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problem09_82

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

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9.82: (a) The kinetic energy of the falling mass after 2.00 m is ( 29 ( 29 . J 100 m/s 00 . 5 kg 00 . 8 2 2 1 2 2 1 = = = mv KE The change in its potential energy while falling is ( 29 ( 29 ( 29 J 8 . 156 m 00 . 2 m/s 8 . 9 kg 00 . 8 2 = = mgh The wheel must have the “missing” 56.8 J in the form of rotational KE. Since its outer rim is moving at the same speed as the falling mass, 5.00 s m : ϖ r v = rad/s 51 . 13 m 370 . 0 m/s 00 . 5 = = = r v ϖ therefore ; 2 1 2 ϖ I KE = ( 29 ( 29 2 2 2 2 m kg 622 . 0 or m kg 6224 . 0 s rad 51 . 13 J 8 . 56 2
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Unformatted text preview: ⋅ = = ϖ = KE I (b) The wheel’s mass is 280 2 s m 8 . 9 N = 28.6 kg. The wheel with the largest possible moment of inertia would have all this mass concentrated in its rim. Its moment of inertia would be ( 29 ( 29 2 2 2 m kg 92 . 3 m 370 . kg 6 . 28 ⋅ = = = MR I The boss’s wheel is physically impossible....
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