Dynamics Solns Ch09

8/17/2019 Dynamics Solns Ch09

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Solutions Manual

Engineering Mechanics: Dynamics1st Edition

Gary L. GrayThe Pennsylvania State University

Francesco CostanzoThe Pennsylvania State University

Michael E. PleshaUniversity of Wisconsin–Madison

With the assistance of:

Chris Punshon

Andrew J. Miller

Justin High

Chris O’Brien

Chandan Kumar

Joseph Wyne

Version: August 10, 2009

The McGraw-Hill Companies, Inc.


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Copyright © 2002–2010

Gary L. Gray, Francesco Costanzo, and Michael E. Plesha

This solutions manual, in any print or electronic form, remains the property of McGraw-Hill, Inc. It

may be used and/or possessed only by permission of McGraw-Hill, and must be surrendered upon

request of McGraw-Hill. Any duplication or distribution, either in print or electronic form, without

the permission of McGraw-Hill, is prohibited.


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Dynamics 1e 3

Important Information about

this Solutions Manual

Even though this solutions manual is nearly complete, we encourage you to visit

http://www.mhhe.com/pgc

often to obtain the most up-to-date version. In particular, as of July 30, 2009, please note the following:

The solutions for Chapters 1 and 2 have been accuracy checked and have been edited by us. They are

in their final form.

The solutions for Chapters 4 and 7 have been accuracy checked and should be error free. We will be

adding some additional detail to these solutions in the coming weeks.

The solutions for Chapters 3, 6, 8, and 9 are being accuracy checked and the accuracy checked versionsshould be available by the end of August 2009. We will be adding some additional detail to these

solutions in the coming weeks.

The solutions for Chapter 10 should be available in their entirety by the end of August 2009.

All of the figures in Chapters 6–10 are in color. Color will be added to the figures in Chapters 1–5 over the

coming weeks.

Contact the Authors

If you find any errors and/or have questions concerning a solution, please do not hesitate to contact the

authors and editors via email at:

[email protected]

We welcome your input.

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4 Solutions Manual

Accuracy of Numbers in Calculations

Throughout this solutions manual, we will generally assume that the data given for problems is accurate to

3 significant digits. When calculations are performed, all intermediate numerical results are reported to 4significant digits. Final answers are usually reported with 3 significant digits. If you verify the calculations in

this solutions manual using the rounded intermediate numerical results that are reported, you should obtain

the final answers that are reported to 3 significant digits.

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Dynamics 1e 1221

Chapter 9 Solutions

Problem 9.1

Show that Eq. (9.15) is equivalent to Eq. (9.3) if C D p A2 C B2 and tan D A=B .

Solution

Start with Eq. (9.15), which is given by

x.t/ D xi cos !nt C vi!n

sin !nt D A cos !nt C B sin !nt;

where we have let xi

DA and vi=!n

DB . Factoring out B , we obtain

x.t/ D B

A

B cos !nt C sin !nt

:

Letting A=B D tan , this becomes

x.t/ D B.tan cos !nt C sin !nt / D Bcos

.sin cos !nt C cos sin !nt /:

Since sin cos !nt C cos sin !nt D sin.!nt C /, this equation becomes

x.t/ D Bcos

sin.!nt C /:

This is Eq. (9.3) as long as we can show that B= cos D p A2 C B2. To show this, we writeB

cos D B

Bp A2CB2

Dp

A2 C B2 D C;

where we have use the identity that if tan D A=B , then cos D B=p A2 C B2. We have thus shown thatEq. (9.15) is equivalent to Eq. (9.3) for the proper definition of C and .

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1222 Solutions Manual

Problem 9.2

Derive the formula for the mass moment of inertia of an arbitrarily shaped rigid body

about its mass center based on the body’s period of oscillation when suspended as a

pendulum. Assume that the mass of the body m is known and that the location of the

mass center G is known relative to the pivot point O .

Solution

Using the FBD shown at the right, we can sum moments about the pivot O to

obtain

XM O W mgL sin D I OR :

Rearranging this equation and assuming small oscillations, we obtain

R C mgLI O

sin D 0 ) R C mgLI O

D 0:

This is the standard form of the harmonic oscillator, so the natural frequency can be written as

!2n D mgL

I O) I O D

mgL

!2n:

Since !n D 2= , we have that

I O D mgL 2

42 :

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Dynamics 1e 1223

Problem 9.3

The thin ring of radius R and mass m is suspended by the pin at O . Determine its

period of vibration if it is displaced a small amount and released.

Solution

Referring to the FBD at the right, we can sum moments about point O to obtainXM O W mgR sin D I O˛G ; (1)

where ˛ D R is the angular acceleration of the thin ring. Using the parallelaxis theorem, we find that I O D mR

2

C mR2

D 2mR2

. Substituting this intoEq. (1), we find the equation of motion to be

mgR sin D 2mR2 R ) R C g2R

sin D 0:

Assuming small angles , we can use the approximation sin so that the equation of motion becomes

R C g2R

D 0:

Therefore, the natural frequency and period are found to be

!n D r g2R

) D 2!n

D 2s 2Rg

.

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Problem 9.4

The thin square hoop has mass m and is suspended by the pin at O . Determine its

period of vibration if it is displaced a small amount and released.

Solution

Using the FBD shown at the right, we can sum moments about the pivot O to

obtain XM O W mg

L

2 sin D I O R:

Rearranging this equation and assuming small oscillations, we obtain

R C mgL2I O

sin D 0 ) R C mgL2I O

D 0:

This is the standard form of the harmonic oscillator, so the natural frequency and period can be written as

!2n D mgL

2I Oand D 2

!nD 2

s 2I O

mgL; (1)

respectively. Noting that each of the four segments of the thin square hoop has mass m=4, computing I O we

obtain

I O D 1

12

m4

L2„ ƒ‚ …

segment AB

C 2

1

12

m4

L2 C m

4

L2

4 C L

2

4

„ ƒ‚ …

segments BC and AD

C

1

12

m4

L2 C m

4 L2

„ ƒ‚ …

segment CD

D 712

mL2:

Substituting this expression for I O into the expression for the period given in Eq. (1), we obtain

D 2s

7L

6g:

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Dynamics 1e 1225

Problem 9.5

The swinging bar and the vibrating mass are made to vibrate on

Earth, and their respective natural frequencies are measured. The

two systems are then taken to the Moon and are again allowed to

vibrate at their respective natural frequencies. How will the naturalfrequency of each system change when compared with that on the

Earth, and which of the two systems will experience the larger

change in natural frequency?

Solution

As can be seen in Example 9.1, the natural frequency of a rigid body that swings like a pendulum under the

action of gravity depends on the value of the acceleration due to gravity g such that !n / p g. On the otherhand, as can be seen in the mini-example on p. 675, the natural frequency of a mass at the end of a spring that

oscillates under the action of gravity does not depend on g. Therefore, on the Moon, where the acceleration

due to gravity gmoon is less than g on Earth, the natural frequency of the swinging bar will decrease and thenatural frequency of the vibrating mass will stay the same.

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Problems 9.6 and 9.7

A rigid body of mass m, mass center at G , and mass moment of inertia I G is pinned at

an arbitrary point O and allowed to oscillate as a pendulum.

Problem 9.6 By writing the Newton-Euler equations, determine the distance ` from

G to the pivot point O so that the pendulum has the highest possible natural frequency

of oscillation.

Problem 9.7 Using the energy method, determine the distance ` from G to the pivot

point O so that the pendulum has the highest possible natural frequency of oscillation.

Solution to 9.6

We will write the equation of motion using the Newton-Euler equations, find

!n as a function of `, and then find .!n/max by letting d !n=d ` D 0.Referring to the FBD at the right and summing moments about O, we

obtain XM O W mg` sin D I O R ; (1)

where, using the parallel axis theorem, I O D I G C m`2. Substituting I O intoEq. (1), we obtain

I G C m`2 R C mg` sin D 0 ) R C mg`

I G C m`2 D 0;

where the second equation was obtained from the first by dividing through by the coefficient of R andassuming small . Therefore, the natural frequency is given by

!n D s mg`I G C m`2 :Differentiating !n with respect to ` and setting it equal to zero, we find that

d!n

d ` D d

d `

mg`

I G C m`21=2

D 12

mg`

I G C m`21=2 "mgI G C m`2 2m`.mg`/

I G C m`22

#D 0;

which implies that

mg

I G C m`2 2m`.mg`/ D 0 ) I G m`2 D 0 ) ` D

r I G

m

where we have used the fact that ` needs to be positive and real. Since !n as a function of ` has the shape

shown below, we can see that this expression for ` gives a maximum value of !n and not a minimum.

Ωn

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Dynamics 1e 1227

Solution to 9.7

We will write the equation of motion using the energy method, find !nas a function of `, and then find .!n/max by letting d!n=d ` D 0.

Referring to the FBD at the right we see that the only force that does

work as the body swings is the weight force mg. Since this system is

conservative, we can use the energy method to determine the equation

of motion.

The kinetic energy can be written as

T D 12

I O!2G D 12

I G C m`2

P 2;where !G is the angular velocity of the rigid body and we have used the parallel axis theorem to find I O . In

addition, the potential energy can be written as

V D mg` cos mg`

1 2

2

;

where we have assumed small in approximating the cosine function. Applying the energy method, we

obtaind

dt.T C V / D I G C m`2 P R C mg` P D 0 ) I G C m`2 R C mg` D 0;

which implies that

!n Ds

mg`

I G C m`2:

Differentiating !n with respect to ` and setting it equal to zero, we

find that

d!n

d ` D d

d ` mg`

I G C m`21=2 D 12 mg`I G C m`2

1=2 "mgI G C m`2 2m`.mg`/I G C m`2

2 # D 0;which implies that

mg

I G C m`2 2m`.mg`/ D 0 ) I G m`2 D 0 ) ` D

r I G

m

where we have used the fact that ` needs to be positive and real. Since !n as a function of ` has the shape

shown below, we can see that this expression for ` gives a maximum value of !n and not a minimum.

Ωn

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Problem 9.8

The uniform disk of radius R and thickness t is attached to the thin shaft of radius

r , length L, and negligible mass. The end A of the shaft is fixed. From mechanics

of materials, it can be shown that if a torque M ́ is applied to the free end of the

shaft, then it can be related to the twist angle via

D M ́ LGJ

;

where G is the shear modulus of elasticity of the shaft and J D 2

r4 is the polar

moment of inertia of the cross-sectional area of the shaft. Letting be the mass

density of the disk and using the given relationship between M ́ and , determine

the natural frequency of vibration of the disk in terms of the given dimensions and

material properties when it is given a small angular displacement in the plane of

the disk.

Solution

If the disk is displaced through a positive angle then the FBD (as viewed

down the ´ axis) is as shown on the right. The moment M ́ is the restoring

moment due to the thin shaft of radius r and length L. Summing moments

about O gives XM O W M ́ D I O R : (1)

From the given relationship between M ́ and , we can find that

M ´ D

GJ

L

) GJ

L

DI O R

: (2)

where we have made use of Eq. (1). Since

I O D 12mR2; m D R2t; and J D 2 r4;

Eq. (2) becomes

G r4=2

L D 1

2R4t R ) R C

Gr 4

LR4t

D 0:

Therefore, the natural frequency of oscillation is

!n Ds Gr 4LR4t

:

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Dynamics 1e 1229

Problem 9.9

A construction worker C is standing at the midpoint of a 14 ft long

pine board that is simply supported. The board is a standard 212,so its cross-sectional dimensions are as shown. Assuming the

worker weighs 180 lb and he flexes his knees once to get the boardoscillating, determine his vibration frequency. Neglect the weight

of the beam and use the fact that a load P applied to a simply

supported beam will deflect the center of the beam PL3=.48EI cs/,

where L is the length of the beam, E is its modulus of elasticity,

and I cs is the area moment of inertia of the cross section of the

beam. The elastic modulus of pine is 1:8106 psi.

Solution

Referring to the FBD of the construction worker shown at the right,

we see that the elastic restoring force due to the board F b

as well as

the weight of the worker mg both do work on the worker. Since the

force due to the board is assumed to be linear elastic with the force

law

F b D 48EI cs

L3 ı D keqı; where keq D 48EI cs

L3

and where ı is the deflection of the center of the board due to the

force F b as shown on the right and keq is the equivalent linear spring

constant of the deflected board. Since both forces doing work are

conservative, we can apply the energy method to determine the

equation of motion of the construction worker on the board. The

kinetic energy of worker isT D 1

2mv2G D 12m Py2:

The potential energy is given by

V D mg.y C ıst/ C 12keq.y C ıst/2;where y is measured from the static equilibrium position of the worker standing on the board. Applying the

energy method, we obtain

d

dt.T C V / D m Py Ry mg Py C keq.y C ıst/ Py D 0 ) m Ry mg C keq.y C ıst/ D 0:

Using the fact that mg D keqıst, this equation becomes the standard harmonic oscillator equation

Ry C 48EI csmL3

y D 0 ) f D !n2

D 2

r 3EI csmL3

;

where f is the vibration frequency. Substituting in given values

f D 2

"3

1:8106 lbin:2

3:164 in:4

0:4658 lbs2in:

168 in:

3#1=2

) f D 1:77 Hz,

where I cs D 112bh3 in which b D 11:25 in: and h D 1:5 in:, L D 14 ft D 168 in:, and m D 5:590 slug D0:4658 lbs2=in:

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Problems 9.10 through 9.13

The L-shaped bar lies in the vertical plane and is pinned at O . One end of the

bar has a linear elastic spring with constant k attached to it, and attached at

the other end is a mass m of negligible size. The angle is measured from the

equilibrium position of the system and it is assumed to be small.

Problem 9.10 Assuming that the L-shaped bar has negligible mass, determine

the natural period of vibration of the system by writing the Newton-Euler

equations.

Problem 9.11 Assuming that the L-shaped bar has negligible mass, determine

the natural period of vibration of the system via the energy method.

Problem 9.12 Assuming that the L-shaped bar has mass per unit length ,

determine the natural period of vibration of the system by writing the Newton-

Euler equations.

Problem 9.13 Assuming that the L-shaped bar has mass per unit length ,

determine the natural period of vibration of the system via the energy method.

Solution to 9.10

Ignoring the mass of the L-shaped bar, the FBD of the system is as shown on the

right. Summing moments about point O , we find thatXM O W F sh mgd D I O R ; (1)

where the minus sign on the right side is due to the fact that is positive in

the clockwise direction and positive ´ is in the counterclockwise direction. The

force in the spring F s is due to the static deflection ıst at equilibrium as well as

any additional deflection during vibration. Therefore the spring force is

F s D k.ıst C h/:

In addition, if we neglect the mass of the bar, then I O D md 2, which means that Eq. (1) becomes

k.ıst C h/h mgd D md 2 R;

Since kısth D mgd , this equation becomes md 2 R C kh2 D 0, which means that the natural frequency andperiod of vibration are given by

!n Ds

kh2

md 2 D h

d

r k

m ) D 2d

h

r m

k .

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Dynamics 1e 1231

Solution to 9.11

Ignoring the mass of the L-shaped bar, the FBD of the system is as shown on

the right. From this FBD we can see that the spring force F s and the weight

force mg both do work and thus the system is conservative. Applying the

energy method, the kinetic energy of the system is

T D 12

I O P 2 D 12md 2 P 2;where we have used the parallel axis theorem to obtain I O D md 2.Changes in the potential energy of the system are due to the deflection

of the spring as well as changes in height of the mass m. The deflection

of the spring is due to the static deflection ıst at equilibrium as well as any

additional deflection during vibration. Therefore, the potential energy of the system can be written as

V D 12

k.ıst C h/2 mgd sin D 12k.ıst C h/2 mgd;where we have approximated sin as . Applying the energy method, we obtain

d

dt .T C V / D md 2

P R C k.ıst C h/h P mgd P D 0 ) md 2

R C k.ıst C h/h mgd D 0:Since kısth D mgd , this equation becomes md 2 R C kh2 D 0, which means that the natural frequency andperiod of vibration are given by

!n Ds

kh2

md 2 D h

d

r k

m ) D 2d

h

r m

k .

Solution to 9.12

Including the mass of the L-shaped bar, the FBD of the system is as shown

on the right. Summing moments about point O , we find thatXM O W F sh mgd dg d 2 D I O R; (2)

where the minus sign on the right side is due to the fact that is positive in

the clockwise direction and positive ´ is in the counterclockwise direction.

The force in the spring F s is due to the static deflection ıst at equilibrium as

well as any additional deflection during vibration. Therefore the spring force

is

F s D k.ıst C h/:In addition, if we include the mass of the bar, then

I O D md 2 C 13.h/h2 C 13 .d/d 2;which means that Eq. (2) becomes

k.ıst C h/h mgd dg d 2 D

md 2 C 13

.h/h2 C 13

.d/d 2 R;

Since kısth D mgd C 12gd 2, this equation becomes13

.h3 C d 3/ C md 2 R C kh2 D 0, which meansthat the natural frequency and period of vibration are given by

!n Ds

kh2

13

.h3 C d 3/ C md 2 ) D 2s

13

.h3 C d 3/ C md 2kh2

.

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Solution to 9.13

Including the mass of the L-shaped bar, the FBD of the system is as

shown on the right. From this FBD we can see that the spring force

F s and the weight forces mg and gd do work and thus the system is

conservative. Applying the energy method, the kinetic energy of the

system is

T D 12

I O P 2 D 1213

h3 C d 3C md 2 P 2;where we have used the parallel axis theorem to find I O D md 2 C13

.h/h2 C 13

.d/d 2 D 13

h3 Cd 3Cmd 2. Changes in the potentialenergy of the system are due to the deflection of the spring as well as changes in height of the mass m and

the centers of mass of each of the two segments of the L-shaped bar. The deflection of the spring is due to

the static deflection ıst at equilibrium as well as any additional deflection during vibration. Therefore, the

potential energy of the system can be written as

V D 1

2k.ıst C h/2

mgd sin dgd

2 sin C hgh

2 cos D 1

2k.ıst C h/2 mgd dg d 2 C hg h2

1 2

2

;

where we have approximated sin as and cos as 1 2=2. Applying the energy method, we obtain

d

dt.T C V / D 1

3

h3 C d 3C md 2 P R C k.ıst C h/h P mgd P 12d 2g P 12h2g P D 0) 1

3

h3 C d 3C md 2 R C k.ıst C h/h mgd 12gd 2 h2 D 0:Since kısth D mgd C 12g

d 2Ch2 , this equation becomes 1

3.h3 C d 3/ C md 2 R Ckh2 D 0, which

means that the natural frequency and period of vibration are given by

!n Ds

kh2

13

.h3 C d 3/ C md 2 ) D 2s

13

.h3 C d 3/ C md 2kh2

.

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Dynamics 1e 1233

Problem 9.14

An off-highway truck drives onto a concrete deck scale to be

weighed, thus causing the truck and scale to vibrate vertically

at the natural frequency of the system. The empty truck weighs

74;000 lb, the scale platform weighs 51;000 lb, and the platformis supported by eight identical springs (four of which are shown),

each with constant k D 3:6105 lb=ft. Modeling the truck, itscontents, and the concrete deck as a single particle, if a vibration

frequency of 3:3 Hz is measured, what is the weight of the payload

being carried by the truck?

Solution

Since the frequency of vibration f is 3:3 Hz, the natural frequency !n is given by

!n D 2f ) s keqmtot D 2f; (1)where keq is the equivalent spring constant of the eight springs supporting the platform and mtot is the total

mass that is vibrating (i.e., the mass of the truck, the scale platform, and the payload of the truck). The

equivalent spring constant is equal to the sum of the eight individual spring constants so that

keq D 8k D 2:88106 lb=ft;

and the total vibrating mass is given by

mtot

D

W tot

g D

W scale C W truck C W payloadg

D

51;000 lb C 74;000 lb C W payload32:2 ft=s2

;

where W tot is the total vibrating weight. Substituting f D 3:3 Hz, keq, and mtot into Eq. (1), we obtain

W payload D 90;700 lb:

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Problem 9.15

A mass m of 3 kg is in equilibrium when a hammer hits it, imparting a

velocity v0 of 2 m=s to it. If k is 120 N=m, determine the amplitude of

the ensuing vibration and find the maximum acceleration experienced

by the mass.

Solution

Using the FBD at the right and summing forces in the x direction, we obtainXF x W F s D max D m Rx ) m Rx C kx D 0; (1)

where F s D kx is the force in the spring and x is measured from the undeformedposition of the spring. Since the initial velocity of the mass is Px.0/ D v0 D 2 m=sand its initial position is x.0/ D x0 D 0 and since !2n D k= m, the amplitude of vibration is

C Ds

v20!2n

C x20 Ds

v20k= m

Ds

mv20k

) C D 0:316 m.

To find the maximum acceleration of the mass, we write the position as a function of time in the form

x.t/ D C sin.!nt C /;

and then differentiate it twice to obtain

Rx.t/ D C !2n sin.!nt C /; ) Rxmax D C !2n ) Rxmax D 12:6 m=s2.

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Dynamics 1e 1235

Problem 9.16

The buoy in the photograph can be modeled as a circular cylinder

of diameter d and mass m. If the buoy is pushed down in the

water, which has density , it will oscillate vertically. Determine

the frequency of oscillation. Evaluate your result for d D 1:2 m,m D 900 kg, and surface seawater, which has a density of D1027 kg=m3. Hint: Use Archimedes’ principle, which states that

a body wholly or partially submerged in a fluid is buoyed up by a

force equal to the weight of the displaced fluid.

Solution

Referring to the FBD of the buoy at the right, mg is it’s weight and F b is the added

buoyancy force due to the displacement of the buoy by an amount y from it’s static

equilibrium position. Summing forces in the y direction, we obtain

XF y W mg .mg C F b/ D m Ry ) m Ry C F b D 0: (1)

The added buoyancy force is equal to the weight of the fluid displaced when the

buoy is deflected by the distance y . Therefore, it must be

F b D ˝g D "

d

2

2y

#g;

where ˝ is the volume of the fluid displaced. Therefore, Eq. (1) becomes

m

Ry

C 4

gd 2y

D0

) Ry

C gd 2

4m

y

D0:

This equation is in standard form, so we can write the natural frequency !n and the frequency f as

!n D d 2

r g

m ) f D !n

2 D d

4

r g

m D 0:566 Hz,

where, to obtain the numerical result, we have used d D 1:2 m, D 1027 kg=m3, m D 900 kg, andg D 9:81 m=s2.

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Problem 9.17

For the silicon nanowire in Example 9.2, use the lumped mass model

shown, in which a point mass m is connected to a rod of negligible

mass and length L that is pinned at O , to determine the natural fre-

quency !n and frequency f of the nanowire. Use the values givenin Example 9.2 for the mass of the lumped mass, the length of the

massless rod, and the parameters used to determine the spring constant

k D 3EI cs=L3. You may use either ı or as the position variable inyour solution. Assume that the displacement of m is small so that it

moves vertically.

Solution

Referring to the FBD of the nanowire at the right, which shows the

nanowire displaced from its static equilibrium position, we let F sbe the elastic force in the spring. Assuming that both ı and are

small and summing moments about point O , we obtainXM O W .mg F s/L D I O˛n D I O R ; (1)

where ˛n D R from the angular acceleration of the nanowire,I O D mL2 by the parallel axis theorem, and and ı are measured from the static equilibrium position of the nanowire. Noting that the force in the spring is given by F s D k.ı C ıst/ D kL. C st/, we can writeEq. (1) as

mL2 R C kL2. C st/ mgL D 0 ) R C km

D 0;where we have used the fact that kL st D mg.

From Example 9.2, we know that the cross section of the Si nanowire is circular and that it has thefollowing properties:

L D 9:8106 m; d D 2r D 330109 m; D 2330 kg=m3; and E D 152 GPa;where d is the diameter of the wire, r is its radius, is its density, and E is its modulus of elasticity. Since

the centroidal area moment of inertia of the cross section is I cs D 14 r4, we can compute k as

k D 3.152109 N=m2/

14

.165109 m/4.9:8106 m/3 D 0:2820 N=m:

The mass m of the lumped mass at the end of the wire is simply the mass of the wire, which is the density of

Si multiplied by the volume of the wire, that is,

m D r2L D 1:9531015 kg:Using the values of k and m found above, we find that the natural frequency !n and frequency f are

!n Dr

k

m D 1:20107 rad=s and f D !n

2 D 1:91106 Hz: (2)

The frequency of vibration goes down when all the mass is lumped at the end of the wire since it now has

a larger mass moment of inertia with respect to point O when compared with rigid body model used in

Example 9.2.

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Dynamics 1e 1237

Problem 9.18

The small sphere A has mass m and is fixed at the end of the arm OA of negligible mass, which is pinned

at O . If the linear elastic spring has stiffness k , determine the equation of motion for small oscillations,

using

(a) the vertical position of the mass A as the position coordinate,

(b) the angle formed by the arm OA with the horizontal as the position coordinate.

Solution

When the nanowire is displaced from its equilibrium position,the FBD is as shown on the right, where F s is the elastic

restoring force due to the spring. Summing moments about

point O and using the fact that both y and are small, we

obtain XM O W mgL F sL D I O˛OA; (1)

where I O D mL2 by the parallel axis theorem and ˛OA is the angular acceleration of the nanowire.Part (a). With y as the vertical position of m measured from the static equibrium position of the nanowire,

the force in the spring can be written as

F s D

k.ıstC

y/;

where ıst is the deflection of the spring in the static equilibrium position. Using this force law, along with the

kinematics relation ˛OA D Ry=L, the equation of motion (Eq. (1)) in terms of y becomesmgL k.ıst C y/L D mL2 Ry=L:

Since mg D kıst, we havem Ry C ky D 0:

Part (b). Using as the coordinate measuring the position of the nanowire from its static equibrium

position, the force in the spring can be written as

F s D k.ıst C L/;where ıst is the deflection of the spring in the static equilibrium position. Using this force law, along with the

kinematics relation ˛OA D R , the equation of motion (Eq. (1)) in terms of y becomesmgL k.ıst C L/L D mL2 R:

Since mg D kıst, we havem R C k D 0:

Of course both equations of motion give the same vibration frequency.

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Grandfather clocks keep time by advancing the hands a set amount per oscillation of the pendulum.

Therefore, the pendulum needs to have a very accurate period for the clock to keep time accurately. As

a fine adjustment of the pendulum’s period, many grandfather clocks have an adjustment nut on a bolt

at the bottom of the pendulum disk. By screwing this nut inward or outward, the mass distribution of the pendulum can be changed and its period adjusted. In the following problems, model the pendulum

as a uniform disk of radius r and mass mp, which is at the end of a rod of negligible mass and length

L r . Model the adjustment nut as a particle of mass mn, and let the distance between the bottom of thependulum disk and the nut be d .

Problem 9.19 If the adjustment is initially at a distance d D 9 mm from the bottom of the pendulumdisk, how much would the period of the pendulum change if the nut were screwed 4 mm closer to the

disk? In addition, how much time would the clock gain or lose in a 24 h period if this were done? Let

mp D 0:7 kg, r D 0:1 m, mn D 8 g, and L D 0:85 m.Problem 9.20 The clock is running slow so that it is losing 1 minute every 24 hours (i.e., the clock

takes 1441 minutes to complete a 1440 minute day). If the adjustment nut is at d D 2 cm, what would itsmass need to have to correct the pendulum’s period if the nut is moved to d D 0 cm? Let mp D 0:7 kg,r D 0:1 m, and L D 0:85 m.

Solution to 9.19

Referring to the FBD at the right, we see that as the pendulum swings, on

the weight of the pendulum disk and the adjustment nut do work. Since the

system is conservative, we apply the energy method to find the equation

of motion and thus the natural frequency of oscillation. The kinetic energy

of the system can be written as

T

D 12

.I O/d !2p

C 12

.I O/n!2p

D 12

Œ.I O/d

CI O/n !

2p ;

where .I O/d is the mass moment of inertia of the pendulum disk with

respect to point O , .I O/n is the mass moment of inertia of the adjustment

nut with respect to point O, and !p D P is the angular velocity of thependulum. The moments of inertia are given by

.I O/p D 12mpr2 C mpL2 and .I O/n D mn.L C r C d /2D mp.12r2 C L2/:

August 10, 2009


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Dynamics 1e 1239

The potential energy of the system can be written as

V D mpgL cos mng.L C r C d / cos D g mpL C mn.L C r C d / cos g mpL C mn.L C r C d / 1

2

2 ;where we have used the approximation cos 1 2=2. Applying the energy method, we find

d

dt.T C V / D mp12r2 C L2C mn.L C r C d /2 P R C g mpL C mn.L C r C d / P D 0

) mp12r2 C L2C mn.L C r C d /2 R C g mpL C mn.L C r C d / D 0;and therefore the natural frequency is

!2n D g

mpL C mn.L C r C d /

mp12r2 C L2C mn.L C r C d /2:

Using g D 9:81 m=s2, mp D 0:7 kg, L D 0:85 m, mn D 0:008 kg, r D 0:1 m, and the fact that theperiod is D 2=!n, with the initial position of the adjustment nut at d D 9 mm D 0:009 m, the period 9is (note that we keep more than four significant figures since the difference in period is very small)

9 D 2!n

ˇ̌̌ˇd D0:009

D 1:85731339 s;

and with the nut 4 mm closer at d D 5 mm D 0:005 m, the period 5 is

5 D 2!n ˇ̌̌̌d D0:005 D 1:85725259 s;

so that the change in period is

9 5 D 0:00006080 s:To determine how much time the clock would gain or lose in a 24 hr period, we write

9

5D 1:00003274 D 1440 min

tnew) tnew D 1439:953 min:

so that the day would be shorter by

1440

1439:953

D0:047 min

) 0:047 min 60 smin D 2:82 s

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Solution to 9.20

Referring to the FBD at the right, we see that as the pendulum swings, on

the weight of the pendulum disk and the adjustment nut do work. Since the

system is conservative, we apply the energy method to find the equation

of motion and thus the natural frequency of oscillation. The kinetic energy

of the system can be written as

T D 12

.I O/d !2p C 12 .I O/n!2p D 12 Œ.I O/d C I O/n !2p ;

where .I O/d is the mass moment of inertia of the pendulum disk with

respect to point O , .I O/n is the mass moment of inertia of the adjustment

nut with respect to point O, and !p D P is the angular velocity of thependulum. The moments of inertia are given by

.I O/p D 12mpr2 C mpL2 and .I O/n D mn.L C r C d /2D mp.12r2 C L2/:

The potential energy of the system can be written as

V D mpgL cos mng.L C r C d / cos D g mpL C mn.L C r C d / cos g mpL C mn.L C r C d / 1 2

2

;

where we have used the approximation cos 1 2=2. Applying the energy method, we find

d

dt.T C V / D mp

12

r2 C L2C mn.L C r C d /2

P R C g mpL C mn.L C r C d /

P D 0) mp12r2 C L2C mn.L C r C d /2 R C g mpL C mn.L C r C d / D 0;

and therefore the natural frequency is

!2n D g

mpL C mn.L C r C d /

mp

12

r2 C L2C mn.L C r C d /2 :Now that we have the natural frequency, the period can be calculated using D 2=!n. Since the clock

is running slow and loses one minute every 24 hours, it takes 1441 minutes to complete a 1440 minute day.

That is, correct D 1440 min and actual D 1441 min. Therefore correct

actual D 1440

1441 ) correct.d

D0 m/

actual.d D 0:02 m/ D 1440

1441 ; (1)

that is, the actual period is when the nut is at d D 0:02 m and the correct period is when the nut is at d D 0 m.Substituting in the expression for the period in terms of the natural frequency into Eq. (1), we obtain

2.!n/correct

2.!n/actual

D

s gŒmpLCmn.LCrC0:02/

mp

12r2CL2

Cmn.LCrC0:02/2s

gŒmpLCmn.LCr/mp

12r

2CL2Cmn.LCr/2

D 14401441

; (2)

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Dynamics 1e 1241

where we have substituted d D 0 m into .!n/correct and d D 0:02 m into .!n/actual. When we substituteg D 9:81 m=s2, mp D 0:7 kg, L D 0:85 m, and r D 0:1 m into Eq. (2), both sides are squared, and it issimplified, it becomes the following quadratic equation in the unknown mn

m2n C 0:6524mn 0:02446 D 0: (3)

Solving Eq. (3) for mn, we obtain

mn D 0:688 kg and mn D 0:0356 kg ) mn D 35:6 g,

where the positive root has been chosen for the mass.

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The uniform cylinder rolls without slipping on a flat surface. Let k1 Dk2 D k and r D R=2. Assume that the horizontal motion of G issmall.

Problem 9.21 Determine the equation of motion for the cylinder by

writing its Newton-Euler equations. Use the horizontal position of the

mass center G as the degree of freedom.

Problem 9.22 Determine the equation of motion for the cylinder using

the energy method. Use the horizontal position of the mass center G as

the degree of freedom.

Solution to 9.21

Referring to the FBD on the right, we can sum moments about point O

to obtain XM O W 2RF A C .R r/F B D I O˛G; (1)

where F A is the force in the spring of constant k1, F B is the force in the

spring of constant k2, ˛G is the angular acceleration of the cylinder, and

where

I O D I G C mR2 D 12mR2 C mR2 D 32mR2:For the spring forces, we can write

F A D k1xA and F B D k2xB ;where x

A and x

B are the horizontal deflection of the springs at A and B ,

respectively. Since the cylinder rolls without slip at O , we can write the velocity of points A and B as

EvA D EvO C !G ErA=O D !G Ok 2R O| D 2R!G O{;EvB D EvO C !G ErB=G D !G Ok .R r/ O| D .R r/!G O{ D 12R!G O{;

where we have used the fact that r D R =2. Since we are assuming that the horizontal motions are small,these expressions for velocity imply that

xA D 2R G and xB D 12R G ;where G is the rotation of the cylinder. Since xG D R G , we can write the above two expressions for themotion of points A and B as

xA D 2xG and xB D 12xG;

and so the spring force equations become (also using k1 D k2 D k)F A D 2kxG and F B D 12kxG:

Substituting everything into Eq. (1) and noting that ˛G D RxG=R, we obtain

4kRxG C12

R12

kxG D 32mR2 RxG

R

) RxG C

17k

6m xG D 0.

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Dynamics 1e 1243

Solution to 9.22

Referring to the FBD on the right, we that only F A and F B do work as the

wheel undergoes small horizontal motion, where F A is the force in the

spring of constant k1, F B is the force in the spring of constant k2. Since

both of those spring forces are conservative, we can apply the energy

method to determine the equation of motion of the cylinder. With this

in mind, the kinetic energy of the cylinder is

T D 12

mv2G C 12I G!2G;

here !G is the angular speed of the cylinder as it rolls without slip at

point O . Since I G D 12mR2, vG D PxG , and !G D PxG=R, the kineticenergy can be written as

T D 12

m Px2G C 1212

mR2

PxG

R

2D 3

4m Px2G:

For the potential energy of the springs, we can write

V D 12

k1x2A C 12k2x2B D 12kx2A C 12kx2B ;

where xA and xB are the horizontal deflection of the springs at A and B , respectively, and we have used the

fact that k1 D k2 D k. Since the cylinder rolls without slip at O , we can write the velocity of points A andB as

EvA D EvO C !G ErA=O D !G Ok 2R O| D 2R!G O{;EvB D EvO C !G ErB=G D !G Ok .R r/ O| D .R r/!G O{ D 12R!G O{;

where we have used the fact that r D R =2. Since we are assuming that the horizontal motions are small,these expressions for velocity imply that

xA D 2R G and xB D 12R G ;

where G is the rotation of the cylinder. Since xG D R G , we can write the above two expressions for themotion of points A and B as

xA D 2xG and xB D 12xG;and so the potential energy becomes

V D 12

k.2xG/2 C 1

2k

12

xG

2 D 17

8 k x2G:

Substituting T and V into d dt

.T C V / D 0, we obtain

32

m PxG RxG C 174 kxG PxG D 0 ) RxG C 17k

6m xG D 0.

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The uniform cylinder of mass m and radius R rolls without slipping on the

inclined surface. The spring with constant k wraps around the cylinder as it

rolls.

Problem 9.23 Determine the equation of motion for the cylinder by writing

its Newton-Euler equations. Determine the numerical value of the period of

oscillation of the cylinder using k D 30 N=m, m D 10 kg, R D 30 cm, and D 20ı.Problem 9.24 Determine the equation of motion for the cylinder using the

energy method. Determine the numerical value of the period of oscillation of

the cylinder using k D 30 N=m, m D 10 kg, R D 30 cm, and D 20ı.

Solution to 9.23

Referring to the FBD at the right and summing moments about point O , weobtain X

M O W .mg sin /R C 2RF s D I G˛G C mRaGx ; (1)

where F s is the force in the spring, ˛G is the angular acceleration of the cylinder,

F is the force of friction between the cylinder and the ground, and aGx is the x

component of the acceleration of the mass center of the cylinder. If the center of

the cylinder at G moves a distance xG down the incline, then point A must move a distance 2xG and so the

total deflection of the spring must consist of its deflection when the system is in static equilibrium ıst plus its

deflection away from static equilibrium 2xG , which implies that

F s D k.ıst C 2xG/;

In addition, since the cylinder rolls without slip at point O , we can write the angular acceleration of the

cylinder as ˛G D RxG=R. Substituting this kinematics equation and the force law shown above into Eq. (1)and using I G D 12mR2, we obtain

mgR sin C 2kR.ıst C 2xG/ D 12mR2 RxG

R

mR RxG:

Using the fact that mgR sin D 2kRıst and then simplifying this equation, we obtain the equation of motionas

RxG C 8k

3m xG D 0: (2)

From Eq. (2), we can see that the natural frequency and period are given by

!n Dr

8k

3m ) D 2

!nD 2

r 3m

8k D 2:22 s,

where we have substituted in m D 10 kg and k D 30 N=m.

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Dynamics 1e 1245

Solution to 9.24

Referring to the FBD at the right we see that F s and mg are the only two forces

that do work (since the cylinder rolls without slip on a stationary surface, the

friction force F does no work). Since both of these forces are conservative, we

can apply the energy method to obtain the equation of motion for the system.

The kinetic energy of the system is

T D 12

I O!2G D 12

12

mR2 C mR2!2G D 34mR2!2G D 34mR Px2G;where !G is the angular velocity of the cylinder, we have used the parallel axis theorem to find I O , and

we have used the kinematic relation that PxG D R!G to write !G since the cylinder rolls without slip atpoint O . If the center of the cylinder at G moves a distance xG down the incline, then point A must move

a distance 2xG and so the total deflection of the spring must consist of its deflection when the system is in

static equilibrium ıst plus its deflection away from static equilibrium 2xG , which implies that

V D 12

k.ıst C 2xG/2 mgxG sin ;

Substituting the kinetic and potential energies into the energy method, we obtain

d

dt.T C V / D 3

2m PxG RxG C k.ıst C 2xG/.2 PxG/ mg PxG sin D 0;

Canceling PxG , using the fact that mg sin D 2kıst, and then simplifying this equation, we obtain the equationof motion as

RxG C 8k

3mxG D 0: (3)

From Eq. (3), we can see that the natural frequency and period are given by

!n Dr

8k

3m ) D 2

!nD 2

r 3m

8k D 2:22 s,

where we have substituted in m D 10 kg and k D 30 N=m.

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Problem 9.25

A uniform bar of mass m is placed off-center on two counter-rotating

drums A and B . Each drum is driven with constant angular speed !0,

and the coefficient of kinetic friction between the drums and the bar

is k. Determine the natural frequency of oscillation of the bar onthe rollers. Hint: Measure the horizontal position of G relative to the

midpoint between the two drums, and assume that the drums rotate

sufficiently fast so that the drums are always slipping relative to the bar.

Solution

Referring to the FBD at the right, notice that the coordinate

system we are using measures the horizontal position of

the mass center of the bar xG relative to the midpoint

between the two drums. Since drum A rotates clockwise,

the friction force it exerts on the bar F A is to the right andsince drum B rotates counterclockwise, the friction force

it exerts on the bar F B is to the left. With this in mind, the

balance laws for the bar shown can be written asXF x W F A F B D maGx ; (1)XF y W N A C N B mg D maGy; (2)XM G W N B.h=2 xG/ N A.h=2 C xG/ D I G˛bar: (3)

Since the counter-rotating drums must be slipping relative to the bar, the force laws are given by

F A

DkN A and F B

DkN B :

Since the bar does not rotate or move vertically, the kinematics equations are given by

aGx D RxG; aGy D 0; and ˛bar D 0:Substituting the force laws and kinematics into Eq. (1), we obtain

kN A kN B D m RxG: (4)Substituting the kinematics into Eqs. (2) and (3) and then solving the resulting two equations for N A and N Bgives

N A D mg

1

2 xG

h

and N B D mg

1

2 C xG

h

:

Substituting the two normal forces into Eq. (4), we obtain the equation of motion as

kmg

1

2 xG

h

kmg

1

2 C xG

h

D m RxG ) RxG C

2kg

h xG D 0:

Therefore, the natural frequency of oscillation is

!n Dr

2kg

h :

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Dynamics 1e 1247

Problem 9.26

The uniform cylinder A of radius r and mass m is released from

a small angle inside the large cylinder of radius R. Assuming

that it rolls without slipping, determine the natural frequency and

period of oscillation of A.

Solution

Referring to the FBD shown at the right, as long as the cylinder rolls

without slipping, only the weight force mg will do work on it. Since

the weight force is conservative, we can apply the energy method

to find the equation of motion. With this in mind, the kinetic energy

can be written as

T D 12

mv2G C 12I G!2G;

where vG is the speed of the mass center of the cylinder, !G is the

angular speed of the cylinder, and I G is its mass moment of inertia.

Since point G is moving in a circle centered at O and since the

cylinder A of radius r is rolling without slip inside the larger cylinder B , we can write the speed of G as

vG D .R r/ P D r!G ) !G D R r

rP;

where !G is the angular speed of the cylinder A. Using these results, the kinetic energy becomes

T D 12

m.R r/2 P 2 C 12

12

mr2R r

r

2 P 2 D 34

m.R r/2 P 2; (1)

where we have used I G D 12mr 2 for a uniform cylinder. In addition, using the datum line shown in the figureabove, the potential energy of the cylinder can be written as

V D mg.R r/ cos :

Adding the potential energy to the kinetic energy in Eq. (1) and applying the energy method, we obtain

d

dt.T C V / D 3

2m.R r/2 P R C mg.R r/ P sin D 0 ) R C 2g

3.R

r/ sin D 0:

For small values of , we can approximate sin by and we obtain

R C 2g3.R r/ D 0 ) !n D

s 2g

3.R r/ and D 2s

3.R r/2g

.

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Problem 9.27

The uniform sphere A of radius r and mass m is released from

a small angle inside the large cylinder of radius R. Assuming

that it rolls without slipping, determine the natural frequency and

period of oscillation of the sphere.

Solution

Referring to the FBD shown at the right, as long as the sphere rolls

without slipping, only the weight force mg will do work on it. Since

the weight force is conservative, we can apply the energy method

to find the equation of motion. With this in mind, the kinetic energy

can be written as

T D 12

mv2G C 12I G!2G;

where vG is the speed of the mass center of the sphere, !G is the

angular speed of the sphere, and I G is its mass moment of inertia.

Since point G is moving in a circle centered at O and since the

sphere A of radius r is rolling without slip inside the larger cylinder B , we can write the speed of G as

vG D .R r/ P D r!G ) !G D R r

rP;

where !G is the angular speed of the sphere A. Using these results, the kinetic energy becomes

T D 12

m.R r/2 P 2 C 12

25

mr 2R r

r

2 P 2 D 710

m.R r/2 P 2; (1)

where we have used I G D 25mr 2 for a uniform cylinder. In addition, using the datum line shown in the figureabove, the potential energy of the sphere can be written as

V D mg.R r/ cos :

Adding the potential energy to the kinetic energy in Eq. (1) and applying the energy method, we obtain

d

dt.T C V / D 7

5m.R r/2 P R C mg.R r/ P sin D 0 ) R C 5g

7.R

r/ sin D 0:

For small values of , we can approximate sin by and we obtain

R C 5g7.R r/ D 0 ) !n D

s 5g

7.R r/ and D 2s

7.R r/5g

.

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Dynamics 1e 1249

Problem 9.28

The U-tube manometer lies in the vertical plane and contains a fluid of density

that has been displaced a distance y and oscillates in the tube. If the cross-

sectional area of the tube is A and the total length of the fluid in the tube is L,

determine the natural period of oscillation of the fluid, using the energy method. Hint: As long as the curved portion of the tube is always filled with liquid (i.e.,

the oscillations don’t get large enough to empty part of it), the contribution of

the liquid in the curved portion to the potential energy is constant .

Solution

To apply the energy method to find the equation of motion of the oscillating

fluid, we need to write the kinetic and potential energy of the fluid in the

tube. Referring to the top figure on the right and assuming the all the fluid

moves as a unit, we can write the kinetic energy of the fluid as

T D 12

m Py2 D 12

AL Py2;where we have computed the mass of the fluid in the tube as m D AL, Pyis the speed of each element of the fluid, and the coordinate y is measure

downward from the equilibrium level of the surface of the fluid. Referring

the bottom figure on the right, the total potential energy of the fluid can

be computed as the sum of the individual potential energies of the four

segments of fluid B , D , E , and F , that is,

V D V B C V D C V E C V F

D AgL C 2 y„ ƒ ‚ …

length of B

y C 1

2L C

2 y„ ƒ‚ …

dist. from datum to center of B

CV D

Ag

L C 2

y

„ ƒ ‚ …length of E

y C 1

2

L C

2 y

„ ƒ‚ …

dist. from datum to center of E

CV F ;

where C is the constant length of the fluid in the curved portion of the tube and V F is always equal to zero.

Simplifying this expression for V , we obtain

V D 14

Ag

.C L/2 4y2:Applying the energy method to this system, we obtain

d

dt.T C V / D 0 ) AL Py Ry C 2Agy Py D 0 ) Ry C 2g

L y D 0 ) !n D

r 2g

L ;

where we have used the fact that V D, C , and L are all constant. Therefore, the period of oscillation of the

fluid is

D 2!n

D 2s

L

2g:

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Problem 9.29

The uniform semicylinder of radius R and mass m rolls without slip on

the horizontal surface. Using the energy method, determine the period

of oscillation for small .

Solution

Referring to the FBD on the right and noting the semicylinder

rolls with slip on the surface at point O , we see that the only

force that does work on the semicylinder as it rocks back and

forth is the weight force mg. Since that force is conservative, we

can apply the energy method to find the equation of motion as

instructed. Using the datum indicated, the potential energy of thesemicylinder is

V D mg

4R

3

cos mg

4R

3

1 2=2; (1)

where we have used the small angle approximation for cos .

Since the semicylinder is rolling without slip at point O, that point must be the IC of the semicylinder.

Therefore, we can write the kinetic energy as

T D 12

I O P 2; (2)

where P is the angular velocity of the semicylinder. To find I O we first note that

I G D

1

2 16

92

mR2 and then I O D I G C m`2 I G C m

R 4R

3

2;

where ` is the distance from O to G as seen in the figure above and we have approximated ` by R 4R3

for

small . Simplifying I O , we obtain

I O D 9 16

6 mR2: (3)

Substituting Eq. (3) into Eq. (2) and then using Eqs. (1) and (2) to apply the energy method to the semicylinder,

we obtain the following equation of motion:

d

dt.T C V / D 0 )

9 166

mR2 P R C mg

4R3

P D 0 ) R C 8g

.9 16/R D 0:

Therefore, the natural frequency and period of oscillation are

!n Ds

8g

.9 16/R ) D s

.9 16/R2g

.

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Dynamics 1e 1251

Problem 9.30

The thin shell semicylinder of radius R and mass m rolls without slip

on the horizontal surface. Using the energy method, determine the

period of oscillation for small .

Solution

Referring to the FBD on the right and noting the thin shell

semicylinder rolls with slip on the surface at point O , we see

that the only force that does work on the thin shell semicylinder

as it rocks back and forth is the weight force mg. Since that

force is conservative, we can apply the energy method to find

the equation of motion as instructed. Before doing so, we needto find the distance d from point A to the center of mass of

the thin shell semicylinder at G . Since the total mass of the

thin shell semicylinder is m and that the length of the shell

is 2R C R, its mass per unit length is D m=.2R C R/.Therefore, applying the definition of the center of mass, the

distance d is given by

md D 0 C R

2R

D R

m

2R C R

2R

) d D 2R

2 C :

Now that we have d , using the datum indicated, the potential energy of the thin shell semicylinder is

V D mg

2R

2 C

cos mg

2R

2 C

1 2=2; (1)where we have used the small angle approximation for cos . Since the thin shell semicylinder is rolling

without slip at point O , that point must be the IC of the thin shell semicylinder. Therefore, we can write the

kinetic energy as

T D 12

I O P 2; (2)where P is the angular velocity of the thin shell semicylinder. To find I O we must first find I G and thenapply the parallel axis theorem to obtain as I O D I G C m`2, where ` is the distance from G to O . SinceI G for the body we are considering does not appear in the inside back cover of the books, we will first find

I A by treating the thin shell semicylinder as a composite body made up of the straight segment B C andsemicircular segment BO C . Therefore

I A D .I A/BC C .I A/BOC D 112mBC .2R/2 C mBOC R2:

Since the linear density of the thin shell semicylinder is as given above, we can write the mass of each

segment as

mBC D .2R/ D 2m

2 C and mBOC D .R/ D m

2 C ;

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and therefore I A becomes

I A D 1

12

2m

2 C

.2R/2 C m2 C R

2 D 2 C 33.2 C / mR

2:

We can now find I G using the parallel axis theorem as

I A D I G C md 2 ) I G D I A md 2 D 2 C 33.2 C / mR

2 m

2R

2 C 2

D 32 C 8 8

3.2 C /2 mR2:

Finally, we can find I O by again using the parallel axis theorem as

I O D I G C m`2 I G C m.R d /2 D 32 C 8 8

3.2 C /2 mR2 C m

R 2R

2 C 2

D 2.3 2/3.2 C / mR

2;

where we have approximated ` by R d for small . Substituting I O into Eq. (2) and then using Eqs. (1)and (2) to apply the energy method to the thin shell semicylinder, we obtain the following equation of motion:

d dt

.T C V / D 0 ) 2.3 2/3.2 C /

mR2 P R C mg 2R2 C

P D 0 ) R C 3g.3 2/R D 0:

Therefore, the natural frequency and period of oscillation are

!n Ds

3g

.3 2/R ) D 2s

.3 2/R3g

.

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Dynamics 1e 1253

Problem 9.31

The magnification factor for a forced (undamped) harmonic oscillator is measured to be equal to 5.

Determine the driving frequency of the forcing if the natural frequency of the system is 100 rad=s.

Solution

The MF for a harmonic oscillator depends on the forcing frequency and the natural frequency of a system

according to

MF D 11 .!0=!n/2 :

Since we know that MF D 5 and !n D 100 rad=s, we can substitute these values in to the above equationand solve for the forcing frequency !0. Doing so, we obtain

5 D 11 .!0=100/2 ) !0 D 89:4 rad=s.

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Problem 9.32

Suppose that equation of motion of a forced harmonic oscillator is

given by Rx C !2nx D .F 0=m/ cos !0t . Obtain the expression for theresponse of the oscillator, and compare it to the response presented in

Eq. (9.42) (which is for a forced harmonic oscillator with the equationof motion given in Eq. (9.37)).

Solution

The response is given by general solution to the equation of motion, that is, it is given by

x.t/ D xc C xp;

where xc is the complementary solution and xp is the particular solution. The complementary solution for

the given equation of motion is still given by

xc D A sin !nt C B cos !nt;where A and B are constants determined by the initial conditions. To find a particular solution, we try a

solution of the form xp D D cos !0t , where D is a constant to be determined. Substituting this assumedsolution into the equation of motion, we obtain

D!20 cos !0t C !2nD cos !0t D F 0

m cos !0t ) D

!2n !20

D F 0m

) D D F 0=m!2n !20

) D DF 0m

1

!2n

1 .!0=!n/2 D

F 0m

mk

1 .!0=!n/2 D

F 0=k

1 .!0=!n/2

Therefore, the response is given by

x.t/ D A sin !nt C B cos !nt C F 0=k1 .!0=!n/2

cos !0t:

Notice that we can find the response due to cos !0t forcing by replacing the “sin” with a “cos” in the response

due to sin !0t forcing.

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Dynamics 1e 1255

Problem 9.33

Derive the equations of motion for the unbalanced motor introduced

in this section by applying Newton’s second law to the center of

mass of the system shown in Fig. 9.12.

Solution

The FBD of the mass center G of the system is shown in the top

figure on the right, where N is the net horizontal force acting on

the system, m is the total mass of the system (i.e., m D mm C mp),and F s is the force on the platform due to the springs. Applying

Newton’s second law to the center of mass and summing forces in

the vertical direction, we findXF y W mg F s D m RyG; (1)

where, referring to the middle figure on the right, yG measures the

vertical position of the mass center. The force law for the springs

is given by

F s D keq.ym ıst/;where, referring to the middle figure on the right, ym is the vertical

position of the motor and platforms (they move together), keq is

the equivalent spring constant of the six springs supporting the

platform, and ıst is the static deflection of the spring when the

system is in static equilibrium. Substituting the force law into thebalance law in Eq. (1) we find

mg keq.ym ıst/ D m RyG ) m RyG C keqym D 0; (2)where we have used the fact that mg D keqıst. Notice that Eq. (2) contains two dependent variables ym andyG and so we need to eliminate one of them. To show that this equation is equivalent to Eq. (9.36) in the

textbook, we will need to eliminate yG by using the definition of the mass center. Referring to the bottom

figure above, we find

myG D muyu C .m mu/ym ) myG D mu.ym C h C " sin / C .m mu/ym:Canceling muym and noting that sin D sin !r t , this equation becomes

myG D mu.h C " sin !r t / C mym ) m RyG D mu"!2r sin !r t C m Rym;where we have used the fact that h is constant in taking the second derivative. Substituting m RyG from thisexpression into Eq. (2), we obtain the equation of motion as

.mm C mp/ Rym C keqym D mu"!2r sin !r t ) Rym C keq

mm C mpym D mu"!

2r

mm C mpsin !r t ,

where we have made the substitution m D mm C mp.August 10, 2009


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Problem 9.34

Determine the amplitude of vibration of the unbalanced motor we

studied in Example 9.4 if the forcing frequency of the motor is

0:95!n.

Solution

Referring to Example 9.4, the response if the unbalance motor was found to be

ym D"

vm0

!n !r

!n

mu"!2r =keq

1 .!r=!n/2#

sin !nt C mu"!

2r =keq

1 .!r=!n/2 sin !r t; (1)

where ym in defined in the figure on the right, vm0 D Pym.0/, andthe natural frequency was found to be

!n Ds

keq

mm C mp:

The goal then is to find the amplitude of ym in Eq. (1). We begin by

writing Equation (1) as

ym D P sin !nt C Q sin !r t;where

P D vm0!n

!r!n

mu"!2r = keq

1 .!r=!n/2 and Q D mu"!

2r =keq

1 .!r=!n/2:

We now let ! D !r !n, which means that ym can be written asym D P sin !nt C Q sin.! C !n/t D P sin !nt C Q.sin !t cos !nt C cos !t sin !nt /

D Q sin !t„ ƒ‚ …A

cos !nt C .P C Q cos !t/„ ƒ ‚ …B

sin !nt

Dp

A2 C B2 sin.!nt C /;where the angle can be determined, but is not needed to find the amplitude of vibration. Thus we see that

the amplitude is

jymj Dp

A2 C B2 Dq

.Q sin !t/2 C .P C Q cos !t/2 Dq

P 2 C Q2 C 2PQ cos.!r !n/t ;where we have substituted !r !n for ! . Notice that the amplitude of vibration is itself a function of time. We can now apply the known quantities from Example 9.4, which were mm

D 40 kg, mp

D 15 kg,

keq D 6ks D 420;000 N=m, " D 15 cm, !r D 0:95!n, vm.0/ D 0:4 m=s, and three different values of mu:10 g, 100 g, and 1000 g. Using these, we find that the amplitude of vibration for each of the three unbalanced

masses is

mu D 0:01 kg ) jymj Dq

2:19106 cos.4:37t/ C 1:89105 m;

mu D 0:1 kg ) jymj Dq

1:10105 cos.4:37t/ C 1:11105 m;

mu D 1 kg ) jymj Dq

9:80104 cos.4:37t/ C 1:01103 m;August 10, 2009


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Dynamics 1e 1257

A Closer Look. A plot showing each of these three amplitudes as

a function of time is shown at the right. The green curve corresponds

to mu D 10 g, the blue curve corresponds to mu D 100 g, and the redcurve corresponds to mu D 1000 g. Note that these curves representthe amplitude of the response at any time t . These curves envelope the

actual response ym as can be seen in the three figures below. The first

figure shows the response as well as the amplitude of the response for

mu D 10 g.

The second plot shows the response as well as the amplitude of the response for mu D 100 g. The thirdplot shows the response as well as the amplitude of the response for mu D 1000 g.

Notice that in each case the amplitude of the response jymj that we found forms an envelope over the responseym. When the forcing frequency is very close to the natural frequency as it is in this problem, the amplitude

of the response increases and decreases as we have seen in this problem. If the vibrations were sound waves,the volume would alternately increase and decrease; this is referred to as the beat phenomenon.

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Problem 9.35

A uniform bar of mass m and length L is pinned to a slider at O . The slider is

forced to oscillate horizontally according to y.t/ D Y sin !st . The system liesin the vertical plane.

(a) Derive the equation of motion of the bar for small angles .

(b) Determine the amplitude of steady-state vibration of the bar.

Solution

Referring to the FBD on the right, to obtain the equation of motion of the bar,

we can sum moments about point O using the equation

EM O D I O Ębar C ErG=O mEaO :Doing so, we obtainX

M O W mgL

2 sin Ok D I O R Ok C

L

2 .cos O{ C sin O| / m RyO O| ;

where RyO O| is the acceleration of point O . Carrying out the cross product,noting that I O D 13mL2, and that RyO D Y !2s sin !st , the equation of motionbecomes

1

3mL2 R C mg L

2 sin m L

2 Y !2s sin !st cos D 0;

which, for small can be written as

1

3mL2 R C mg L

2 D m L

2 Y !2s sin !st ) R C

3g

2L D 3Y

2L !2s sin !st .

To find the amplitude of steady-state vibration of the bar, we need to find a particular solution and then the

amplitude of that solution will be what we seek. We assume a particular solution of the form

p D D sin !st;where D is a constant to be determined. Substituting this particular solution into the equation of motion, we

obtain

D!2

s sin !st C 3g

2L D sin !st D 3Y

2L !2

s sin !st ) D!2n !2s D 3Y 2L !2s) D D 3Y=.2L/.!s=!n/

2

1 .!s=!n/2 ;

where we have used !2n D 3g=.2L/ from the equation of motion. Therefore, the amplitude of steady-statevibration is

amp D 3Y=.2L/.!s=!n/2

1 .!s=!n/2 :

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Dynamics 1e 1259

Problem 9.36

Consider a sign mounted on a circular hollow steel pole of length

L D 5 m, outer diameter d o D 5 cm, and inner diameter d i D 4 cm.Aerodynamic forces due to wind provide a harmonic torsional excitation

with frequency f 0 D 3 Hz and amplitude M 0 D 10 Nm about the ´axis. The mass center of the sign lies on the central axis ´ of the pole.The mass moment of inertia of the sign is I ́ D 0:1 kgm2. The torsionalstiffness of the pole can be estimated as k t D Gst

d 4o d 4i

=.32L/,

where Gst is the shear modulus of steel, which is 79 GPa. Neglecting the

inertia of the pole, calculate the amplitude of vibration of the sign.

Solution

Referring to the FBD of the sign on the right, T e is the exciting torque on the

sign due to the wind and T p is the restoring torque due to the steel pole of

length L. Summing moments about the ´ axis, we obtainXM ́ W T e T p D I ́ R; (1)

where is the angle of rotation of the sign about the ´ axis from its static

equilibrium position. The force law for the excitation torque and restoring

torque are given by, respectively,

T e D

M 0 sin.2f 0t / and T p D

kt

D "Gstd

4o d 4i 32L #;

where kt is the torsional stiffness of the pole and we have substituted in the given relationship for kt .

Substituting the force laws in to Eq. (1), we obtain

M 0 sin.2f 0t / "

Gst

d 4o d 4i

32L

# D I ́ R ) I ́ R C

"Gst

d 4o d 4i

32L

# D M 0 sin.2f 0t /;

so we see that using d o D 0:05 m, d i D 0:04 m, L D 5 m, Gst D 79109 Pa, and I ́ D 0:1 kgm2, we obtain

!n

Ds Gstd

4o d 4i 32LI ́ D 239:2 rad=s; !0 D 2f 0 D 18:85 rad=s

keff D Gst

d 4o d 4i

32L

D 5724 Nm; .F 0/eff D M 0 D 10 Nm;

where !0 is the forcing frequency, keff is the effective spring constant, and .F 0/eff is the effective forcing

amplitude. Applying Eq. (9.43) from the text, we find the following amplitude

amp D .F 0/eff = keff 1 .!0=!n/2 ) amp D 0:00176 rad=s.

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One of the propellers on the Beech King Air 200 is unbalanced such that the eccentric mass mu is a

distance R from the spin axis of the propeller. The propellers spin at a constant rate !r , and the mass of

each engine is me (this includes the mass of the propeller). Assume that the wing is a uniform beam that is

cantilevered at A, has mass mw and bending stiffness EI , and whose mass center is at G. For each problem,evaluate your answers for mu D 3 oz, me D 450 lb, R D 5:1 ft, !r D 2000 rpm, EI D 1:131011 lbin:2,d D 8:7 ft, and h D 10:9 ft.Problem 9.37 Neglect the mass of the wing and model the wing as done in Example 9.2. Determine the

resonance frequency of the system, and find the MF for the given parameters.

Problem 9.38 Let the mass of the wing be mw D 350 lb, and model the wing as done in Example 9.2.Determine the resonance frequency of the system and find the MF for the given parameters.

Solution to 9.37

The figure at the right shows an FBD of the wing

and engine with the unbalanced mass removed as

well as an FBD of just the unbalanced mass mu.

Summing forces in the x and y directions on theunbalanced mass mu givesX

F x W F x D muaux; (1)XF y W F y mug D muauy; (2)

where aux and auy are the x and y components,

respectively, of the acceleration of the unbalanced

mass. Next, summing moments about point A on

the wing, we obtain

XM A W megd F xR sin ˇ F y.d C R cos ˇ/ M t D I A˛wing: (3)where M t is the lumped torsional stiffness of the wing at point A, I A D med 2, and ˛wing D R . Note that wehave treated the engine as a point mass. The force law for the torsional stiffness is given by

M t D kt. C st/ D 3EI d

. C st/;

where k t is the lumped torsional spring constant, st is the rotation of the wing in its static equilibrium

position, and kt was determined using the result from Example 9.2. The last thing we need to do is determine

the acceleration of the unbalanced mass. Noting that the propeller is rotating with a constant angular velocity

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Dynamics 1e 1261

and neglecting the rotation of the wing when computing the angular velocity and acceleration of the propeller,

we find the acceleration of the unbalanced mass to be

Eau D Eae C Ęprop Eru=e !2propEru=e D d R O| !2p.R cos ˇ O{ C R sin ˇ O| /;

so that

aux D R!2p cos ˇ and auy D d R R!2p sin ˇ:Substituting the kinematics equations into Eqs. (1) and (2) gives

F x D muR!2p cos ˇ and F y D mud R C muR!2p sin ˇ mug:

Substituting F x and F y into Eq. (3), we obtain the following equation of motion

megd C muR2!2p sin ˇ cos ˇ mu.d C R cos ˇ/

d R C R!2p sin ˇ g 3EI

d . C st/ D med 2 R :

Canceling terms and rearranging, we obtain

.mu C me/d 2 R C muRd cos ˇ R .mu C me/dg C mudR!2p sin ˇ mugR cos ˇ C 3EI d . C st/ D 0:

Since .3EI=d/ st D megd C mug.d C R cos ˇ/, this becomes

.mu C me/d 2 C muRd cos !pt„ ƒ‚ …time-dependent inertia

R C 3EI d

D mudR!2p sin !pt;

where we have substituted !pt for ˇ. Notice that the inertia term (i.e., the coefficient of R ) is time-dependentsince it contains cos !pt , which means that this equation of motion is not in our standard form, which required

that the coefficients be constant. On the other hand, also notice that

ˇ̌.mu C me/d 2ˇ̌ D 1058 slugft2 and jmuRd cos !pt j 0:2584 slug ft2;

and therefore, the inertia term is dominated by the constant part, which means that we can approximate the

equation of motion as

.mu C me/d 2 R C 3EI d

D mudR!2p sin !pt; (4)and so using

EI D 1:131011 lbin:2 D 7:847108 lbft2;mu D 3 oz

1 lb

16 oz 1

32:2 ft=s2D 5:823103 slug;

me D 450 lb 1

32:2 ft=s2

D 13:98 slug; and

d D 8:7 ft;

we find that the resonance frequency of the wing is

!n Ds

3EI

.mu C me/d 3 D 506 rad=s or f D !n

2 D 80:5 Hz.

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As for the MF, we need to find a particular solution p to the equation of motion in Eq. (4), and then use its

amplitude to find determine the MF. The most direct way to find a particular solution is to assume a solution

of the form p D D sin !pt and then substitute it into the equation of motion. Doing so, we obtain

D.mu C me/d 2!2p sin !pt C 3EI

d D sin !pt D mudR!2p sin !pt;

which, upon canceling sin !pt and solving for D gives

D D mudR!2p

3EI d

.mu C me/d 2!2pD mud

2R!2p=.3EI /

1 .mu C me/d 3!2p=.3EI / D mud

2R!2p=.3EI /

1 .!p=!n/2

D muR.mu C me/d

.!p=!n/2

1 .!p=!n/2:

Therefore,

j p

j D muR

.mu C me/d .!p=!n/

2

1 .!p=!n/2

) MF D j pj.mu C me/d muR

D .!p=!n/2

1 .!p=!n/2 D 0:207,

where we have used !n D 505:7 rad=s and !p D 2000 rpm D 209:4 rad=s.

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Dynamics 1e 1263

Solution to 9.38

The figure at the right shows an FBD of the wing

and engine with the unbalanced mass removed as

well as an FBD of just the unbalanced mass mu.

Summing forces in the x and y directions on the

unbalanced mass mu givesXF x W F x D muaux; (5)XF y W F y mug D muauy; (6)

where aux and auy are the x and y components,

respectively, of the acceleration of the unbalanced

mass. Next, summing moments about point A on

the wing, we obtain

XM A W mwgh C megd F xR sin ˇ F y.d C R cos ˇ/ M t D I A˛wing: (7)where M t is the lumped torsional stiffness of the wing at point A and ˛wing D R . Note that we have treatedthe engine as a point mass. The mass moment of inertia with respect to point A is

I A D med 2 C 112mw.2h/2 C mwh2 D med 2 C 43mwh2;where we have assumed that the wing is a uniform beam of length 2h. The force law for the torsional stiffness

is given by

M t D kt. C st/ D 3EI d

. C st/;where k t is the lumped torsional spring constant, st is the rotation of the wing in its static equilibrium

position, and kt was determined using the result from Example 9.2. The last thing we need to do is determine

the acceleration of the unbalanced mass. Noting that the propeller is rotating with a constant angular velocity

and neglecting the rotation of the wing when computing the angular velocity and acceleration of the propeller,

we find the acceleration of the unbalanced mass to be

Eau D Eae C Ęprop Eru=e !2propEru=e D d R O| !2p.R cos ˇ O{ C R sin ˇ O| /;so that

aux D R!2p cos ˇ and auy D d R R!2p sin ˇ:Substituting the kinematics equations into Eqs. (5) and (6) gives

F x

D muR!

2p cos ˇ and F y

Dmud R

CmuR!

2p sin ˇ

mug:

Substituting F x and F y into Eq. (7), we obtain the following equation of motion

mwgh C megd C muR2!2p sin ˇ cos ˇ mu.d C R cos ˇ/

d R C R!2p sin ˇ g 3EI

d . C st/

D med 2 C 43mwh2 R :Canceling terms and rearranging, we obtain

.mu C me/d 2 C 43mwh2 C muRd cos ˇ

R .mu C me/dg mwgh C mudR!2p sin ˇAugust 10, 2009


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mugR cos ˇ C 3EI d

. C st/ D 0;

or, since .3EI=d/ st D mwgh C megd C mug.d C R cos ˇ/, this becomes

.mu Cme/d

2

C 43

mwh2

CmuRd cos !pt„ ƒ‚ …

time-dependent inertia R C

3EI

d

D mudR!

2p sin !pt;

where we have substituted !pt for ˇ. Notice that the inertia term (i.e., the coefficient of R ) is time-dependentsince it contains cos !pt , which means that this equation of motion is not in our standard form, which required

that the coefficients be constant. On the other hand, also notice thatˇ̌.mu C me/d 2 C 43mwh2

ˇ̌ D 2780 slug ft2 and jmuRd cos !pt j 0:2584 slug ft2;and therefore, the inertia term is dominated by the constant part, which means that we can approximate the

equation of motion as

.mu C me/d 2 C 43mwh2 R C 3EI

d

D mudR!

2

p

sin !pt: (8)

Therefore, the resonance frequency of the wing is

!n Ds

3EI

.mu C me/d 3 C 43mwh2d D 312 rad=s or f D !n

2 D 49:7 Hz,

where we have used

EI D 1:131011 lbin:2 D 7:847108 lbft2;mu

D3 oz 1 lb16 oz

1

32:2 ft=s2 D 5:82310

3 slug;

me D 450 lb

1

32:2 ft=s2

D 13:98 slug;

mw D 350 lb

1

32:2 ft=s2

D 10:87 slug

d D 8:7 ft; andh D 10:9 ft:

As for the MF, we need to find a particular solution p to the equation of motion in Eq. (8), and then use its

amplitude to find determine the MF. The most direct way to find a particular solution is to assume a solution

of the form p

DD sin !pt and then substitute it into the equation of motion. Doing so, we obtain

D.mu C me/d 2 C 43mwh2!2p sin !pt C 3EI d D sin !pt D mudR!2p sin !pt;which, upon canceling sin !pt and solving for D gives

D D mudR!2p

3EI d

.mu C me/d 2 C 43mwh2!2p Dmud 2R!2p=.3EI /

1 .mu C me/d 3 C 43mwh2d !2p=.3EI /D mud

2R!2p=.3EI /

1 .!p=!n/2 D muR

.mu C me/d .!p=!n/

2

1 .!p=!n/2:

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Dynamics 1e 1265

Therefore,

j pj D muR.mu C me/d

.!p=!n/2

1 .!p=!n/2

) MF D j p

j.mu

Cme/d

muR D

.!p=!n/2

1 .!p=!n/2 D 0:820,

where we have used !n D 312:0 rad=s and !p D 2000 rpm D 209:4 rad=s.

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Problem 9.39

An unbalanced motor is mounted at the tip of a rigid beam of mass mb and length L. The beam is restrained

by a torsional spring of stiffness kt and an additional support of stiffness k located at the half length of the

beam. In the static equilibrium position, the beam is horizontal and the torsional spring does not exert any

moment on the beam. The mass of the motor is mm, and the unbalance results in a harmonic excitationF.t/ D F sin !0t in the vertical direction. Derive the equation of motion for the system assuming that is small.

Solution

Referring to the�

Dynamics Solns Ch09

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Transcript of Dynamics Solns Ch09