Question

The total kinetic energy of rolling solid sphere having translational velocity $\nu$ is

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

Answer 1

The correct option is A $\frac{7}{10}$m${\nu }^2$

Total kinetic energy = $\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2$

For pure rotation, $\omega =\frac{v}{r}$ and $I=\frac{2}{5}m{r}^2$

So.Total kinetic energy = $\frac{1}{2}m{v}^2+\frac{1}{2}\left(\frac{2}{5}m{r}^2\right)(\frac{v}{r}{)}^2$

$=\frac{1}{2}m{v}^2+\frac{2}{10}m{v}^2=\frac{7}{10}m{v}^2$

Therefore, A is correct option.