ch15-p046 - 46. To use Eq. 15-29 we need to locate the...

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46. To use Eq. 15-29 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.

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