A body having a moment of inertia about its a | vedclass

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(A) Given: Moment of inertia $I = 3 \ kg \ m^{2}$,angular velocity $\omega = 3 \ rad \ s^{-1}$,and mass $m = 27 \ kg$.The rotational kinetic energy of

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(A) Given: Moment of inertia $I = 3 \ kg \ m^{2}$,angular velocity $\omega = 3 \ rad \ s^{-1}$,and mass $m = 27 \ kg$.The rotational kinetic energy of the body is given by $K_{rot} = \frac{1}{2} I \omega^{2}$.The translational kinetic energy of the second body is given by $K_{trans} = \frac{1}{2} m v^{2}$.According to the problem,$K_{rot} = K_{trans}$.Therefore,$\frac{1}{2} I \omega^{2} = \frac{1}{2} m v^{2}$.$I \omega^{2} = m v^{2}$.$v^{2} = \frac{I \omega^{2}}{m}$.$v = \omega \sqrt{\frac{I}{m}}$.Substituting the values: $v = 3 	imes \sqrt{\frac{3}{27}} = 3 	imes \sqrt{\frac{1}{9}} = 3 	imes \frac{1}{3} = 1 \ m \ s^{-1}$.

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