Solved Problems
Chapter 3: Mechanical Systems
Problem A-3-8-
In Figure 3-23, the simple pendulum shown consists of a sphere of mass m suspended by a string of negligible mass. Neglecting the elongation of the string, find a mathematical model of the pendulum. In addition, find the natural frequency of the system when θ is small. Assume no friction.
Solution
Assumptions:
1. Mass of the rope is neglected compared to the mass of the bob.
2. Motion occurs in one plane only. 3. Friction in the pivot is neglected. 4. Angle
θθ
θ
θ
is small .l
θ
m
mg
T
θ
sin
l
θ
cos
l
h
l
Figure 3-23 Simple Pendulum
Using Newton’s second law
Taking CCW moment about the pivot. The only force that is producing moment about the pivot is the weight
mg
. This force is producing a CW moment. Therefore, the moment produced bymg
ismgl sin
θθ
θ
θ
−
−
−
−
.J
mgl sin
J
θ
θθ
θ
θ
θ
θ
θ
θ
θ
θ
θ
=
=
=
=
−
=
−
=
−
=
−
=
∑
∑
∑
∑
&&
&&
M
M
M
M
WhereJ
=
=
=
=
ml
2. Therefore, 20
ml
θ
θ
θ
θ
&&
+
+
+
+
mgl sin
θ
θ
θ
θ
=
=
=
=
For small
θθ
θ
θ
,sin
θ
θ
θ
θ
≅
≅
≅
≅
θ
θ
θ
θ
, and the above equation of motion simplifies to0
g
l
θ
θ
θ
θ
θ
θ
θ
&&
+
+
+
+
θ
=
=
=
=
which is of the form
2
0
nθ
ω θ
θ
ω θ
θ
ω θ
θ
&&
+
+
+
+
ω θ
=
=
=
=
From the above, the natural frequency of the pendulum is
g
ω
ω
ω
ω
=
=
=
=
Using The Conservation of Energy Method
The potential energy is
((((
1
))))
V
=
=
=
=
mgh
=
=
=
=
mgl
−
−
−
−
cos
θ
θθ
θ
The kinetic energy is
(((( ))))
2 2 21
1
1
2
2
2
T
=
=
=
=
mv
=
=
=
=
ms
&
=
=
=
=
m l
θ
θ
θθ
&
(((( ))))
2((((
))))
1
1
2
T
+
+
+
+
V
=
=
=
=
m l
θ
θ
θ
θ
&
+
+
+
+
mgl
−
−
−
−
cos
θ
θ
θ
θ
((((
))))
((((
))))
21
2
0
2
d T
V
ml
mgl
sin
dt
θθ
θθ
θθ
θθ
θ
θ
θ
θ
θ
θ
θ
θ
+
+
+
+
=
+
=
=
=
+
+
=
=
=
&&&
+
&
=
((((
))))
((((
2))))
0
d T
V
ml
mgl sin
dt
θ
θ
θ
θ
θ θ
θ θ
θ θ
θ θ
+
+
+
+
=
+
=
=
+
=
=
+
=
=
&&
+
&
=
Since
θθ
θ
θ
&
≠
≠
≠
≠
0
, then the terms in brackets must be equal to zero, Therefore((((
))))
0
ml l
θ
θ
θ
θ
&&
+
+
+
+
g sin
θ
θ
θ
θ
=
=
=
=
Or0
g
l
θ
θ
θ
θ
θ
θ
θ
&&
+
+
+
+
θ
=
=
=
=
Which is similar to the differential equation obtained before.
Problem A-3-12
For the spring-mass-pulley system of Figure 3-27, the moment of inertia of the pulley about the axis of rotation is
J
and the radius isR
. Assume that the system is initially at equilibrium. The gravitational force of massm
causes a static deflection of the spring such thatk
δ
st=
mg
. Assuming that the displacementx
of massm
is measured from the equilibrium position, obtain a mathematical model of the system. In addition, find the natural frequency of the system.Solution
Assumptions:
1. Lever is rigid and massless.
2. Reactions at lever P are neglected. 3. Displacement
x
is small.4. The wire is inextensible.
k
δ
m
mg
R
x
θ
T
T
Figure 3-27F
=
=
=
=
mx
∑
∑
∑
∑
&&
orT
mx
−
=
−
=
−
=
−
=
&&
(1)Where
T
is the tension in the wire (Notice that sincex
is measured from the static equilibrium position the termmg
does not enter into the equation).Applying Newton’s second law for the pulley
T
=
=
=
=
J
θ
θθ
θ
∑
∑
∑
∑
&&
orTR
−
−
−
−
kxR
=
=
=
=
J
θ
θθ
θ
&&
(2)Eliminating
T
from equations (1) and (2) givesmxR
kxR
J
θ
θθ
θ
−
−
=
−
−
=
−
−
=
−
&&
−
=
&&
(3)Noting that
x
=
=
=
=
R
θ
θθ
θ
, one can simplify the above equation((((
2))))
20
J
+
+
+
+
mR
θ
θ
θ
θ
&&
+
+
+
+
kR
θ
θ
θ
θ
=
=
=
=
(4) or 2 20
kR
J
mR
θ
θ
θ
θ
θ
θ
θ
+
+
+
+
θ
=
=
=
=
+
+
+
+
&&
(5)which represents the mathematical model for the system. The natural frequency of the system is given by 2 2 n
kR
J
mR
ω
ω
ω
ω
=
=
=
=
+
+
+
+
(6)Using The Energy Method
The kinetic energy of the system is
2 2
1
1
2
2
Due to Rotation Due to TranslationT
=
=
=
=
mx
&
+
+
+
+
J
θ
θθ
θ
&
123
123
(7)The potential energy of the system is
{
2
1
2
Noting that
x
=
=
=
=
R
θ
θθ
θ
, one can simplify the above equation((((
))))
2 2 2 2 2 2 2 2 2 2 2 21
1
1
2
2
2
1
1
1
2
2
2
1
1
2
2
T
V
mx
J
kx
T
V
mR
J
kR
T
V
mR
J
kR
θ
θθ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
+
=
+
+
+
=
+
+
+
=
+
+
+
=
+
+
+
=
+
+
+
=
+
+
+
=
+
+
+
=
+
+
+
=
+
+
+
=
+
+
+
=
+
+
+
=
+
+
&
&
&
&
&
(9)((((
))))
1
((((
2))))
1
20
2
2
0
2
2
d
T
V
mR
J
kR
dt
+
+
+
+
=
=
=
=
⇒
⇒
⇒
⇒
+
+
+
+
θθ
θθ
θθ
θθ
+
+
+
+
θθ
θθ
θθ
θθ
=
=
=
=
& &&
&
(10)Or
((((
2))))
20
mR
+
+
+
+
J
θ
θ
θ
θ
&&
+
+
+
+
kR
θ
θ
θ
θ
=
=
=
=
(11)which is similar to the one obtained before
Problem A-3-13
In the mechanical system of Figure 3-28, one end of the lever is connected to a spring and a damper, and a force
f t
(((( ))))
is applied to the other end of the lever. Derive a mathematical model of the system. Assume that the displacementx
is small and the lever is rigid and massless.1
l
l
2k
b
x
P
(((( ))))
f t
Figure 3-28Solution
Assumptions:1. Lever is rigid and massless. 2. Reactions at lever P are neglected. 3. Displacement
x
is small.Free Body Diagram (FBD): The FBD is shown in the Figure below 1
l
l
2P
k= −
k x
F
b= −
b x
F
&
x
(((( ))))
f t
Figure 3-28 Equation of motion:Applying Newton’s second law, for a system in rotation about point P gives
(((( ))))
1 2 20
0
P=
=
=
=
⇒
⇒
⇒
⇒
f t l
−
−
−
−
b x l
−
−
−
−
k x l
=
=
=
=
∑
∑
∑
∑
M
&
DID YOU ASK YOURSELF
WHY THE RHS of the above equation is ZERO
?
Rearranging, one can write:(((( ))))
1((((
))))
20
f t l
−
−
−
−
b x
&
+
+
+
+
k x l
=
=
=
=
or(((( ))))
1 2l
b x
k x
f t
l
+
=
+
+
=
=
+
=
&
which is the mathematical model of the system. It is clear that this differential equation represents is a first order system since it is represented by a first order ordinary differential equation with constant coefficients. The RHS of the above differential equation is not zero.
System Response:
In the above system, the input is the force
F
while the output is the displacementx
. We will try to find a relationship between the input and the output.Taking Laplace transform of both sides of the above equations provided we have zero initial conditions gives
((((
))))
(((( )))) 1 (((( )))) 2l
b s
k X s
F s
l
+
=
+
=
+
=
+
=
where
X s
( )
andF
( )
s
are Laplace transform ofx
andf t
(((( ))))
, respectively. The above equation can be written as(((( )))) (((( )))) (((( ))))
((((
)))) ((((
))))
1 21
Output
Input
X s
l
C
G s
F s
l
b s
k
b s
k
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
+
+
+
+
+
+
+
+
where
C
=
=
=
=
(((( ))))
l l
1 2 . The above relation can be written in the standard form as: (((( )))) (((( )))) (((( ))))1
Output
Input
C b
X s
C *
G s
k
F s
s
s
b
ττττ
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
+
+
+
+
+
+
+
+
whereC
*
=
C b
andτ
=
b k
(
1
)
*C
s
+
τ
X s
( )
( )
s
F
Input14243
Output14
4244
3
Transfer Function
144
42444
3
The response
x t
( )
will depend on the input forcef t
(((( ))))
. Iff t
(((( ))))
is a step input of magnitude 1:In this case
F s
(((( ))))=
=
=
=
1
/ s
and( )
( )
*
*
1
1
1
1
C
C
X s
s
s
s
s
s
s
α
β
τ
τ
τ
=
=
=
+
+
+
+
F
where1
s
α
=
s
0*
*
1
sC
C
s
τ
τ
==
+
and1
1
s
s
τ
β
+
=
1
*
s
C
τ
+
1*
sC
ττ
= −= −
Therefore,( )
1
1
1
X s
C
s
s
τ
τ
=
−
+
and hence( )
*1
t
