316_Physics ProblemsTechnical Physics

316_Physics ProblemsTechnical Physics - 318 P10.82 Rotation...

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318 Rotation of a Rigid Object About a Fixed Axis P10.82 Conservation of energy between apex and the point where the grape leaves the surface: mg y mv I mgR mv mR v R ff f f ∆= + −= + F H G I K J F H G I K J 1 2 1 2 1 1 2 1 2 2 5 22 2 ω θ cos af which gives g v R f 1 7 10 2 F H G I K J cos a f (1) Consider the radial forces acting on the grape: mg n mv R f cos 2 . At the point where the grape leaves the surface, n 0. Thus, mg mv R f cos = 2 or v R g f 2 =c o s . Substituting this into Equation (1) gives gg g cos cos θθ 7 10 or cos = 10 17 and 54 0 .. R i y = R—R cos f n mg sin mg cos FIG. P10.82 P10.83 (a) There are not any horizontal forces acting on the rod, so the center of mass will not move horizontally. Rather, the center of mass drops straight downward (distance h /2) with the rod rotating about the center of mass as it falls. From conservation of energy:
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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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