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465_Physics ProblemsTechnical Physics

# 465_Physics ProblemsTechnical Physics - Chapter 15 P15.69(a...

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Chapter 15 467 P15.69 (a) ∆∆ KU += 0 Thus, KUKU top top bot bot +=+ where top bot == 0 Therefore, mgh I = 1 2 2 ω , but hRR R v R =− = − = cos cos θθ 1 af and I MR mr mR =+ + 22 2 Substituting we find v M θ R FIG. P15.69 mgR MR mr mR v R mgR Mm r R m v 1 1 2 1 4 4 2 2 2 2 2 2 2 −= + + F H G I K J + + L N M O Q P cos cos a f a f and vg R M m r R 2 4 1 2 2 2 = ++ cos a f ej so v Rg M m r R = 2 1 2 2 2 cos (b) T I mg d T = 2 π CM mm M T d mR M mM CM = + + 0 T MR mr mR mgR = 2 1 2 2 1 2 P15.70 (a) We require Ae A bt m = 2 2 e bt m + = 2 2 or bt m 2 2 = ln or 0100 20375 0693 . . . kg s kg bg t = ∴= t 520 . s The spring constant is irrelevant. (b) We can evaluate the energy at successive turning points, where cos ωφ t ± 1 and the energy is 1 2 1 2 2 kx kA e bt m = . We require 1 2 1 2 1 2 2 kA e kA bt m = F H G
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