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hrly_exam_ii_vc_solution.20070919.46f178ae17e7f3.84817784

hrly_exam_ii_vc_solution.20070919.46f178ae17e7f3.84817784 -...

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TAM212 Hour Exam II - Spring 2005 Solution- Version C A1) v C = 20 i ω C k × r I C = 20 i ω C k × 1 j = 20 i - ω C i = 20 i ω C = - 20 rad/s v A = ω C k × r I A = - 20 k × 2 j = 40 i m/s v B = v A + ω B k × r AB v B i = 40 i + ω B k × ( r AB,x i + r AB,y j ) v B i = 40 i + ω B k × ( r AB,x i + r AB,y j ) v B i = 40 i + ω B r AB,x j - ω B r AB,y i j components: 0 = ω B r AB,x ω B = 0 i components: v B = 40 - ω B r AB,y v B = 40 v B = 40 i m/s A2) From previous problem, we saw that ω B = 0 . a C = 0 C i = 0 α C = 0 a A = a C + > 0 α C k × r CA - ω 2 C r CA = - ( - 20) 2 1 j = - 400 j a B = a A + α B k × r AB - > 0 ω 2 B r AB a B i = - 400 j + α B k × ( r AB,x i + r AB,y j ) a B i = - 400 j + α B r AB,x j - α B r AB,y i j components: 0 = - 400 + α B r AB,x α B = 400 r AB,x 6 = 0 ω B = 0 , α B 6 = 0 . A3) Full Sphere: I C zz,S = 2 5 m S R 2 = 1 kg - m 2 Removed Sphere: I C zz,s = 2 5 m s R 2 2 1
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TAM212 Hour Exam II - Spring 2005 Solution- Version C = 2 5 ρV s R 2 4 = 2 5 ρ 4 3 π R 2 3 R 2 4 = 2 5 ρ 4 3 πR 3 R 2 32 = 1 32 2 5 m S R 2 = 1 32 2 5 m S R 2 = 1 32 kg - m 2 Hollow Sphere: I C zz,H = I C zz,S - I C zz,s = 31 32 kg - m 2 A4) Symmetry is about the yz plane. So the following statement is true
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