problem10_32

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

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10.32: 2 2 2 1 2 2 1 m kg 2 . 42 m) kg)(2.08 117 ( = = = mL I 10 a) . rad/s 2 . 46 m kg 42.2 m N 1950 2 2 = = = I τ α b) rad/s. 9 . 53 rev) 2 rev 0 . 5 )( rad/s 2 . 46 ( 2 2 2 = × = = π αθ ϖ c) From either (10.24), Eq. or 2 1 2 = = K W J. 10 6.13 rad/rev) 2 rev N.m)(5.00 1950 ( 4 × = × = θ = π τ W d), e) The time may be found from the angular acceleration and the total angle, but the
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Unformatted text preview: instantaneous power is also found from hp). kW(141 105 = = P The average power is half of this, or kW. 6 . 52...
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This document was uploaded on 02/04/2008.

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