Question

A circular disc rolls down an inclined plane without slipping. What fraction of its total energy is translational energy?

Answer 1

Rotational kinetic energy is,

$K{E}_r=\frac{1}{2}I{\omega }^2=\frac{1}{4}M{R}^2{\omega }^2$

$(\because$ For the disc, $I=\frac{1}{2}M{R}^2)$

Translational kinetic energy is,

$K{E}_t=\frac{1}{2}M{v}^2=\frac{1}{2}M(R\omega {)}^2=\frac{1}{2}M{R}^2{\omega }^2$

$(\because$ For pure rolling, $v=R\omega )$

Now, total kinetic energy is,

$KE=K{E}_r+K{E}_t=\frac{3}{4}M{R}^2{\omega }^2$

So,

$\frac{K{E}_t}{KE}=\frac{\frac{1}{2}M{R}^2{\omega }^2}{\frac{3}{4}M{R}^2{\omega }^2}=\frac{2}{3}$

Hence, option $\left(D\right)$ is the correct answer.