problem05_49

University Physics with Modern Physics with Mastering Physics (11th Edition)

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5.49: a) Setting rad a g = in Eq. (5.16) and solving for the period T gives s, 1 . 40 s m 9.80 m 400 2 2 2 = = = π g R π T so the number of revolutions per minute is min rev 1.5 s) 1 . 40 ( min) s 60 ( = . b) The lower acceleration corresponds to a longer period, and hence a lower rotation
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Unformatted text preview: rate, by a factor of the square root of the ratio of the accelerations, min. rev 92 . 9.8 3.70 min) rev 5 . 1 ( = × = ′ T ....
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This document was uploaded on 02/04/2008.

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