Question

A quarter disc of radius $R$ and mass $m$ is rotating about an axis $O{O}^{\prime }$ which is perpendicular to the plane of the disc as shown in figure. Rotational kinetic energy of the disc is

• A. $\frac{1}{2}m{R}^2{\omega }^2$
• B. $\frac{1}{4}m{R}^2{\omega }^2$
• C. $\frac{1}{8}m{R}^2{\omega }^2$
• D. $\frac{1}{16}m{R}^2{\omega }^2$

Answer 1

The correct option is B $\frac{1}{4}m{R}^2{\omega }^2$

Rotational kinetic energy $K{E}_{\text{Rot}}=\frac{1}{2}I{\omega }^2...\left(i\right)$
For the complete disc, the mass will be $4m$.
By symmetry of quarter disc, its moment of inertia about axis $O{O}^{\prime }$ will be:
$I=\frac{\text{Moment of inertia of complete disc}}{4}$
$I=\frac{\left[\frac{\left(4m\right)R^2}{2}\right]}{4}$
$\therefore I=\frac{m{R}^2}{2}...\left(i\right)$
From Eq $\left(i\right)\&\left(ii\right)$
$K{E}_{\text{Rot}}=\frac{1}{2}\left(\frac{m{R}^2}{2}\right){\omega }^2$
$\therefore K{E}_{\text{Rot}}=\frac{1}{4}m{R}^2{\omega }^2$