Question

A solid sphere rolls on a horizontal table. Then the ratio of its rotational kinetic energy to its total energy is:

• A. 1:2
• B. 1:1
• C. 2:5
• D. 2:7

Answer 1

The correct option is D 2:7

$foranonslippingrotatingsphere-v=r\omega rotationalKE=\frac{1}{2}\times \frac{2}{5}M{r}^2\times {\omega }^2=\frac{1}{5}M{r}^{2}totalKE=rotationalKE+translationalKEtotalKE=\{\frac{1}{2}\times \frac{2}{5}M{r}^2\times {\omega }^2\}+\left\{\frac{1}{2}M{v}^2\right\}=\frac{7}{10}M{r}^{2}ratio=\frac{\frac{1}{5}}{\frac{7}{10}}=\frac{2}{7}$