Mechanical Vibrations: Theory and Applications 1st Edition Solution Manual

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Solutions Manual to Accompany
MECHANICAL VIBRATIONS: THEORY AND APPLICATIONS,
1ST EDITION
S. GRAHAM KELLY

Contents
Preface vii
Chapter 1 1
Chapter 2 55
Chapter 3 134
Chapter 4 220
Chapter 5 369
Chapter 6 424
Chapter 7 526
Chapter 8 660
Chapter 9 727
Chapter 10 765
Chapter 11 853
Chapter 12 921
Chapter 13 993

1
CHAPTER 1: INTRODUCTION
Short Answer Problems
1.1 True: The earth is taken to be non-accelerating for purposes of modeling systems on
the surface of the earth.
1.2 False: Systems undergoing mechanical vibrations are not subject to nuclear reactions is
an example of an implicit assumption.
1.3 True: Basic laws of nature can only be observed and postulated.
1.4 False: The point of application of surface forces is on the surface of the body.
1.5 False: The number of degree of freedom necessary to model a mechanical system is
unique.
1.6 False: Distributed parameter systems are another name for continuous systems.
1.7 True: The Buckingham Pi theorem states that the number of dimensionless variables
required in the formulation of a dimensional relationship is the number of dimensional
variables, including the dependent variable, minus the number of dimensions involved in
the dimensional variables.
1.8 True: The displacement of its mass center (x and y coordinates) and the rotation about
an axis perpendicular to the mass center are degrees of freedom the motion of an
unconstrained rigid body undergoing planar motion.
1.9 False: A particle traveling in a circular path has a velocity which is tangent to the
circle.
1.10 False: The principle of work and energy is derived from Newton’s second law by
integrating the dot product of the law with a differential displacement vector as the particle
moves from one location to another.
1.11 The continuum assumption treats all matter as a continuous material and implies that
properties are continuous functions of the coordinates used in modeling the system.
1.12 An explicit assumption must be stated every time it is used, whereas an implicit
assumption is taken for granted.
1.13 Constitutive equations are used to model the stress-strain relationships in materials.
They are used in vibrations to model the force-displacement relationships in materials that
behave as a spring.
1.14 A FBD is a diagram of a body abstracted from its surroundings and showing the
effects of the surroundings as forces. They are drawn at an arbitrary time.

Chapter 1: Introduction
2
1.15 The equation represents simple harmonic motion
1.16 (a) X is the amplitude of motion; (b)߱ is the frequency at which the motion occurs
(c) ߶ is the phase between the motion and a pure sinusoid.
1.17 The phase angle is positive for simply harmonic motion. Thus the response lags a
pure sinusoid.
1.18 A particle has mass that is concentrated at a point. A rigid body has a distribution of
mass about the mass center.
1.19 A rigid body undergoes planar motion if (1) the path of its mass center lies in a plane
and (2) rotation occurs only about an axis perpendicular to the plane of motion of the mass
center.
1.20 The acceleration of a particle traveling in a circular path has a tangential component
that is the radius of the circle times the angular acceleration of the particle and a centripetal
acceleration which is directed toward the center of the circle which is the radius time the
square of the angular velocity.
1.21 An observer fixed at A observes, instantaneously that particle B is moving in a
circular path of radius ܚห஻/஺ ห about A.
1.22 It is applied to the FBD of the particle.
1.23 The effective forces for a rigid body undergoing planar motion are a force applied at
the mass center equal to݉ ത ܉ and a moment equal to ߙҧܫ.
1.24 The two terms of the kinetic energy of a rigid body undergoing planar motion are
ଵ
݉ଶ ݒ ҧ ଶ , the translational kinetic energy, and ଵ
ଶ ߱ҧܫ ଶ , the rotational kinetic energy.
1.25 The principle of impulse and momentum states that a body’s momentum (linear or
angular) momentum at ݐଵ plus the external impulses applied to the body (linear or angular)
between ݐଵ and ݐଶ is equal to the system’s momentum (linear or angular) at ݐଶ .
1.26 One, let ߠ be the angular rotation of the bar, measured positive counterclockwise,
from the system’s equilibrium position.
1.27 Four, let ݔଵ be the absolute displacement of the cart, ݔଶ the displacement of the
leftmost block relative to the cart, ݔଷ the displacement of the rightmost block away from
the cart and ߠ the counterclockwise angular rotation of the bar.
1.28 Four, let ݔଵ represent the displacement of the center of the disk to the right, ݔଶ the
downward displacement of the hanging mass, ݔଷ the displacement of the sliding mass to
the left and ߠ the counterclockwise angular rotation of the rightmost pulley.

Chapter 1: Introduction
3
1.29 Two, let ߠ be clockwise the angular displacement of the bar and x the downward
displacement of the hanging mass.
1.30 Three, let x be the downward displacement of the middle of the upper bar, ߠ its
clockwise angular rotation and ߶ the clockwise angular rotation of the lower bar.
1.31 Three, let ߠ represent the clockwise angular rotation of the leftmost disk, ߶ the
clockwise angular rotation of the rightmost disk and x the upward displacement of the
leftmost hanging mass.
1.32 Infinite, let x be a coordinate measured along the neutral axis of the beam measured
for the fixed support. Then the displacement is a continuous function of x and t, w(x,t).
1.33 Three, let ݔଵ be the downward displacement of the hand, ݔଶ the downward
displacement of the palm and ݔଷ the displacement of the fingers.
1.34 Given: Uniform acceleration, a=2 m/s. (a) ݒ ൅ ݐ ܽൌ ሻݐሺݒ ଴ ฺ ݒ ሺ5ሻ ൌ ቀ2 ௠
௦ మ ቁ ሺ5 sሻ ൅
ቀ0 ୫
ୱ ቁ ൌ 10 m (b) ܽൌ ሻݐሺݔ ௧ మ
ଶ ݒ ൅଴ ݔ ൅ ݐ଴ ฺ ݔ ሺ5ሻ ൌ ଵ
ଶ ቀ2 ୫
ୱమ ቁ ሺ5 sሻଶ ൌ 25 m
1.35 Given: ܞൌ 2 cos 2 ܑݐ൅ 3 sin 2 ܒ ݐ൅ 0.4 m/s. (a) ൌ ܉ ܞௗ
ௗ௧ ൌ െ4 sin 2ݐ ܑ ൅
6 cos 2 ܒ ݐm/s ଶฺܽ ሺߨሻ ൌ െ4 sin 2ߨ ܑ ൅ 6 cos 2ߨ ܒ ୫
ୱమ ൌ 6 ܒm/s ଶ (b) ൌ ݐ݀ܞ ׬ ൌ ܚ
ቂሺsin 2 ݐ൅ ܥଵ ሻ ܑ൅ ቀെ ଷ
ଶ cos 2 ݐ൅ ܥଶ ቁ ܒ൅ ሺ0.4 ݐ൅ ܥଷ ቃܓሻ m. The particle starts at the origin
at t = 0. Application of this condition leads to) ܚሺtሻ ൌ ቂሺsin 2 ݐሻ ܑ൅ ቀെ ଷ
ଶ cos 2ݐ ൅ ଷ
ଶ
ቁ ܒ ൅
ܓݐ0.4 m. Evaluation at
ߨ leads to ሺܚπሻ
ൌsin
൅ ܑ ߨ2−
32cos
ܒ൅32ߨ2
ܓߨ൅0.4 m =
ܓߨ0.4 m.
1.36 Given: v=2 m/s, r=3 m, ߠሺ0ሻ ൌ 0 (a) ൌ ݒ ௗ௦
ௗ௧ฺ ൌ ݏ ݐ2 ൌ ݐ݀ݒ ׬ at t=2 s the particle
has traveled 4 m. But ߠݎ ൌ ݏ thus ൌ ߠ ସ ୫
ଷ ୫ ൌ 1.33 rad ൌ 76.2°. (b) The acceleration of a
particle traveling on a circular path has two components. One is ௗ௩
ௗ௧ which is tangent to the
circle and is zero for this problem. The other component is ௩ మ
௥ ൌ ሺଶ ୫/ୱሻ మ
ଷ ୫ ൌ 1.33 m directed
toward the center of the circle from the position of the particle.
1.37 Given: m=2 kg,ܫҧ ൌ 0.5 kg · mଶ ܉ ,ത ൌ ሺ5 ܑ൅ 3 ܒሻ m/s ଶ, ߙൌ 10 rad/s ଶ. Effective
forces are݉ ܉ത ൌ ሺ2 kgሻ ቂሺ5 ܑ൅ 3ܒ ሻ ୫
ୱమ ቃ ൌ ሺ10 ܑ൅ 15ܒ ሻ ୫
ୱమ applied at the mass center and a
couple ܫҧ ߙൌ ሺ0.5 kg · mଶ ሻሺ10 rad/s ଶ ሻ ൌ 5 N · m.
1.38 Given: m = 0.1 kg, ܞൌ ሺ9 ܑ൅ 11j ሻ m/s. The kinetic energy of the particle isܶ ൌ
ଵ
݉ଶ |ܞ| ଶ ൌ ଵ
ଶ ሺ0.1 kgሻ൫√9 ଶ ൅ 11 ݉ ଶ ൯ݏ/ ଶ
ൌ 0.711 J.

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Chapter 1: Introduction
4
1.39 Given: m=3 kg, ܞത ൌ ሺ3 ܑ൅ 4 ܒሻ m/s, d=0.2 m The angular velocity is calculated from
ܞ|ฺ߱߱݀ൌ |ത ൌ ܞ|ത|
ௗ =ହ ୫/ୱ
଴.ଶ ୫=20 rad/s.
1.40 Given: ൌ 100 J , ܫൌ 0.03 kg · mଶ The kinetic energy of a rigid body which rotates
about its centroidal axis isܶ ൌ ଵ
ଶ ߱ܫ ଶ. Thus 100 J ൌ ଵ
ଶ ሺ0.03 kg · mଶ ߱ሻ ଶ which leads to߱
ൌ 81.65 ୰ୟୢ
ୱୣୡ .
1.41 Given: m = 5 kg, ݒҧ ൌ 4 m/s,߱ ൌ 20 rad/s , ܫҧ ൌ 0.08 kg · mଶ. The kinetic energy of
a rigid body undergoing planar motion isܶ ൌ ଵ
݉ଶ ݒ ҧ ଶ ൅ ଵ
ଶ ߱ҧܫ ଶ ൌ ଵ
ଶ ሺ5 kgሻሺ4 m/sሻଶ ൅
ଵ
ଶ ሺ0.08 kg · mଶ ሻሺ20 rad/sሻ ଶ ൌ 56 J.
1.42 Given: F=12,000 N, ∆ ݐൌ 0.03 s. The impulse applied to the system is ൌ ݐ∆ܨൌ ܫ
ሺ12,000 Nሻሺ0.03 sሻ ൌ 360 N · s.
1.43 Given: m = 3 kg, ݒଵ ൌ 0 m/s, force as given in Figure (a) The impulse imparted to
the particle is ׬ ൌ ܫ ݐ݀ܨ
ଷ ୱ
଴ ൌ ଵ
ଶ ሺ1ሻሺ100ሻ ൅ 2ሺ100ሻ ൅ ଵ
ଶ ሺ1ሻሺ100ሻ ൌ 300 N · s (b) The
velocity at t=2 s is given by the principle of impulse and momentum݉ ൌ ݒ ׬ ݐ݀ܨ
ଶ ୱ
଴ฺ
ൌ ݒ ׬ ிௗ௧
మ ౩
బ
௠ ൌ ଶହ଴ N·ୱ
ଷ ୩୥ ൌ 83.3 m/s. (c) The velocity after 5 s is ൌ ݒ ׬ ிௗ௧
ఱ ౩
బ
௠ ൌ ଷ଴଴ N·ୱ
ଷ ୩୥ ൌ
100 m/s.
1.44 Given: m = 2 kg, F=6 N, t=10 s, ݒଵ ൌ 4 m/s. The principle of work and energy is
used to calculate how far the particle travelsܶ ଵ ܷ൅ ଵ՜ଶ ܶൌ ଶ after the velocity ݒଶis
calculated from the principle of impulse and momentum݉ ݒ ଵ ݒ ݉ൌ ܫ ൅ ଶฺ ݒ ଶ ൌ
௠௩భ ାூ
௠ ൌ ሺଶ ୩୥ሻሺସ ୫/ୱሻାሺ଺ Nሻሺଵ଴ ୱሻ
ଶ ୩୥ ൌ 34 m/s. Then letting x be the distance traveled
application of work and energy gives ଵ
ଶ ሺ2 kgሻሺ4 m/sሻଶ ൅ ሺ6 Nሻݔ ൌ ଵ
ଶ ሺ2 kgሻሺ34 m/
s2which is solved to yield x=190 m.
1.45 (a) -(ii) (b)-(iv) (c)-(i) (d)-(v) (e)-(i) (f)-(v) (g)-(vi) (h)-(iii) (i)-(ix)

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Chapter 1: Introduction
5
Chapter Problems
1.1 The one-dimensional displacement of a particle is
ݔሺݐሻ ൌ 0.5݁ି ଴.ଶ௧ sin 5ݐ m
(a) What is the maximum displacement of the particle? (b) What is the maximum velocity
of the particle? (c) What is the maximum acceleration of the particle?
Given: x(t)
Find:ݔሻ ܽሺ௠௔௫ ݒሻ ܾሺ௠௔௫ ܽሻ ܿሺ ௠௔௫
Solution: (a) The maximum displacement occurs when the velocity is zero. Thus
ݔሶሺݐሻ ൌ 0.5݁ି ଴.ଶ௧ ሺെ0.2 sin 5 ݐ൅ 5 cos 5ݐሻ
Setting the velocity to zero leads to
െ0.2 sin 5 ݐ൅ 5 cos 5 ݐൌ 0
or tan 5 ݐൌ 25 . The first time that the solution is zero is t=0.3062. Substituting this value
of t into the expression for x(t) leads to
ݔ௠௔௫ ൌ 0.4699 m
(b) The maximum velocity occurs when the acceleration is zero
ݔሷሺݐሻ ൌ 0.5݁ି ଴.ଶ௧ ሾെ0.2ሺെ0.2 sin 5 ݐ൅ 5 cos 5ݐሻ െ cos 5 ݐെ 25 sin 5ݐሿ
ൌ 0.5݁ି ଴.ଶ௧ ሺെ24.96 sin 5 ݐെ 6 cos 5ݐሻ
The acceleration is zero when 24.96 sin 5 ݐെ 6 cos 5 ݐൌ 0ฺ tan 5 ݐൌ െ0.240 .
The first time that this is zero is t=0.5812 which leads to a velocity of
ݒ௠௜௡ ൌ െ2.185 m/s
(c) The maximum acceleration occurs when ݔഺ ൌ 0,
ݔഺ ൌ 0.5݁ି ଴.ଶ௧ ሾെ0.2ሺെ24.96 sin 5 ݐെ 6 cos 5ݐሻ െ ሺ24.96ሻሺ5ሻ cos 5 ݐ൅ 30 sin 5ݐሿ
ൌ 0.5݁ି ଴.ଶ௧ ሺ34.992 sin 5 ݐെ 123.6 cos 5ݐሻ
The maximum acceleration occurs when 34.992 sin 5 ݐെ 123.6 cos 5 ݐൌ 0ฺ
tan 5 ݐൌ 3.53. The time at which the maximum acceleration occurs is t=0.2589 s
which leads toܽ
௠௔௫ ൌ െ12.18 m/s ଶ
Problem 1.1 illustrates the relationships between displacement, velocity and acceleration.

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Chapter 1: Introduction
6
1.2 The one-dimensional displacement of a particle is
msin.0 0.24)+(5te0.5=x(t) 2t- (1)
(a) What is the maximum displacement of the particle? (b) What is the maximum velocity of
the particle? (c) What is the maximum acceleration of the particle?
Given: x(t)
Find:ݔሻ ܽሺ௠௔௫ ݒሻ ܾሺ௠௔௫ ܽሻ ܿሺ ௠௔௫
Solution: (a) The maximum displacement occurs when the velocity is zero. Thus
ݔሶሺݐሻ ൌ 0.5݁ି ଴.ଶ௧ ሾെ0.2 sinሺ5 ݐ൅ 0.24ሻ ൅ 5 cosሺ5 ݐ൅ 0.24ሻሿ
Setting the velocity to zero leads to
െ0.2 sinሺ5 ݐ൅ 0.24ሻ ൅ 5 cosሺ5 ݐ൅ 0.24ሻ ൌ 0
or tanሺ5 ݐ൅ 0.24ሻ ൌ 0.2582. The first time that the solution is zero is t=0.3062.
Substituting this value of t into the expression for x(t) leads to
ݔ௠௔௫ ൌ 0.4745 m
(b) The maximum velocity occurs when the acceleration is zero
ݔሷሺݐሻ ൌ 0.5݁ି ଴.ଶ௧ ሼെ0.2ሾሺെ0.2 sinሺ5 ݐ൅ 0.24ሻ ൅ 5 cosሺ5 ݐ൅ 0.24ሻሻሿ
െ cosሺ5 ݐ൅ 0.24ሻ െ 25 sinሺ5 ݐ൅ 0.24ሻሽ
ൌ 0.5݁ି ଴.ଶ௧ ሾെ24.96 sinሺ5 ݐ൅ 0.24ሻ െ 6 cosሺ5 ݐ൅ 0.24ሻሿ
The acceleration is zero when
െ24.96 sinሺ5 ݐ൅ 0.24ሻ െ 6 cosሺ5 ݐ൅ 0.24ሻ ൌ 0ฺ tan ሺ5 ݐ൅ 0.24ሻ ൌ െ0.240.
The first time that this is zero is t = 0.5332 which leads to a velocity of
ݒ௠௜௡ ൌ െ2.0188 m/s
(c) The maximum acceleration occurs when ݔഺ ൌ 0,
ݔഺ ൌ 0.5݁ି ଴.ଶ௧ ሼെ0.2ሾെ24.96 sinሺ5 ݐ൅ 0.24ሻ െ 6 cosሺ5 ݐ൅ 0.24ሻሿ
െ ሺ24.96ሻሺ5ሻ cosሺ5 ݐ൅ 0.24ሻ ൅ 30 sinሺ5 ݐ൅ 0.24ሻሽ
ൌ 0.5݁ି ଴.ଶ௧ ሾ34.992 sinሺ5 ݐ൅ 0.24ሻ െ 123.6 cosሺ5 ݐ൅ 0.24ሻሿ
The maximum acceleration occurs when
34.992 sinሺ5 ݐ൅ 0.24ሻ െ 123.6 cosሺ5 ݐ൅ 0.24ሻ ൌ 0ฺ tan ሺ5 ݐ൅ 0.24ሻ ൌ 3.53.
The time at which the maximum acceleration occurs is t=0.2109 s which leads toܽ
௠௔௫ ൌ െ12.30 m/s ଶ
Problem 1.2 illustrates the relationships between displacement, velocity and acceleration.

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Chapter 1: Introduction
7
42m/sec
75m/sec
2
2
1.3 At the instant shown in Figure P1.3, the slender
rod has a clockwise angular velocity of 5 rad/sec and a
counterclockwise angular acceleration of 14 rad/sec2
.
At the instant shown, determine (a) the velocity of
point P and (b) the acceleration of point P.
Given: ω = 5 rad/sec, α = 14 rad/sec2
, θ = 10°
Find: ݒ௣ , aP
Solution: The particle at the pin support, call it O, is fixed. Hence its velocity and acceleration
are zero. Using the relative velocity and acceleration equations between two particles on a
rigid body
ܞ௉ ܞ ൌܗ ൅ ૑ ൈ ܚ௉/ை ൌ െ5 ܓൈ ሺ3 cos 10°ܑ െ 3 sin 10 ° ܒሻ ൌ െ15 sin 10°ܑ െ 15 cos 10 ° ܒ
ൌ െ2.604 ܑെ 14.772 ܒ
and
OPOPOP +)(+= // αxrωxrxωaa
s
m
9.85
s
m
2
2
=P
P )54.3+(-66.5=
a
jia
Alternate solution: The bar is rotating about a fixed point. Thus any point on the bar moves on
a circular arc about the point of support. The particle P has two components of acceleration,
one directed between P and O (the normal acceleration), and one tangent to the path of P
whose direction is determined using the right hand rule (the tangential component).
The component normal to the path of P is
s
m
s
rad
5m 2
75=)(3=a 2
n
and is directed between P and O. The tangential acceleration is
s
m
s
rad
m 22 42=))(14(3=a t
The normal and tangential components of acceleration are illustrated on the diagram below.

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Chapter 1: Introduction
8
Problem 1.3 illustrates the use of the relative acceleration equation of rigid body kinetics.
1.4 At t = 0, a particle of mass 1.2 kg is traveling with a speed of 10 m/s that is increasing
at a rate of 0.5 m/s2 . The local radius of curvature at this instant is 50 m. After the particle
travels 100 m, the radius of curvature of the particle's path is 50 m.
(a) What is the speed of the particle after it travels 100 m?
(b) What is the magnitude of the particle’s acceleration after it travels 100 m?
(c) How long does it take the particle to travel 100 m?
(d) What is the external force acting on the particle after it travels 100 m?
Given: m = 1.2 kg, v(t=0) = 10 m/s, dv/dt= 0.5 m/s2 , and r = 25 m when s = 100 m
Find: (a) v when s = 100 m, (b) a when s = 3 m, (c) t when s = 3 m
Solution: Let s(t) be the displacement of the particle, measured from t = 0. The particle’s
velocity is
∫∫ +==+=
t
t
tdtvdt
dt
dv
tv
0
0 105.05.0)0()(
By definition v=ds/dt. Thus the displacement of the particle is obtained as
( )∫ ∫ +=+=+=
t t
ttdttsdtvts
0 0
2 1025.0105.0)0()(
When s = 100 m,
s28.81025.0m100 2 =⇒+= ttt
(a) The velocity when s = 100 m is
m/s14.1410)28.8(5.0 =+=v
(b) Since the particle is traveling along a curved path, its acceleration has two components:
a tangential component equal to the rate of change of the velocity
2
m/s5.0== dt
dv
a t
and a normal component directed toward the center of curvature
2
22
m/s00.4
m50
)m/s14.14( === r
v
a n
The magnitude of the acceleration at this instant is

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Chapter 1: Introduction
9
2
222222
m/s03.4a
)m/s00.4()m/s5.0(a
=
+=+= nt aa
(c) The time for the particle to travel 100 m is previously calculated as t = 8.28 s
(d) The external force equation written in terms of magnitudes is
෍|۴| ൌ݉ | ܽ|
which upon application to the particle gives
෍|۴| ൌ ሺ1.2 kgሻ ቀ4.03 m
s ଶ ቁ ൌ 4.84 N
Problem 1.4 illustrates the kinematics of a particle traveling along a curved path.
1.5 The machine of Figure P1.15 has a vertical displacement,
x(t). The machine has component which rotates with a constant
angular speed,
ω. The center of mass of the rotating component
is a distance e from its axis of rotation. The center of mass of the
rotating component is as shown at t = 0. Determine the vertical
component of the acceleration of the rotating component.
Given: e, ω, x (t)
Find: a y
Solution: The particle of interest is on a component that moves
relative to the machine. From the relative acceleration equation,
MGMG aaa +=
where
( ) ja txM &&−=
and
( )jia
θ
θ
ω sincos2 −−= eMG
Since the angular velocity of the rotating component is constant and θ = 0 when t = 0,
t
ω
θ =
Hence the vertical acceleration of the center of mass of the rotating component is

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Chapter 1: Introduction
10
( ) tetxa 2
y
ω
ω sin−−= &&
Problem 1.5 illustrates application of the relative acceleration equation. Vibrations of
machines subject to a rotating unbalance are considered in Chapter 4.
1.6 The rotor of Figure P1.6 consists of a disk mounted
on a shaft. Unfortunately, the disk is unbalanced, and
the center of mass is a distance e from the center of the
shaft. As the disk rotates, this causes a phenomena
called “whirl”, where the disk bows. Let r be the
instantaneous distance from the center of the shaft to
the original axis of the shaft and ߠ be the angle made
by a given radius with the horizontal. Determine the
acceleration of the mass center of the disk.
Given: e, r
Find: ܉ത
Solution: The position vector from the origin to the center of the disk is ܑݎ ௥ where r varies
with time. The mass center moves in a circular path about the center of the disk. The
relative acceleration equation gives
܉ ௖ ܉ ൌ ௢ ൅ હ ൈ ܑݎ ௥ ൅ ૑ ൈ ሺ૑ ൈ ܑݎ ௥ ሻ ൅ ݎሷܑ ௥ ൅ 2૑ ൈ ݎሶܑ ௥
܉ ௖ ൌ ൫ݎሷ െ ߠݎሶ ଶ ܑ൯ ௥ ܑሶ൯ߠሶݎ2 ൅ ሷߠݎ൫ ൅ ఏ
The acceleration of the mass center is then
܉ത ൌ ൫ݎሷ െ ߠݎሶ ଶ ܑ൯ ௥ ܑሶ൯ߠሶݎ2 ൅ ሷߠݎ൫ ൅ ఏ ߱݁െ ଶ ሾcosሺ ݐ߱െ ߠ ܑሻ ௥ ൅ sinሺ ݐ߱െ ߠ ܑሻ ఏ ሿ
Problem 1.6 illustrates application of the relative acceleration equation.
1.7 A 2 ton truck is traveling down an
icy, 10º hill at 50 mph when the driver
sees a car stalled at the bottom of the
hill 250 ft away. As soon as he sees the
stalled car, the driver applies his brakes,
but due to the icy conditions, a braking
force of only 2000 N is generated. Does
the truck stop before hitting the car?
Given: W = 4000 lb.,
θ = 10 o , d = 250 ft., Fb = 2000 N = 449.6 lb,
vo = 50 mph = 73.33 ft/sec
10º
250 ft

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Chapter 1: Introduction
11
N
=
W
W
g a
F b
EXTERNAL FORCES EFFECTIVE FORCES
Find: v = 0 before x = 250 ft.
Solution: Application of Newton’s Law to the free body diagram of the truck at an
arbitrary instant
( ) ( )
2
0
2
..
sec
ft
973.1
10sin
lb4000
lb449.6
sec
ft
2.32
sin
sin
=
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛ +−=
⎟
⎠
⎞
⎜
⎝
⎛ +−=
=+−
= ∑∑
a
a
W
F
ga
a
g
W
WF
FF
b
b
effxextx
θ
θ
Since the acceleration is constant, the velocity and displacement of the truck are
tttv
t
ax
tvatv
33.73986.0
2
33.73973.1
2
0
2
0
+=+=
+=+=
The acceleration is positive thus the vehicle speeds up as it travels down the incline. The
truck does not stop before hitting the car.
Problem 1.7 illustrates application of Newton’s Law to a particle and kinematics of
constant acceleration.
1.8 The contour of a bumpy road is approximated by y(x) = 0.03 sin(0.125x) m. What is the
amplitude of the vertical acceleration of the wheels of an automobile as it travels over this
road at a constant horizontal speed of 40 m/s?

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Chapter 1: Introduction
12
Given: y(x) = 0.03sin(0.125x) m, v = 40 m/s
Find: A
Solution: Since the vehicle is traveling at a constant horizontal speed its horizontal distance
traveled in a time t is x = vt. Thus the vertical displacement of the vehicle is
[ ] m)5sin(03.0)40(125.0sin03.0)( ttty ==
The vertical velocity and acceleration of the vehicle are calculated as
2
m/s)5sin(75.0)5sin()5(15.0)(
m/s)5cos(15.0)5cos()5(03.0)(
ttta
tttv
−=−=
==
Thus the amplitude of acceleration is A=0.75 m/s2 .
Problem 1.8 illustrates the relationship between displacement, velocity, and acceleration
for the motion of a particle.
1.9 The helicopter of Figure P1.9 has a horizontal speed of 110 ft/s and a horizontal
acceleration of 3.1 ft/s2 . The main blades rotate at a constant speed of 135 rpm. At the
instant shown, determine the velocity and acceleration of particle A.
Given: v h = 110 ft/s, a h =3ft/s 2 , ω = 135 rpm = 14.1 rad/s, r = 2.1 ft
Find: v A , aA
Solution: Construct a x-y coordinate system in the horizontal plane
As illustrated. Using this coordinate system
2
ft/s3ft/s,110 iaiv −=−=
The position vector of A relative to the helicopter at this instant is
[ ] jijir hA 48.148.1)4/sin()4/cos(/ −=−=
π
πr
The relative velocity equation is used to determine the velocity of particle A as
ft/s9.201.89
)48.148.1(1.14110
/
jiv
jikiv
rkvv
A
A
hAhA
+−=
−×+−=
×+=
ω
The relative acceleration equation is used to determine the acceleration of particle A as

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Chapter 1: Introduction
13
2
//
ft/s6.2944.297
)87.2087.20(1.141.3
)(
jia
jikia
rkkrkaa
A
A
hAhAhA
+−=
+×+−=
××+×+=
ω
ω
α
Problem 1.9 illustrates the use of the relative velocity and relative acceleration equations.
1.10 For the system shown in Figure P1.10,
the angular displacement of the thin disk is
given by ߠሺݐሻ ൌ 0.03 sin ቀ30ݐ ൅ గ
ସ ቁ rad. The
disk rolls without slipping on the surface.
Determine the following as functions of time.
(a) The acceleration of the center of the disk.
(b) The acceleration of the point of contact
between the disk and the surface. (c) The
angular acceleration of the bar. (d) The
vertical displacement of the block. (Hint:
Assume small angular oscillations ߶ of the
bar. Then sin ߶ൎ ߶.)
Given: ሻݐሺߠ, ݎௗ ൌ 0.1 m, ℓ ൌ 0.3 m,݀ ൌ 0.2 m
Find: (a)ܽ ത ௗ (b)ܽ ത ௖ (c) ߙ (d) x
Solution: (a) The angular acceleration of the disk is
ߠሷሺݐሻ ൌ െሺ30ሻ ଶ 0.03 sin ቀ30 ݐ൅ ߨ
4ቁ ൌ െ27 sin ቀ30 ݐ൅ ߨ
4ቁ rad
s ଶ
Since the disk rolls without slip the acceleration of the mass center isܽ
ௗ ൌ ߙݎൌ ሺ0.1 mሻ ൬െ27 sin ቀ30 ݐ൅ ߨ
4ቁ rad
s ଶ ൰ ൌ െ2.7 sin ቀ30 ݐ൅ ߨ
4ቁ m
s ଶ
(b) Since the disk rolls without slip the horizontal acceleration of the point of contact is
zero. The vertical acceleration is ሶߠݎ ଶ towards the center
܉ ௖ ൌ ሺ0.1 mሻ ൤ሺ30ሻሺ0.03ሻ cos ቀ30 ݐ൅ ߨ
4ቁ rad
s ൨
ଶ
ܒ

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Mechanical Vibrations: Theory and Applications 1st Edition Solution Manual

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