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180_Physics ProblemsTechnical Physics

180_Physics ProblemsTechnical Physics - 182 P6.60 Circular...

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182 Circular Motion and Other Applications of Newton’s Laws P6.60 For the block to remain stationary, F y = 0 and F ma x r = . n m m g p b 1 = + e j so f n m m g s s p b = + µ µ 1 1 1 e j . At the point of slipping, the required centripetal force equals the maximum friction force: + = + m m v r m m g p b s p b e j e j max 2 1 µ or v rg s max . . . . = = = µ 1 0 750 0 120 9 80 0 939 a fa fa f m s . For the penny to remain stationary on the block: F n m g y p = = 0 0 2 or n m g p 2 = and F ma f m v r x r p p = = 2 . When the penny is about to slip on the block, f f n p p s = = , max µ 2 2 or µ s p p m g m v r 2 2 = max v rg s max . . . . = = = µ 2 0 520 0 120 9 80 0 782 a fa fa f m s m g b m g p m g b m g p n 1 f p f m g p n 2 f p FIG. P6.60 This is less than the maximum speed for the block, so the penny slips before the block starts to slip. The maximum rotation frequency is Max rpm = = L N M O Q P F H G I K J = v r max . . . 2 0 782 1 2 0 120 60 62 2 π π m s rev m s 1 min rev min b g a f . P6.61 v r T = = = 2 2 9 00 15 0 3 77 π π . . . m s m s a f a f (a) a v r r = = 2 1 58 . m s 2 (b) F m g a r low N = + = b g 455 (c) F m g a r high
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