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Sample Problem 8/2
The 8-kg body is moved 0.2 m to the right of the equilibrium position andreleased from rest at time t � 0. Determine its displacement at time t � 2 s. Theviscous damping coefficient c is 20 and the spring stiffness k is 32 N/m.
Solution. We must first determine whether the system is underdamped, criti-cally damped, or overdamped. For that purpose, we compute the damping ratio �.
Since � � 1, the system is underdamped. The damped natural frequency is �d � � � 1.561 rad/s. The motion is given by Eq. 8/12and is
The velocity is then
Evaluating the displacement and velocity at time t � 0 gives
Solving the two equations for C and � yields C � 0.256 m and � � 0.896 rad.Therefore, the displacement in meters is
Evaluation for time t � 2 s gives x2 � �0.01616 m. Ans.
Sample Problem 8/3
The two fixed counterrotating pulleys are driven at the same angular speed�0. A round bar is placed off center on the pulleys as shown. Determine the nat-ural frequency of the resulting bar motion. The coefficient of kinetic friction be-tween the bar and pulleys is �k.
Solution. The free-body diagram of the bar is constructed for an arbitrary dis-placement x from the central position as shown. The governing equations are
Eliminating NA and NB from the first equation yields
We recognize the form of this equation as that of Eq. 8/2, so that the natural fre-quency in radians per second is �n � and the natural frequency in cy-cles per second is
Ans.ªn � 12�
�2�k g/a
�2�k g/a
x � 2�k g
a x � 0
aNB� �a2
� x�mg � 0[ΣMA � 0]
NA � NB � mg � 0[ΣFy � 0]
�k NA � �k NB � mx[ΣFx � mx]
x � 0.256e�1.25t sin (1.561t � 0.896)
x0 � C sin � � 0.2 x0 � �1.25C sin � � 1.561C cos � � 0
x � �1.25Ce�1.25t sin (1.561t � �) � 1.561Ce�1.25t cos (1.561t � �)
x � Ce���nt sin (�dt � �) � Ce�1.25t sin (1.561t � �)
2�1 � (0.625)2�n�1 � �2
�n � �k/m � �32/8 � 2 rad/s � � c2m�n
� 202(8)(2)
� 0.625
N � s/m,
610 Chapter 8 Vibrat ion and T ime Response
k
xc
8 kg
x
Equilibriumposition
mg
N
cx = 20x··
kx = 32x
a0ω 0ω
Centralposition
y
a––2
a––2x
A BG
NA NBmg
k NAµ k NBµ
�
�
�
Helpful Hint
� We note that the exponential factore�1.25t is 0.0821 at t � 2 s. Thus, � � 0.625 represents severe damp-ing, although the motion is still oscillatory.
Helpful Hints
� Because the bar is slender and doesnot rotate, the use of a momentequilibrium equation is justified.
� We note that the angular speed �0
does not enter the equation of mo-tion. The reason for this is our as-sumption that the kinetic frictionforce does not depend on the relativevelocity at the contacting surface.
c08.qxd 6/28/06 4:48 PM Page 610
Sample Problem 8/7
A simplified version of a pendulum used in impact tests is shown in the figure.Derive the equation of motion and determine the period for small oscillationsabout the pivot. The mass center G is located a distance � 0.9 m from O, and theradius of gyration about O is kO � 0.95 m. The friction of the bearing is negligible.
Solution. We draw the free-body diagram for an arbitrary, positive value of theangular-displacement variable , which is measured counterclockwise for the coor-dinate system chosen. Next we apply the governing equation of motion to obtain
or Ans.
Note that the governing equation is independent of the mass. When is small,sin � , and our equation of motion may be written as
The frequency in cycles per second and the period in seconds are
Ans.
For the given properties: Ans.
Sample Problem 8/8
The uniform bar of mass m and length l is pivoted at its center. The springof constant k at the left end is attached to a stationary surface, but the right-endspring, also of constant k, is attached to a support which undergoes a harmonicmotion given by yB � b sin �t. Determine the driving frequency �c which causesresonance.
Solution. We use the moment equation of motion about the fixed point O toobtain
Assuming small deflections and simplifying give us
The natural frequency should be recognized from the now-familiar form of theequation to be
Thus, �c � �n � will result in resonance (as well as violation of our small-angle assumption!). Ans.
�6k/m
�n � �6k/m
� 6km
� 6kbml
sin �t
��k l2
sin � l2
cos � k� l2
sin � yB� l2
cos � 112
ml2
� gr
kO
2 � 0
� gr
kO
2 sin � 0
�mgr sin � mkO
2 [ΣMO � IO ]
r
636 Chapter 8 Vibrat ion and T ime Response
G
O
Oy
Ox
θ
r–
mg
θ
r–
yB = b sin ωt
O
B
mk
l—2
l—2
k
Oy
mg
Oxθ
k )( l—2
sin θk – y
B)( l—2
sin θ
Helpful Hints
� With our choice of point O as themoment center, the bearing reac-tions Ox and Oy never enter theequation of motion.
� For large angles of oscillation, deter-mining the period for the pendulumrequires the evaluation of an ellipticintegral.
Helpful Hints
� As previously, we consider only thechanges in the forces due to a move-ment away from the equilibriumposition.
� The standard form here is � �
where M0 � and IO �
The natural frequency �n of asystem does not depend on the exter-nal disturbance.
112
ml2.
klb2
M0 sin �tIO
,
�n
2
�
�
�
�
ƒn �1
2� � gr
kO
2 � �
1ƒn
� 2��kO
2
gr
� � 2�� (0.95)2
(9.81)(0.9)� 2.01 s
c08.qxd 6/28/06 4:48 PM Page 636
Sample Problem 8/9
Derive the equation of motion for the homogeneous circular cylinder, whichrolls without slipping. If the cylinder mass is 50 kg, the cylinder radius 0.5 m,the spring constant 75 N/m, and the damping coefficient 10 determine
(a) the undamped natural frequency
(b) the damping ratio
(c) the damped natural frequency
(d) the period of the damped system.
In addition, determine x as a function of time if the cylinder is released from restat the position x � �0.2 m when t � 0.
Solution. We have a choice of motion variables in that either x or the angulardisplacement of the cylinder may be used. Since the problem statement in-volves x, we draw the free-body diagram for an arbitrary, positive value of x andwrite the two motion equations for the cylinder as
The condition of rolling with no slip is � Substitution of this conditioninto the moment equation gives F � Inserting this expression for thefriction force into the force equation for the x-direction yields
Comparing the above equation with that for the standard dampedoscillator, Eq. 8/9, allows us to state directly
(a) Ans.
(b) Ans.
Hence, the damped natural frequency and the damped period are
(c) Ans.
(d) Ans.
From Eq. 8/12, the underdamped solution to the equation of motion is
At time t � 0, x and become
The solution to the two equations in C and � gives
Thus, the motion is given by
Ans.x � �0.200e�0.0667t sin (0.998t � 1.504) m
C � �0.200 m � � 1.504 rad
x0 � �0.0667C sin � � 0.998C cos � � 0
x0 � C sin � � �0.2
x
� 0.998Ce�0.0667t cos (0.998t � �)
x � �0.0667Ce�0.0667t sin (0.998t � �)The velocity is
x � Ce���nt sin (�dt � �) � Ce�(0.0667)(1)t sin (0.998t � �)
�d � 2�/�d � 2�/0.998 � 6.30 s
�d � �n�1 � �2 � (1)�1 � (0.0667)2 � 0.998 rad/s
2��n � 23
cm
� � 13
cm�n
� 103(50)(1)
� 0.0667
�n
2 � 23
km
�cx � kx � 12
mx � mx or x � 23
cm
x � 23
km
x � 0
�12
mx.r .x
�Fr � 12
mr2 [ΣMG � I ]
�cx � kx � F � mx[ΣFx � mx]
N � s/m,
Art ic le 8/4 Vibrat ion of Rig id Bodies 637
�
�
x
mk cr
F
O
N
mg
Equilibriumposition
kx cx·
x+θ
Helpful Hints
� The angle is taken positive clock-wise to be kinematically consistentwith x.
� The friction force F may be as-sumed in either direction. We willfind that the actual direction is tothe right for x � 0 and to the leftfor x � 0; F � 0 when x � 0.
�n � �23
km
� �23
7550
� 1 rad/s
c08.qxd 6/28/06 4:48 PM Page 637
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