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) mg T ma 222 −= Tmg a Tm g m a T 22 11 1 1 20 0 2 00 156 37 0 15 0 9 80 37 0 2 00 118 =− = = −° = + = bg ej a f .. sin . s i n . . kg 9.80 m s m s N kg m s N 2 (b) TTRI I a R 21 = F H G I K J α I TTR a = = =⋅ 2 2 156 118 0 250 200 117 a f a f N N m ms kg m 2 2 . . . FIG. P10.71 P10.72 For the board just starting to move, τα = I : mg m A A 2 1 3 2 F H G I K J = F H G I K J cos θα αθ = F H G I K J 3 2 g A cos The tangential acceleration of the end is ag t == A 3 2 cos The vertical component is aa g yt cos cos θθ 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 gg cos θ gives cos 2 2 3 so cos 2 3 and ≤° 35 3 . (b) When 35 3 . , the cup will land underneath the release-point of the ball if r c = A cos When A = 100 . m, and 35 3 . r c 2 3 0816 m m so the cup should be 1 00 0 816 0 184 . m m f r o m t h e m o v i n g e n d af P10.73 At t = 0, ωω 350 0 0 . r a d s e . Thus, ω 0 = r a d s At t = 930 s , σ 0 . . rad s s e , yielding −− 602 10 s (a) ωσ d dt de
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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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