181_Physics ProblemsTechnical Physics

181_Physics ProblemsTechnical Physics - Chapter 6 P6.63(a...

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Chapter 6 183 P6.63 (a) The mass at the end of the chain is in vertical equilibrium. Thus T mg cos θ = . Horizontally T ma mv r r sin θ = = 2 r r = + = °+ = 2 50 4 00 2 50 28 0 4 00 5 17 . sin . . sin . . . θ a f a f m m m Then a v r = 2 5 17 . m . By division tan . θ = = a g v g r 2 5 17 v g v 2 5 17 5 17 9 80 28 0 5 19 = = ° = . tan . . tan . . θ a fa fa f m s m s 2 2 (b) T mg cos θ = T mg = = ° = cos . . cos . θ 50 0 9 80 28 0 555 kg m s N 2 b g e j T R = 4.00 m θ l = 2.50 m r mg FIG. P6.63 P6.64 (a) The putty, when dislodged, rises and returns to the original level in time t . To find t , we use v v at f i = + : i.e., = + v v gt or t v g = 2 where v is the speed of a point on the rim of the wheel.
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