ch10-p056 - 2 , then Eq. 10-45 gives | | = 2.955 3.0 rad/s...

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56. (a) In this case, the force is mg = (70 kg)(9.8 m/s 2 ), and the “lever arm” (the perpendicular distance from point O to the line of action of the force) is 0.28 m. Thus, the torque (in absolute value) is (70 kg)(9.8 m/s 2 )(0.28 m). Since the moment-of-inertia is I = 65 kg·m
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Unformatted text preview: 2 , then Eq. 10-45 gives | | = 2.955 3.0 rad/s 2 . (b) Now we have another contribution (1.4 m 300 N) to the net torque, so | net | = (70 kg)(9.8 m/s 2 )(0.28 m) + (1.4 m)(300 N) = (65 kgm 2 ) | | which leads to | | = 9.4 rad/s 2 ....
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