Question

A solid sphere of mass M and radius roots on a horizontal surface without slipping. The ratio of rotational $K.E$ to $K.E$ is:

• A. $1/2$
• B. $3/7$
• C. $2/7$
• D. $2/10$

Answer 1

The correct option is C $2/7$

$RotationalKineticEnergy$ $\left(K_r\right)=\frac{1}{2}I{\omega }^2$

where, $I=$ Moment of Inertia of solid sphere $=\frac{2}{5}m{r}^2$

$\omega =$ Angular Velocity $=\frac{v}{r}(\because forpurerolling)$

So, $K_r=\frac{1}{2}\left(\frac{2}{5}m{r}^2\right)(\frac{v}{r}{)}^2$

$=\frac{1}{5}m{v}^2$

$TotalKineticEnergy$ $\left(K_T\right)=\frac{1}{2}m{v}^2$ $+K_r$

$=\frac{1}{2}m{v}^2$ $+\frac{1}{5}m{v}^2$

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

Therefore, $\frac{K_r}{K_T}=\frac{\frac{1}{5}m{v}^2}{\frac{7}{10}m{v}^2}$

$=\frac{2}{7}$

So, correct option will be $\left(C\right).$