PASCO ME-9833 Physical Pendulum Set - Manual (Page 11)

Physical Pendulum Set Model No. ME-9833
11
®
Experiment 2: Minimum Period for an Oscillating Bar
Equipment
The first section outlines the calculation of the distance from the pivot point to the center of mass of
a long bar that produces the minimum period of oscillation for the bar at small amplitudes. The next
section describes the measurements needed to determine the minimum period.
Calculating Distance for Minimum Period
Recall that the moment of inertia of a “long rod” about its center of gravity is: where M
is the mass of the long rod and L is the overall length of the rod. A more precise physical model of
the 28-cm Pendulum Bar as a rectangular-type rod has a moment of inertia about its center of
gravity that is:
where M is the mass of the rectangular bar, “a” is the length, and
“b” is the thickness. However, if a>>b, can be used as a very good approximation.
The parallel axis theorem enables us to write the moment of inertia of the bar about a pivot point
not at the center of gravity as:
where L
cg
is the distance from the pivot point to the
center of gravity.
The period of a physical pendulum depends on its moment of inertia, its mass, and the distance
from the pivot point to the center of gravity. The period is: . For the 28-cm Pendulum
Bar, the period becomes:
Use calculus to find the derivative of the period, T, with respect to L
cg
, the distance between the
pivot and the center of gravity.
Set the derivative equal to zero and solve for L
cg
. Confirm that the minimum distance is:
Item Item
PASCO Interface and DataStudio software Base and Support Rod
Rotary Motion Sensor Metric ruler
28-cm Bar from Physical Pendulum Set
I
cg
1
12
------ML
2
=
I
cg
1
12
------M a
2
b
2
+ ( ) =
I
cg
1
12
------ML
2
=
I
pivot
I
cg
ML
cg
2
+ =
T 2 π
I
pivot
mgL
cg
----------------- =
T 2 π
1
12
------ML
2
ML
cg
2
+
MgL
cg
---------------------------------------- 2 π
1
12
------L
2
L
cg
2
+
gL
cg
---------------------------- = =
L
cg
1
12
----------L =