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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 f f 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 2 2 2 2 2 ω θ cos a f which gives g v R f 1 7 10 2 = F H G I K J cos θ a f (1) Consider the radial forces acting on the grape: mg n mv R f cos θ = 2 . At the point where the grape leaves the surface, n 0. Thus, mg mv R f cos θ = 2 or v R g f 2 = cos θ . Substituting this into Equation (1) gives g g g = cos cos θ θ 7 10 or cos θ = 10 17 and θ = ° 54 0 . . R θ i y = R—R cos θ f n mg sin θ mg cos θ 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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