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, its rotational kinetic energy when it comes to the bottom of track is

• A. $mgh$
• B. $\frac{10}{7}mgh$
• C. $\frac{5}{7}mgh$
• D. $\frac{2}{7}mgh$

Answer 1

Answer: (D)

$\frac{2}{7}mgh$

By applying conservation of energy

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

$\frac{1}{2}m{v}^2+\frac{1}{2}\left(\frac{2}{5}M{R}^2\right)\frac{v^2}{R^2}=mgh$;

$\frac{1}{2}m{v}^2\left[\frac{7}{5}\right]=mgh\Rightarrow \text{Translational KE}=\frac{5mgh}{7}$

Rotational KE $=\frac{2}{5}$ [Translational KE] $\Rightarrow =\frac{2}{5}\left(\frac{5mgh}{7}\right)=\frac{2}{7}mhg$