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Unformatted text preview: energy of E final = mgH + 1 2 I 2 = E initial = mgH . We see that since some of the initial energy is stuck in the rotational motion, it cannot go as high. Problems 1. A rod with a Fxed pivot falls down (a) KE final = 1 2 MgL (b) = r 3 g L (into the page) (c) v top = 3 gL 2. The bowling balls Fnal speed is: v 2 cm = 10 7 gL sin . 3. Various objects roll down a ramp: (a) The block reaches the bottom Frst. (b) The bowling ball and the marble reach the bottom at the same time. (c) The bowling ball reaches the bottom Frst. 4. A cylinder rolls down a track. 1 (a) v 2 f = 4 3 g ( H-h-R ) (b) KE linear = 1 2 mv 2 f < U = Mg ( H-h ) (c) D 2 = 8 3 h ( H-h-R ) 5. A sphere rolls down a circular ramp. (a) v 2 = 4 3 gR (1-cos ) (b) cos c = 4 7 55 . 2 o 2...
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