A ring of radius 0.5, m and mass 10 ,kg is ro | vedclass

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(D) The rotational kinetic energy $K$ of a body rotating about an axis is given by $K = \frac{1}{2}I\omega^2$.For a ring of mass $M$ and radius $R$ ro

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$250$

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(D) The rotational kinetic energy $K$ of a body rotating about an axis is given by $K = \frac{1}{2}I\omega^2$.For a ring of mass $M$ and radius $R$ rotating about its diameter,the moment of inertia $I$ is given by $I = \frac{1}{2}MR^2$.Given: $M = 10 \,kg$,$R = 0.5 \,m$,and $\omega = 20 \,rad/s$.Substituting the values into the formula:$I = \frac{1}{2} 	imes 10 	imes (0.5)^2 = 5 	imes 0.25 = 1.25 \,kg \cdot m^2$.Now,calculate the kinetic energy:$K = \frac{1}{2} 	imes 1.25 	imes (20)^2$$K = \frac{1}{2} 	imes 1.25 	imes 400$$K = 1.25 	imes 200 = 250 \,J$.

$A$ ring of radius $0.5\, m$ and mass $10 \,kg$ is rotating about its diameter with an angular velocity of $20 \,rad/s.$ Its kinetic energy is .......... $J$.

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