Question

A small spherical ball is released from a point at a height h on a rough track shown in figure (10 - E13). Assuming that it does not slip anywhere, find its linear speed when it rolls on the horizontal part of the track.

Answer 1

A small spherical ball is released from a point at a height h on rough track and the sphere does not slip.

Therefore, potential energy it has gained w.r.t. the surface will be converted to angular kinetic energy about the centre and linear kinetic energy.

Therefore, mgh = $\frac{1}{2}I{\omega }^2=m{v}^2$

$\Rightarrow mgh=\frac{1}{2}\times \frac{2}{5}m{R}^2{\omega }^2+\frac{1}{2}m{v}^2$

$\Rightarrow 9h=\frac{1}{5}{v}^2+\frac{1}{2}{v}^2$

$\Rightarrow v^2=\left(\frac{10}{7}\right)gh$

$\Rightarrow v=\sqrt{\left(\frac{10gh}{7}\right)}$