problem10_52

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

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10.52: a) The net torque must be m. N 60 . 2 s) 00 . 9 ( rev/min rad/s 60 2 rev/min 120 ) m kg 86 . 1 ( 2 = π × = = = t ω I τ This torque must be the sum of the applied force FR and the opposing frictional torques f τ at the axle and nr μ fr k = due to the knife. Combining, ) ( 1 k f nr μ τ τ R F + + = ( 29 N. 1 . 68 m) N)(0.260 (0.60)(160 m) N 50 . 6 ( m) N 60 . 2 ( m 500 . 0 1 = + + = b) To maintain a constant angular velocity, the net torque τ is zero, and the force is F N. 62.9 m) N 24.96 m N 50 . 6 ( m 500 .
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Unformatted text preview: 1 = ⋅ + ⋅ = ′ F c) The time t needed to come to a stop is found by taking the magnitudes in Eq. (10.27), with f τ τ = constant; ( 29 ( 29 ( 29 s. 6 . 3 m N 6.50 m kg 86 . 1 rev/min 120 2 rev/min rad/s 60 2 f f = ⋅ ⋅ × = = = π τ ωI τ L t Note that this time can also be found as ( 29 . s 00 . 9 m 6.50N m N 60 . 2 ⋅ ⋅ = t...
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