Question

A ball (solid sphere) of mass $m$ is rolling on a smooth horizontal surface as shown in figure. At an instant the magnitude of the velocity of the centre of mass is $v_0$ and its angular velocity is $\omega \frac{v_0}{2R}$, where $R$ is the radius of the ball. The total kinetic energy of the rolling ball at this instant is:-

• A. $\frac{7}{5}m{R}^2{\omega }_0^2$
• B. $\frac{11}{5}m{R}^2{\omega }_0^2$
• C. $\frac{11}{20}m{v}_0^2$
• D. $\frac{7}{20}m{v}_0^2$

Answer 1

The correct option is C $\frac{11}{20}m{v}_0^2$

$KE=\frac{1}{2}m{v_0}^2+\frac{1}{2}\left(\frac{2m{R}^2}{5}\right)w^2$

$=\frac{1}{2}m{v_0}^2+\frac{m{R}^2}{5}\cdot \frac{{v_0}^2}{4R^2}$

$=\frac{1}{2}m{v_0}^2+\frac{m{v_0}^2}{20}$

$=\frac{11m{v_0}^2}{20}$

Hence,

option $\left(C\right)$ is correct answer.