208322 Mechanical Vibrations
Lesson 4-2
1 Equations of Motion 1: Newton’s Method
☻ Example 1: [2] The rod has no mass. Derive the equation of motion
using the angular impulse - momentum principle.
Solution
From the principle, we have
M o = H o ,
− mgL sin θ =
(
d mL2θ
dt
) = mL θ,
2
Lθ + g sin θ = 0.
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Copyright  2007 by Withit Chatlatanagulchai
208322 Mechanical Vibrations
Lesson 4-2
☻ Example 2: [2] Find the equation of motion for the system shown
below. The rod has inertia I RG about its mass center G and has mass
mr .
Solution
From the Newton’s law of motion,
∑M
o
= ∑ I α,
− mc gLc sin θ − mr gL sin θ = ( I RG + mr L2 + mc L2c )θ.
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Copyright  2007 by Withit Chatlatanagulchai
208322 Mechanical Vibrations
Lesson 4-2
☻ Example 3: [2] The pendulum shown below consists of a slender rod of
mass 1.4 kg and a block of mass 4.5 kg.
a) Determine the location of the center of mass.
b) Derive the equation of motion in terms of θ .
Solution
a) Let L be the distance from the pivoted point O to the center of
mass G. From the definition of the center of mass, we have
( 4.5 + 1.4 ) L = (1.4 )( 0.015) + ( 4.5 )( 0.09 + 0.015) ,
L = 0.084.
b) Use the center of mass and the combined mass 1.4 + 4.5 = 5.9,
we have
− mgL sin θ = I oθ,
2 
1
 12 (1.4 )( 0.06 + 0.09 ) 


2
θ,
−5.9 ( 9.81) ( 0.084 ) sin θ =  + (1.4 )( 0.015 )


2
 + ( 4.5 )( 0.09 + 0.015 ) 




θ + 92.61sin θ = 0.
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Copyright  2007 by Withit Chatlatanagulchai
208322 Mechanical Vibrations
Lesson 4-2
☻ Example 4: [2] A slender rod 1.4 m long and of mass 20 kg is attached
to a wheel of radius 0.05 m and negligible mass, as shown below. A
horizontal force f is applied to the wheel axle. Derive the equation of
motion in terms of θ . Assume that the wheel does not slip.
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Copyright  2007 by Withit Chatlatanagulchai
208322 Mechanical Vibrations
Lesson 4-2
The coordinates of the mass center of the rod are
xG = xP − ( L / 2 ) sin θ ,
yG = − ( L / 2 ) cosθ .
Therefore, we have
L
L
xG = xP − θcos θ + θ 2 sin θ ,
2
2
L
L
yG = θsin θ + θ 2 cos θ .
2
2
Summing forces in the x direction, we have
mxG = f ,
L
L


m  xP − θcos θ + θ 2 sin θ  = f .
2
2


(1)
Summing forces in the y direction, we have
Solution
myG = R − mg ,
L
L

m  θsin θ + θ 2 cos θ  = R − mg .
2
2

(2)
Summing moments about the mass center of the rod, we have
I Gθ = ( fL / 2 ) cos θ − ( RL / 2 ) sin θ .
(3)
Use (2) to eliminate R in (3). The resulting equation together with
(1) constitute the two equations of motion for this two DOFs system.
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Copyright  2007 by Withit Chatlatanagulchai
208322 Mechanical Vibrations
Lesson 4-2
☻ Example 5: [2] An object of mass 0.005 kg moving with a velocity 365
m/s strikes and becomes embedded in the pendulum as shown below,
where L = 1.2 m. Assume that the pendulum is a slender rod whose mass
is 4.5 kg. Determine the angular velocity of the pendulum immediately
after the impact.
Solution
From conservation of angular momentum,
H o1 = H o 2 ,
2
m1v1 ( 2 L ) =  I G + m2 L2 + m1 ( 2 L )  ω ,


ω = 0.505 rad / s.
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Copyright  2007 by Withit Chatlatanagulchai
208322 Mechanical Vibrations
Lesson 4-2
2 Equations of Motion 2: Energy Method
We can apply the principle of work and energy
∑U
1− 2
= T2 − T1 ,
dU dT
=
,
dt
dt
with problems in Vibrations to obtain the equations of motion of the
system.
☻ Example 6: [2] The rod has no mass. Derive the equation of motion
using the conservation of energy method.
Solution
The work from gravitational force is
U = −mg ( L − L cos θ ) ,
dU
= −mgL sin θ θ.
dt
The kinetic energy is
T=
( )
2
1
m Lθ ,
2
dT
.
= mL2θθ
dt
From the principle of work and energy, we have
Lθ + g sin θ = 0.
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Copyright  2007 by Withit Chatlatanagulchai
208322 Mechanical Vibrations
Lesson 4-2
☻ Example 7: [1] Determine the natural frequency of the system below
where θ is measured from the equilibrium position.
k
r2
r1
J
θ
Solution
Because θ is measured from the equilibrium position, we need not
compute work done by gravitational force. We then have
m
T=
( )
2
1 2 1
Jθ + m r1θ ,
2
2
dT
+ mr 2θθ
,
= Jθθ
1
dt
1
2
U = − k ( r2θ ) ,
2
dU
= −kr22θθ.
dt
Using the principle of work and energy, we have
dU dT
=
,
dt
dt
+ mr 2θθ
,
− kr 2θθ = Jθθ
2
θ +
1
2
2
kr
θ = 0,
J + mr12
ωn =
8
kr22
.
J + mr12
Copyright  2007 by Withit Chatlatanagulchai
208322 Mechanical Vibrations
Lesson 4-2
☻ Example 8: [1] A cylinder of weight w rolls without slipping on a
cylindrical surface as shown below. Determine its differential equation of
motion for small oscillations about the lowest point and find its natural
frequency.
Solution
Since the cylinder rolls without slipping, we have that the cylinder
both translates and rotates. We have
θ
R
T=
r
1
1
mG vG2 + J Gω 2
2
2
2
1w
1 w r 2  R  
( R − r )θ  +
=
 − 1 θ 
2g
2 g 2  r
 
3w
2
=
( R − r ) θ2 .
4g
φ
2
The work done from the gravitational force is
U = − mgh
= − w ( R − r )(1 − cos θ ) .
Using the principle of work and energy, we have
dU dT
=
,
dt
dt
3 w

2
 2 g ( R − r ) θ + w ( R − r ) sin θ  θ = 0.


Letting sin θ = θ for small angles, we have
θ +
Therefore,
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ωn =
2g
θ = 0.
3( R − r )
2g
.
3( R − r )
Copyright  2007 by Withit Chatlatanagulchai
208322 Mechanical Vibrations
Lesson 4-2
☻ Example 9: [2] Derive the equation of motion of the system below. The
cylinder has mass M and inertia I G . The link has mass m and inertia I o
about its C.G. The rolling is without slipping.
Solution
We have
2
2
2
1 2 1
1  L  1 
 L   2


T = I Gφ + M ( r − R )θ  + m  θ  +  I o + m    θ ,
2
2
2  2  2 
 2  
L

U = − Mg ( r − R )(1 − cos θ ) + mg  r −  (1 − cos θ ) .
2

Using the principle of work and energy, we have
dU dT
=
.
dt
dt
Similar to the previous example, we finally obtain
  r 2
L2
L2  2
I
+
M
r
−
R
+
m
+
I
+
m
(
)
 G 
θ
o
4
4 
  R 

L 

+  Mg ( r − R ) + mg  r −   sin θ = 0.
2 


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Copyright  2007 by Withit Chatlatanagulchai
208322 Mechanical Vibrations
Lesson 4-2
☻ Example 10: [2] The disk has mass m and inertia I about its center.
The free length of the spring equals L and its stiffness is k . The disk is
released from rest in the position shown and rolls without slipping.
Calculate its angular velocity when its center is directly below point O.
Solution
Using the principle of work and energy, we have
U1− 2 = T2 − T1 ,
1 
k
2 
2
1
1
1
2
2
2
2

D
−
R
+
D
−
L
− k ( D1 − R − L ) = m (ω R ) + I ω 2 .
( 1 )
2
 2
2
2
From the equation above, we can find
11
ω.
Copyright  2007 by Withit Chatlatanagulchai
208322 Mechanical Vibrations
Lesson 4-2
Lesson 4 Homework Problems
2.15, 2.45, 2.46, 2.71, 2.73, 2.82, 2.83.
Homework problems are from the required textbook (Mechanical
Vibrations, by Singiresu S. Rao, Prentice Hall, 2004)
References
[1]
[2]
Theory of Vibration with Applications, by William T. Thomson and
Marie Dillon Dahleh, Prentice Hall, 1998
Mechanical Vibration, by William J. Palm III, Wiley, 2007
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Copyright  2007 by Withit Chatlatanagulchai