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8. A torque of magnitude \( 400 N - m \), acting on a body of mass \( 40 kg \), produces an angular acceleration of \( 20 rad / s ^{2} \). Calculate the M.I. and radius of gyration of the body. |
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Answer» \(\tau\) = 400 N-m M = 40 kg α = 20 rad/s2 I = ? k = ? We know that \(\tau\) = Iα ∵ I = Mk2 \(\tau\) = Mk2α k = \(\sqrt{\frac{\tau}{M\alpha}}\) where k = radius of gyration. k = \(\sqrt{\frac{400}{40 \times 20}}\) k = \(\sqrt{\frac{1}{2}}\) Radius of gyration k = 0.70 m Moment of inertia I = Mk2 I = 40 × \(\frac{1}{2}\) I = 20 kg m2 |
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