Solutions Manual
Engineering Mechanics: Dynamics
1st Edition
Gary L. Gray
The Pennsylvania State University
Francesco Costanzo
The Pennsylvania State University
Michael E. Plesha
University 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.
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.
3
Dynamics 1e
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 versions
should 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:
dyn_solns@email.esm.psu.edu
We welcome your input.
August 10, 2009
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 4
significant 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.
August 10, 2009
1221
Dynamics 1e
Chapter 9 Solutions
Problem 9.1
Show that Eq. (9.15) is equivalent to Eq. (9.3) if C D
p
A2 C B 2 and tan D A=B.
Solution
Start with Eq. (9.15), which is given by
x.t / D xi cos !n t C
vi
sin !n t D A cos !n t C B sin !n t;
!n
where we have let xi D A and vi =!n D B. Factoring out B, we obtain
A
x.t / D B
cos !n t C sin !n t :
B
Letting A=B D tan , this becomes
x.t / D B.tan cos !n t C sin !n t / D
B
.sin cos !n t C cos sin !n t /:
cos Since sin cos !n t C cos sin !n t D sin.!n t C /, this equation becomes
B
sin.!n t C /:
cos p
This is Eq. (9.3) as long as we can show that B= cos D A2 C B 2 . To show this, we write
x.t / D
B
D
cos B
p B
A2 CB 2
D
p
A2 C B 2 D C;
p
where we have use the identity that if tan D A=B, then cos D B= A2 C B 2 . We have thus shown that
Eq. (9.15) is equivalent to Eq. (9.3) for the proper definition of C and .
August 10, 2009
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
X
MO W
mgL sin D IO R :
Rearranging this equation and assuming small oscillations, we obtain
mgL
sin D 0
R C
IO
)
mgL
R C
D 0:
IO
This is the standard form of the harmonic oscillator, so the natural frequency can be written as
!n2 D
mgL
IO
)
IO D
mgL
:
!n2
Since !n D 2=, we have that
IO D
mgL 2
:
4 2
August 10, 2009
1223
Dynamics 1e
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 obtain
X
MO W
mgR sin D IO ˛G ;
(1)
where ˛ D R is the angular acceleration of the thin ring. Using the parallel
axis theorem, we find that IO D mR2 C mR2 D 2mR2 . Substituting this into
Eq. (1), we find the equation of motion to be
mgR sin D 2mR2 R
)
g
R C
sin D 0:
2R
Assuming small angles , we can use the approximation sin so that the equation of motion becomes
g
R C
D 0:
2R
Therefore, the natural frequency and period are found to be
r
!n D
g
2R
)
2
D
D 2
!n
s
2R
.
g
August 10, 2009
1224
Solutions Manual
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
X
MO W
mg
L
sin D IO R :
2
Rearranging this equation and assuming small oscillations, we obtain
mgL
sin D 0
R C
2IO
)
mgL
R C
D 0:
2IO
This is the standard form of the harmonic oscillator, so the natural frequency and period can be written as
s
mgL
2
2IO
2
!n D
and D
D 2
;
(1)
2IO
!n
mgL
respectively. Noting that each of the four segments of the thin square hoop has mass m=4, computing IO we
obtain
L2
7
1 m 2
1 m 2 m L2
1 m 2 m 2
L C2
L C
C
C
L C L D mL2 :
IO D
12 ƒ‚
4 …
12 4
4 4
4
12 4
4
12
„
„
ƒ‚
… „
ƒ‚
…
segment AB
segments BC and AD
segment CD
Substituting this expression for IO into the expression for the period given in Eq. (1), we obtain
s
D 2
7L
:
6g
August 10, 2009
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 natural
frequency 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
p
action of gravity depends on the value of the acceleration due to gravity g such that !n / g. On the other
hand, 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 the
natural frequency of the vibrating mass will stay the same.
August 10, 2009
1226
Solutions Manual
Problems 9.6 and 9.7
A rigid body of mass m, mass center at G, and mass moment of inertia IG is pinned at
an arbitrary point O and allowed to oscillate as a pendulum.
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 9.6
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.
Problem 9.7
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
X
MO W
mg` sin D IO R ;
(1)
where, using the parallel axis theorem, IO D IG C m`2 . Substituting IO into
Eq. (1), we obtain
mg`
IG C m`2 R C mg` sin D 0 ) R C
D 0;
IG C m`2
where the second equation was obtained from the first by dividing through by the coefficient of R and
assuming small . Therefore, the natural frequency is given by
s
mg`
:
!n D
IG C m`2
Differentiating !n with respect to ` and setting it equal to zero, we find that
#
1=2
1=2 "
mg IG C m`2
2m`.mg`/
d!n
1
d
mg`
mg`
D
D
D 0;
2
d`
d ` IG C m`2
2 IG C m`2
IG C m`2
which implies that
r
mg IG C m`
2
2m`.mg`/ D 0
)
IG
2
m` D 0
)
`D
IG
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
!
August 10, 2009
1227
Dynamics 1e
Solution to 9.7
We will write the equation of motion using the energy method, 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 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
2
T D 21 IO !G
D 12 IG 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 IO . In
addition, the potential energy can be written as
2
V D mg` cos mg` 1
;
2
where we have assumed small in approximating the cosine function. Applying the energy method, we
obtain
d
.T C V / D IG C m`2 P R C mg` P D 0 )
IG C m`2 R C mg` D 0;
dt
which implies that
s
!n D
mg`
:
IG C m`2
Differentiating !n with respect to ` and setting it equal to zero, we
find that
#
1=2
1=2 "
mg IG C m`2
2m`.mg`/
d!n
d
1
mg`
mg`
D
D
D 0;
2
d`
d ` IG C m`2
2 IG C m`2
IG C m`2
which implies that
r
mg IG C m`
2
2m`.mg`/ D 0
)
IG
m`2 D 0
)
`D
IG
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
!
August 10, 2009
1228
Solutions Manual
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´ L
;
GJ
where G is the shear modulus of elasticity of the shaft and J D 2 r 4 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
X
MO W
M´ D IO R :
(1)
From the given relationship between M´ and , we can find that
M´ D
GJ
L
)
GJ
D IO R :
L
(2)
where we have made use of Eq. (1). Since
IO D 12 mR2 ;
m D R2 t;
and J D
4
2r ;
Eq. (2) becomes
G r 4 =2
1
D R4 t R
L
2
)
R C
Gr 4
LR4 t
D 0:
Therefore, the natural frequency of oscillation is
s
!n D
Gr 4
:
LR4 t
August 10, 2009
1229
Dynamics 1e
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 2 12,
so its cross-sectional dimensions are as shown. Assuming the
worker weighs 180 lb and he flexes his knees once to get the board
oscillating, 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 =.48EIcs /,
where L is the length of the beam, E is its modulus of elasticity,
and Ics 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 Fb 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
48EIcs
48EIcs
ı D keq ı; where keq D
Fb D
3
L
L3
and where ı is the deflection of the center of the board due to the
force Fb 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 is
2
T D 21 mvG
D 12 myP 2 :
The potential energy is given by
V D
mg.y C ıst / C 12 keq .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
.T C V / D myP yR mg yP C keq .y C ıst /yP D 0 ) myR mg C keq .y C ıst / D 0:
dt
Using the fact that mg D keq ıst , this equation becomes the standard harmonic oscillator equation
r
48EIcs
!n
2 3EIcs
yR C
yD0 ) f D
D
;
mL3
2
mL3
where f is the vibration frequency. Substituting in given values
"
#1=2
lb
2 3 1:8106 in:2 3:164 in:4
f D
)
f D 1:77 Hz,
3
2
0:4658 lbs
168
in:
in:
1
where Ics D 12
bh3 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 D
2
0:4658 lbs =in:
August 10, 2009
1230
Solutions Manual
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.
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 9.10
Assuming that the L-shaped bar has negligible mass, determine
the natural period of vibration of the system via the energy method.
Problem 9.11
Assuming that the L-shaped bar has mass per unit length ,
determine the natural period of vibration of the system by writing the NewtonEuler equations.
Problem 9.12
Assuming that the L-shaped bar has mass per unit length ,
determine the natural period of vibration of the system via the energy method.
Problem 9.13
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 that
X
MO W Fs h mgd D IO 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 Fs is due to the static deflection ıst at equilibrium as well as
any additional deflection during vibration. Therefore the spring force is
Fs D k.ıst C h /:
In addition, if we neglect the mass of the bar, then IO D md 2 , which means that Eq. (1) becomes
k.ıst C h /h
mgd D
md 2 R ;
Since kıst h D mgd , this equation becomes md 2 R C kh2 D 0, which means that the natural frequency and
period of vibration are given by
s
!n D
kh2
h
D
2
md
d
r
k
m
)
2d
D
h
r
m
.
k
August 10, 2009
1231
Dynamics 1e
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 Fs 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 IO P 2 D 12 md 2 P 2 ;
where we have used the parallel axis theorem to obtain IO 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 21 k.ıst C h /2
mgd;
where we have approximated sin as . Applying the energy method, we obtain
d
.T C V / D md 2 P R C k.ıst C h /hP
dt
mgd P D 0
md 2 R C k.ıst C h /h
)
mgd D 0:
Since kıst h D mgd , this equation becomes md 2 R C kh2 D 0, which means that the natural frequency and
period of vibration are given by
s
r
r
h k
2d m
kh2
D
)
D
.
!n D
md 2
d m
h
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 that
X
MO W Fs h mgd dg d2 D IO 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 Fs is due to the static deflection ıst at equilibrium as
well as any additional deflection during vibration. Therefore the spring force
is
Fs D k.ıst C h /:
In addition, if we include the mass of the bar, then
IO D md 2 C 13 .h/h2 C 13 .d /d 2 ;
which means that Eq. (2) becomes
2 1
md C 3 .h/h2 C 31 .d /d 2 R ;
Since kıst h D mgd C 12 gd 2 , this equation becomes 13 .h3 C d 3 / C md 2 R C kh2 D 0, which means
that the natural frequency and period of vibration are given by
k.ıst C h /h
s
!n D
mgd
dg d2 D
kh2
1
3
3 .h
C d 3 / C md 2
s
)
D 2
1
3
3 .h
C d 3 / C md 2
.
kh2
August 10, 2009
1232
Solutions Manual
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
Fs 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 IO P 2 D 21 13 h3 C d 3 C md 2 P 2 ;
where we have used the parallel axis
to find IO D md 2 C
theorem
1
1
1
2
2
3
3
2
C md . Changes in the potential
3 .h/h C 3 .d /d D 3 h C d
energy 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 21 k.ıst C h /2
D 21 k.ıst C h /2
dg d2 sin C hg h2 cos 2
dg d2 C hg h2 1 2 ;
mgd sin mgd
where we have approximated sin as and cos as 1
2 =2. Applying the energy method, we obtain
d
.T C V / D 13 h3 C d 3 C md 2 P R C k.ıst C h /hP mgd P
dt
) 13 h3 C d 3 C md 2 R C k.ıst C h /h
2 P
1
2 d g mgd
2
1
P
2 h g 1
2 g
d2
D0
h2 D 0:
Since kıst h D mgd C 12 g d 2 C h2 , this equation becomes 13 .h3 C d 3 / C md 2 R C kh2 D 0, which
means that the natural frequency and period of vibration are given by
s
!n D
kh2
1
3
3 .h
C d 3 / C md 2
s
)
D 2
1
3
3 .h
C d 3 / C md 2
.
kh2
August 10, 2009
1233
Dynamics 1e
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 platform
is supported by eight identical springs (four of which are shown),
each with constant k D 3:6 105 lb=ft. Modeling the truck, its
contents, 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
s
keq
D 2f;
!n D 2f
)
mtot
(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
Wscale C Wtruck C Wpayload
51;000 lb C 74;000 lb C Wpayload
Wtot
D
D
;
g
g
32:2 ft=s2
where Wtot is the total vibrating weight. Substituting f D 3:3 Hz, keq , and mtot into Eq. (1), we obtain
Wpayload D 90;700 lb:
August 10, 2009
1234
Solutions Manual
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 obtain
X
Fx W
Fs D max D mxR ) mxR C kx D 0;
(1)
where Fs D kx is the force in the spring and x is measured from the undeformed
position of the spring. Since the initial velocity of the mass is x.0/
P
D v0 D 2 m=s
2
and its initial position is x.0/ D x0 D 0 and since !n D k=m, the amplitude of vibration is
s
s
s
2
v02
v
mv02
0
2
C D
C
x
D
D
)
C D 0:316 m.
0
!n2
k=m
k
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.!n t C /;
and then differentiate it twice to obtain
x.t
R /D
C !n2 sin.!n t C /;
)
xR max D C !n2
)
xR max D 12:6 m=s2 .
August 10, 2009
1235
Dynamics 1e
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 D
1027 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 Fb 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
X
Fy W mg .mg C Fb / D myR ) myR C Fb 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
" #
d 2
Fb D ˝g D y g;
2
where ˝ is the volume of the fluid displaced. Therefore, Eq. (1) becomes
myR C
2
4 gd y
D0
)
yR C
gd 2
y D 0:
4m
This equation is in standard form, so we can write the natural frequency !n and the frequency f as
d
!n D
2
r
g
m
)
!n
d
f D
D
2
4
r
g
D 0:566 Hz,
m
where, to obtain the numerical result, we have used d D 1:2 m, D 1027 kg=m3 , m D 900 kg, and
g D 9:81 m=s2 .
August 10, 2009
1236
Solutions Manual
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 frequency !n and frequency f of the nanowire. Use the values given
in 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 3EIcs =L3 . You may use either ı or as the position variable in
your 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 Fs
be the elastic force in the spring. Assuming that both ı and are
small and summing moments about point O, we obtain
X
MO W .mg Fs /L D IO ˛n D IO R ;
(1)
where ˛n D R from the angular acceleration of the nanowire,
IO 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 Fs D k.ı C ıst / D kL. C st /, we can write
Eq. (1) as
k
mL2 R C kL2 . C st / mgL D 0 ) R C D 0;
m
where we have used the fact that kLst D mg.
From Example 9.2, we know that the cross section of the Si nanowire is circular and that it has the
following properties:
L D 9:810
6
d D 2r D 33010
m;
9
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 Ics D 14 r 4 , we can compute k as
3.152109 N=m2 / 14 .16510 9 m/4
kD
D 0:2820 N=m:
.9:810 6 m/3
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 r 2 L D 1:95310
15
kg:
Using the values of k and m found above, we find that the natural frequency !n and frequency f are
r
!n D
k
D 1:20107 rad=s
m
and f D
!n
D 1:91106 Hz:
2
(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.
August 10, 2009
1237
Dynamics 1e
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 Fs 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
X
MO W mgL Fs L D IO ˛OA ;
(1)
where IO 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
Fs D k.ıst C 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 y=L,
R
the equation of motion (Eq. (1)) in terms of y becomes
mgL
k.ıst C y/L D mL2 y=L:
R
Since mg D kıst , we have
myR 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
Fs 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
R the equation of motion (Eq. (1)) in terms of y becomes
kinematics relation ˛OA D ,
mgL
k.ıst C L /L D mL2 R :
Since mg D kıst , we have
mR C k D 0:
Of course both equations of motion give the same vibration frequency.
August 10, 2009
1238
Solutions Manual
Problems 9.19 and 9.20
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 the
pendulum disk and the nut be d .
If the adjustment is initially at a distance d D 9 mm from the bottom of the pendulum
disk, 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 9.19
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 its
mass 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 .IO /d !p2 C 12 .IO /n !p2 D
1
2
Œ.IO /d C IO /n  !p2 ;
where .IO /d is the mass moment of inertia of the pendulum disk with
respect to point O, .IO /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 the
pendulum. The moments of inertia are given by
.IO /p D 21 mp r 2 C mp L2
and .IO /n D mn .L C r C d /2
D mp . 12 r 2 C L2 /:
August 10, 2009
1239
Dynamics 1e
The potential energy of the system can be written as
V D
mp gL cos mn g.L C r C d / cos g mp L C mn .L C r C d / cos 2
;
g mp L C mn .L C r C d / 1
2
D
where we have used the approximation cos 1
2 =2. Applying the energy method, we find
d
.T C V / D mp 12 r 2 C L2 C mn .L C r C d /2 P R C g mp L C mn .L C r C d / P D 0
dt
)
mp 21 r 2 C L2 C mn .L C r C d /2 R C g mp L C mn .L C r C d / D 0;
and therefore the natural frequency is
!n2
g mp L C mn .L C r C d /
:
D
mp 12 r 2 C L2 C 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 the
period is D 2=!n , with the initial position of the adjustment nut at d D 9 mm D 0:009 m, the period 9
is (note that we keep more than four significant figures since the difference in period is very small)
ˇ
2 ˇˇ
9 D
D 1:85731339 s;
!n ˇd D0:009
and with the nut 4 mm closer at d D 5 mm D 0:005 m, the period 5 is
ˇ
2 ˇˇ
5 D
D 1:85725259 s;
!n ˇd D0:005
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
1440 min
D 1:00003274 D
5
tnew
)
tnew D 1439:953 min:
so that the day would be shorter by
1440
1439:953 D 0:047 min
)
s
0:047 min 60 min
D 2:82 s
August 10, 2009
1240
Solutions Manual
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 .IO /d !p2 C 12 .IO /n !p2 D
1
2
Œ.IO /d C IO /n  !p2 ;
where .IO /d is the mass moment of inertia of the pendulum disk with
respect to point O, .IO /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 the
pendulum. The moments of inertia are given by
.IO /p D 21 mp r 2 C mp L2
and .IO /n D mn .L C r C d /2
D mp . 12 r 2 C L2 /:
The potential energy of the system can be written as
V D
D
mp gL cos mn g.L C r C d / cos g mp L C mn .L C r C d / cos 2
;
g mp L C mn .L C r C d / 1
2
where we have used the approximation cos 1
2 =2. Applying the energy method, we find
d
.T C V / D mp 12 r 2 C L2 C mn .L C r C d /2 P R C g mp L C mn .L C r C d / P D 0
dt
)
mp 21 r 2 C L2 C mn .L C r C d /2 R C g mp L C mn .L C r C d / D 0;
and therefore the natural frequency is
!n2
g mp L C mn .L C r C d /
:
D
mp 12 r 2 C L2 C 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
1440
D
actual
1441
)
correct .d D 0 m/
1440
D
;
actual .d D 0:02 m/
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
s
g Œmp LCmn .LCrC0:02/
2
.!n /correct
2
.!n /actual
D
1
mp 2 r 2 CL2 Cmn .LCrC0:02/2
s
g Œmp LCmn .LCr/
1
mp 2 r 2 CL2 Cmn .LCr/2
D
1440
;
1441
(2)
August 10, 2009
1241
Dynamics 1e
where we have substituted d D 0 m into .!n /correct and d D 0:02 m into .!n /actual . When we substitute
g 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 is
simplified, 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.
August 10, 2009
1242
Solutions Manual
Problems 9.21 and 9.22
The uniform cylinder rolls without slipping on a flat surface. Let k1 D
k2 D k and r D R=2. Assume that the horizontal motion of G is
small.
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 9.21
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.
Problem 9.22
Solution to 9.21
Referring to the FBD on the right, we can sum moments about point O
to obtain
X
MO W 2RFA C .R r/FB D IO ˛G ;
(1)
where FA is the force in the spring of constant k1 , FB is the force in the
spring of constant k2 , ˛G is the angular acceleration of the cylinder, and
where
IO D IG C mR2 D 12 mR2 C mR2 D 32 mR2 :
For the spring forces, we can write
FA D k1 xA
and
FB D k2 xB ;
where xA and xB 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
vEA D vEO C !G rEA=O D !G kO 2R |O D 2R!G {O;
vEB D vEO C !G rEB=G D !G kO .R r/ |O D .R r/!G {O D
1
2 R!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
2RG
and xB D
1
2 RG ;
where G is the rotation of the cylinder. Since xG D RG , we can write the above two expressions for the
motion of points A and B as
xA D 2xG and xB D 12 xG ;
and so the spring force equations become (also using k1 D k2 D k)
FA D 2kxG
and FB D 12 kxG :
Substituting everything into Eq. (1) and noting that ˛G D
4kRxG C
1
1
2 R 2 kxG
D
2
3
2 mR
xR G
R
xR G =R, we obtain
)
xR G C
17k
xG D 0.
6m
August 10, 2009
1243
Dynamics 1e
Solution to 9.22
Referring to the FBD on the right, we that only FA and FB do work as the
wheel undergoes small horizontal motion, where FA is the force in the
spring of constant k1 , FB 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
2
2
T D 12 mvG
C 21 IG !G
;
here !G is the angular speed of the cylinder as it rolls without slip at
point O. Since IG D 12 mR2 , vG D xP G , and !G D xP G =R, the kinetic
energy can be written as
T D
2
1
PG
2 mx
C
2
1 1
2 2 mR
xP G 2
2
:
D 34 mxP G
R
For the potential energy of the springs, we can write
V D 12 k1 xA2 C 12 k2 xB2 D 12 kxA2 C 21 kxB2 ;
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 and
B as
vEA D vEO C !G rEA=O D !G kO 2R |O D 2R!G {O;
vEB D vEO C !G rEB=G D !G kO .R r/ |O D .R r/!G {O D
1
2 R!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
2RG
1
2 RG ;
and xB D
where G is the rotation of the cylinder. Since xG D RG , we can write the above two expressions for the
motion of points A and B as
xA D 2xG and xB D 12 xG ;
and so the potential energy becomes
V D 12 k.2xG /2 C 12 k
Substituting T and V into
d
.T
dt
2
1
2 xG
D
2
17
8 kxG :
C V / D 0, we obtain
3
P G xR G
2 mx
C
17
PG
4 kxG x
D0
)
xR G C
17k
xG D 0.
6m
August 10, 2009
1244
Solutions Manual
Problems 9.23 and 9.24
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.
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 9.23
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 .
Problem 9.24
Solution to 9.23
Referring to the FBD at the right and summing moments about point O, we
obtain
X
MO W
.mg sin /R C 2RFs D IG ˛G C mRaGx ;
(1)
where Fs 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
Fs 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 xR G =R. Substituting this kinematics equation and the force law shown above into Eq. (1)
and using IG D 12 mR2 , we obtain
xR G
2
1
mgR sin C 2kR.ıst C 2xG / D 2 mR
mRxR G :
R
Using the fact that mgR sin D 2kRıst and then simplifying this equation, we obtain the equation of motion
as
8k
xR G C
(2)
xG D 0:
3m
From Eq. (2), we can see that the natural frequency and period are given by
r
!n D
8k
3m
)
2
D
D 2
!n
r
3m
D 2:22 s,
8k
where we have substituted in m D 10 kg and k D 30 N=m.
August 10, 2009
1245
Dynamics 1e
Solution to 9.24
Referring to the FBD at the right we see that Fs 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
2
2
2
2
T D 12 IO !G
D 12 21 mR2 C mR2 !G
D 34 mR2 !G
D 43 mRxP G
;
where !G is the angular velocity of the cylinder, we have used the parallel axis theorem to find IO , and
we have used the kinematic relation that xP G D R!G to write !G since the cylinder rolls without slip at
point 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
.T C V / D 32 mxP G xR G C k.ıst C 2xG /.2xP G /
dt
mg xP G sin D 0;
Canceling xP G , using the fact that mg sin D 2kıst , and then simplifying this equation, we obtain the equation
of motion as
8k
xG D 0:
xR G C
(3)
3m
From Eq. (3), we can see that the natural frequency and period are given by
r
!n D
8k
3m
)
2
D
D 2
!n
r
3m
D 2:22 s,
8k
where we have substituted in m D 10 kg and k D 30 N=m.
August 10, 2009
1246
Solutions Manual
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 on
the 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 FA is to the right and
since drum B rotates counterclockwise, the friction force
it exerts on the bar FB is to the left. With this in mind, the
balance laws for the bar shown can be written as
X
Fx W
FA FB D maGx ;
X
Fy W
NA C NB mg D maGy ;
X
MG W NB .h=2 xG / NA .h=2 C xG / D IG ˛bar :
(1)
(2)
(3)
Since the counter-rotating drums must be slipping relative to the bar, the force laws are given by
FA D k NA
and
FB D k NB :
Since the bar does not rotate or move vertically, the kinematics equations are given by
aGx D xR G ;
aGy D 0;
and
˛bar D 0:
Substituting the force laws and kinematics into Eq. (1), we obtain
k NA
k NB D mxR G :
(4)
Substituting the kinematics into Eqs. (2) and (3) and then solving the resulting two equations for NA and NB
gives
1 xG
1 xG
C
NA D mg
and NB D mg
:
2
h
2
h
Substituting the two normal forces into Eq. (4), we obtain the equation of motion as
1 xG
1 xG
2k g
k mg
k mg
C
D mxR G ) xR G C
xG D 0:
2
h
2
h
h
Therefore, the natural frequency of oscillation is
r
!n D
2k g
:
h
August 10, 2009
1247
Dynamics 1e
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
2
2
T D 21 mvG
C 21 IG !G
;
where vG is the speed of the mass center of the cylinder, !G is the
angular speed of the cylinder, and IG 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
1
2 m.R
2 P2
r/ C
1
2
2
1
2 mr
R
r
2
r
P 2 D 34 m.R
r/2 P 2 ;
(1)
where we have used IG D 12 mr 2 for a uniform cylinder. In addition, using the datum line shown in the figure
above, 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
.T C V / D 23 m.R
dt
r/2 P R C mg.R
r/P sin D 0
)
R C
2g
sin D 0:
3.R r/
For small values of , we can approximate sin by and we obtain
2g
R C
D0
3.R r/
s
)
!n D
2g
3.R r/
s
and
D 2
3.R r/
.
2g
August 10, 2009
1248
Solutions Manual
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
2
2
T D 21 mvG
C 21 IG !G
;
where vG is the speed of the mass center of the sphere, !G is the
angular speed of the sphere, and IG 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
1
2 m.R
2 P2
r/ C
1
2
2
2
5 mr
R
r
r
2
P 2 D
r/2 P 2 ;
7
10 m.R
(1)
where we have used IG D 25 mr 2 for a uniform cylinder. In addition, using the datum line shown in the figure
above, 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
.T C V / D 57 m.R
dt
r/2 P R C mg.R
r/P sin D 0
)
R C
5g
sin D 0:
7.R r/
For small values of , we can approximate sin by and we obtain
5g
R C
D0
7.R r/
s
)
!n D
5g
7.R r/
s
and
D 2
7.R r/
.
5g
August 10, 2009
1249
Dynamics 1e
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 crosssectional 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 myP 2 D 21 ALyP 2 ;
where we have computed the mass of the fluid in the tube as m D AL, yP
is 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 VB C VD C VE C VF
L C
1 L C
D Ag
y yC
2
2
2
„
ƒ‚
…„
ƒ‚
length of B
Ag
L
C
ƒ‚
CVD
y
…
dist. from datum to center of B
2
„
length of E
1 L C
y yC
2
2
…„
ƒ‚
CVF ;
y
…
dist. from datum to center of E
where C is the constant length of the fluid in the curved portion of the tube and VF is always equal to zero.
Simplifying this expression for V , we obtain
V D 14 Ag .C L/2 4y 2 :
Applying the energy method to this system, we obtain
d
.T C V / D 0
dt
)
ALyP yR C 2Agy yP D 0
)
2g
yR C
yD0
L
r
)
!n D
2g
;
L
where we have used the fact that VD , C , and L are all constant. Therefore, the period of oscillation of the
fluid is
s
2
L
D
D 2
:
!n
2g
August 10, 2009
1250
Solutions Manual
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 the
semicylinder is
4R
4R
V D mg
cos mg
1 2 =2 ; (1)
3
3
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 IO P 2 ;
(2)
where P is the angular velocity of the semicylinder. To find IO we first note that
IG D
1
2
16
mR2
9 2
and then IO
D IG C m` IG C m R
2
4R
3
2
;
where ` is the distance from O to G as seen in the figure above and we have approximated ` by R 4R
3 for
small . Simplifying IO , we obtain
9 16
IO D
mR2 :
(3)
6
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:
9 16
4R
d
8g
2PR
.T C V / D 0 )
mR C mg
P D 0 ) R C
D 0:
dt
6
3
.9 16/R
Therefore, the natural frequency and period of oscillation are
s
!n D
8g
.9 16/R
s
)
D
.9
16/R
.
2g
August 10, 2009
1251
Dynamics 1e
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 need
to 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
2R
m
2R
md D 0 C R
D R
2R C R
)
dD
2R
:
2C
Now that we have d , using the datum indicated, the potential energy of the thin shell semicylinder is
2R
2R
cos mg
1 2 =2 ;
V D mg
2C
2C
(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 21 IO P 2 ;
(2)
where P is the angular velocity of the thin shell semicylinder. To find IO we must first find IG and then
apply the parallel axis theorem to obtain as IO D IG C m`2 , where ` is the distance from G to O. Since
IG for the body we are considering does not appear in the inside back cover of the books, we will first find
IA by treating the thin shell semicylinder as a composite body made up of the straight segment BC and
semicircular segment BOC . Therefore
IA D .IA /BC C .IA /BOC D
2
1
12 mBC .2R/
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
2m
m
mBC D .2R/ D
and mBOC D .R/ D
;
2C
2C
August 10, 2009
1252
Solutions Manual
and therefore IA becomes
1
m
2 C 3
2m
IA D
.2R/2 C
R2 D
mR2 :
12 2 C 2C
3.2 C /
We can now find IG using the parallel axis theorem as
IA D IG C md
2
)
2R 2
3 2 C 8 8
m
mR2 :
D
2C
3.2 C /2
2 C 3
mR2
md D
3.2 C /
2
IG D IA
Finally, we can find IO by again using the parallel axis theorem as
2
IO D IG C m` IG C m.R
3 2 C 8 8
2
d/ D
mR C m R
3.2 C /2
2
2R
2C
2
D
2.3 2/
mR2 ;
3.2 C /
where we have approximated ` by R d for small . Substituting IO 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:
2R
3g
d
2.3 2/
2PR
mR C mg
P D 0 ) R C
.T C V / D 0 )
D 0:
dt
3.2 C /
2C
.3 2/R
Therefore, the natural frequency and period of oscillation are
s
!n D
s
3g
.3
2/R
)
D 2
.3
2/R
3g
.
August 10, 2009
1253
Dynamics 1e
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
1
MF D
:
1 .!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 equation
and solve for the forcing frequency !0 . Doing so, we obtain
5D
1
1
.!0 =100/2
)
!0 D 89:4 rad=s.
August 10, 2009
1254
Solutions Manual
Problem 9.32
Suppose that equation of motion of a forced harmonic oscillator is
given by xR C !n2 x D .F0 =m/ cos !0 t . Obtain the expression for the
response of the oscillator, and compare it to the response presented in
Eq. (9.42) (which is for a forced harmonic oscillator with the equation
of 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 !n t C B cos !n t;
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 !0 t , where D is a constant to be determined. Substituting this assumed
solution into the equation of motion, we obtain
D!02 cos !0 t C !n2 D cos !0 t D
F0
cos !0 t
m
)
F0
D !n2 !02 D
)
m
)
DD
F0
m
1
1
!n2
.!0 =!n /2
D
F0
m
1
m
k
.!0 =!n /2
DD
D
1
F0 =m
!n2 !02
F0 =k
.!0 =!n /2
Therefore, the response is given by
x.t / D A sin !n t C B cos !n t C
1
F0 =k
cos !0 t:
.!0 =!n /2
Notice that we can find the response due to cos !0 t forcing by replacing the “sin” with a “cos” in the response
due to sin !0 t forcing.
August 10, 2009
1255
Dynamics 1e
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 Fs 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 find
X
Fy W
mg Fs D myRG ;
(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
Fs 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 the
balance law in Eq. (1) we find
mg
keq .ym
ıst / D myRG
)
myRG C keq ym D 0; (2)
where we have used the fact that mg D keq ıst . Notice that Eq. (2) contains two dependent variables ym and
yG 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 mu yu C .m
mu /ym
)
myG D mu .ym C h C " sin / C .m
mu /ym :
Canceling mu ym and noting that sin D sin !r t , this equation becomes
myG D mu .h C " sin !r t / C mym
)
myRG D
mu "!r2 sin !r t C myRm ;
where we have used the fact that h is constant in taking the second derivative. Substituting myRG from this
expression into Eq. (2), we obtain the equation of motion as
.mm C mp /yRm C keq ym D mu "!r2 sin !r t
)
yRm C
keq
mu "!r2
ym D
sin !r t ,
mm C mp
mm C mp
where we have made the substitution m D mm C mp .
August 10, 2009
1256
Solutions Manual
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
#
"
mu "!r2 =keq
vm0 !r mu "!r2 =keq
sin
!
t
C
sin !r t;
ym D
n
!n
!n 1 .!r =!n /2
1 .!r =!n /2
(1)
where ym in defined in the figure on the right, vm0 D yPm .0/, and
the natural frequency was found to be
s
keq
!n D
:
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 !n t C Q sin !r t;
where
mu "!r2 =keq
vm0 !r mu "!r2 =keq
and
Q
D
:
!n
!n 1 .!r =!n /2
1 .!r =!n /2
!n , which means that ym can be written as
P D
We now let ! D !r
ym D P sin !n t C Q sin.! C !n /t D P sin !n t C Q.sin !t cos !n t C cos !t sin !n t /
D Q sin !t cos !n t C .P C Q cos !t / sin !n t
„ ƒ‚ …
„
ƒ‚
…
A
D
p
A2
C
B
B2
sin.!n t C /;
where the angle can be determined, but is not needed to find the amplitude of vibration. Thus we see that
the amplitude is
q
q
p
jym j D A2 C B 2 D .Q sin !t /2 C .P C Q cos !t /2 D 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
q
mu D 0:01 kg
)
jym j D 2:1910 6 cos.4:37t / C 1:8910 5 m;
q
mu D 0:1 kg
)
jym j D 1:1010 5 cos.4:37t / C 1:1110 5 m;
q
mu D 1 kg
)
jym j D 9:8010 4 cos.4:37t / C 1:0110 3 m;
August 10, 2009
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 red
curve corresponds to mu D 1000 g. Note that these curves represent
the 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 third
plot 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 jym j that we found forms an envelope over the response
ym . 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.
August 10, 2009
1258
Solutions Manual
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 !s t. The system lies
in 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
E O D IO ˛Ebar C rEG=O mE
M
aO :
Doing so, we obtain
X
MO W
mg
L
L
sin kO D IO R kO C .cos {O C sin |O/ myRO |O;
2
2
where yRO |O is the acceleration of point O. Carrying out the cross product,
noting that IO D 13 mL2 , and that yRO D Y !s2 sin !s t , the equation of motion
becomes
1
L
L
mL2 R C mg sin m Y !s2 sin !s t cos D 0;
3
2
2
which, for small can be written as
1
L
L
mL2 R C mg D m Y !s2 sin !s t
3
2
2
3g
3Y 2
R C
D
! sin !s t .
2L
2L s
)
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 !s t;
where D is a constant to be determined. Substituting this particular solution into the equation of motion, we
obtain
D!s2 sin !s t C
3g
3Y 2
D sin !s t D
! sin !s t
2L
2L s
)
D !n2
3Y 2
!s2 D
!
2L s
3Y =.2L/.!s =!n /2
;
) DD
1 .!s =!n /2
where we have used !n2 D 3g=.2L/ from the equation of motion. Therefore, the amplitude of steady-state
vibration is
amp D
3Y =.2L/.!s =!n /2
:
1 .!s =!n /2
August 10, 2009
1259
Dynamics 1e
Problem 9.36
Consider a sign mounted on a circular hollow steel pole of length
L D 5 m, outer diameter do D 5 cm, and inner diameter di D 4 cm.
Aerodynamic forces due to wind provide a harmonic torsional excitation
with frequency f0 D 3 Hz and amplitude M0 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 torsional
stiffness of the pole can be estimated as k t D Gst do4 di4 =.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, Te is the exciting torque on the
sign due to the wind and Tp is the restoring torque due to the steel pole of
length L. Summing moments about the ´ axis, we obtain
X
M´ W Te Tp 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,
"
#
Gst do4 di4
;
Te D M0 sin.2f0 t / and Tp D k t D
32L
where k t is the torsional stiffness of the pole and we have substituted in the given relationship for k t .
Substituting the force laws in to Eq. (1), we obtain
"
"
#
#
Gst do4 di4
Gst do4 di4
D I´ R ) I´ R C
D M0 sin.2f0 t /;
M0 sin.2f0 t /
32L
32L
so we see that using do D 0:05 m, di D 0:04 m, L D 5 m, Gst D 79109 Pa, and I´ D 0:1 kgm2 , we obtain
s
Gst do4 di4
!n D
D 239:2 rad=s;
!0 D 2f0 D 18:85 rad=s
32LI´
Gst do4 di4
keff D
D 5724 Nm;
.F0 /eff D M0 D 10 Nm;
32L
where !0 is the forcing frequency, keff is the effective spring constant, and .F0 /eff is the effective forcing
amplitude. Applying Eq. (9.43) from the text, we find the following amplitude
amp D
.F0 /eff =keff
1 .!0 =!n /2
)
amp D 0:00176 rad=s.
August 10, 2009
1260
Solutions Manual
Problems 9.37 and 9.38
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.
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 9.37
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.
Problem 9.38
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 the
unbalanced mass mu gives
X
Fx W
Fx D mu aux ;
(1)
X
Fy W
Fy mu g D mu auy ;
(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
X
MA W me gd Fx R sin ˇ
Fy .d C R cos ˇ/
M t D IA ˛wing :
(3)
R Note that we
where M t is the lumped torsional stiffness of the wing at point A, IA D me d 2 , and ˛wing D .
have treated the engine as a point mass. The force law for the torsional stiffness is given by
M t D k t . C st / D
3EI
. C st /;
d
where k t is the lumped torsional spring constant, st is the rotation of the wing in its static equilibrium
position, and k t 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
August 10, 2009
1261
Dynamics 1e
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
aEu D aEe C ˛Eprop rEu=e
2
!prop
rEu=e D
d R |O
!p2 . R cos ˇ {O C R sin ˇ |O/;
so that
aux D R!p2 cos ˇ
d R
and auy D
R!p2 sin ˇ:
Substituting the kinematics equations into Eqs. (1) and (2) gives
Fx D
mu R!p2 cos ˇ
and Fy D mu d R C mu R!p2 sin ˇ
mu g:
Substituting Fx and Fy into Eq. (3), we obtain the following equation of motion
me gd C mu R2 !p2 sin ˇ cos ˇ
mu .d C R cos ˇ/ d R C R!p2 sin ˇ
g
3EI
. C st / D me d 2 R :
d
Canceling terms and rearranging, we obtain
.mu C me /d 2 R C mu Rd cos ˇ R
.mu C me /dg C mu dR!p2 sin ˇ
mu gR cos ˇ C
3EI
. C st / D 0:
d
Since .3EI =d /st D me gd C mu g.d C R cos ˇ/, this becomes
3EI
D
.mu C me /d 2 C mu Rd cos !p t R C
d
ƒ‚
…
„
mu dR!p2 sin !p t;
time-dependent inertia
where we have substituted !p t for ˇ. Notice that the inertia term (i.e., the coefficient of R ) is time-dependent
since it contains cos !p t, 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
ˇ
ˇ
ˇ.mu C me /d 2 ˇ D 1058 slugft2 and jmu Rd cos !p t j 0:2584 slugft2 ;
and therefore, the inertia term is dominated by the constant part, which means that we can approximate the
equation of motion as
3EI
.mu C me /d 2 R C
D mu dR!p2 sin !p t;
(4)
d
and so using
EI D 1:131011 lbin:2 D 7:847108 lbft2 ;
1 lb
1
mu D 3 oz
D 5:82310
16 oz
32:2 ft=s2
1
me D 450 lb
D 13:98 slug; and
32:2 ft=s2
d D 8:7 ft;
3
slug;
we find that the resonance frequency of the wing is
s
!n D
3EI
D 506 rad=s
.mu C me /d 3
or
f D
!n
D 80:5 Hz.
2
August 10, 2009
1262
Solutions Manual
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 !p t and then substitute it into the equation of motion. Doing so, we obtain
D.mu C me /d 2 !p2 sin !p t C
3EI
D sin !p t D
d
mu dR!p2 sin !p t;
which, upon canceling sin !p t and solving for D gives
DD
D
mu dR!p2
3EI
d
.mu C me /d 2 !p2
D
mu d 2 R!p2 =.3EI /
1
.mu C me /d 3 !p2 =.3EI /
D
mu d 2 R!p2 =.3EI /
1
.!p =!n /2
.!p =!n /2
mu R
:
.mu C me /d 1 .!p =!n /2
Therefore,
jp j D
mu R
.!p =!n /2
.mu C me /d 1 .!p =!n /2
)
MF D
jp j.mu C me /d
.!p =!n /2
D 0:207,
D
mu R
1 .!p =!n /2
where we have used !n D 505:7 rad=s and !p D 2000 rpm D 209:4 rad=s.
August 10, 2009
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Dynamics 1e
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 gives
X
Fx W
Fx D mu aux ;
(5)
X
Fy W
Fy mu g D mu auy ;
(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
X
MA W mw gh C me gd Fx R sin ˇ
Fy .d C R cos ˇ/
M t D IA ˛wing :
(7)
where M t is the lumped torsional stiffness of the wing at point A and ˛wing D R . Note that we have treated
the engine as a point mass. The mass moment of inertia with respect to point A is
IA D me d 2 C
2
1
12 mw .2h/
C mw h2 D me d 2 C 34 mw h2 ;
where we have assumed that the wing is a uniform beam of length 2h. The force law for the torsional stiffness
is given by
3EI
M t D k t . C st / D
. C st /;
d
where k t is the lumped torsional spring constant, st is the rotation of the wing in its static equilibrium
position, and k t 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
aEu D aEe C ˛Eprop rEu=e
2
!prop
rEu=e D
d R |O
!p2 . R cos ˇ {O C R sin ˇ |O/;
so that
aux D R!p2 cos ˇ
and auy D
d R
R!p2 sin ˇ:
Substituting the kinematics equations into Eqs. (5) and (6) gives
Fx D
mu R!p2 cos ˇ
and Fy D mu d R C mu R!p2 sin ˇ
mu g:
Substituting Fx and Fy into Eq. (7), we obtain the following equation of motion
mw gh C me gd C mu R2 !p2 sin ˇ cos ˇ
mu .d C R cos ˇ/ d R C R!p2 sin ˇ
3EI
. C st /
d
D me d 2 C 43 mw h2 R :
g
Canceling terms and rearranging, we obtain
.mu C me /d 2 C 43 mw h2 C mu Rd cos ˇ R
.mu C me /dg
mw gh C mu dR!p2 sin ˇ
August 10, 2009
1264
Solutions Manual
mu gR cos ˇ C
3EI
. C st / D 0;
d
or, since .3EI =d /st D mw gh C me gd C mu g.d C R cos ˇ/, this becomes
3EI
.mu C me /d 2 C 34 mw h2 C mu Rd cos !p t R C
D
d
ƒ‚
…
„
mu dR!p2 sin !p t;
time-dependent inertia
where we have substituted !p t for ˇ. Notice that the inertia term (i.e., the coefficient of R ) is time-dependent
since it contains cos !p t, 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
ˇ
ˇ
ˇ.mu C me /d 2 C 4 mw h2 ˇ D 2780 slugft2 and jmu Rd cos !p tj 0:2584 slugft2 ;
3
and therefore, the inertia term is dominated by the constant part, which means that we can approximate the
equation of motion as
3EI
D
.mu C me /d 2 C 43 mw h2 R C
d
mu dR!p2 sin !p t:
(8)
Therefore, the resonance frequency of the wing is
s
!n D
3EI
.mu C me
/d 3
C
4
2
3 mw h d
D 312 rad=s
or
f D
!n
D 49:7 Hz,
2
where we have used
EI D 1:131011 lbin:2 D 7:847108 lbft2 ;
1
1 lb
D 5:82310
mu D 3 oz
16 oz
32:2 ft=s2
1
D 13:98 slug;
me D 450 lb
32:2 ft=s2
1
mw D 350 lb
D 10:87 slug
32:2 ft=s2
d D 8:7 ft; and
3
slug;
h 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 D D sin !p t and then substitute it into the equation of motion. Doing so, we obtain
3EI
D .mu C me /d 2 C 34 mw h2 !p2 sin !p t C
D sin !p t D
d
mu dR!p2 sin !p t;
which, upon canceling sin !p t and solving for D gives
DD
D
3EI
d
mu dR!p2
D
1
.mu C me /d 2 C 34 mw h2 !p2
mu d 2 R!p2 =.3EI /
1
.!p =!n /2
D
mu d 2 R!p2 =.3EI /
.mu C me /d 3 C 34 mw h2 d !p2 =.3EI /
mu R
.!p =!n /2
:
.mu C me /d 1 .!p =!n /2
August 10, 2009
1265
Dynamics 1e
Therefore,
jp j D
mu R
.!p =!n /2
.mu C me /d 1 .!p =!n /2
)
MF D
jp j.mu C me /d
.!p =!n /2
D
D 0:820,
mu R
1 .!p =!n /2
where we have used !n D 312:0 rad=s and !p D 2000 rpm D 209:4 rad=s.
August 10, 2009
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Solutions Manual
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 k t 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 excitation
F .t/ D F sin !0 t in the vertical direction. Derive the equation of motion for the system assuming that is small.
Solution
Referring to the FBD at the right, we can sum moments about
point O to obtain the equation of motion of the system. Doing
so, we obtain
X
MO W
M t C .mb g Fs / L
2
C Œmm g
F .t /L D IO ˛beam :
(1)
The force laws for the torsional spring and the spring of stiffness k are given by
M t D k t and Fs D k L
2 C ıst ;
where is measured from the static equilibrium position of the beam and ıst is the deflection of the midpoint
R
of the beam when it is in static equilibrium. Substituting these force laws into Eq. (1), noting that ˛beam D ,
1
2
2
that F .t/ D F sin !0 t, and that IO D 3 mb L C mm L , the equation of motion becomes
k t C mb g
k
L
2
C ıst
L
2
C .mm g
F sin !0 t /L D
2
1
3 mL
C mm L2 R :
Now, if we realize that mb gL=2 C mm gL D kıst L=2, we are left with the following equation of motion
1
3 mb
C mm L2 R C k t C 14 kL2 D FL sin !0 t:
August 10, 2009
Dynamics 1e
1267
Problem 9.40
Revisit Example 9.4 and discuss whether it is possible to obtain the equation of motion of the system via
the energy method.
Solution
No, it is not possible since the system is not conservative. The system is not conservative since the amount of
energy the motor adds or takes away from the system is not known and the efficiency of the motor is not
known.
August 10, 2009
1268
Solutions Manual
Problem 9.41
A fatigue-testing machine for electronic components consists of a
platform with an unbalanced motor. Assume that the rotor in the
motor spins at !0 D 3000 rpm, the mass of the platform is mp D
20 kg, the mass of the motor is mm D 15 kg, the eccentric mass is
mu D 0:5 kg, and the equivalent stiffness of the platform suspension
is k D 5106 N=m. 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 2 mm.
Solution
This system is identical to that found in Example 9.4. Therefore, using k D 5106 N=m, mm D 15 kg, and
mp D 20 kg, we know that the natural frequency of the system is given by
!n2 D
k
D 1:429105 rad2 =s2 ;
mm C mp
The amplitude of the particular or steady-state solution is then given by
jymp j D
mu "!r2 =k
D .0:03191"/ m;
1 .!r =!n /2
in which we used !r D 3000 rpm D 314:2 rad=s is the angular velocity of the rotor and we have substituted
in mu D 0:5 kg, k D 5106 N=m, and !n2 D 1:429105 rad2 =s2 . Therefore,
for jymp j D 0:0001 m
) " D 0:00313 m D 3:13 mm;
for jymp j D 0:002 m
) " D 0:0627 m D 62:7 mm;
and therefore the range of " is
3:13 mm < " < 62:7 mm:
August 10, 2009
1269
Dynamics 1e
Problem 9.42
At time t D 0, a forced harmonic oscillator occupies position x.0/ D
0:1 m and has a velocity x.0/
P
D 0. The mass of the oscillator is m D
10 kg, and the stiffness of the spring is k D 1000 N=m. Calculate the
motion of the system if the forcing function is F .t / D F0 sin !0 t,
with F0 D 10 N and !0 D 200 rad=s.
Solution
As we have seen, the equation of motion for this system is given by
xR C !n2 x D
F0
sin !0 t;
m
where !n is the natural frequency of the system, which is given by !n D
the response or motion of this system to be
x.t / D A sin !n t C B cos !n t C
1
p
k=m D 10:00 rad=s. We found
F0 =k
sin !0 t;
.!0 =!n /2
(1)
where A and B are constants that depend on the given initial conditions. Evaluating the response at t D 0,
we obtain
x.0/ D B D 0:1 ) B D 0:1 m;
and
F0 =k
!0 cos !0 t
1 .!0 =!n /2
F0 =k
x.0/
P
D A!n C
!0 ) A D
1 .!0 =!n /2
xP D A!n cos !n t
)
B!n sin !n t C
1
F0 =k
!0
D 0:0005013 m;
2
.!0 =!n / !n
where we have used F0 D 10 N, k D 1000 N=m, !0 D 200 rad=s, and !n D 10:00 rad=s. Substituting A
and B into Eq. (1), we find that
x.t / D 0:000501 sin 10t C 0:100 cos 10t
2:5110
5
sin 200t m;
where we have used F0 D 10 N, k D 1000 N=m, !0 D 200 rad=s, and !n D 10:00 rad=s to obtain the
coefficient of the sin 200t term.
August 10, 2009
1270
Solutions Manual
Problem 9.43
The forced harmonic oscillator shown has a mass m D 10 kg. In
addition the harmonic excitation is such that F0 D 150 N and !0 D
200 rad=s. If all sources of friction can be neglected, determine the
spring constant k such that the magnification factor MF D 5.
Solution
Treating m as a point mass, we obtain the FBD shown on the right.
Summing forces in the x direction gives
X
Fx W F .t / Fs D max D mx;
R
(1)
where Fs is the force on m due to the springs and x is measured
from the location of m when the springs are undeformed. The
force law for the springs is
Fs D keq x;
where keq is the single spring constant that is equivalent to the
four springs shown. To find keq we refer to the bottom figure shown above, where we have assigned to each
spring a different constant to illustrate how to find the equivalent spring constant. We note that the total
deformation of the springs is given by
ıtot D ı1 C ı23 C ı4 ;
where ı1 is the deformation of spring k1 , ı23 is the deformation of springs k2 and k3 , and ı4 is the deformation
of spring k4 . Since the force in each of the three segments is equal to Fs , the displacements can be written in
terms of Fs as
Fs
Fs
Fs
Fs
1
1
1
1
D
C
C
)
D
C
C ;
(2)
keq
k1
k2 C k3
k4
keq
k1
k2 C k3
k4
where the expression for ı23 was found using
Fs
Fs D F2 C F3 D k2 ı23 C k3 ı23 ) ı23 D
;
k2 C k3
and where F2 is the force in k2 and F3 is the force in k3 . Solving Eq. (2) for keq , we obtain
k1 k4 .k2 C k3 /
D 25 k;
k1 k2 C k1 k3 C k1 k4 C k2 k4 C k3 k4
where the last result was obtained by letting k1 D k2 D k3 D k4 D k. Substituting Fs and F .t / D
F0 sin !0 t into Eq. (1), we obtain the equation of motion as
keq D
mxR C 25 kx D F0 sin !0 t:
The MF for this system was found to be (see Eq. (9.44))
MF D
where !n2 D
for k gives
2k
5m
1
1
;
.!0 =!n /2
from the equation of motion. Substituting !n2 into the equation above for the MF and solving
5m!02 MF
)
k D 1:25106 N=m,
2.MF 1/
where we have used MF D 5, m D 10 kg, and !0 D 200 rad=s to obtain the numerical result.
kD
August 10, 2009
1271
Dynamics 1e
Problem 9.44
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/ D F0 sin !0 t and w.t / D
0 (i.e., the support is stationary). In addition, find the response of the
system for y.0/ D 0 and yP D 0.
Solution
Referring to the FBD on the right, we can sum forces in the y direction
to obtain
X
Fy W F .t / 2Fs sin ˇ D may D my;
R
(1)
where Fs is the force in each of the elastic cords. The force law for each
cord is given by
q
Fs D k
y 2 C L2 L0 :
Kinematically, we can write ˇ in terms of y if we note that
y
:
sin ˇ D p
y 2 C L2
Substituting the kinematics and the force law into the balance law in Eq. (1), we obtain
myR C 2k
q
y 2 C L2
L0 p
y
y 2 C L2
D F0 sin !0 t
)
myR C 2k 1
L0
p
y 2 C L2
!
y D F0 sin !0 t:
Looking at the second term in parentheses, we note that using a Taylor series expansion about y D 0 (this is
sometimes called a binomial expansion) we can write it as
"
#
4
1 1
L0
1
L0
L0
1 y 2
y
L0
2 2 C1
D
D
1
C
::: ;
p
p
2
2
2
L
L
2
L
2Š
L
L
y CL
1 C .y=L/
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
myR C 2k 1
L0
y D F0 sin !0 t:
L
We have already seen that the response is given by Eq. (9.42), which is
y D A sin !n t C B cos !n t C
F0 =keq
sin !0 t;
1 .!0 =!n /2
(2)
August 10, 2009
1272
Solutions Manual
where A and B are constants determined by enforcing the initial conditions, keq is the equivalent spring
constant, and !n is the natural frequency, which are given by
s 2k
L0
L0
keq D 2k 1
and !n D
1
:
(3)
L
m
L
Evaluating Eq. (2) at y.0/ D 0, we obtain
y.0/ D 0 D B
)
B D 0:
Evaluating Eq. (2) at y.0/
P
D 0, we find
y.0/
P
D A!n C
F0 =keq
!0 D 0
1 .!0 =!n /2
)
AD
F0 =keq
!0
;
2
1 .!0 =!n / !n
so that the response of the system for the given initial conditions is
F0 =keq
!0
sin
!
t
C
sin
!
t
;
n
0
1 .!0 =!n /2 !n
s L0
2k
L0
D 2k 1
and !n D
1
:
L
m
L
y.t / D
where keq
August 10, 2009
1273
Dynamics 1e
Problem 9.45
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 / D 0 and
w.t/ D w0 sin !t . In addition, find the response of the system for
y.0/ D 0 and y.0/
P
D 0.
Solution
Referring to the FBD at the right, we can sum forces in the y direction
to obtain the following balance law
X
Fy W
2Fs sin ˇ D may ;
(1)
where Fs is the force in each of the elastic cords, y is measured from
the point of attachment of the elastic cords to the support, and ay is the
inertial acceleration of the ring in the y direction. The force law for
each cord is given by
q
Fs D k
y 2 C L2 L0 :
Kinematically, we can write ˇ in terms of y if we note that
y
:
sin ˇ D p
y 2 C L2
In addition, the acceleration of the ring can be found using
ay D
d2
Œw.t / C y.t / D
dt 2
! 2 w0 sin !t C y;
R
where we note that yR is not the inertial acceleration of the ring. Substituting the kinematics and the force law
into the balance law in Eq. (1), we obtain
myR C 2k
q
y 2 C L2
L0 p
y
y2
C
L2
D m! 2 w0 sin !t
)
myR C 2k 1
L0
!
y D m! 2 w0 sin !t:
p
2
2
y CL
Looking at the second term in parentheses, we note that using a Taylor series expansion about y D 0 (this is
sometimes called a binomial expansion) we can write it as
"
#
4
1 1
C
1
L0
L0
1
L0
1 y 2
y
L0
D
D
1
C 2 2
::: ;
p
p
L 1 C .y=L/2
L
2 L
2Š
L
L
y 2 C L2
August 10, 2009
1274
Solutions Manual
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
myR C 2k 1
L0
y D m! 2 w0 sin !t:
L
We have already seen that the general solution to this equation is given by Eq. (9.42), which is
y D A sin !n t C B cos !n t C
F0 =keq
sin !t;
1 .!=!n /2
(2)
where A and B are constants determined by enforcing the initial conditions, keq is the equivalent spring
constant, !n is the natural frequency, and F0 is the amplitude of the forcing, which are given by
s L0
L0
2k
keq D 2k 1
; !n D
1
; and F0 D m! 2 w0 :
(3)
L
m
L
Evaluating Eq. (2) at y.0/ D 0, we obtain
y.0/ D 0 D B
)
B D 0:
Evaluating Eq. (2) at y.0/
P
D 0, we find
y.0/
P
D A!n C
F0 =keq
!D0
1 .!=!n /2
)
AD
F0 =keq
!
;
1 .!=!n /2 !n
so that the response of the system, which is given by w.t / C y.t /, for the given initial conditions is
F0 =keq
!
w.t / C y.t / D w0 sin !t
sin
!
t
C
sin
!t
;
n
1 .!=!n /2 !n
or
F0 =keq
F0 =keq
!
w.t / C y.t / D
sin !t;
sin !n t C w0
1 .!=!n /2 !n
1 .!=!n /2
s L0
2k
L0
where keq D 2k 1
; !n D
1
; and F0 D m! 2 w0 :
L
m
L
August 10, 2009
1275
Dynamics 1e
Problem 9.46
Modeling the beam as a rigid uniform
thin bar, ignoring the inertia of the pulleys, assuming that the system is in
static equilibrium when the bar is horizontal, and assuming that the cord is
inextensible and does not go slack, determine the linearized equation of motion of the system in terms of x, which
is the position of A. Finally, determine
the amplitude of the steady-state vibration of block A.
Solution
An FBD of the block A, an FBD of the bar B, and the
positive coordinate directions are shown at the right.
Summing forces in the x direction on the block, we
obtain the following balance law
X
Fy W mA g C F .t / T D mA aAx
D mA x:
R
(1)
where x measures the vertical position of the block
A. Summing moments about point O for bar B gives
X
MO W T L
Fs L mB g L
2
2 D IO ˛B ; (2)
where T is the cord tension, ˛B is the angular acceleration of the bar, Fs is the force in the spring, and IO D 13 mL2 is the mass moment of inertia of the bar B
about point O. The force laws are
F .t / D F0 sin !0 t and Fs k ıB .ıB /st ;
where .ıB /st is the deflection of the spring when the system is in static equilibrium. For the kinematics
relations, we need to relate the angular acceleration of the bar (˛B ) to the acceleration of the mass A (x)
R and
we need to relate the deflection of point B (ıB ) to the position of mass A (x). Assuming small displacements
and noting that the vertical displacement of A is equal and opposite to that of G, we can write
ıB D LB
and
xD
L
2 B
)
ıB D 2x
and RB D ˛B D
2
R
L x:
Solving Eq. (1) for T and substituting the result into Eq. (2), we obtain
L
2
ŒmA g C F .t /
mA x
R
kL ŒıB
.ıB /st 
2R
1
mB g L
2 D 3 mL B :
Now substituting in the force laws and kinematics relations, this becomes
L
2
ŒmA g C F0 sin !0 t
mA x
R
kL Œ2x
.ıB /st 
2
1
mB g L
2 D 3 mL
2
R
Lx
August 10, 2009
1276
Solutions Manual
Canceling an L and noting that 12 g.mA
2
3 mB
mB / D k.ıB /st , we find the equation of motion to be
C 21 mA xR C 2kx D 12 F0 sin !0 t:
In Eq. (9.43) we found that the amplitude of steady-state vibration is
xamp D
.F0 /eq =keq
;
1 .!0 =!n /2
where .F0 /eq is the amplitude of the forcing, keq is the equivalent spring constant, and !n is the natural
frequency of vibration. Referring to the above equation of motion, these three quantities are given by
.F0 /eq D 12 F0 ;
keq D 2k;
and
!n2 D
2k
2
3 mB
C 12 mA
:
Substituting in these quantities, xamp becomes
xamp D
4k
F0
4
3 mB C
:
mA !02
August 10, 2009
1277
Dynamics 1e
Problem 9.47
For the system in Prob. 9.46 determine
the maximum forcing frequency !0 for
steady state motion such that the cord
does not go slack.
Solution
An FBD of the block A, an FBD of the bar B, and the
positive coordinate directions are shown at the right.
Summing forces in the x direction on the block, we
obtain the following balance law
X
Fy W mA g C F .t / T D mA aAx
D mA x:
R
(1)
where x measures the vertical position of the block
A and T is the tension in the cord. To determine the
maximum !0 such that the cord does not go slack, we
need to make sure that T 0 in Eq. (1). Summing
moments about point O for bar B gives
X
MO W T L
Fs L
2
mB g L
2 D IO ˛B ;
(2)
where ˛B is the angular acceleration of the bar, Fs is the force in the spring, and IO D 13 mL2 is the mass
moment of inertia of the bar B about point O. The force laws are
F .t / D F0 sin !0 t and Fs k ıB .ıB /st ;
where .ıB /st is the deflection of the spring when the system is in static equilibrium. For the kinematics
relations, we need to relate the angular acceleration of the bar (˛B ) to the acceleration of the mass A (x)
R and
we need to relate the deflection of point B (ıB ) to the position of mass A (x). Assuming small displacements
and noting that the vertical displacement of A is equal and opposite to that of G, we can write
ıB D LB
and
xD
L
2 B
)
ıB D 2x
and RB D ˛B D
2
R
L x:
Solving Eq. (1) for T and substituting the result into Eq. (2), we obtain
L
2
ŒmA g C F .t /
mA x
R
kL ŒıB
.ıB /st 
2R
1
mB g L
2 D 3 mL B :
Now substituting in the force laws and kinematics relations, this becomes
L
2
ŒmA g C F0 sin !0 t
mA x
R
kL Œ2x
.ıB /st 
2
1
mB g L
2 D 3 mL
2
R
Lx
August 10, 2009
1278
Solutions Manual
Canceling an L and noting that 12 g.mA
2
3 mB
mB / D k.ıB /st , we find the equation of motion to be
C 21 mA xR C 2kx D 12 F0 sin !0 t:
In Eq. (9.43) we found that the amplitude of steady-state vibration is
.F0 /eq =keq
;
1 .!0 =!n /2
xamp D
where .F0 /eq is the amplitude of the forcing, keq is the equivalent spring constant, and !n is the natural
frequency of vibration. Referring to the above equation of motion, these three quantities are given by
.F0 /eq D 12 F0 ;
keq D 2k;
!n2 D
and
2k
2
3 mB
C 12 mA
:
Substituting in these quantities, xamp becomes
xamp D
F0
4
3 mB C
4k
:
mA !02
and the steady-state vibration response is
x.t / D xamp sin !0 t D
F0
4
3 mB C
4k
sin !0 t:
mA !02
Taking two time derivatives of x, substituting the result into Eq. (1), and then solving for T , we obtain
mA F0 !02
T D mA g C F0 sin !0 t
Since we want T 0, we can write
"
mA g C F0
4
3 mB
4k
sin !0 t:
C mA !02
#
mA F0 !02
4k
4
3 mB
sin !0 t 0:
C mA !02
Since sin !0 t oscillates between 1 and 1, for this expression to be true, we must satisfy
mA g F0
mA F0 !02
4k
4
3 mB
C mA !02
or
!02 4k.F0
4
3 mB .F0
mA g/
mA g/
2
mA
g
.
August 10, 2009
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Dynamics 1e
Problem 9.48
In the design of a MacPherson strut suspension, what would you choose for
the damping ratio ? Explain your answer in terms of automotive ride and
comfort.
shock
absorber
and spring
steering
link
lower
control arm
car frame
Solution
You would want to choose a damping ratio as close to 1 as possible. The reason is that this provides the level
of damping that returns the system to equilibrium in the minimum time and without any vibration.
August 10, 2009
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Solutions Manual
Problem 9.49
For identical systems, one with damping and the other without, would you expect the period of damped
vibration to be greater, less than, or equal to the period of undamped vibration? Explain your answer.
Solution
Equation (9.68) tells us that frequency of damped vibration is always less than the frequency of undamped
vibration. Since the period of vibration is inversely proportional to the frequency, the period of vibration of a
damped system should always be greater than the period of an identical undamped system.
August 10, 2009
Dynamics 1e
1281
Problem 9.50
A vibration test is performed on a structure, in which both the magnification factor MF and the phase
angle are recorded as a function of excitation frequency !0 . After the test, it is discovered that, for some
unfortunate reason, the recording of the magnification factor data is corrupted so that only the phase angle
data is available for analysis. Is it possible to determine the resonant frequency from the available data?
What can be inferred about the amount of damping in the system from the phase data?
Solution
Yes, the resonance frequency can be closely approximated from the phase angle data. As can be seen in
Fig. 9.32, the phase angle is approximately equal to =2 at resonance, so knowing the forcing frequency
when D =2 will tell us the resonance frequency. As for the amount of damping, Fig. 9.32 also demonstrates
that the steeper the slope at resonance (i.e., at D =2), the smaller the amount of damping. Therefore, we
can infer something about the amount of damping by the slope of the phase angle plot at resonance.
August 10, 2009
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Solutions Manual
Problem 9.51
Suppose that equation of motion of a damped forced harmonic oscillator is given by xR C 2!n xP C !n2 x D .F0 =m/ cos !0 t , where
x is measured from the equilibrium position of the system. Obtain
the expression for the amplitude of the steady-state response of the
oscillator, and compare it with the expression presented in Eq. (9.78)
(which is for a system with equation of motion xR C 2!n xP C !n2 x D
.F0 =m/ sin !0 t ).
Solution
To find the steady-state response for cos !0 t forcing, we will follow the same procedure we used to find the
steady-state response for sin !0 t forcing.
We will assume a particular solution of either of the form
xp D D cos.!0 t
/;
(1)
where D and are constants to be determined and D is assumed to be a positive quantity. Substituting this
expression for xp into the given equation of motion, xR C 2!n xP C !n2 x D .F0 =m/ cos !0 t , we obtain
D!02 cos.!0 t
/
2D!n !0 sin.!0 t
/ C D!n2 cos.!0 t
Using the trigonometric identities sin.˛ ˇ/ D sin ˛ cos ˇ
sin ˛ cos ˇ and then collecting terms, we obtain
D
!n2
!02 sin 2!0 !n cos sin !0 t
C D !n2
/ D
cos ˛ sin ˇ and cos.˛
F0
cos !0 t:
m
(2)
ˇ/ D cos ˛ cos ˇ C
F0
!02 cos C 2!0 !n sin cos !0 t D
cos !0 t:
m
Since this equation must be true for all time, we can equate the coefficients of sin !0 t and cos !0 t to obtain
two equations for the unknowns D and . Doing this for sin !0 t allows us to solve for tan as
tan D
2!0 !n
2!0 =!n
:
D
2
2
1 .!0 =!n /2
!n !0
(3)
Equating the coefficients of cos !0 t, we obtain
DD
!n2
2
!0
F0 =m
D
1
cos C 2!0 !n sin F0 =.m!n2 cos /
;
.!0 =!n /2 C 2.!0 =!n / tan 2
2
where we have divided
pthe numerator and denominator by !n cos . Now noting that !n D k=m, that we
can write 1= cos D 1 C tan2 , and that tan is given by Eq. (3), D becomes
r
h
i2
q
2
2 .!0 =!n /
F0
F
0
1
C
1 .!0 =!n /2 C Œ2.!0 =!n /2
k
1 .!0 =!n /2
k
h
iD DD
2
1 .!0 =!n /2 C Œ2.!0 =!n /2
1 .!0 =!n /2 C 2.!0 =!n / 2 .!0 =!n /2
1 .!0 =!n /
August 10, 2009
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Dynamics 1e
This is easily seen to simplify to
DDp
Œ1
F0 =k
.!0 =!n /2 2 C .2!0 =!n /2
:
Thus, we see that both the amplitude and the phase of the steady-state response are the same for cos !0 t
forcing as they are for sin !0 t forcing.
August 10, 2009
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Solutions Manual
Problem 9.52
Differentiate Eq. (9.79) with respect to !0 =!n and set the result equal to zero to determine the frequency
!0 at which peaks in the MF curve occur as a function of and !n . Use this result to show that the peak
always occurs at !0 =!n 1. Finally, determine the value of for which the MF has no peak.
Solution
Equation (9.79) is given by
MF D p
Œ1
1
;
.!0 =!n /2 2 C .2!0 =!n /2
where !0 is the forcing frequency, !n is the natural frequency, and is the damping ratio. Letting x D !0 =!n
and then differentiating this equation with respect to x, we obtain
d MF
D
dx
1
2
h
1
x2
2
C .2x/2
x2
1=2 2 1
x 2 . 2x/ C 8 2 x
2
C .2x/2
8 2 x 4x 1 x 2
i3=2 D 0
h
2
2
2
2 1 x
C .2x/
1
)
i
)
xD0
)
D0
8 2 x
or x D
q
1
4x 1
2 2
x2 D 0
)
!0
D
!n
q
1
2 2 :
p
Therefore, the peak in the MF curve occurs at !0 =!n D 1 2 2 . Since 0, we see that 1 2 2 1
and so !0 =!n 1.
For the MF to have no peak, the quantity under the square root in the equation above must be negative.
This occurs when
p
no peak in MF for 1=2:
August 10, 2009
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Dynamics 1e
Problem 9.53
Calculate the response described by the equations listed below, in which x is measured in feet and time is
measured in seconds.
(a) 5xR C 10xP C 100x D 0, with x.0/ D 0:1 and x.0/
P
D
0:1
(b) 3xR C 15xP C 12x D 0, with x.0/ D 0 and x.0/
P
D 0:5
(c) xR C 10xP C 25x D 0, with x.0/ D 0:15 and x.0/
P
D0
(d) 25xR C 200xP C 1500x D 0, with x.0/ D 0:01 and x.0/
P
D0
Solution
In each case, before finding the response, we will first need to determine the character of the damping. That
is, is the system overdamped, critically damped, or underdamped.
Part (a).
For this system, m D 5, c D 10, and k D 100. Therefore the damping ratio is
D
c
c
D p
D 0:2236 < 1
cc
2 km
)
underdamped:
Therefore the response is given by
!n t
x.t / D De
sin.!d t C /;
where D and are constants determined by the initial conditions and !n and !d are the natural frequency
and damped natural frequency, respectively, which are given by
q
p
!n D k=m D 4:472 rad=s and !d D !n 1 2 D 4:359 rad=s:
Therefore, we can write the response as
t
x.t / D De
sin.4:359t C /:
Applying x.0/ D 0:1 gives
0:1 D D sin :
Since xP D DΠe
t
sin.4:359t C / C 4:359e
0:1 D
t
(1)
cos.4:359t C /, we can apply x.0/
P
D
D sin C 4:359D cos :
0:1 to obtain
(2)
Substituting Eq. (1) into Eq. (2), we find that
0:1 D
0:1 C 4:359D cos )
cos D 0
)
D =2;
where we have used the fact that D cannot be zero. Since D =2, Eq. (1) gives D D 0:1 and so the
response is
x.t / D 0:1e
t
sin.4:36t C =2/:
August 10, 2009
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Solutions Manual
Part (b). For this system, m D 3, c D 15, and k D 12. Therefore the damping ratio is
D
c
c
D p
D 1:25 > 1
cc
2 km
)
overdamped:
Therefore the response is given by
!n t
x.t / D e
p
Ae
2 1 !n t
p
C Be
2 1 !n t
where A and B are constants determined by the initial conditions and where !n is the natural frequency,
which is given by
p
!n D k=m D 2:000 rad=s:
Therefore, we can write the response as
2:5t
x.t / D e
Ae 1:5t C Be
1:5t
t
D Ae
C Be
4t
:
Applying x.0/ D 0 gives
ACB D0
Since xP D
Ae
t
A
4B D 0:5
4Be
4t ,
)
AD
B:
(3)
we can apply x.0/
P
D 0:5 to obtain
)
B
4B D 0:5
)
BD
0:1667
)
A D 0:1667:
(4)
Therefore, the response is
x.t / D 0:167e
t
0:167e
4t
:
Part (c). For this system, m D 1, c D 10, and k D 25. Therefore the damping ratio is
c
c
D1
D p
cc
2 km
D
)
critically damped:
Therefore the response is given by
x.t / D .A C Bt /e
!n t
D .A C Bt /e
5t
;
where A and B are constants determined from the initial conditions and !n D
natural frequency. Applying x.0/ D 0:15, we obtain
p
k=m D 5 rad=s is the
A D 0:15:
Since xP D Be
5t
5.A C Bt /e
5t ,
then applying x.0/
P
D 0 gives
B
5A D 0
)
B D 0:75;
which means that the response is
x.t / D .0:15 C 0:75t /e
5t
:
August 10, 2009
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Dynamics 1e
Part (d). For this system, m D 25, c D 200, and k D 1500. Therefore the damping ratio is
D
c
c
D p
D 0:5164 < 1
cc
2 km
)
underdamped:
Therefore the response is given by
!n t
x.t / D De
sin.!d t C /;
where D and are constants determined by the initial conditions and !n and !d are the natural frequency
and damped natural frequency, respectively, which are given by
q
p
!n D k=m D 7:746 rad=s and !d D !n 1 2 D 6:633 rad=s:
Therefore, we can write the response as
x.t / D De
4t
sin.6:633t C /:
Applying x.0/ D 0:01 gives
D sin D 0:01:
Since xP D DΠ4e
t
sin.6:633t C / C 6:633e
4t
(5)
cos.6:633t C /, we can apply x.0/
P
D 0 to obtain
4D sin C 6:633D cos D 0:
(6)
Substituting Eq. (5) into Eq. (6), we find that
0:04 C 6:633D cos D 0
)
D sin 0:01
D
6:633D cos 0:04
tan D 1:658 ) D 58:91ı D 1:028 rad:
6:633D cos D 0:04
)
)
Since D 58:91ı D 1:028 rad, Eq. (5) gives D D 0:01168 and so the response is
x.t / D 0:0117e
4t
sin.6:63t C 1:03/:
August 10, 2009
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Solutions Manual
Problem 9.54
Derive the equation of motion given in Eq. (1) of Example 9.6 for the
system in that example. The independent variable y is measured from the
equilibrium position of the system, m is the mass of the motor and platform,
c is the total damping coefficient of the dashpots, k is the total constant
of the linear elastic springs, !r is the angular velocity of the unbalanced
rotor, " is the distance of the eccentric mass from the rotor axis, and mu
is the eccentric mass. Note that m includes the eccentric mass so that the
nonrotating mass is equal to m mu .
Solution
Referring to the figure at the right, we have drawn an FBD of the unbalanced mass mu and an FBD of the system with the unbalanced mass
removed. Summing forces in the y direction on the unbalanced mass (top
figure) we obtain
X
Fy W
Ry mu g D mu auy D mu yRu ;
(1)
where yu is defined in the bottom figure on the right. Summing forces in
the y direction on the motor and platform (middle figure), we obtain
X
Fy W Ry .m mu /g Fs Fd D .m mu /amy
D .m
mu /y;
R
(2)
where Fs is the force in the spring, Fd is the force in the dashpot, and y is
the vertical position of the motor and platform as shown on the right. The
force laws for the spring and dashpot forces are given by
Fs D k.y
ıst / and
Fd D c y;
P
where ıst is the deflection of the spring when the system is in static equilibrium. Kinematically, we need to write yRu in terms of y and/or its time
derivatives. We can do this using the relation
yu D y C " sin )
yRu D yR
"P 2 sin D yR
"!r2 sin !r t;
where we have used the fact that P D !r , !r is constant, and D !r t .
Eliminating Ry from Eqs. (1) and (2) and then wubstituting the kinematics
equation and the force laws into the result, we obtain
mu g
mu .yR
"!r2 sin !r t /
.m
mu /g
k.y
ıst /
c yP D .m
mu /y:
R
Since mg D kıst , the equation of motion becomes
myR C c yP C ky D mu "!r2 sin !r t
)
yR C
c
k
mu "!r2
yP C y D
sin !r t;
m
m
m
or, using c=m D 2!n and k=m D !n2 , this becomes the equation of motion in Example 9.6, that is
yR C 2!n yP C !n2 y D
mu "!r2
sin !r t:
m
August 10, 2009
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Dynamics 1e
Problem 9.55
The mass m is coupled to the support A, which is displacing harmonically
according to y D Y sin !0 t , by the linear elastic spring of constant k and
the dashpot with constant c.
(a) Derive its equation of motion, using x as the independent variable, and
explain in what way the resulting equation of motion is not in the form
of Eq. (9.71).
(b) Next, let ´ D x y and substitute it into the equation of motion found
in part (a). After doing so, show that you obtain an equation of motion
in ´ that is of the same form as Eq. (9.71).
(c) Find the steady-state solution to the equation of motion found in part (b)
and then using that, determine the steady-state solution for x.
Solution
Part (a). Referring to the FBD on the right, if we sum forces in the p
direction, we obtain the following balance law
X
Fp W
Fs Fd D mx:
R
The force laws for the spring and dashpot forces are, respectively,
Fs D k.x
y/ and
Fd D c.xP
y/:
P
Substituting the force laws into the balance law, we obtain the equation of motion as
k.x
y/
c.xP
y/
P D mxR
)
mxR C c xP C kx D c yP C ky
)
mxR C c xP C kx D c!0 Y cos !0 t C kY sin !0 t ,
where we have used y D Y sin !0 t and yP D !0 cos !0 t . This equation is not in form of Eq. (9.71) since the
forcing term on the right hand side of the form R cos !0 t C S sin !0 t and not just R cos !0 t or S sin !0 t.
Therefore, we cannot solve this equation with what we have learned in this section.
Part (b). Letting ´ D x y and substituting it into the first form of the equation of motion from Part (a),
we obtain
k´ c Ṕ D m. Ŕ C y/
R
) m Ŕ C c Ṕ C k´ D myR D m!02 Y sin !0 t;
which means that the equation of motion in terms of ´ is
m Ŕ C c Ṕ C k´ D m!02 Y sin !0 t;
which is of the same form as Eq. (9.71).
Since
the system is linear, it turns out that it is actually easy to solve as the superposition of a particular solution for R cos !0 t
and one for S sin !0 t .
August 10, 2009
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Solutions Manual
Part (c). To determine the steady-state solution for ´, that is to the equation of motion found in Part (b), we
can either apply the results derived in Section 9.3, or we can assume a particular solution of the form
´p D A sin !0 t C B cos !0 t
or ´p D D sin.!0 t
/:
Choosing the former and substituting it into the equation of motion in ´, we obtain
m!02 .A sin !0 t C B cos !0 t / C c!0 .A cos !0 t
B sin !0 t / C k.A sin !0 t C B cos !0 t / D m!02 Y sin !0 t;
Since this expression must be true for all time, we can equate the coefficients of sin !0 t and cos !0 t to obtain
two equations whose solution gives A and B. These equations are:
m!02 A c!0 B C kA D m!02 Y
)
k m!02 A c!0 B D m!02 Y;
m!02 B C c!0 A C kB D 0
)
c!0 A C k m!02 B D 0:
Solving, we obtain the following for B
k
m!02
c!0
2
c!0 B D m!02 Y
B
)
BD
k
m!02 Y
D
2
m!02 =c!0 C c!0
k
mc!03 Y
;
2
m!02 C c 2 !02
and for A, we obtain
AD
k
mc!03 Y
2
m!02 C c 2 !02
k
m!02
c!0
!
m!02 Y k m!02
D
:
2
k m!02 C c 2 !02
Substituting A and B into the assumed form of the particular solution, we obtain the steady-state response for
´ as
m!02 Y k m!02
mc!03 Y
´p D
sin
!
t
cos !0 t
0
2
2
k m!02 C c 2 !02
k m!02 C c 2 !02
or
m!02 Y
2
´p D
k
m!
sin
!
t
C
c!
cos
!
t
:
0
0
0
0
2
k m!02 C c 2 !02
Therefore, since x D ´ C y, the steady-state solution for x must be
x.t / D
k
m!02 Y
k
2
m!02 C c 2 !02
m!02 sin !0 t C c!0 cos !0 t C Y sin !0 t:
August 10, 2009
1291
Dynamics 1e
Problem 9.56
A module with sensitive electronics is mounted on a panel that vibrates
due to excitation from a nearby diesel generator. To prevent fatigue
failure, the module is placed on vibration-absorbing mounts. The
displacement of the panel is measured to be yp .t / D y0 sin !0 t , where
y0 D 0:001 m, !0 D 300 rad=s, and the time t is measured in seconds.
Letting the mass of the electronic module be m D 0:5 kg, calculate the
amplitude of the vibration of the module if the equivalent stiffness and
damping coefficients for all the mounts combined are k D 10;000 N=m
and c D 40 Ns=m, respectively.
Solution
The figure at the right shows an FBD of just the electronics module.
Summing forces in the y direction, we find that
X
Fy W
mg Fd Fs D myRm ;
(1)
where Fs is the force in the equivalent spring and Fd is the force in the
equivalent damper. The forces laws for the spring and damper forces are,
respectively,
Fs D k.ym yp ıst / and Fd D c.yPm yPp /;
where ıst is the deflection of the spring in the static equilibrium position of the system. Substituting the force
laws into the equation of motion, we obtain
myRm C c.yPm
yPp / C k.ym
yp
ıst / C mg D 0
)
myRm C c yPm C kym D c yPp C kyp ;
(2)
where we have used the fact that mg D kıst . Substituting in yp D y0 sin !0 t and yPp D !0 y0 cos !0 t, we
obtain the equation of motion as
myRm C c yPm C kym D c!0 y0 cos !0 t C ky0 sin !0 t;
which is not in the form of Eq. (9.69). Letting ´ D ym
m. Ŕ C yRp / C c. Ṕ C yPp / C k.´ C yp / D c yPp C kyp
yp , we can write Eq. (2) as
)
m Ŕ C c Ṕ C k´ D
myRp D my0 !02 sin !0 t;
which is in the form of Eq. (9.69). Assuming a particular solution of the form
´ D D sin.!0 t
/;
and then substituting it into the equation of motion in ´, we obtain
mD!02 sin.!0 t
/ C cD!0 cos.!0 t
/ C kD sin.!0 t
Applying the trigonometric identities sin.˛ ˇ/ D sin ˛ cos ˇ
sin ˛ cos ˇ and then collecting terms, we obtain
/ D my0 !02 sin !0 t;
cos ˛ sin ˇ and cos.˛
D. m!02 cos C c!0 sin C k cos / sin !0 t C D.m!02 sin C c!0 cos ˇ/ D cos ˛ cos ˇ C
k sin / cos !0 t
D my0 !02 sin !0 t:
August 10, 2009
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Solutions Manual
Since this equation must be true for all time, we can equate the coefficients of sin !0 t and cos !0 t to obtain
two equations for the unknowns D and . Doing this for cos !0 t allows us to solve for tan as
c!0
c!0
1
) D tan
D 18:92ı D 0:3303 rad;
(3)
tan D
2
2
k m!0
k m!0
where we have used c D 40 Ns=m, !0 D 300 rad=s, k D 10;000 N=m, and m D 0:5 kg. Equating the
coefficients of sin !0 t, we obtain
DD
my0 !02 = cos :
m!02 C c!0 tan my0 !02
D
k
m!02 cos C c!0 sin k
Since tan is given by Eq. (3), we can write cos as
1
cos D r
1C
and so the expression for D becomes
r
my0 !02 1 C
DD
k
m!02 C c!0
c!0
k m!02
2
c!0
k m!02
D
my0 !02
k
q
c!0
k m!02
2 ;
2
m!02 C c 2 !02
2
m!02 C c 2 !02
k
Dq
k
my0 !02
D 0:001216 m; (4)
2 2
2
2
m!0 C c !0
where we have used c D 40 Ns=m, !0 D 300 rad=s, k D 10;000 N=m, y0 D 0:001 m, and m D 0:5 kg.
Now that we have D and , a particular solution for ´ is
´ D 0:001216 sin.300t C 0:3303/;
and so the solution for ym is
ym D ´ C yp D 0:001216 sin.300t C 0:3303/ C 0:001 sin 300t:
To calculate the amplitude of this vibration, we will write ym as
ym D y0 sin !0 t C D sin.!0 t
D .y0 C D cos / sin !0 t
D .y0 C D cos /.sin !0 t
D
y0 C D cos sin.!0 t
cos
/ D y0 sin !0 t C D sin !0 t cos D cos !0 t sin D sin D sin cos !0 t D .y0 C D cos / sin !0 t
cos !0 t
y C D cos „ 0 ƒ‚
…
tan
y0 C D cos tan cos !0 t / D
.sin !0 t cos
cos !0 t sin /
cos
/;
D D sin =.y0 C D cos /. Therefore, the amplitude of vibration is given by
s
2
D sin y0 C D cos jym j D
D .y0 C D cos / 1 C
)
jym j D 0:00219 m,
cos
y0 C D cos p
where we have used the fact that if tan ˇ D q then cos ˇ D 1= 1 C q 2 .
We could have saved a little time with this solution by noting that Eq. (3) above could have been directly
obtained from Eq. (9.75) in the textbook and Eq. (4) above could have been obtained from Eq. (9.76).
where we have let tan
August 10, 2009
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Dynamics 1e
Problem 9.57
A hard drive arm undergoes flow-induced vibration caused by the
vortices of air produced by a platter that rotates at !0 D 10;000 rpm.
The arm has length L D 0:037 m and mass m D 0:00075 kg, and it is
made from aluminum with a modulus of elasticity E D 70 GPa. In
addition, assume that the cross section of the arm has an area moment
of inertia Ics D 8:510 14 m4 . Following the steps in Example 9.2 on
p. 682, the arm can be modeled as a rigid rod that is pinned at one end
and is restrained by a torsional spring with equivalent spring constant
k t D 3EIcs =L. In addition to the torsional spring, assume that the
arm’s motion is affected by a torsional damper with torsional damping
coefficient c t . Assuming that the damping ratio is D 0:02 and that
the vortices produce an aerodynamic force with the same frequency as
the rotation of the platter, determine the amplitude of the aerodynamic
force needed to cause a steady-state vibration amplitude of 0:0001 m
at the tip of the arm. Assume that the aerodynamic force is applied
at the midpoint B of the hard drive. What vibration amplitude will
result if the same excitation is applied to a hard drive arm assembly
with the damping ratio of 0.05?
Solution
Referring to the FBD on the right and summing moments
about point O, we find that
X
(1)
MO W
Mk Mc C F .t / L
2 D IO ˛arm ;
where Mk is the moment due to the torsional spring, Mc
is the moment due to the torsional damper, F .t / is the
aerodynamic force on the arm, and ˛arm is the angular
acceleration of the arm. The force laws are given by
Mk D k t D k t
y
;
L
yP
Mc D c t P D c t ;
L
and
F .t / D F0 sin !0 t;
where F0 is the desired amplitude of the aerodynamic forcing and and y are defined as shown on the FBD.
Since ˛arm D R D y=L
R
and IO D 13 mL2 , we can write the equation of motion in Eq. (1) as
kt
y
L
ct
yP
L
1
yR
C F0 sin !0 t D mL2
L
2
3
L
)
2
1
3 mL yR
C c t yP C k t y D 21 L2 F0 sin !0 t:
(2)
Defining the following “equivalent” quantities
meq D 13 mL2
and
.F0 /eq D 12 L2 F0 ;
we see that Eq. (2) becomes
meq yR C c t yP C k t y D .F0 /eq sin !0 t:
Therefore the amplitude of steady-state vibration is given by
jyj D p
.F0 /eq =k t
Œ1
.!0 =!n /2 2 C .2!0 =!n /2
;
(3)
August 10, 2009
1294
Solutions Manual
p
p
where !n D k t =meq is the natural frequency, !0 is the forcing frequency, and D c t = 2 k t meq is the
damping ratio. Computing the numerical values of the quantities in Eq. (3), we find that
.F0 /eq D 21 L2 F0 D .0:0006845F0 / N;
3EIcs
kt D
D 0:4824 Nm=rad;
L
!0 D 10;000 rpm D 1047 rad=s;
meq D 31 mL2 D 3:42210 7 kg;
s
kt
!n D
D 1187 rad=s:
meq
Substituting jyj D 0:0001 m as well as the values above into Eq. (3) and then solving for F0 , we obtain
0:0001 D p
Œ1
0:0006845F0 =0:4824
.1047=1187/2 2 C Œ2.0:02/1047=11872
)
F0 D 0:01582 N
)
F0 D 0:0158 N,
where we have reported the result to both 3 and 4 significant figures. Applying this level of excitation to a
drive arm assembly with damping ratio D 0:05, we obtain
jyj D p
Note
0:0006845.0:01582/=0:4824
Œ1
.1047=1187/2 2 C Œ2.0:05/1047=11872
)
jyj D 0:0000941 m.
that this definition for would allow us to find c t if we desired, but c t is not explicitly needed.
August 10, 2009
1295
Dynamics 1e
Problem 9.58
The mechanism consists of a disk D pinned at G, which is both the
geometric center of the disk and its mass center. The outer circumference of the disk has radius ro D 0:1 m and is connected to an
element consisting of a linear spring with stiffness k1 D 100 N=m
in parallel with a dashpot with damping coefficient c D 50 Ns=m.
The disk has a hub of radius ri D 0:05 m that is connected to a linear
spring with constant k2 D 350 N=m. Knowing that for D 0 the
disk is in static equilibrium and that the mass moment of inertia of the
disk is IG D 0:001 kgm2 , derive the linearized equation of motion
of the disk in terms of . In addition, calculate the resulting vibrational motion if the system is released from rest with an initial angular
displacement i D 0:05 rad.
Solution
Referring to the FBD at the right, we can obtain the equation of motion by
summing moments about the mass center G to obtain
X
MG W
.Fc C F1 /ro F2 ri D IG ˛D ;
(1)
where Fc is the for due to the dashpot, F1 is the force due to spring k1 , F2 is
the force due to spring k2 , and the reactions due to the pin at G are designated
by Rx and Ry . Using as the displacement variable and assuming small
displacements, the force laws can be written as
F1 D k1 ro ;
F2 D k2 ri ;
and Fc D cro P :
R Eq. (1) becomes
Using these force laws, along with the fact that ˛D D ,
IG R C cro2 P C k1 ro2 C k2 ri2 D 0:
We now compute the motion of the system when it is released with the initial conditions .0/ D 0:05 rad
P
D 0 rad=s. Before we can determine the response, we need to determine character of the damping.
and .0/
Computing the critical damping for this system, we find
q
q
cc D 2 keq meq D 2 k1 ro2 C k2 ri2 IG D 0:08660 Ns=m;
where we are using the definitions
meq D IG D 0:001 kgm2 ;
ceq D cr02 D 0:5000 Nms;
and keq D k1 ro2 C k2 ri2 D 1:875 Nm:
Since c > cc , the system is overdamped, D c=cc D 5:774, and the solution is given by
p
p 2
2 1 !n t
D e !n t Ae 1 !n t C Be
;
where !n is the natural frequency and is given by
q
!n D keq =meq D 43:30 rad=s;
August 10, 2009
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Solutions Manual
and A and B are constants determined from the initial conditions. Substituting the computed quantities into
the solution for , we obtain
D e 250:0t Ae 246:2t C Be 246:2t :
Applying the initial condition .0/ D 0:05 rad, we obtain
0:05 D A C B:
(2)
P we obtain
Finding ,
P D
250:0e
250:0t
Ae 246:2t C Be
246:2t
Ce
250:0t
246:2Ae 246:2t
D
3:800Ae
246:2Be
3:800t
246:2t
496:2Be
496:2t
;
and then applying the initial condition P .0/ D 0, we get
0D
3:800A
496:2B:
(3)
Solving Eqs. (2) and (3) for A and B, we obtain
A D 0:05039 rad
and B D
0:0003859 rad;
which means that the vibrational motion is given by
.t/ D e
250:0t
0:05039e 246:2t
0:0003859e
246:2t
)
.t / D 0:0504e
3:80t
0:000386e
496t
.
August 10, 2009
1297
Dynamics 1e
Problem 9.59
A box of mass 0:75 kg is thrown on a scale, causing both the scale and the box
to move vertically downward with an initial speed of 0:5 m=s. Before the box
lands on the scale, the scale is in equilibrium. The total mass of the scale’s
moving platform and the box is m D 1:25 kg. Modeling the platform’s support
as a spring and dashpot with stiffness k D 1000 N=m and damping coefficient
c D 70:7 Ns=m, find the response of the scale. Hint: Place the origin of
the y axis at the position of the platform corresponding to the equilibrium
configuration of the platform and box together.
Solution
Referring to the FBD of the box and the platform on the right, Fs is the force due
to the spring and Fd is the force due to the dashpot. Summing forces in the y
direction gives
X
Fy W mg Fd Fs D my;
R
(1)
Writing the force laws for the spring and dashpot forces, we obtain
Fs D k.y C yst / and
Fd D c y;
P
where yst is the static deflection of the spring when the box is on the platform. Substituting the force laws
into Eq. (1) gives
mg c yP k.y C yst / D myR ) myR C c yP C ky D 0;
where we have used the fact that mg D kyst . To determine the nature of the response, we need to determine
the nature of the damping. Doing so, we find
c
c
D 0:9998 1;
D
D p
cc
2 km
where we have used c D 70:7 Ns=m, k D 1000 N=m, and m D 1:25 kg. Since the damping ratio is so
close to 1, can consider the system to be critically damped. The response of a critically damped system is
given by
y D .A C Bt /e !n t ;
where A and B are constants determined from the initial conditions and !n is the natural frequency, which is
given by
r
k
!n D
D 28:28 rad=s ) y D .A C Bt /e 28:28t ;
m
where we have used k D 1000 N=m and m D 1:25 kg and we have substituted !n into the response.
Evaluating the response at the initial condition y.0/ D 0, we obtain
y.0/ D A
)
A D 0:
Differentiating the response with respect to time and then evaluating it at the initial condition y.0/
P
D 0:5 m=s,
we obtain
yP D Be
28:28t
28:28.A C Bt /e
28:28t
)
y.0/
P
DB
28:28A D 0:5
)
B D 0:5 m=s;
where we have used the fact that A D 0. Putting everything together, the response is
y.t / D 0:5t e
28:3t
:
August 10, 2009
1298
Solutions Manual
Problem 9.60
Consider a simple viscously damped harmonic oscillator governed by
Eq. (9.51), and analyze the case in which the damping coefficient c is negative.
Calculate the general expression for the response (without taking into account
specific initial conditions), using m D 1 kg, c D 1 Ns=m, and k D 10 N=m.
Comment on the system’s response.
Solution
We are told that the equation of motion for this system is given by Eq. (9.51), which is given by
mxR C c xP C kx D 0:
The solution of this equation was found by assuming a solution of the form
x D e t
m2 e t C ce t C ke t D 0
)
The roots of this characteristic equation we found to be
r
c
c 2 k
C
;
2 D
1 D
2m
2m
m
c
2m
)
m2 C c C k D 0:
r
c 2
2m
k
:
m
For m D 1 kg, c D 1 Ns=m, and k D 10 N=m, the term under the square root has the value 9:75, which
is less than zero and so the response is given by
xDe
.c=2m/t
.A sin !d t C B cos !d t /;
where !d is the damped natural frequency given by
r
k c 2
D 3:122 rad=s:
!d D
m
2m
Substituting in m D 1 kg, c D
respectively,
xR
1 Ns=m, and k D 10 N=m, the equation of motion and response become,
xP C 10x D 0
and
x.t / D e 0:5t .A sin 3:12t C B cos 3:12t /.
Since the exponent on the exponential is positive, the response will grow without limit and so the system is
unstable.
August 10, 2009
1299
Dynamics 1e
Problem 9.61
The MF for a harmonically excited spring-mass-damper system at
!0 =!n 1 is equal to 5. Calculate the damping ratio of the system.
What would the damping ratio be if the MF were equal to 10? Sketch the
magnification factor at !0 =!n 1 as a function of the damping ratio.
Solution
The equation for the magnification factor for this system is given by Eq. (9.79), which is
MF D p
Œ1
1
.!0 =!n
/2 2
C .2!0 =!n /2
where is the damping ratio, !n is the natural frequency, and !0 is the forcing frequency. Noting that
MF D 5, when !0 =!n D 1, we have
5D p
1
Œ1
1
1
Dp
D
2
.1/2 2 C .2/2
.2/2
)
D
1
D 0:1
10
for MF = 5.
For MF D 10, the damping ratio would be
10 D
1
2
)
D
1
D 0:05 for MF = 10.
20
A plot of MF as a function of for !0 =!n D 1 is shown below.
Ω0 !Ωn # 1
3.0
2.5
MF
2.0
1.5
1.0
0.5
0.0
0.0
0.5
1.0
1.5
2.0
Ζ
August 10, 2009
1300
Solutions Manual
Problem 9.62
A slider moves in the horizontal plane under the action of the harmonic forcing
F .t/ D F0 sin !0 t . The slider is connected to two identical linear springs, each
of which has constant k. When t D 0, x.0/ D 0, the springs are unstretched,
D 45ı , and L D L0 . The slider is also connected to a damper with damping
coefficient c. Treating F0 , k, c, and L0 as known quantities, neglecting friction,
and letting x.0/
P
D vi , (a) derive the equations of motion of the system, (b) derive
the linearized equations of motion about the initial position, and (c) determine the
amplitude of the steady-state vibrations for the linearized equations of motion.
Solution
Referring to the FBD at the right and summing forces in the x direction,
we find that
X
Fy W F .t / 2Fs cos Fd D mx;
R
(1)
where Fs is the force in each spring and Fd is the force in the dashpot. The
forces in the spring and dashpot can be written as, respectively,
Fs D k.Lx
and Fd D c x;
P
Lu /
where Lx is the length of the spring in its deformed configuration (i.e., for
arbitrary x) and Lu is the undeformed length of the spring. Referring to
the bottom figure on the right, we find Lx and Lu to be, respectively,
q
p
Lu D 2L and Lx D L2 C .L C x/2 :
We can write cos as
cos D p
LCx
L2
C .L C x/2
:
Substituting the force laws and the geometric relations into Eq. (1), we obtain the equation of motion
F .t/
2k
q
L2
C .L C
)
x/2
p
2L p
mxR C c xP C 2k
q
LCx
L2
C .L C x/2
L2
x/2
C .L C
c xP D mxR
p
2L p
LCx
L2
C .L C x/2
D F0 sin !0 t ,
where we have replaced F .t / with F0 sin !0 t. To linearize the equation of motion, we only need to linearize
the term containing k since it is the only term that is nonlinear. Copying that term and then rearranging gives
2k
q
L2 C .L C x/2
p
2L p
LCx
L2 C .L C x/2
p
"
D 2k.L C x/ 1
p
„
2L
#
L2 C .L C x/2
ƒ‚
…
f .x/
D 2k.L C x/Œ1
f .x/: (2)
August 10, 2009
1301
Dynamics 1e
We can now use the Taylor series for f .x/ about x D 0 to linearize the stiffness term and thus the equation
of motion. Writing the Taylor series, we have
f .x/ D f .0/ C f 0 .0/x C 12 f 00 .0/x 2 C ˇ
ˇ
p
p
ˇ
ˇ
2L
2L.L
C
x/
ˇ
ˇ
x C Dp
ˇ
3=2 ˇˇ
2
2
ˇ
2
2
L C .L C x/ xD0
L C .L C x/
xD0
p 2
x
2L
D1
x C D 1
C :::;
2L
.2L2 /3=2
(3)
where the prime .0 / indicates differentiation with respect to x and we have omitted the x 2 term in the Taylor
series expansion since it is nonlinear. Substituting Eq. (3) into Eq. (2) and then substituting the result into the
equation of motion, we obtain
h
mxR C c xP C 2k.L C x/ 1
x i
D F0 sin !0 t
2L
1
)
mxR C c xP C kx C
k 2
x D F0 sin !0 t
L
mxR C c xP C kx D F0 sin !0 t ,
)
where we have ignored terms of the order x 2 and higher. Since the linearized equation of motion for this
system is identical to Eq. (9.69), the amplitude of steady-state vibration is given by Eq. (9.78), which is
DDp
r
F0 =k
Œ1
.!0 =!n
/2 2
C .2!0 =!n
/2
;
where !n D
k
m
c
and D p
:
2 km
August 10, 2009
1302
Solutions Manual
Problem 9.63
The mechanism shown is a pendulum consisting of a pendulum bob B
with mass m and a T bar, which is pinned at O and has negligible mass.
The horizontal portion of the T bar is connected to two supports, each
of which has an identical spring and dashpot system, each with spring
constant k and damping coefficient c. The springs are unstretched when
B is vertically aligned with the pin at O. Modeling B as a particle,
derive the linearized equations of motion of the system. In addition,
assuming that the system is underdamped, derive the expression for the
damped natural frequency of vibration of the system.
Solution
Referring to the FBD on the right and assuming that the
angle is small (so the horizontal and vertical dimensions
are as shown on the FBD), we can sum moments about
point O to obtain the equation of motion
X
MO W
mgL 2Fd .0:3L/ 2Fs .0:6L/
D IO R ;
(1)
where Fs is the force in each of the two springs and Fd is
the force in each of the two dashpots. The mass moment
of inertia IO D mL2 and the forces Fs and Fd can be
written as
Fs D k.0:6L / and Fd D c 0:3LP :
Substituting the force laws and the expression for the moment of inertia into Eq. (1), we obtain
mgL
0:72kL2 0:18cL2 P D mL2 R ;
or, rearranging and canceling an L, we get the linear equation of motion as
mLR C 0:18cLP C .mg C 0:72kL/ D 0:
Assuming the system is underdamped, the damped natural frequency, which is given by Eq. (9.61), is
s
!d D
keq
meq
ceq
2meq
s
2
D
mg C 0:72kL
mL
0:18cL
2mL
2
s
)
!d D
g
k
C 0:72
L
m
0:0081
c2
.
m2
August 10, 2009
1303
Dynamics 1e
Problem 9.64
The engine in the rocket shown is supposed to provide a constant
thrust of 5000 kN. The turbopump unit in the engine nominally
operates at 7000 rpm, and as a result of a design issue, the actual thrust provided by the engine oscillates harmonically with
an amplitude of 10 kN at the same rotational frequency of the
turbopump unit. The mass of the engine is m D 5000 kg. The rest
of the rocket is much heavier than the engine and can be treated as
being fixed. The engine is mounted to the rocket via two structural
members, each of which can be modeled as consisting of a linear
spring of stiffness k in parallel with a dashpot with linear viscous
damping coefficient c. Determine the smallest values of k and c
such that the static deflection due to the constant component of the
thrust is less than 0:01 m and so that the vibration amplitude in the
nominal operating regime is less than 0:001 m. Ignore the stiffness
and damping due to the piping. Hint: If x is measured from the
equilibrium position of the engine that results from the combined
effect of the thrust and gravity, then the engine is subject to an
externally applied forcing equal to .10 kN/ sin !0 t , where !0 is
the rotational frequency of the turbopump.
Solution
Referring to the figure at the right and summing forces in the x direction,
we obtain
X
Fx W
2Fs 2Fd C F .t / D mx;
R
(1)
or, noting that Fs D kx, Fd D c x,
P and F .t / D F0 sin !0 t , the balance
law becomes the following equation of motion
mxR C 2c xP C 2kx D F0 sin !0 t;
where F0 D 10;000 N and !0 D 7000 rpm D 733:0 rad=s. Writing the
static deflection as ıst , we note that we can write it as
ıst D
Fst
Fst
D
< 0:01 m
keq
2k
)
k > 50Fst
)
k > 2:5108 N=m;
where Fst D 5000103 N is the constant thrust static force. Note that we also want the amplitude of the
deflection during vibration to not exceed 0:001 m. The steady-state amplitude for damped system is
DDp
Œ1
where
r
!n D
keq
D
m
r
F0 =keq
.!0 =!n /2 2 C .2!0 =!n /2
2k
m
and D p
< 0:001 m;
ceq
2c
c
D p
Dp
:
2keq m
2 2km
2km
August 10, 2009
1304
Solutions Manual
As a limiting case, we can ask whether or not we can satisfy this limit on D with c D 0? Let’s check. Since
c D 0 implies that D 0, the steady-state amplitude becomes
DD
F0 =keq
F0
F0 =.2k/
D
< 0:001 m
D
2
2
1 .!0 =!n /
1 !0 m=.2k/
2k m!02
)
k > 1:348109 N=m;
where we have used F0 D 10;000 N, m D 5000 kg, and !0 D 733:0 rad=s. Therefore, we can satisfy the
given constraints with
k > 1:35109 N=m
and c D 0:
Of course, with c D 0, the amplitude would be infinite at resonance, which is generally undesirable.
August 10, 2009
1305
Dynamics 1e
Problem 9.65
A simple model for a ship rolling on waves treats the waves as sinusoids.
Using this model, it can be shown that a linear model for the roll angle
is given by
IG R C c P C mg D
IG k!02 sin !0 t;
where G denotes the mass center of the ship, IG is the ship’s mass moment of inertia, c is a rotational viscous damping constant coefficient, m
is the mass of the ship, A is the wave amplitude, and k is the wavelength
of the waves.
(a) What is the natural frequency of the system?
(b) Find the magnification factor for the system.
(c) Assuming that the damping is negligible (i.e., c 0), if the maximum amplitude of oscillation that the ship can undergo without
capsizing is max D 1 rad, find the maximum A so that the crew
remains safe.
Solution
Note: The given equation of motion is missing an A in the forcing term so that it should be written as
IG R C c P C mg D
IG Ak!02 sin !0 t:
This does not affect parts (a) and (b) of the problem, but it does impact part (c).
Part (a).
The natural frequency is
s
!n D
keq
meq
r
)
!n D
mg
.
IG
Part (b). For an equation of motion of the given form, the MF is given by
MF D p
Œ1
1
.!0 =!n /2 2 C .2!0 =!n /2
;
where !n is given above and is given by
c
c
:
D p
D p
2 mgIG
2 keq meq
Substituting !n and into the MF, we obtain
MF D r
1
!02 IG
mg
1
2
C
4!02 c 2 IG
4mgIG mg
D r
1
1
!02 IG
mg
2
C
!02 c 2
m2 g 2
)
MF D q
mg
mg
IG !02
2
.
C c 2 !02
August 10, 2009
1306
Solutions Manual
Part (c). If c D 0, then the amplitude of the response is given by
amp D
IG Ak!02 =.mg/
IG Ak!02
.F0 /eq =keq
< 1 rad;
D
D
IG !02
1 .!0 =!n /2
mg IG !02
1
mg
where we have used the fact that we want the amplitude of vibration to be less than 1 rad. Solving for A, we
obtain
IG Ak!02 < mg
IG !02
)
A<
mg IG !02
.
IG k!02
August 10, 2009
1307
Dynamics 1e
Problem 9.66
A delicate instrument of mass m must be isolated from excessive vibration of the ground, which is described by the function u.t / D A sin !0 t .
To do so, we need to design a vibration isolating mount, modeled by the
spring and dashpot system shown.
(a) Find the equation of motion of the instrument and reduce it to standard form.
(b) Find the steady-state response y.t /.
(c) Find the displacement transmissibility, i.e., the response amplitude
D divided by A, where A is the amplitude of the ground’s vibration.
Solution
Part (a). Referring to the FBD on the right and summing forces in the
y direction, we obtain
X
Fy W
mg Fs Fd D my;
R
(1)
where Fs is the force in the spring and Fd is the force in the dashpot. The
force laws for the spring and dashpot are, respectively,
Fs D kŒy.t /
u.t /
ıst  D kŒy.t /
Fd D cŒy.t
P /
u.t
P / D cŒy.t
P /
A sin !0 t
ıst ;
A!0 cos !0 t :
Substituting Fs and Fd into Eq. (1), we obtain the equation of motion as
mg
kŒy.t /
A sin !0 t
cŒy.t
P /
ıst 
A!0 cos !0 t  D myR
)
myR C c yP C ky D kA sin !0 t C cA!0 cos !0 t; (2)
where we have used the fact that mg D kıst . Equation (2) has a forcing function that is not in standard form.
We can put it in standard form if we let s.t / D y.t / u.t /. Doing so, the force laws become
Fs D k.s
Noting that yR D sR C uR D sR
mg
k.s
ıst /
ıst /
and Fd D c sP :
A!02 sin !0 t , Eq. (1) becomes
c sP D m.Rs
A!02 sin !0 t /
)
mRs C c sP C ks D mA!02 sin !0 t ,
where we still have mg D kıst .
Part (b). The steady-state response for this system is given by
s D B sin !0 t C C cos !0 t;
August 10, 2009
1308
Solutions Manual
where B and C are constants to be determined. Substituting this assumed solution into the equation of
motion, we obtain
m!02 .B sin !0 t CC cos !0 t /Cc!0 .B cos !0 t C sin !0 t /Ck.B sin !0 t CC cos !0 t / D mA!02 sin !0 t:
Because of the orthogonality of sin !0 t and cos !0 t, we can equate the coefficients of each to obtain the
following two equations for the unknowns B and C
m!02 B c!0 C C kB D mA!02
)
k m!02 B c!0 C D mA!02 ;
m!02 C C c!0 B C kC D 0
)
c!0 B C k m!02 C D 0:
Solving these two equations for B and C , we obtain
mA!02 k m!02
and
BD
2
k m!02 C c 2 !02
C D
Therefore, the solution for s.t / is given by
"
#
mA!02 k m!02
s.t / D
sin !0 t
2
k m!02 C c 2 !02
k
"
mcA!03
:
2
m!02 C c 2 !02
#
mcA!03
cos !0 t;
2
m!02 C c 2 !02
k
and therefore the solution for the steady-state response y.t / is
"
"
#
mA!02 k m!02
sin !0 t
y.t/ D s.t/ C u.t / D
2
k m!02 C c 2 !02
k
#
mcA!03
cos !0 t C A sin !0 t;
2
m!02 C c 2 !02
or,
"
y.t / D
#
mA!02 k m!02
C A sin !0 t
2
k m!02 C c 2 !02
"
k
#
mcA!03
cos !0 t:
2
m!02 C c 2 !02
Part (c). To find the displacement transmissibility, we divide the amplitude of the response by the amplitude
of the ground’s vibration. To find the amplitude of the response, we first define E and F as follows
"
#
"
#
mA!02 k m!02
mcA!03
y.t / D
C A sin !0 t
cos !0 t;
2
2
k m!02 C c 2 !02
k m!02 C c 2 !02
„
ƒ‚
…
„
ƒ‚
…
E
F
and then, since sin !0 t and cos !0 t are orthogonal, the response amplitude D is given by
v"
#2 "
#2
u
3
u mA! 2 k m! 2
p
mcA!
0
0
t
0
D D E2 C F 2 D
CA C
2
2
k m!02 C c 2 !02
k m!02 C c 2 !02
r
h
i2
2
A
2
2
2 2
2!2
D
m!
k
m!
C
k
m!
C
c
C mc!03
0
0
0
0
2
k m!02 C c 2 !02
q
2
2
A
D
k m!02 m!02 C k m!02 C c 2 !02 C mc!03
2
k m!02 C c 2 !02
q
2
2
A
D
k k m!02 C c 2 !02 C mc!03 :
2 2
2
k m!0 C c 2 !0
August 10, 2009
1309
Dynamics 1e
Calling the term under the square root Q and then expanding it, we obtain
Q D k2 k
D k2 k
m!02
2
C c 2 !02 c 2 !02 C 2k k m!02 C m2 !04
2
m!02 C c 2 !02 c 2 !02 C k 2 C k 2 2mk!02 C m2 !04
ƒ‚
…
„
2
k m!02
2
2
m!02 C c 2 !02 C c 2 !02 k 2 C c 2 !02 k
2
D k m!02 k 2 C c 2 !02 C c 2 !02 k 2 C c 2 !02
i
h
2
D k 2 C c 2 !02 k m!02 C c 2 !02 :
D k2 k
m!02
2
Substituting Q back into the expression for D, we obtain
v
u
u
k 2 C c 2 !02
:
D D At
2
k m!02 C c 2 !02
Finally, since the displacement transmissibility, DT, becomes
D
DT D
A
)
v
u
u
DT D t
k
k 2 C c 2 !02
.
2
m!02 C c 2 !02
August 10, 2009
1310
Solutions Manual
Problem 9.67
When the connecting rod shown is suspended from the knife-edge at point O and displaced
slightly so that it oscillates as a pendulum, its period of oscillation is 0:77 s. In addition, it
is known that the mass center G is located a distance L D 110 mm from O and that the
mass of the connecting rod is 661 g. Using the energy method, determine the mass moment
of inertia of the connecting rod IG .
Solution
Referring to the FBD on the right, the kinetic energy of the connecting rod can
be written as
1
1
T D IO P 2 D IG C mL2 P 2 :
2
2
Using the datum shown, the potential energy of the connecting rod can be
written as
2
V D mgL cos mgL 1
;
2
where we have approximated the cosine function using Eq. (9.28). Applying
the energy method to the kinetic and potentials energies shown above, we
obtain
d
d 1
2
2 P2
D0
.T C V / D
IG C mL mgL 1
dt
dt 2
2
D IG C mL2 P R C mgL P D 0:
P we obtain
Canceling ,
s
IG C mL2 R C mgL D 0
)
!n D
mgL
IG C mL2
)
2
D
D 2
!n
s
IG C mL2
:
mgL
Solving for IG
2
IG C mL2
D
4 2
mgL
g 2
IG D mL
4 2
)
L
)
IG D 0:00271 kgm2 .
August 10, 2009
1311
Dynamics 1e
Problem 9.68
Derive the equation of motion for the system, in which the springs with constants
k1 and k2 connecting m to the wall are joined in series. Neglect the mass of the
small wheels, and assume that the attachment point A between the two springs
has negligible mass. Hint: The force in the two springs must be the same; use
this fact, along with the fact that the total deflection of the mass must equal the
sum of the deflections of the springs to find an equivalent spring constant keq .
Solution
Referring to the FBD at the right and summing forces in the x direction, we obtain
the following balance law for the mass m
X
Fx W
Fs D mx:
R
(1)
To find the force law for Fs , we begin by noting that the total displacement of the
mass m, which is given by x, is the sum of the deflection of k1 and the deflection of k2 , that is
x D ı1 C ı2 ;
where ı1 is the deflection of k1 and ı2 is the deflection of k2 . Since we are neglecting the mass of the springs,
the force in each spring must equal Fs so that
Fs D k1 ı1 D k2 ı2 D keq x;
where keq is the equivalent spring constant of the two springs in series. Therefore
xD
Fs
Fs
Fs
C
D
k1
k2
keq
)
1
1
1
C
D
k1
k2
keq
)
keq D
k1 k2
:
k1 C k2
Therefore, since Fs D keq x, the equation of motion becomes
mxR C
k1 k2
x D 0:
k1 C k2
August 10, 2009
1312
Solutions Manual
Problem 9.69
Revisit Example 9.2 and compute the natural frequency of the
silicon nanowire, using the energy method. Use a uniform Si
nanowire with a circular cross section that is 9:8 m long and
330 nm in diameter and with all its flexibility lumped in a torsional
spring at the base of the wire. In addition, use D 2330 kg=m3
for the density of silicon and E D 152 GPa for its modulus of
elasticity.
Solution
Referring to the FBD at the right, we see that only the moment due
to the torsional spring does work. With this in mind, we can write
the kinetic energy as
1
1 1
1
2
2 P2
P
T D IO D
mL D mL2 P 2 ;
2
2 3
6
where IO is the mass moment of inertia of the bar with respect to
point O. The potential energy of the torsional spring can be written as
1
1 3EIcs
2
V D kt D
2;
2
2
L
where we have used the fact that k t D 3EIcs =L from Eq. (3) in Example 9.2. Applying the energy method,
we obtain
1
d
d 1
1 3EIcs
3EIcs
2
2 P2
2PR
D mL C
P D 0:
.T C V / D
mL C
dt
dt 6
2
L
3
L
Canceling P and we obtain the equation of motion and the natural frequency as
1
mL2 R C
3
3EIcs
D0
L
)
R C
9EIcs
D0
mL3
r
)
!n D 3
EIcs
:
mL3
To obtain a numerical value for !n , we find that the volume of the nanowire is r 2 L, where r D 165 nm D
16510 9 m and L D 9:810 6 m. Therefore, the mass of the wire is m D r 2 L D 1:95310 15 kg,
where D 2330 kg=m3 . Referring back to Example 9.2, the area moment of inertia is Ics D 41 r 4 D
5:82110 28 m4 . Using these results, along with E D 152109 N=m2 , we find that
!n D 2:08107 rad=s:
August 10, 2009
1313
Dynamics 1e
Problem 9.70
Structural health monitoring technology detects damage in civil, aerospace, and other structures. Structural
damage is usually comprised of cracking, delaminations, or loose fasteners, which result in the reduction of
stiffness. Many structural health monitoring methods are based on tracking changes in natural frequencies.
Modeling a structure as a one DOF harmonic oscillator, calculate the change in stiffness needed to cause a
3% reduction in the natural frequency of the structure being monitored.
Solution
With a 3% reduction in frequency, the ratio of the new !1 frequency to the old !0 can be written as
r
r
!1
k1
k0
k1
k0
D 0:97 ) !1 D 0:97!0 )
D 0:97
)
D 0:9409 :
!0
m
m
m
m
Therefore, since k1 D 0:9409k0 , the percent change in stiffness is
k D
5:91%:
August 10, 2009
1314
Solutions Manual
Problem 9.71
The harmonic oscillator shown has a mass m D 5 kg, a spring with
constant k D 4000 N=m, and a dashpot with a damping coefficient
c D 20 Ns=m. Calculate the amplitude F0 of the sinusoidal excitation force that is necessary to produce a steady-state vibration with a
velocity amplitude of 10 m=s at resonance. What is the corresponding
amplitude of the acceleration?
Solution
Referring to the FBD at the right and summing forces in the x direction, we obtain
the following balance law
X
Fx W
Fd Fs C F0 sin !0 t D mx;
R
(1)
where Fd is the force in the dashpot and Fs is the force in the spring. Writing the
force laws as follows,
Fd D c xP and Fs D kx;
Eq. (1) becomes
c xP
kx C F0 sin !0 t D mxR
)
mxR C c xP C kx D F0 sin !0 t:
Now that we have the equation of motion, we know that we can write its steady-state response and its time
derivative as
x D D sin.!0 t
/
)
xP D !0 D cos.!0 t
)
/
jxj
P D !0 D D 10 m=s;
(2)
where we have noted that we want the velocity jxj
P to be 10 m=s (at resonance). The position amplitude D
can be written as
F0 =k
;
(3)
DDp
Œ1 .!0 =!n /2 2 C .2!0 =!n /2
p
where !n D k=m is the natural frequency. Combining Eqs. (2) and (3) and solving for the amplitude of
the forcing F0 , we obtain
q
!0 F0 =k
jxjk
P
Œ1 .!0 =!n /2 2 C .2!0 =!n /2 :
D jxj
P
) F0 D
p
!0
Œ1 .!0 =!n /2 2 C .2!0 =!n /2
Since we are interested in finding F0 at resonance, we note that
!0
D1
!n
)
!0
D1
p
k=m
)
!0 D
p
k=m D 28:28 rad=s;
where we have used k D 4000 N=m and m D 5 kg. In addition, calculating the damping ratio , we find
D
c
c
D p
D 0:07071;
cc
2 km
where we used c D 20 Ns=m, k D 4000 N=m, and m D 5 kg. Substituting into the F0 we found above, we
obtain
August 10, 2009
1315
Dynamics 1e
jxjk
P
F0 D
!0
q
Œ1
.!0 =!n
/2 2
C .2!0 =!n
/2
10.4000/
D
28:28
q
1
.1/2
2
C Œ2.0:07071/.1/2
)
F0 D 200 N.
To find the corresponding amplitude of the acceleration, we differentiate x in Eq. (2) a second time to obtain
xR D
!02 D sin.!0 t
/
)
jxj
R D !02 D D p
!02 F0 =k
Œ1
.!0 =!n /2 2 C .2!0 =!n /2
)
jxj
R D 283 m=s2 ,
where, to obtain the final numerical result, we used !0 D 28:28 rad=s, F0 D 200:0 N, k D 4000 N=m,
!0 =!n D 1, and D 0:07071.
August 10, 2009
1316
Solutions Manual
Problem 9.72
Revisit Example 9.4 and derive the equations of motion of the motor,
using the following equation, called Lagrange’s equation,
d
@T
@T
@V
C
D 0;
dt @yPm
@ym
@ym
where T and V are the kinetic and potential energies of the system,
respectively.
Solution
Referring to the FBD at the right, we see that the spring force Fs as well
as all three weight forces do work as the system oscillates vertically.
On the other hand, we have also seen that as long as we measure the
position of the system ym from its static equilibrium position, we do
not have to include the potential energy of gravity. Doing so, we find
that the potential energy can be written as simply
2
V D 12 keq ym
;
where, as we saw in Example 9.4, keq is the single spring constant
equivalent to all the springs supporting the motor and platform. To find
the kinetic energy T , we note that there are two contributions: one from
motor and platform, which move together, and one from the rotating
mass. Therefore, we can write T as
T D 21 .mm
2
mu C mp /yPm
C 12 mu vu2 ;
where vu is the speed of the unbalanced mass. To determine vu , we can write its position as
rEu D " cos {O C .ym C " sin / |O D " cos !r t {O C .ym C " sin !r t / |O;
where we have let D !r t since the unbalanced mass spins at a constant angular velocity !r . Differentiating
rEu with respect to time and noting that {O and |O do not rotate, we have that
vEu D
"!r sin !r t {O C .yPm C "!r cos !r t / |O:
Since we need vu2 , we sum the square of the components of vEu to obtain
2
2
vu2 D "2 !r2 sin2 !r t C yPm
C 2yPm "!r cos !r t C "2 !r2 cos2 !r t D "2 !r2 C yPm
C 2yPm "!r cos !r t:
Therefore, the kinetic energy becomes
T D 21 .mm
2
2
mu C mp /yPm
C 21 mu "2 !r2 C yPm
C 2yPm "!r cos !r t :
Carrying out the derivatives in the given Lagrange’s equation, we obtain
@T
D .mm
@yPm
mu C mp /yPm C mu yPm C mu "!r cos !r t D .mm C mp /yPm C mu "!r cos !r t;
August 10, 2009
1317
Dynamics 1e
and then
d
dt
@T
@yPm
D .mm C mp /yRm
mu "!r2 sin !r t;
where we have used the fact that !r is constant. Taking the remaining two derivatives, we obtain
@T
D 0 and
@ym
@V
D keq ym :
@ym
Substituting these derivatives into the given Lagrange’s equation, we obtain
.mm C mp /yRm
mu "!r2 sin !r t C keq ym D 0
)
.mm C mp /yRm C keq ym D mu "!r2 sin !r t ,
which agrees with the equation of motion used in Example 9.4.
August 10, 2009
1318
Solutions Manual
Problem 9.73
Modeling the beam as a uniform thin bar, ignoring the inertia of the pulleys, assuming that the system is
in static equilibrium when the bar is horizontal, and assuming that the cord is inextensible and does not
go slack, determine the linearized equation of motion of the system. In addition, determine the system’s
natural frequency of vibration. Treat the parameters shown in the figure as known.
Solution
An FBD of the block A, an FBD of the bar B, and the
positive coordinate directions are shown at the right.
Summing forces in the x direction on the block, we
obtain the following balance law
X
Fy W mA g T D mA aAx D mA x:
R
(1)
where x measures the vertical position of the block
A. Summing moments about point O on bar B gives
X
Fs L mB g L
MO W T L
2
2 D IO ˛B ; (2)
where T is the cord tension, ˛B is the angular acceleration of the bar, Fs is the force in the spring, and
IO D 13 mB L2 is the mass moment of inertia of the bar B about point O. The only force law is for Fs and it
can be written as
Fs D k ıB .ıB /st ;
where .ıB /st is the deflection of the spring when the system is in static equilibrium. For the kinematics
relations, we need to relate the angular acceleration of the bar (˛B ) to the acceleration of the mass A (x)
R and
we need to relate the deflection of point B (ıB ) to the position of mass A (x). Assuming small displacements
and noting that the vertical displacement of A is equal and opposite to that of G, we can write
ıB D LB
and
xD
L
2 B
)
ıB D 2x
and RB D ˛B D
2
R
L x:
Solving Eq. (1) for T and substituting the result into Eq. (2), we obtain
L
2
.mA g
mA x/
R
kL ŒıB
.ıB /st 
2R
1
mB g L
2 D 3 mB L B :
Now substituting in the force laws and kinematics relations, this becomes
L
2 mA .g
x/
R
kL Œ2x
.ıB /st 
2
1
mB g L
2 D 3 mB L
2
R
Lx
August 10, 2009
1319
Dynamics 1e
Canceling an L and noting that 12 g.mA
mB / D k.ıB /st , we find the equation of motion to be
2
3 mB
C 21 mA xR C 2kx D 0:
Therefore, the natural frequency of vibration is
s
!n D
2k
2
3 mB
C 12 mA
.
August 10, 2009
1320
Solutions Manual
Problem 9.74
Revisit Example 9.6 and obtain the expression for the force transmitted to
the floor, using the expression for the steady-state response of the unbalanced motor.
Solution
Referring to the diagram at the right, we see that the force transmitted to
the floor F t is equal to the force in the springs Fs plus the force in the
dashpots Fd , i.e.,
F t D Fs C Fd D kyss C c yPss ;
where yss is the steady-state response and is measured from the static
equilibrium position of the system and we have ignored the static force
transmitted to the floor. In Example 9.6 we found the steady-state response
to be
yss D 0:00352mu sin.125:7t C 0:842/ m;
in which the following values were used for k and c
k D 420;000 N=m
and c D 4000 Ns=m:
Differentiating yss with respect to time and then substituting yss and yPss into the equation for F t , we obtain
yPss D 0:442mu cos.125:7t C 0:842/ m=s
)
F t D 1480mu sin.125:7t C 0:842/ C 1770mu cos.125:7t C 0:842/ N:
Therefore, for each of the values of the unbalanced mass mu D 0:01 kg, mu D 0:1 kg, and mu D 1 kg, we
obtain
mu D 0:01 kg W
F t D 14:8mu sin.125:7t C 0:842/ C 17:7mu cos.125:7t C 0:842/ N;
mu D 0:1 kg W
F t D 148mu sin.125:7t C 0:842/ C 177mu cos.125:7t C 0:842/ N;
mu D 1 kg W
F t D 1480mu sin.125:7t C 0:842/ C 1770mu cos.125:7t C 0:842/ N:
August 10, 2009
1321
Dynamics 1e
Problem 9.75
The system shown is released from rest when both springs
are unstretched and x D 0. Neglecting the inertia of the
pulley P and assuming that the disk rolls without slip,
derive the equation of motion of the system in terms of x.
Assume that point G is both the mass center of the disk
and its geometric center. Treat the quantities k1 , k2 , c,
m1 , m2 , and IG as known, where IG is the mass moment
of inertia of the disk. Finally, assuming that the system
is underdamped, derive an expression for the damped
natural frequency of the system.
Solution
The figure at the right shows an FBD of the mass m1 as
well as an FBD of the disk of mass m2 , where T is the
tension in the connecting cable, F is the friction force
between the disk and the ground, Fs1 is the force in spring
k1 , Fs2 is the force in spring k2 , and Fd is the force in the
dashpot. Summing forces in the x direction on mass m1 ,
we obtain
X
Fx W m1 g T Fs2 D m1 x:
R
(1)
Summing moments about point O on the disk of mass m2 , we obtain
X
MO W 2ro Fs1 C ro Fd .ro C ri /T D IO ˛disk ;
(2)
where ˛disk is the angular acceleration of the disk and IO D IG C m2 ro2 via the parallel axis theorem. For
the force laws, we have that
Fs2 D k2 x;
Fs1 D k1 ık1 ;
and
Fd D c yPG ;
where ık1 is the deflection of spring k1 and yPG is velocity of point G. In addition, we need to relate the
position of mass m1 , that is x, to ık1 and yPG . For small displacements, this is done via the relations
ık1
x
D
2ro
ro C ri
yG
x
D
ro
ro C ri
yG
x
D
ro
ro C ri
)
)
)
2ro
x;
ro C ri
ro
yPG D
x;
P
ro C ri
ık1 D
yRG D
ro ˛disk D
ro
xR
ro C ri
)
˛disk D
xR
;
ro C ri
where we have used the fact that point O is the instantaneous center of rotation for the disk. Substituting the
kinematics relations into the force laws and then substituting the force laws into Eqs. (1) and (2), we obtain
m1 g
T
k2 x D m1 xR
and
cro2
4ro2 k1
xC
xP
ro C ri
ro C ri
.ro C ri /T D
IG C m2 ro2
xR
:
ro C ri
August 10, 2009
1322
Solutions Manual
Solving the first equation for T and substituting the result into the second, we obtain the equation of motion
as
xR
4ro2 k1
cro2
xC
xP .ro C ri /.m1 g k2 x m1 x/
R D IG C m2 ro2
ro C ri
ro C ri
ro C ri
or, rearranging
2
4ro k1
IG C m2 ro2
cro2
xP C
C m1 .ro C ri / xR C
C k2 .ro C ri / x D .ro C ri /m1 g
ro C ri
ro C ri
ro C ri
IG C m2 ro2 C m1 .ro C ri /2 xR C cro2 xP C 4ro2 k1 C k2 .ro C ri /2 x D .ro C ri /2 m1 g.
)
Notice that because we measured the position of mass m1 from the unstretched state of the springs, we have
the constant forcing term .ro C ri /2 m1 g on the right-hand side of the equation of motion. Notice that the
static solution (i.e., for xR D 0 and xP D 0 is given by
xst D
.ro C ri /2 m1 g
;
4ro2 k1 C k2 .ro C ri /2
which is clearly a constant. As we have seen many times, if we were to measure x from this position instead,
we would add this constant to x to obtain xnew D x C xst , then the right-hand side would disappear, and the
coefficients of xR and xP would not change since xP new D xPpand xR new D x.
R
2
The damped natural frequency is given by !d D !n 1 , where
s
s
keq
4ro2 k1 C k2 .ro C ri /2
!n D
D
;
meq
IG C m2 ro2 C m1 .ro C ri /2
ceq
cro2
D p
D q
:
2 keq meq
2 4ro2 k1 C k2 .ro C ri /2 IG C m2 ro2 C m1 .ro C ri /2
Therefore, we can write the damped natural frequency as
s
!d D
"
4ro2 k1 C k2 .ro C ri /2
1
IG C m2 ro2 C m1 .ro C ri /2
c2r 4
o
4 4ro2 k1 C k2 .ro C ri /2 IG C m2 ro2 C m1 .ro C ri /2
#1=2
August 10, 2009
1323
Dynamics 1e
Problem 9.76
A ring of mass m is attached by two linear elastic cords to the
vertical supports as shown. The cords have elastic constant k
and unstretched length L0 < L. Assuming that the pretension
in the cords is large enough that the deflection of the cords due
to the ring’s weight can be neglected, find the nonlinear equation
of motion for the mass m.
Solution
Referring to the FBD on the right, we can sum forces in the y direction
to obtain
X
Fy W
2Fs sin ˇ D may D my;
R
(1)
where Fs is the force in each of the elastic cords. The force law for
each cord is given by
q
Fs D k
y 2 C L2 L0 :
Kinematically, we can write ˇ in terms of y if we note that
y
:
sin ˇ D p
y 2 C L2
Substituting the kinematics and the force law into the balance law in Eq. (1), we obtain
q
myR C 2k
y 2 C L2
L0 p
y
y 2 C L2
D0
)
myR C 2k 1
L0
p
y 2 C L2
!
y D 0.
August 10, 2009
1324
Solutions Manual
Problem 9.77
A ring of mass m is attached by two linear elastic cords to the
vertical supports as shown. The cords have elastic constant k and
unstretched length L0 < L. Assuming that the pretension in the
cords is large enough that the deflection of the cords due to the
ring’s weight can be neglected, use Newton’s second law to find
the linearized equation of motion about y D 0 for the mass m. In
addition, determine the natural frequency of the ring’s vibration.
Solution
Referring to the FBD on the right, we can sum forces in the y direction
to obtain
X
Fy W
2Fs sin ˇ D may D my;
R
(1)
where Fs is the force in each of the elastic cords. The force law for
each cord is given by
q
Fs D k
y 2 C L2 L0 :
Kinematically, we can write ˇ in terms of y if we note that
y
sin ˇ D p
:
2
y C L2
Substituting the kinematics and the force law into the balance law in Eq. (1), we obtain
q
myR C 2k
y 2 C L2
L0 p
y
y 2 C L2
D0
)
myR C 2k 1
L0
p
y 2 C L2
!
y D 0:
Looking at the second term in parentheses, we note that using a Taylor series expansion about y D 0 (this is
sometimes called a binomial expansion) we can write it as
"
#
4
1 1
L0
L0
1
L0
1 y 2
y
L0
2 2 C1
D
D
1
C
::: ;
p
p
2
2
2
L 1 C .y=L/
L
2 L
2Š
L
L
y CL
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
myR C 2k 1
L0
y D 0:
L
The natural frequency of the ring’s vibration is given by !n D
keq D 2k 1
L0
;
L
p
keq =meq , and so we can write
s
meq D m;
and
!n D
2k
1
m
L0
.
L
August 10, 2009
1325
Dynamics 1e
Problem 9.78
Solve Prob. 9.77 by finding the linearized equations of motion via
the energy method.
Solution
Referring to the FBD at the right, we see that only the forces due to
the elastic cords do work on the ring. Since they are linear elastic, the
system is conservative and we can apply the energy method to find
the equation of motion. The kinetic energy can be written as
T D 12 myP 2 :
In addition, the potential energy of the two elastic can be written as
" q
#
2
V D2
1
2k
y2
C
L2
L0
2
Dk y CL
2
2L0
q
y2
C
L2
C
L20
:
Since V is nonlinear, to apply the energy method, we need to expand it in a Taylor series about y D 0 and
keep up through quadratic terms (see Section 9.1). Rewriting V and expanding in a Taylor series, we obtain
q
V D k L2 C L20 C ky 2 2kL0 y 2 C L2
D k L2 C L20 C ky 2
8
9
ˇ
#ˇ
"
ˇ
<
ˇ
=
2
y
1
y
1
ˇ
ˇ
2
2kL0 L C p
y
C
yC
p
ˇ
ˇ
3=2 ˇ
:
;
2
y 2 C L2 ˇyD0
y 2 C L2
y 2 C L2
yD0
y2
C D k L2 C L20 C ky 2 2kL0 L C
2L
L0
2
2
k L C L0 2L0 L C k 1
y2
L
„
ƒ‚
…
constant
where we have used the Taylor series p
expansion about y D 0, which is given by f .y/ D f .0/ C f 0 .0/y C
1 00
2
y 2 C L2 . Notice that the first term in V is a constant and so we can
2 f .0/y C : : : and where f .y/ D
ignore it when applying the energy method. Doing so, we obtain
d
.T C V / D myP yR C 2k 1
dt
L0
y yP D 0
L
)
myR C 2k 1
L0
y D 0.
L
August 10, 2009