A disc rolls without slipping with a constant | vedclass

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(B) The total kinetic energy $(K_{total})$ of a rolling body is the sum of its translational kinetic energy $(K_t)$ and rotational kinetic energy $(K_

Vedclass · Accepted Answer

$1/3$

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(B) The total kinetic energy $(K_{total})$ of a rolling body is the sum of its translational kinetic energy $(K_t)$ and rotational kinetic energy $(K_r)$.$K_{total} = K_t + K_r = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$For a disc,the moment of inertia about its central axis is $I = \frac{1}{2}mr^2$. Since it rolls without slipping,$v = r\omega$,so $\omega = v/r$.Substituting these into the rotational kinetic energy formula:$K_r = \frac{1}{2} (\frac{1}{2}mr^2) (\frac{v}{r})^2 = \frac{1}{4}mv^2$Now,calculate the total kinetic energy:$K_{total} = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2$The fraction of total kinetic energy that is rotational is:$\frac{K_r}{K_{total}} = \frac{\frac{1}{4}mv^2}{\frac{3}{4}mv^2} = \frac{1}{3}$

$A$ disc rolls without slipping with a constant velocity. What fraction of its total kinetic energy is in the form of rotational kinetic energy?

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