Question

A circular disc rolls down an inclined plane. The ratio of the rotational kinetic energy to the total kinetic energy is

• A. $\frac{1}{2}$
• B. $\frac{1}{3}$
• C. $\frac{2}{3}$
• D. $\frac{3}{4}$

Answer 1

Answer: (B)

Rotational kinetic energy, $K_R=\frac{1}{2}/{\omega }^2$
$K_R=\frac{1}{2}\times \frac{M{R}^2}{2}{\omega }^2=\frac{1}{4}M{v}^2$
$(\because \vee =R\omega )$
Translational kinetic energy
$K_T=\frac{1}{2}M{v}^2$
Total kinetic energy $=K_T+K_R$
$=\frac{1}{2}M{v}^2+\frac{1}{4}M{v}^2$
$=\frac{3}{4}M{v}^2$
$\therefore \frac{\text{ Rotational kinetic energy }}{\text{ Total kinetic energy }}$
$=\frac{\frac{1}{4}M{v}^2}{\frac{3}{4}M{v}^2}$
$=\frac{1}{3}$