Mechanical Engineering Chapter 21 Problem The Mass Each Slender Bar The Frequency Small Vibrations The System

Problem 21.34 The mass of each slender bar is 1 kg.

If the frequency of small vibrations of the system is

0.935 Hz, what is the mass of the object A?

350

mm

280

mm

A

0.280 m, m=1 kg, and Mthe mass of A. The moments of inertia

about the ﬁxed point is the same for the two vertical bars. The kinetic

energy is

T=1

2I˙

θ2+1

2I˙

θ2+1

2mv2+1

2Mv2

A,

+MgLA(1−cos θ),

V=(MgLA+2mgL)(1−cos θ).

=const.

A

θ

is I=mL2

3, from which



g(2+η)

Solve:

Problem 21.35* The 4-kg slender bar is 2 m in length.

It is held in equilibrium in the position θ0=35◦by a tor-

sional spring with constant k. The spring is unstretched

when the bar is vertical. Determine the period and fre-

quency of small vibrations of the bar relative to the

equilibrium position shown.

θ

k

0

Solution: The total energy is

Solving for k,

k=mgL

sin θ0

sin 35◦

(4)(2)2−(3)(9.81)

(2)(2)cos 35◦

so τ=2π

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Problem 21.36 The mass m=2 slug, the spring con-

stant is k=72 lb/ft, and the damping constant is c=

8 lb-s/ft. The spring is unstretched when x=0. The

mass is displaced to the position x=1 ft and released

from rest.

(a) If the damping is subcritical, what is the frequency

of the resulting damped vibrations?

(b) What is the value of xat t=1s?

k

c

x

m

Problem 21.37 The mass m=2 slug, the spring con-

stant is k=72 lb/ft, and the damping constant is c=

32 lb-s/ft. The spring is unstretched with x=0. The

mass is displaced to the position x=1 ft and released

from rest.

(a) If the damping is subcritical, what is the frequency

of the resulting damped vibrations?

(b) What is the value of xat t=1s?

k

c

x

m

750

x=0.

(a) What value of the damping constant ccauses the

system to be critically damped?

(b) Suppose that chas the value determined in part (a).

At t=0,x=1 ft and dx/dt =4 ft/s. What is the

value of xat t=1s?

k

c

x

m

Problem 21.39 The mass m=2 kg, the spring con-

stant is k=8 N/m, and the damping coefﬁcient is c=

12 N-s/m. The spring is unstretched when x=0. At

t=0, the mass is released from rest with x=0. Deter-

mine the value of xat t=2s.

k

c

x

Problem 21.40 The mass m=0.15 slugs, the spring

constant is k=0.5 lb/ft, and the damping coefﬁcient is

c=0.8 lb-s/ft. The spring is unstretched when x=0.

We identify

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Problem 21.41 A 2570-lb test car moving with velo-

city v0=5 mi/h collides with a rigid barrier at t=0. As

a result of the behavior of its energy-absorbing bumper,

the response of the car to the collision can be simulated

by the damped spring-mass oscillator shown with k=

8000 lb/ft and c=3000 lb-s/ft. Assume that the mass

is moving to the left with velocity v0=5 mi/h and the

spring is unstretched at t=0. Determine the car’s posi-

tion (a) at t=0.04 s and (b) at t=0.08 s.

k

c

x

y

Car colliding with a rigid barrier

Simulation model

v0

Solution: The equation of motion is

Problem 21.42 A 2570-lb test car moving with velo-

city v0=5 mi/h collides with a rigid barrier at t=0. As

a result of the behavior of its energy-absorbing bumper,

the response of the car to the collision can be simulated

by the damped spring-mass oscillator shown with k=

8000 lb/ft and c=3000 lb-s/ft. Assume that the mass

is moving to the left with velocity v0=5 mi/h and the

spring is unstretched at t=0. Determine the car’s decel-

eration (a) immediately after it contacts the barrier; (b) at

t=0.04 s and (c) at t=0.08 s.

x

y

Car colliding with a rigid barrier

v0

Problem 21.43 The motion of the car’s suspension

shown in Problem 21.42 can be modeled by the damped

spring–mass oscillator in Fig. 21.9 with m=36 kg, k=

22 kN/m, and c=2.2 kN-s/m. Assume that no external

forces act on the tire and wheel. At t=0, the spring is

unstretched and the tire and wheel are given a velocity

dx/dt =10 m/s. Determine the position xas a function

of time.

Shock absorber

Coil spring

x

kc

m

x

Solution: Calculating ωand d, we obtain

The time derivative is

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Problem 21.44 The 4-kg slender bar is 2 m in length.

Aerodynamic drag on the bar and friction at the sup-

port exert a resisting moment about the pin support of

magnitude 1.4(dθ/dt ) N-m, where dθ/dt is the angular

velocity in rad/s.

(a) What are the period and frequency of small vibra-

tions of the bar?

(b) How long does it take for the amplitude of vibration

to decrease to one-half of its initial value?

θ

Solution:

−1.4dθ

dt −mg L

2sin θ=1

3mL2α.

The (linearized) equation of motion is

d2θ

dt2+4.2

mL2

dθ

dt +3g

2Lθ=0.

This is of the form of Eq. (21.16) with

d=4.2

2mL2=4.2

2(4)(2)2=0.131 rad/s,

ω=3g

2L=3(9.81)

2(2)=2.71 rad/s.

From Eq. (21.18), ωd=√ω2−d2=2.71 rad/s, and from

Eqs. (21.20),

τd=2π

ωd=2.32 s,

fd=1

τd=0.431 Hz.

(b) Setting e−dt =e−0.131t=0.5, we obtain t=5.28 s.

mg

θ

O

Problem 21.45 The bar described in Problem 21.44 is

given a displacement θ=2◦and released from rest at

t=0. What is the value of θ(in degrees) at t=2s?

Problem 21.46 The radius of the pulley is R=

100 mm and its moment of inertia is I=0.1 kg-m2. The

mass m=5 kg, and the spring constant is k=135 N/m.

The cable does not slip relative to the pulley. The

coordinate xmeasures the displacement of the mass

relative to the position in which the spring is unstretched.

Determine xas a function of time if c=60 N-s/m and

the system is released from rest with x=0. x

c

k

R

m

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Problem 21.47 For the system described in Prob-

lem 21.46, determine xas a function of time if c=

120 N-s/m and the system is released from rest with

x=0.

Solution: From the solution to Problem 21.46 the canonical form

of the equation of motion is

The system is supercritically damped, since d2>ω

2. The homogenous

k=0.3633. The solution is

Apply the initial conditions, at t=0, x=0, dx

dt =0, from which

x(t) =mg

Problem 21.48 For the system described in Prob-

lem 21.46, choose the value of cso that the system is

critically damped, and determine xas a function of time

if the system is released from rest with x=0.

R

k=0.3633 m. The solution:

Apply the initial conditions at t=0, x=0, dx

Problem 21.49 The spring constant is k=800 N/m,

and the spring is unstretched when x=0. The mass of

each object is 30 kg. The inclined surface is smooth.

The radius of the pulley is 120 mm and it moment of

inertia is I=0.03 kg-m2. Determine the frequency and

period of vibration of the system relative to its equilib-

rium position if (a) c=0, (b) c=250 N-s/m.

k

x

c

20⬚

Solution: Let T1be the tension in the rope on the left of the

pulley, and T2be the tension in the rope on the right of the pulley.

The equations of motion are

Problem 21.50 The spring constant is k=800 N/m,

and the spring is unstretched when x=0. The damping

0.03 kg-m2.Att=0,x=0 and dx/dt =1 m/s. What

is the value of xat t=2s?

Solution: From Problem 21.49 we know that the damping is sub-

critical and the key parameters are

The solution is

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Problem 21.51 The homogeneous disk weighs 100 lb

and its radius is R=1 ft. It rolls on the plane surface.

The spring constant is k=100 lb/ft and the damping

constant is c=3 lb-s/ft. Determine the frequency of

small vibrations of the disk relative to its equilibrium

position.

c

k

R

Solution: Choose a coordinate system with the origin at the center

Problem 21.52 In Problem 21.51, the spring is

unstretched at t=0 and the disk has a clockwise angular

velocity of 2 rad/s. What is the angular velocity of the

disk when t=3s?

Solution: From the solution to Problem 21.51, the canonical form

of the equation of motion is

Problem 21.53 The moment of inertia of the stepped

disk is I. Let θbe the angular displacement of the disk

relative to its position when the spring is unstretched.

Show that the equation governing θis identical in form

to Eq. (21.16), where d=R2c

2Iand ω2=4R2k

I.

θ

2R

ck

R

Solution: The sum of the moments about the center of the stepped

θ

760

Problem 21.54 In Problem 21.53, the radius R=

250 mm, k=150 N/m, and the moment of inertia of

the disk is I=2 kg-m2.

(a) At what value of cwill the system be critically

damped?

Problem 21.55 The moments of inertia of gears Aand

Bare IA=0.025 kg-m2and IB=0.100 kg-m2. Gear

Ais connected to a torsional spring with constant k=

10 N-m/rad. The bearing supporting gear Bincorporates

a damping element that exerts a resisting moment on

gear Bof magnitude 2(dθB/dt) N-m, where dθB/dt is

the angular velocity of gear Bin rad/s. What is the fre-

quency of small angular vibrations of the gears?

140 mm

200 mm

AB

d2θB

M=6.622 rad/s

762

Problem 21.56 At t=0, the torsional spring in

Problem 21.55 is unstretched and gear Bhas a coun-

terclockwise angular velocity of 2 rad/s. Determine the

Solution: From the solution to Problem 21.55, the canonical form

of the equation of motion for gear Bis

M=6.622 rad/s,

Problem 21.57 For the case of critically damped

motion, conﬁrm that the expression x=Ce−dt +

Solution: Eq. (21.16) is

Problem 21.58 The mass m=2 slug and the spring

constant is k=72 lb/ft. The spring is unstretched when

x=0. The mass is initially stationary with the spring

unstretched, and at t=0 the force F(t) =10 sin 4tlb is

applied to the mass. What is the position of the mass at

t=2s?

k

F(t)

x

m

Problem 21.59 The mass m=2 slug and the spring

F(t) =10 sin 4t+10 cos 4tlb is applied to the mass.

Solution: The equation of motion is

mx=F(t)

m⇒¨x+72

2x=10

2sin 4t+10

2cos 4t

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Problem 21.60 The damped spring–mass oscillator is

initially stationary with the spring unstretched. At t=0,

a constant force F(t) =6 N is applied to the mass.

(a) What is the steady-state (particular) solution?

12 N/m

x

F(t)

dt2+2dx

dt +4x=F(t)

3.(1)

(a) If F(t) =6 N, we seek a particular solution of the form xp=

A0, a constant. Substituting it into Equation (1), we get 4xp=

F(t)

3=2 and obtain the particular solution: xp=0.5m.

(b) Comparing equation (1) with Equation (21.26), we see that d=

1 rad/s and ω=2 rad/s. The system is subcritically damped and

the homogeneous solution is given by Equation (21.19) with

ωd=√ω2−d2=1.73 rad/s. The general solution is

x=xh+xp=e−1(A sin 1.73t+Bcos 1.73t) +0.5m.

Problem 21.61 The damped spring–mass oscillator

shown in Problem 21.60 is initially stationary with the

spring unstretched. At t=0, a force F(t) =6 cos 1.6tN

is applied to the mass.

(a) What is the steady-state (particular) solution?

0.234 cos 1.6t(m).

(b) The system is subcritically damped so the homogeneous solution

−0.234, so the solution is

Problem 21.62 The disk with moment of inertia I=

3 kg-m2rotates about a ﬁxed shaft and is attached to

a torsional spring with constant k=20 N-m/rad. At

t=0, the angle θ=0, the angular velocity is dθ/dt =

4 rad/s, and the disk is subjected to a couple M(t) =

θ

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