231_Dynamics 11ed Manual

# 231_Dynamics 11ed Manual - Engineering Mechanics - Dynamics...

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Engineering Mechanics - Dynamics Chapter 13 Given: M 0.2 kg = θ ' 2 rad s = 1 π 4 = '' 0 rad s 2 = g 9.81 m s 2 = Solution: Σ F r = Ma r ; Mg sin () M r'' r ' 2 = r'' r ' 2 g sin 0 = [1] Σ F = Ma ; cos N s Mr '' 2 r' ' + = 2 M 'r ' N s + cos 0 = (Q.E.D) The solution of the differential equation (Eq.[1] is given by rC 1 e ' t C 2 e 't + g 2 ' 2 sin = r' ' C 1 e ' t 'C 2 e + g 2 ' cos = At t 0 = r 0 = 0 C 1 C 2 + = r' 0 = 0 ' C 1 2 + g 2 ' = Thus C 1 g 4 ' 2 = C 2 g 4 ' 2 = t 1 ' = t 0.39 s = 1 e ' t C 2 e + g 2 ' 2 sin = r 0.198 m = *Problem 13-112 The rocket is in circular orbit about the earth at altitude
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## This note was uploaded on 08/16/2011 for the course EGN 3321 taught by Professor Christianfeldt during the Spring '08 term at University of Central Florida.

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