Analytical Mech Homework Solutions 99

# Analytical Mech Homework Solutions 99 - 1 2 gt g 1 gt ln(1...

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() 2 11 ln 1 1 21 gt g Cg t k t d kk  =− − − −+   t k t 2 2 1 ln 1 ln 1 2 gt gt g t k t k k − + + t = 2 2 1 1l n 1 2 gg t k t k t −+− + t but 0 y = at t so 0 = 0 C = 2 2 1 n 1 2 gt g yg t k t + k t and at tt = D (a) ()( ) 2 2 22 l n 2 gt M m m M M m mM  =+ + +  +  D DD D HM m (b) ( ) ln Mm gt m + D D D vM m (c) letting 1 m M < < D ε = we have ()() ( 2 2 1 l n 1 2 gt H ) εε ++ D 2 2 1 gt + + D 3 3 2 6 gt H D Similarly: n 1 gt v =++ D 23 1 gt + D 1 gt 2 D (d) Hm ; 327 = 1 9.8 vm s = 7.26 or mm m ± k k t D Burn-out occurs at time m k T = D So – the rocket equation (7.7.7) becomes dv dt ± mV (-) since V is oppositely directed to m v ± 15
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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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