Homework Solutions
Physics 121
Set 09
Spring 2008

1

Problem 1
The moment of inertia of a stick of length
L
and mass
M
with respect to the centerofmass is
equal to (1/12)
M
(
L
2
+
W
2
).
The moment of inertia of the stick with respect to the pivot point can
be found using the parallelaxis theorem:
I
O
=
1
12
M L
2
+
W
2
( )
+
Mx
2
In order to determine the period of the pendulum we need to use the equation of the period of a
physical pendulum:
T
=
2
!
I
0
Mgx
=
2
1
12
M L
2
+
W
2
( )
+
Mx
2
Mgx
=
2
g
1
12
L
2
+
W
2
( )
x
+
x
The period has a minimum or maximum value when dT/dx = 0.
This requires that
d
dx
2
g
1
12
L
2
+
W
2
( )
x
+
x
"
#
$
$
%
&
’
’
=
2
g
1
2
(
1
12
L
2
+
W
2
( )
x
2
+
1
"
#
$
%
&
’
1
12
L
2
+
W
2
( )
x
+
x
=
(
1
12
L
2
+
W
2
( )
x
2
+
1
"
#
$
%
&
’
1
12
L
2
+
W
2
( )
x
+
x
=
0
This condition is satisfied when
!
1
12
L
2
+
W
2
( )
x
2
+
1
=
0
or
x
=
1
12
L
2
+
W
2
( )
=
1
2
3
L
2
+
W
2
( )
The period at this distance is equal to
T
=
2
g
1
12
L
2
+
W
2
( )
x
+
x
=
2
g
x
1
12
L
2
+
W
2
( )
x
2
+
1
=
2
g
x
=
2
g
L
2
+
W
2
( )
3
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Physics 121
Set 09
Spring 2008

2

Problem 2
When the cylinder is released, the total mechanical energy of the system is equal to the potential
energy of the spring, which is stretched by a distance
d
:
E
i
=
1
2
kd
2
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 Spring '08
 Wolfs
 Energy, Inertia, Kinetic Energy, Mass, Potential Energy, Work, Moment Of Inertia, Homework Solutions Set

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