ch11-p056 - 56. We denote the cockroach with subscript 1...

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Lm v r I mR ii i i =+ = + 111 2 2 1 0 2 20 2 1 2 ωω ω . After the cockroach has completed its walk, its position (relative to the axis) is rR f 1 2 = so the final angular momentum of the system is R ff f = F H G I K J + 1 2 2 2 2 1 2 . Then from L f = L i we obtain f 1 4 1 2 1 2 1 2 1 2 2 2 + F H G I K J F H G I K J . Thus, 22 12 2 1 00 0 0 2 1 21 ( / ) 2 1.33 . 42 1 / 4 ( / ) 2 1 / 4 2 f m m §· § · ++ + == = = ¨¸ ¨ ¸ + ©¹ © ¹ With 0 = 0.260 rad/s, we have f =0.347 rad/s. (b) We substitute I = L / into KI = 1 2 2 and obtain KL = 1 2 . Since we have L i = L f , the kinetic energy ratio becomes /2 1.33. f L K = (c) The cockroach does positive work while walking toward the center of the disk, increasing the total kinetic energy of the system.
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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