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46. To use Eq. 1529 we need to locate the center of mass and we need to compute the
rotational inertia about
A
. The center of mass of the stick shown horizontal in the figure is
at
A
, and the center of mass of the other stick is 0.50 m below
A
. The two sticks are of
equal mass so the center of mass of the system is
1
2
(0.50 m) 0.25m
h
==
below
A
, as
shown in the figure. Now, the rotational inertia of the system is the sum of the rotational
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics, Center Of Mass, Inertia, Mass

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