hrly_exam_i_solution_vc.20070919.46f177e4b5e1c1.82675612

hrly_exam_i_solution_vc.20070919.46f177e4b5e1c1.82675612 -...

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TAM212 Introductory Dynamics Spring 2005 S. Balachandar Hourly I Exam (Solution - Version C) A1 W = 1 2 k δ 2 i - 1 2 k δ 2 f N-m = 1 2 k ( 4 2 - 2 2 ) N-m = 6 k N-m A2 When a satellite revolves around the Earth in a circular orbit with constant speed it has: radial acceleration pointed towards the center of Earth . A3 y x A B a b d e p q r p c The length of the rope is a constant. a + b + c + d + e = constant ( y - p ) + ( y - p - q ) + ( y - p - q ) + d + ( x + r ) = constant Differentiating wrt t , ˙ y + ˙ y + ˙ y + ˙ x = 0 3 ˙ y = - ˙ x 1
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A4 v dv dt = x dx dt 1 2 dv 2 dt = 1 2 dx 2 dt dv 2 = dx 2 v 2 v f 2 = x 2 4 2 v f = 4 m / s A5 Elastic collision, hence e = 1 e = v B f - v A f v Ai - v Bi v B f - v A f 1 - ( - 2) = 1 v B f - v A f = 3 Conservation of momentum mv Ai + mv Bi = mv A f + mv B f v A f + v B f = 1 + ( - 2) v A f + v B f = - 1 Solving, we get v A f = - 2 i m / s v B f = 1 i m / s 2
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B1 (a) v A = v C + ω 3 × r C A v A j = - 0 . 3 i + ω 3 k × ( - 2 . 15407 i + 2 . 5 j ) v A j = - 0 . 3 i - 2 . 15407 ω 3 j - 2 . 5 ω 3 i Equating the i and j coefficients separately i : 0 = - 0 . 3 - 2 . 5 ω 3 ω 3 = - 0 . 12 rad / s j : v A = - 2 . 15407 ω 3 v A = - 2 . 15407( - 0 . 12) v A = 0 . 258488 j m / s v B = v A + ω 2 × r A B v B i = 0 . 258488 j + ω 2 k × ( - 2 . 44949 i - 2 . 5 j ) v B i = 0 . 258488 j - 2 . 44949 ω 2 j + 2 . 5 ω 2 i Equating the i and j coefficients separately j : 0 = 0 . 258488 - 2 . 44949 ω 2 ω 2 = 0 . 105527 rad / s i : v B = 2 . 5 ω 2 v B = 2 . 5(0 . 105527)
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