Question

A sphere of radius $r$ is rolling without sliding. What is the ratio of rotational K.E. and total K.E. associated with the sphere?

• A. $\frac{1}{2}$
• B. $\frac{2}{7}$
• C. $\frac{2}{5}$
• D. 1

Answer 1

The correct option is B $\frac{2}{7}$

Given: A sphere of radius r is rolling without sliding.

To find the ratio of rotational K.E. and total K.E. associated with the sphere

Solution:

We know,

For an object that is rotating only, the kinetic energy is,

$K{E}_{rotational}=\frac{1}{2}I{\omega }^2⟹K{E}_{rotational}=\frac{1}{2}\times \frac{2}{5}M{R}^2\frac{v_{CM}^2}{R^2}⟹K{E}_{rotational}=\frac{1}{5}M{v}_{CM}^2....\left(i\right)$

For an object that is rolling, i.e., translating and rotating simultaneously, the total kinetic energy of such an object is:

And $K{E}_{total}=\frac{1}{2}M{v}_{CM}^2+\frac{1}{2}I{\omega }^2⟹K{E}_{total}=\frac{1}{2}M{v}_{CM}^2+\frac{1}{2}\times \frac{2}{5}M{R}^2\frac{v_{CM}^2}{R^2}⟹K{E}_{total}=\frac{1}{2}M{v}_{CM}^2+\frac{1}{5}M{v}_{CM}^2⟹K{E}_{total}=\frac{5+2}{10}M{v}_{CM}^2⟹K{E}_{total}=\frac{7}{10}M{v}_{CM}^2....\left(ii\right)$

So, by divinding eqn(i) and eqn(ii), we get

$\frac{K{E}_{rotational}}{K{E}_{total}}=\frac{\frac{1}{5}M{v}_{CM}^2}{\frac{7}{10}M{v}_{CM}^2}=\frac{2}{7}$

is the required ratio of rotational K.E. and total K.E. associated with the sphere