Question

A sphere is rolling down an inclined plane without slipping. The ratio of rotational energy to total kinetic energy is

• A. $\frac{5}{7}$
• B. $\frac{2}{5}$
• C. $\frac{2}{7}$
• D. none of these

Answer 1

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

For solid sphere, $I=\frac{2}{5}M{R}^2$

Also, $v=Rw$ (no slipping)

Rotational K.E, $E_R=\frac{1}{2}I{w}^2=\frac{1}{2}\times \frac{2}{5}M{R}^2{w}^2=\frac{1}{5}M{v}^2$

Total K.E $E=\frac{1}{2}I{w}^2+\frac{1}{2}M{v}^2=\frac{7}{10}M{v}^2$

$\frac{E_R}{E}=\frac{2}{7}$