Question

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

• A. $1/2$
• B. $1/3$
• C. $2/3$
• D. $3/4$

Answer 1

The correct option is B $1/3$

Moment of inertia, I = $\frac{1}{2}m{R}^2$

$K.E{.}_{rotational}=\frac{1}{2}I{w}^2=\frac{1}{4}m{R}^2{w}^2$

$K.E{.}_{translational}=\frac{1}{2}m{v}^2=\frac{1}{2}m{R}^2{w}^2$

$K.E{.}_{total}=\frac{3}{4}m{R}^2{w}^2$

so taking ration,

$\frac{K.E{.}_{Rotational}}{K.E{.}_{Total}}=\frac{\frac{1}{4}m{R}^2{w}^2}{\frac{3}{4}m{R}^2{w}^2}=\frac{1}{3}$