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Unformatted text preview: E is a constant of motion, G = ± m ˙ x 2k x 2 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 ² ˙ G = 2 E± m ˙ x 2k x 2 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 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 ² 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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 Spring '08
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 Physics, mechanics, Derivative, Noether's theorem

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