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(D) For a rolling disc,the total kinetic energy $(K_{total})$ is the sum of translational kinetic energy and rotational kinetic energy.$K_{total} = K_

Vedclass · Accepted Answer

$3:1$

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(D) For a rolling disc,the total kinetic energy $(K_{total})$ is the sum of translational kinetic energy and rotational kinetic energy.$K_{total} = K_{trans} + K_{rot} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$.Since $v = R\omega$ and for a disc $I = \frac{1}{2}mR^2$,we have $K_{rot} = \frac{1}{2}(\frac{1}{2}mR^2)(\frac{v}{R})^2 = \frac{1}{4}mv^2$.Thus,$K_{total} = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2$.The ratio of total kinetic energy to rotational kinetic energy is $\frac{K_{total}}{K_{rot}} = \frac{\frac{3}{4}mv^2}{\frac{1}{4}mv^2} = \frac{3}{1}$.

The ratio of the total kinetic energy to the rotational kinetic energy for a rolling disc is:

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