Analytical Mech Homework Solutions 143

Analytical Mech Homework Solutions 143 - T= 1 21 21 mv I Mx...

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22 111 222 Tm v I M ω =++ ± 2 x For rolling without slipping, x a = ± . 2 2 5 I ma = () 2 11 2c o s 25 x x x x m x M θ ′′ =+ + ++ ±± ± ± ± ± 2 1 2 x sin Vm g x =− , for V at the initial position of the ball. 0 = 17 o s LTV mx x x x =−= + +  ±±± ± 2 1 sin 2 Mx mgx ± 7 cos 5 L mx mx x , 7 cos 5 dL mx mx dt x   ± sin L mg x = 7 cos sin 5 mx mx mg += 5 sin cos 7 xg x cos L mx mx Mx x + ± ± , cos mMxm x dt x + ± 0 L x = cos 0 x = 2 55 sin cos cos 0 77 mMx m g m x θθ = 2 5s i n c o s 5c o s 7 mg x mm M = −+ 10.7 Let x be the slant height of the particle … vxx 2 ± t = x 2 v m x x 2 == + ± sin sin g x m g x t = = 1 sin 2 mxx m g x t + ± L mx x = ± ± , mx dt x = ± 2 sin L mx mg t x 2 sin mx mx mg t 2 sin x
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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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