Unformatted text preview: .wikipedia.org/wiki/Fluidized_bed_reactor.) 3. A pendulum is formed by pivoting a long thin rod of mass M and length L about a point on
the rod. If the pivot is a distance x from the rod’s center, for what x is the period of the pendulum
1
minimum? The moment of inertia for a thin rod about its center of mass is I = 12 ML2 .
Solution: In the end, we only have a physical pendulum, and we already know that the period is
given by T = 2π I
mgh where I is the moment of inertia of the rod (of mass m) about the pivot point, and h is the distance
between the rod’s center of mass and the pivot point. Let the pivot be a distance x from the end
of the rod, making it a distance l/2 − x from the center of mass. The moment of inertia is then
I = Icom + m l
−x
2 2 = 1
ml2 + m
12 l
−x
2 2 The distance between the center of mass and the pivot is h = l/2 − x, so
I= 1
ml2 + mh2
12 The period is thus Problems ions
ured
, reperlope
ring.
tion,
With
ured
lues
ding
and
each
and
eastalue
e for
alue 483 12
l + h2
l2
h
T = 2π 12
= 2π
+
gh
70. Consider a damped o...
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 Fall '08
 StanJones
 Physics, Force, Mass, Simple Harmonic Motion, Work, pivot point

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