mass? (See

xix

k

c

m

TOP VIEW

774

Problem 21.73 In Problem 21.72, determine the posi-

tion xof the mass relative to the base as a function of

time. (See Example 21.7.)

Solution: From Example 21.7 and the solution to Problem 21.70,

the canonical form of the equation of motion is

Apsin ωit+Bpcos ωit, where

Ap=(ω2−ω2

i)ω2

ixi

(ω2−ω2

i)2+4d2ω2

i=4.615.

ixi

Solve:

i)ω2

ixi

ωd=√ω2−d2=3 rad/s. The complete solution is x=xh+xp.

Problem 21.74 The coordinate xmeasures the dis-

placement of the mass relative to the position in which

the spring is unstretched. The mass is given the initial

conditions

t=0





x=0.1m,

dx

dt =0.

(a) Determine the position of the mass as a function

of time.

(b) Draw graphs of the position and velocity of the

mass as functions of time for the ﬁrst 5 s of motion.

90 N/m

x

10 kg

776

Problem 21.75 When t=0, the mass in Prob-

lem 21.74 is in the position in which the spring is

unstretched and has a velocity of 0.3 m/s to the right.

Determine the position of the mass as functions of time

and the amplitude of vibration

(a) by expressing the solution in the form given by

Eq. (21.8) and

(b) by expressing the solution in the form given by

Eq. (21.9)

90 N/m

x

10 kg

Solution: From Eq. (21.5), the canonical form of the equation of

Problem 21.76 A homogenous disk of mass mand

radius Rrotates about a ﬁxed shaft and is attached to

a torsional spring with constant k. (The torsional spring

exerts a restoring moment of magnitude kθ, where θis

the angle of rotation of the disk relative to its position in

which the spring is unstretched.) Show that the period

of rotational vibrations of the disk is

τ=πR2m

k.

k

R

Solution: From the equation of angular motion, the equation of

Problem 21.77 Assigned to determine the moments of

inertia of astronaut candidates, an engineer attaches a

horizontal platform to a vertical steel bar. The moment

of inertia of the platform about Lis 7.5 kg-m2, and

the frequency of torsional oscillations of the unloaded

platform is 1 Hz. With an astronaut candidate in the

position shown, the frequency of torsional oscillations

is 0.520 Hz. What is the candidate’s moment of inertia

about L?

L

778

Problem 21.78 The 22-kg platen Prests on four roller

bearings that can be modeled as 1-kg homogenous

cylinders with 30-mm radii. The spring constant is k=

900 N/m. What is the frequency of horizontal vibrations

of the platen relative to its equilibrium position?

kP

2.

Since the system is conservative, T+V=const. Substitute the

kinematic relations and reduce:

Problem 21.79 At t=0, the platen described in

Problem 21.78 is 0.1 m to the left of its equilibrium

position and is moving to the right at 2 m/s. What are

the platen’s position and velocity at t=4s?

kP

780