Question

A solid spherical ball is rolling on a frictionless horizontal plane surface about its axis of symmetry. The ratio of rotational kinetic energy of the ball to its total kinetic energy is

• A. $\frac{2}{5}$
• B. $\frac{2}{7}$
• C. $\frac{1}{5}$
• D. $\frac{7}{10}$

Answer 1

The correct option is B $\frac{2}{7}$

Rotational kinetic energy,
$K{E}_R=\frac{1}{2}I{\omega }^2$
$=\frac{1}{2}\times \frac{2}{5}\times \left(m{R}^2\right)\times {\omega }^2$

Total kinetic energy,

$K{E}_{total}=\frac{1}{2}m{v}^2[1+\frac{K^2}{R^2}]$

$K{E}_{\text{total}}=\frac{1}{2}\times m\times {\omega }^2{R}^2\times \frac{7}{5}$

$\therefore \frac{K{E}_R}{K{E}_{\text{total}}}=\frac{2}{7}$