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.

Figure

Answer 1

Let r be the radius of the ball.
Let v be the linear speed of the ball when it rolls on the horizontal part of the track.
Let ω be the angular speed of the ball when it rolls on the horizontal part of the track.
Potential energy the ball has gained w.r.t. the surface will be converted to angular kinetic energy about the centre and linear kinetic energy.
Therefore, we have:
$mgh=\frac{1}{2}I{\omega }^2+m{v}^2$
$$\begin{gathered} \Rightarrow mgh=\frac{1}{2}\times \left(\frac{2}{5}m{R}^2\right)\times {\left(\frac{v}{R}\right)}^2+\frac{1}{2}m{v}^2 \\ \Rightarrow gh=\frac{1}{5}{v}^2+\frac{1}{2}{v}^2 \\ \Rightarrow v^2=\left(\frac{10}{7}\right)gh \\ \Rightarrow v=\sqrt{\left(\frac{10gh}{7}\right)} \end{gathered}$$