hrly_exam_i_solution_va.20070919.46f177a61256d1.39911426

hrly_exam_i_solution_va.20070919.46f177a61256d1.39911426 -...

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Unformatted text preview: 1 a+b+c+d+e ( y − p) + ( y − p − q) + ( y − p − q) + d + ( x + r) ˙ ˙ ˙˙ Differentiating wrt t, y + y + y + x ˙ x = = = = constant constant 0 ˙ −3 y The length of the rope is a constant. A ¤¤¤¤¤¤¤ £¡£¡£¡£¡£¡£¡£¡ ¤¡¤¡¤¡¤¡¤¡¤¡¤¡ £¡£¡£¡£¡£¡£¡£¡ ¤¡¤¡¤¡¤¡¤¡¤¡¤¡¤ £¡£¡£¡£¡£¡£¡£¡ ¡¡¡¡¡¡¡£¤£¤£ q   ¡©¡©¡        ©¡©¡©¡©¡¡¡ ¡ ¡©¡©¡©¡©¡©¡©¡©¡ ©¡  ¡¡¡¡©¡¢¢¡¢¡¢¡ ¡¡¡¡¡¡¡ ¡©¡©¡ ©¡©¡©¡©¡¡ ¡¢¡¢¡ ©¡©¡©¡©¡©¡©¡©¡ ¡¡¡¡©¡¡ ¡ ¡¢ ¡¡¡¡¡¡¡¥¦ ©© ©¡¡¡ ¡¢¡¢¡ ¢¡ ¡ ¡ ¢¡¢¡¢¡¢ ¡ ¡ ¡ ¡¢¡¢¡ ¢ ¢¡ ¡ ¡ ¡ ¡ ¡ ¢ ¢¡¢¡ ¦¥¦ ¢ ¢¡¢¡¢¡¢ ¡ ¡ ¡ ¢¡¢¡¢¡¢ ¡ ¡ ¡ ¥¦¥ ¢¡¢¡¢¡¢ ¡ ¡ ¡ ¡¡¡ ¢ ¢ y r a b c e B d x ¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨ §¡§¡§¡§¡§¡§¡§¡§¡§¡§¡§¡§¡§¡§¡§¡§¡§¡§¡§¡ ¨¡¨¡¨¡¨¡¨¡¨¡¨¡¨¡¨¡¨¡¨¡¨¡¨¡¨¡¨¡¨¡¨¡¨¡¨¡ §¡§¡§¡§¡§¡§¡§¡§¡§¡§¡§¡§¡§¡§¡§¡§¡§¡§¡§¡ ¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡§¨§¨ p p A3 When a satellite revolves around the Earth in a circular orbit with constant speed it has: radial acceleration pointed towards the center of Earth. A2 1212 kδ − kδ f N-m 2i 2 1 = k 12 − 22 N-m 2 3 = − k N-m 2 W= A1 Hourly I Exam (Solution - Version A) Spring 2005 S. Balachandar TAM212 Introductory Dynamics A4 dv dt 1 dv2 2 dt dv2 v v2 1 f vf v dx dt 1 dx2 2 dt dx2 5 x2 1 5 m/s =x = = = = A5 Elastic collision, hence e = 1 e= vB f − vA f v Ai − v Bi vB f − vA f =1 2 − (−1) vB f − vA f = 3 Conservation of momentum mv Ai + mv Bi = mv A f + mv B f v A f + v B f = 2 + (−1) vA f + vB f = 1 Solving, we get v A f = −1i m/s vB f = 2i m/s 2 B1 (a) v A = v C + ω 3 × rC A v A j = −0.2i + ω3 k × (−2.15407i + 2.5j) v A j = −0.2i − 2.15407ω3j − 2.5ω3 i Equating the i and j coefficients separately i: j: 0 ω3 vA vA vA = = = = = −0.2 − 2.5ω3 −0.08 rad/s −2.15407ω3 −2.15407(−0.08) 0.172325j m/s vB = vA + ω2 × rA B v B i = 0.172325j + ω2 k × (−2.44949i − 2.5j) v B i = 0.172325j − 2.44949ω2j + 2.5ω2 i Equating the i and j coefficients separately j: i: 0 ω2 vB vB vB = = = = = 0.172325 − 2.44949ω2 0.070351 rad/s 2.5ω2 2.5(0.070351) 0.175879i m/s (b) v A = vC + ω 3 × r C A v A j = −0.2i + ω3 k × 3.3j v A j = −0.2i − 3.3ω3 i Since there is no j coefficient on the RHS, vA = 0 3 B2 B: Block, b: bullet (a) Conservation of momentum m B+b v B+b = mb vbi + m B v Bi (10 + 0.01)v B+b = 0.01(200) + 10(0) v B+b = 0.1998 m/s (b) Kinetic Energy Lost = KEi − KE f 1 1 1 = m b v2 + m B v2 − m B+b v2 +b b B B 2 2 2 1 1 = (0.01)(200)2 + 0 − (10.01)(0.1998)2 2 2 2 2 = 199.8 kgm /s (c) W = ∆T 0 12 1 2 7   − v2 ) − k δ = m B+b (v f i  2 2 1 1 (8)δ 2 = (10.01)(0.1998)2 2 2 δ = 0.223496 m 4 ...
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This note was uploaded on 04/17/2008 for the course TAM 212 taught by Professor Keane during the Fall '08 term at University of Illinois at Urbana–Champaign.

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