Question

If a solid sphere of mass M and radius R is rolling perfectly on a rough horizontal surface, what is the percentage of the rotational kinetic energy

• A. 50%
• B. 71.5%
• C. 28.57%
• D. 10%

Answer 1

The correct option is C 28.57%

Given: if a solid sphere of mass M and radius R is rolling perfectly on a rough horizontal surface,

To find the percentage of the rotational kinetic energy

Solution:

We know

Total energy = Kinetic energy due to translation + Kinetic energy due to rotation

And we also know,

Moment of inertia of solid sphere, $I=\frac{2}{5}M{R}^2$

and $\omega =\frac{v}{R}$

So KE due to translation $K{E}_{tr}=\frac{1}{2}M{v}^2$

KE due to rotation $K{E}_{rot}=\frac{1}{2}I{\omega }^2⟹=\frac{1}{2}\times \frac{2}{5}M{R}^2\times {\omega }^2⟹\frac{1}{5}M{R}^2\times \frac{v^2}{R^2}⟹K{E}_{rot}=\frac{1}{5}M{v}^2$

So the total kinetic energy becomes

$K{E}_{tot}=K{E}_{tr}+K{E}_{rot}⟹K{E}_{tot}=\frac{1}{2}M{v}^2+\frac{1}{5}M{v}^2⟹K{E}_{tot}=\frac{2+5}{10}M{v}^2⟹K{E}_{tot}=\frac{7}{10}M{v}^2$

Contribution of rotational energy to the total energy

$=\frac{\text{Rotational KE}}{\text{Total Energy}}⟹=\frac{K{E}_{rot}}{K{E}_{tot}}⟹=\frac{\frac{1}{5}M{v}^2}{\frac{7}{10}M{v}^2}⟹=\frac{2}{7}$

Percentage is $\frac{2}{7}\times 100=28.57\%$

is the percentage of the rotational kinetic energy