455_Physics ProblemsTechnical Physics

# 455_Physics ProblemsTechnical Physics - Chapter 15 457...

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Chapter 15 457 Additional Problems P15.51 Let F represent the tension in the rod. (a) At the pivot, FM gM g M g =+= 2 A fraction of the rod’s weight Mg y L F H G I K J as well as the weight of the ball pulls down on point P . Thus, the tension in the rod at point P is g y L Mg Mg y L = F H G I K J += + F H G I K J 1. M P pivot L y FIG. P15.51 (b) Relative to the pivot, II I M L M L M L + = rod ball 1 3 4 3 22 2 For the physical pendulum, T I mgd = 2 π where mM = 2 and d is the distance from the pivot to the center of mass of the rod and ball combination. Therefore, d MM L L L = + + = 2 3 4 ch and T ML Mg L g L == 2 2 4 3 2 4 3 2 3 4 af . For L = 200 . m , T 4 3 2200 980 268 . . . m ±ms s 2 . P15.52 (a) Total energy = 1 2 1 2 100
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## This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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