Question

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

Answer 1

we know that,

$K{E}_{Total}=K{E}_{Transnational}+K{E}_{Rotational}$

Let the disc(MR) has moved down a

height 'h' has velocity v and

angular w.

Than for pure Rolling

$RW=V$

$K{E}_{Total}=\frac{1}{2}m{v}^2+\frac{I}{2}{w}^2$

For disc, $I=\frac{1}{2}m{R}^2$

So, $K{E}_{Total}=\frac{1}{2}m{v}^2+\frac{1}{2}\left(\frac{m{R}^2}{2}{w}^2\right)=\frac{3}{4}m{v}^2$

Fraction of total energy that is transnational

$=\frac{\frac{1}{2}m{v}^2}{\frac{3}{4}m{v}^2}=\frac{2}{3}$