Analytical Mech Homework Solutions 100

# Analytical Mech Homework Solutions 100 - dv Vk = dt m kt...

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dv Vk dt m k t =+ D Thus () 1 ln dt k V m k t C mk t == D D vV where C is a constant. + 1 Now, v so 0@ 0 t 1 ln CVm = D Hence ln t m  =−   D D , 0 m t k ε ≤≤ D Let y be the displacement at the time t so 2 ln t yV d tC m + D D Integrating the above expression by parts 2 ln t tdt t V k C mm k t + D DD 2 1 t m Vt V dt C k t  + +   D D ln 2 ln t Vm Vt Vt m k t C + + + D D D D ln since 0 y = at t , 0 = 2 ln Vm m k C = and we have ( ) ln t V t t km D D D At burn-out m k tT D so (a) 1l n 1 mV yD k εε +− D (b) cannot exceed 1.0 although it can approach 1.0 for small payloads Thus max lim 1 yy k D 7.27 From eqn. 7.7.5,
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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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