Question

A small solid sphere of mass $m$ is released from a point $A$ at a height $h$ above the bottom of a rough track as shown in the figure. If the sphere rolls down the track without slipping, find its rotational kinetic energy when it comes to the bottom of track.

Answer 1

$K.E=P.E$

$\frac{1}{2}m{v}^2[1+\frac{k^2}{r^2}]=mgh$

$v^2[1+\frac{2}{3}]=gh$

$v=\sqrt{\frac{3}{5}gh}$

Now $\frac{1}{2}m{v}^2+RotationalK.E=mgh$

$\therefore mgh-\frac{1}{2}m\times \frac{3}{5}gh=$ Rotational K.E

So Rotational Kinetic Energy

$=(1-\frac{3}{10})mgh=\frac{7}{10}mgh$