hrly_exam_ii_vc_solution.20070919.46f178ae17e7f3.84817784

hrly_exam_ii_vc_solution.20070919.46f178ae17e7f3.84817784 -...

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Unformatted text preview: 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: = ω B r AB,x ω B = 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 = rα C i = ∴ α C = a A = a C + > 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- > 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: =- 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 TAM212 Hour Exam II - Spring 2005 Solution- Version C = 2 5 ρV s R 2 4 = 2 5 ρ...
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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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