3_6_Problem_3_119

3_6_Problem_3_119 - e a r r r s n t C 3 1 1 2 1 2 ω − =...

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3.119 Given: B 1 and B 2 are identical cylinders of radius r in the positions shown. They are connected with B 3 through pin connections and roll on the surface shown. The angular velocity for B 1 is constant k 0 1 ω = ω v . Since everything is in plane motion k 2 2 α = α v , k 3 3 α = α v , k 2 2 ω = ω v and k 3 3 ω = ω v . From the diagram, j i r r r CG 2 6 + = , i v C C v = , and i v G G v = . Find: The angular accelerations of B 2 and B 3 2 α v and 3 α v Solution: CG CG C G r r a a 2 3 3 ω × α + = v . (1) To find 3 ω , we could use ) 3 ( 2 ) 3 ( 2 3 3 3 j i i j i k v r v v + ω + = + × ω + = × ω + = r v r C C CG C G v (2) i v r G 0 ω = (equation 3.27) Plugging this into (2) we have ) 3 ( 2 3 0 j i i i + ω + = ω r v r C , so 0 3 = ω and C G v v = j e
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Unformatted text preview: e a r r r s n t C 3 1 1 2 1 2 ω − + = ρ θ + = & & & . j i e e a r r s r s n t G 5 2 2 2 2 2 2 ω − + = ρ θ + = & & & & & Plugging these all into (1): j i j j i k j j i 3 3 2 3 2 2 2 6 2 3 ) 3 ( 2 3 5 α + α − ω = − + × α + ω = ω − r r r r r r s & & j-direction: 45 4 2 3 ω − = α , so i-direction: 3 2 2 α − = r s & & , so 45 8 2 2 ω = r s & & k 45 4 2 3 ω − = α v k 45 8 2 3 ω − = α v...
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This note was uploaded on 04/09/2008 for the course TAM 212 taught by Professor Keane during the Spring '08 term at University of Illinois at Urbana–Champaign.

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