The rotational kinetic energy and translation | vedclass

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(D) Let $m$ be the mass,$v$ be the velocity of the center of mass,$R$ be the radius,and $I$ be the moment of inertia of the body.The translational kin

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ring

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(D) Let $m$ be the mass,$v$ be the velocity of the center of mass,$R$ be the radius,and $I$ be the moment of inertia of the body.The translational kinetic energy is $KE_{	ext{trans}} = \frac{1}{2} mv^2$.The rotational kinetic energy is $KE_{	ext{rot}} = \frac{1}{2} I \omega^2$.Since the body is rolling without slipping,$v = \omega R$,so $\omega = \frac{v}{R}$.Substituting this into the rotational kinetic energy formula: $KE_{	ext{rot}} = \frac{1}{2} I \left(\frac{v}{R}ight)^2 = \frac{1}{2} \left(\frac{I}{R^2}ight) v^2$.Given $KE_{	ext{trans}} = KE_{	ext{rot}}$,we have $\frac{1}{2} mv^2 = \frac{1}{2} \left(\frac{I}{R^2}ight) v^2$.This implies $m = \frac{I}{R^2}$,or $I = mR^2$.The moment of inertia $I = mR^2$ corresponds to a ring (or a thin hollow cylinder) about its central axis.

The rotational kinetic energy and translational kinetic energy of a rolling body are the same. The body is:

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