Analytical Mech Homework Solutions 18

Analytical Mech - 2 2c2 v c1 c1 4mgc2 t 1 = ln 2 2 m c1 4mgc2 2c2 v c1 c1 4mgc2 t 2 c1 4mgc2 m 1 2 2c v c = ln 2c v c 2 v 0(c 4mgc c 2 1 c1 4mgc2 2

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2 21 1 2 22 12 2 1 1 0 24 1 ln 42 4 v cv c c mgc t m cm g c c v ccm g c −− − + = +− + + 2 () ( ) ( ) ( ) ( ) 1 2 1 1 2 2 2 2 2 4 4l n 4 c c c t g c m c c c ++ + + += + + + as t , →∞ 2 211 2 t cv c c + = 0 1 2 2 11 2 t g v ccc    =− + +    Alternatively, when v , t v = 2 0 tt dv mm g c v dt == c v 1 2 2 2 t g v + + 2.16 2 dv k av x dx m 2 0 vx b kdx vdv mx ∫∫ 2 2 k v mx b 1 2 dx k k b x v dt m x b mb x = 1 2 1 1 3 2 2 00 01 1 t b x mb x mb x b dt dx d x kbx k b b Since x b , say 2 sin x b θ = 33 2 sin 2sin cos 2 sin 2c o s d mb mb td kk ππ θθ       1 3 2 8 mb t k π = 7
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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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