978-0073380308 Chapter 9 Solution Manual Part 7

Problem 9.38

Revisit Example 9.6 and discuss whether it is possible to obtain the

equation of motion of the system via the energy method.

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Dynamics 2e 1981

Problem 9.39

A fatigue-testing machine for electronic components consists of a

platform with an unbalanced motor. Assume that the rotor in the

motor spins at

!rD3000 rpm

, the mass of the platform is

mpD

20 kg

, the mass of the motor is

mmD15 kg

, the unbalanced mass is

muD0:5 kg

, and the equivalent stiffness of the platform suspension

is

keq DnksD5106N=m

, where

n

is the number of springs. For

the testing machine, the distance

“

between the spin axis of the rotor

and the location at which

mu

is placed can be varied to obtain the

desired vibration level. Calculate the range of values of

“

that would

provide amplitudes of the particular solution ranging from

0:1 mm

to

2mm.

Solution

This system is identical to that found in Example 9.6. Therefore, using

kD5106N=m

,

mmD15 kg

, and

mpD20 kg, we know that the square of the natural frequency of the system is given by

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Problem 9.40

At time

tD0

, a forced harmonic oscillator occupies position

x.0/ D

0:1

m and has a velocity

Px.0/ D0

. The mass of the oscillator is

mD10 kg

, and the stiffness of the spring is

kD1000 N=m

. Calculate

the motion of the system if the forcing function is

F .t/ DF0sin !0t

,

with F0D10 N and !0D200 rad=s.

Solution

As we saw in Eq. (9.31) on p. 681, the equation of motion for this system is given by

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Dynamics 2e 1983

Problem 9.41

The forced harmonic oscillator shown has a mass

mD10 kg

. In

addition, the harmonic excitation is such that

F0D150

N and

!0D200 rad=s

. If all sources of friction can be neglected, determine

the spring constant ksuch that the magnification factor MF D5.

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Computation. Substituting the force law, the kinematic equation, and F .t/ DF0sin !0tinto Eq. (1), we

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Dynamics 2e 1985

Problem 9.42

A ring of mass

m

is attached by two linear elastic cords with elastic

constant

k

and unstretched length

L0< L

to a support, as shown.

Assuming that the pretension in the cords is large, so that the cords’

deﬂection due to the ring’s weight can be neglected, find the linearized

equation of motion for the case where

F .t/ DF0sin !0t

and

w.t/ D

0

(i.e., the support is stationary). In addition, find the response of the

system for y.0/ D0and Py.0/ D0.

In addition, since we want the equation of motion, we let ayD Ry.

Computation.

Substituting the kinematics equations and the force law into the balance law in Eq. (1), we

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where we obtained the last approximation by noting that

yL

and so

.y=L/2

and higher powers can be

ignored. Therefore the linearized equation of motion becomes

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Dynamics 2e 1987

Problem 9.43

A ring of mass

m

is attached by two linear elastic cords with

elastic constant

k

and unstretched length

L0< L

to a support, as

shown. Assuming that the pretension in the cords is large, so that

the cords’ deﬂection due to the ring’s weight can be neglected,

find the linearized equation of motion for the case where

F .t/ D0

and

s.t/ Ds0sin !t

. In addition, find the response of the system

for y.0/ D0and Py.0/ D0.

In addition, the acceleration of the ring can be found using

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Looking at the second term in parentheses, we note that using a Taylor series expansion about

yD0

(this is

sometimes called a binomial expansion) we can write it as

1

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Dynamics 2e 1989

Problem 9.44

Modeling the beam as a rigid uniform

thin bar, ignoring the inertia of the pul-

leys, assuming that the system is in

static equilibrium when the bar is hor-

izontal, and assuming that the cord is

inextensible and does not go slack, de-

termine the linearized equation of mo-

tion of the system in terms of

x

, which

is the position of

A

. Finally, determine

an expression for the amplitude of the

steady-state vibration of block A.

2FsLmBgL

2DIO˛B;(2)

where

T

is the cord tension,

˛B

is the angular acceleration of the bar,

Fs

is the force in the spring, and

IOD1

3mBL2is the mass moment of inertia of the bar Babout point O.

Force Laws. The force laws are

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