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310_Physics ProblemsTechnical Physics

310_Physics ProblemsTechnical Physics - 312 P10.71 Rotation...

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312 Rotation of a Rigid Object About a Fixed Axis P10.71 (a) m g T m a 2 2 2 = T m g a T m g m a T 2 2 1 1 1 1 20 0 2 00 156 37 0 15 0 9 80 37 0 2 00 118 = = = °= = °+ = b g e j b ga f . . sin . . . sin . . kg 9.80 m s m s N kg m s N 2 2 2 (b) T T R I I a R 2 1 = = F H G I K J b g α I T T R a = = = 2 1 2 2 156 118 0 250 2 00 1 17 b g a fa f N N m m s kg m 2 2 . . . FIG. P10.71 P10.72 For the board just starting to move, τ α = I : mg m A A 2 1 3 2 F H G I K J = F H G I K J cos θ α α θ = F H G I K J 3 2 g A cos The tangential acceleration of the end is a g t = = A α θ 3 2 cos The vertical component is a a g y t = = cos cos θ θ 3 2 2 If this is greater than g , the board will pull ahead of the ball falling: FIG. P10.72 (a) 3 2 2 g g cos θ gives cos 2 2 3 θ so cos θ 2 3 and θ ° 35 3 . (b) When θ = ° 35 3 . , the cup will land underneath the release-point of the ball if r c = A cos θ When A = 1 00 . m, and θ = ° 35 3 . r c = = 1 00 2 3 0 816 . . m m so the cup should be 1 00 0 816 0 184 . . . m m m from the moving end = a f P10.73 At t = 0, ω ω = = 3 50 0 0 . rad s e . Thus, ω 0 3 50 = . rad s At t = 9 30 . s, ω ω σ = = 2 00 0 9 30 . . rad s s e a f
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