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(A) For a rolling disc of mass $M$ and radius $R$ moving with velocity $v$ and angular velocity $\omega = v/R$,the translational kinetic energy is $K_

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$3$ times its rotational kinetic energy

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(A) For a rolling disc of mass $M$ and radius $R$ moving with velocity $v$ and angular velocity $\omega = v/R$,the translational kinetic energy is $K_t = \frac{1}{2} Mv^2$. The rotational kinetic energy about the center of mass is $K_r = \frac{1}{2} I \omega^2$. For a disc,the moment of inertia $I = \frac{1}{2} MR^2$. Substituting $I$ and $\omega = v/R$,we get $K_r = \frac{1}{2} (\frac{1}{2} MR^2) (v/R)^2 = \frac{1}{4} Mv^2$. The total kinetic energy $K = K_t + K_r = \frac{1}{2} Mv^2 + \frac{1}{4} Mv^2 = \frac{3}{4} Mv^2$. Comparing $K$ with $K_r$: $K = \frac{3/4 Mv^2}{1/4 Mv^2} K_r = 3 K_r$. Thus,the total kinetic energy is $3$ times its rotational kinetic energy.

$A$ thin uniform circular disc rolls with a constant velocity without slipping on a horizontal surface. Its total kinetic energy is

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