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Unformatted text preview: E is a constant of motion, G = m x 2k x 2 tmx x = 2 Etmx x G = 2 Em x 2mx x Now substitute x from the equation of motion. G = 2 Em x 2x k x 3 G = 2 E m x 2k x 2 G = 2 E2 E = 0 (c) Calculate E and G at t = 0 . E = 0k 2 R 2 =k 2 R 2 G = 2 E (0)mx (0) x (0) = 0 Thus, for any time t , we have E =k 2 R 2 = 1 2 m x 2k 2 x 2 G = 0 =2 k 2 R 2 tmx x From the second equation, x =kt mxR 2 Substitute this into the energy equation.k 2 R 2 = 1 2 m kt mxR 2 2k 2 x 2k 2 R 2 = k 2 t 2 2 mR 4 x 2k 2 x 21 R 2 = kt 2 mR 4 x 21 x 21 R 2 =1 x 2 1kt 2 mR 4 x ( t ) = R r 1kt 2 mR 4 (d) From previous part, x = 0 when t = R 2 r m k . 2...
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This note was uploaded on 10/06/2011 for the course PHYS 325 taught by Professor Staff during the Spring '08 term at University of Illinois, Urbana Champaign.
 Spring '08
 Staff
 Physics, mechanics

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