Question

When a sphere rolls without slipping, the ratio of its kinetic energy of translation to its total kinetic energy is

• A. 1 : 7
• B. 1 : 2
• C. 1 : 1
• D. 5 : 7

Answer 1

The correct option is

D

5 : 7

We know that,

The kinetic energy of translation motion = $\frac{1}{2}m{v}^2$

The kinetic energy of the rolling sphere = $\frac{1}{2}m{v}^2+\frac{1}{2}I{w}^2$

So the ratio of translation motion and total kinetic energy =

$\frac{T}{E_t}=\frac{\frac{1}{2}m{v}^2}{\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2}=\frac{\frac{1}{2}m{v}^2}{\frac{1}{2}m{v}^2+\frac{1}{2}\left(\frac{2}{5}m{r}^2\right)\left(\frac{v^2}{r^2}\right)}=\frac{5}{7}$