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 Fm a xr = . nm m g pb 1 =+ ej so fn m m g ss p b ≤= + µµ 11 1 . At the point of slipping, the required centripetal force equals the maximum friction force: ∴+ = + mm v r g s max 2 1 µ or vr g s max ... . == = 1 0 750 0 120 9 80 0 939 af a f ms. For the penny to remain stationary on the block: Fn m g yp =⇒ − = 00 2 or g p 2 = and a fm v r p p =⇒ = 2 . When the penny is about to slip on the block, ff n pp s , max 22 or sp p mg m v r 2 2 = max g s max . = 2 0 520 0 120 9 80 0 782 a f ms 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 0782 1 20 1 2 0 60 62 2 ππ rev m s 1 min rev min bg . P6.61 v r T = 2 29 0 0 15 0 377 π . . . m s a f (a) a v r r 2 158 . m s 2 (b) g a r low N = 455 (c) g a r high N =− = 328 (d) g a r mid N upward and
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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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