Question:
A small spherical ball is released from a point at a height $h$ on a rough track shown in figure. Assuming that it does not slip anywhere, find its linear speed when it rolls on the horizontal part of the track.
Solution:
By energy conservation,
$m q h=\frac{1}{2} I \omega^{2}+\frac{1}{2} m v^{2}$
$\mathrm{mgh}=\frac{1}{2}\left(\frac{2}{5} m R^{2}\right)\left(\frac{v^{2}}{R^{2}}\right)+\frac{1}{2} m v^{2}$
$v=\sqrt{\frac{10 h}{7}}$