What is the ratio of total kinetic energy to | vedclass

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(D) The total kinetic energy $(K_{total})$ of a rolling body is given by $K_{total} = K_{translational} + K_{rotational} = \frac{1}{2}mv^2(1 + \frac{K

Vedclass · Accepted Answer

$3:1$

Vedclass · Answer

(D) The total kinetic energy $(K_{total})$ of a rolling body is given by $K_{total} = K_{translational} + K_{rotational} = \frac{1}{2}mv^2(1 + \frac{K^2}{R^2})$.For a disc,the radius of gyration $K$ is given by $K^2 = \frac{1}{2}R^2$,so $\frac{K^2}{R^2} = \frac{1}{2}$.The rotational kinetic energy is $K_{rot} = \frac{1}{2}I\omega^2 = \frac{1}{2}mv^2(\frac{K^2}{R^2})$.The ratio of total kinetic energy to rotational kinetic energy is $\frac{K_{total}}{K_{rot}} = \frac{1 + \frac{K^2}{R^2}}{\frac{K^2}{R^2}}$.Substituting $\frac{K^2}{R^2} = \frac{1}{2}$,we get $\frac{1 + 1/2}{1/2} = \frac{3/2}{1/2} = 3$.Thus,the ratio is $3:1$.

What is the ratio of total kinetic energy to rotational kinetic energy for a rolling disc?

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