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 atA, 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 12(0.50 m) 0.25mh==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.