Analytical Mech Homework Solutions 55

# Analytical Mech Homework Solutions 55 - = k , with = 0 r =...

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ˆ k ω = G , with 0 = G ± G GG r , so ˆ r b e ′ = ( ) 2 ˆ r rb ωω e ××= G G vv ˆ e θ ′′ = , so ˆ r e ×= GGG From eqn 5.2.14, 2 ( ) aa v r =+ ×+×× and putting in terms from above 2 2 2 r v av b b =− For no slipping s Fm µ G g , so s ag G 2 2 2 s v vb g b ++≤ 2 22 20 mm s v b b g ++ = 2 2 ms vbb bb g ′ =− ± + Since v was defined positive, the +square root is used. b vb g + (b) ˆ e G ˆ r e + 2 2 2 r v b b + 2 2 2 s v g b −+≤ b g =+ 5.9 As in Example 5.2.2, ˆ V j ρ = D G and 2 ˆ V Ai = D D G For the point at the front of the wheel: 2 ˆ V b = D G G rj ±± and vV ˆ k D G 0 = ± () ˆ ˆˆ VV rkb j b i ′′′ ×− = DD 2 ˆ bV b rki ρρ × = D G j 0 V vkV k × = D D
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## This note was uploaded on 01/06/2012 for the course PHYSICS 4360C taught by Professor Johnanderson during the Fall '11 term at UNF.

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