Analytical Mech Homework Solutions 116

# Analytical Mech - 14 m l 2 2 1 ml 2 2 l = mg(1 cos 24 2 12 2 2 l = g(1 cos 3 1 3g 2 =(1 cos l 1 1 3g 3g 2 3g =(1 cos sin = sin 2 l 2l l ml 3g 3g Rx

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14 () 22 2 2 1 1c o s 24 2 1 2 2 ml l mg θ += ± ± 2 o s 3 l g =− ± 1 2 3 o s g l θθ    ± 1 2 13 3 3 o s s i n s i n gg ll g l =  ±± ± 33 sin 1 cos cos sin x ml g g R  − +  3 sin 3cos 2 4 x mg R cos 1 cos sin sin y ml g g Rm g + 2 2 3s cos cos y mg g i n + 2 3cos 1 4 y mg R The reaction force constrains the tail of the rocket from sliding backward for : 0 x R > 2 0 −> 1 2 cos 3 < The rocket is constrained from sliding forward for 0 x R < : 1 2 cos 3 > 8.19 sin cos mx mg mg µθ ( ) sin cos xg + x Since acceleration is constant, 2 1 2 x xt =+ D ± : mg G f G θ 2 sin cos 2 gt xv t + D Meanwhile 2 2 cos 5 mg a I ma µ θω == ω 5c o s 2 g a = ± o s 2 g t a = 14
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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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