Question

One quarter sector is cut from a uniform circular disc of radius $R$. This sector has mass $M$. It is made to rotate about a line perpendicular to its plane and passing through the centre of the original disc. Its moment of inertia about the axis of rotation is

• A. $\frac{M{R}^2}{8}$
• B. $\frac{M{R}^2}{2}$
• C. $\frac{M{R}^2}{4}$
• D. $M{R}^2$

Answer 1

The correct option is B $\frac{M{R}^2}{2}$

For complete disc with mass ${}^{\prime }4M^{\prime }$, moment of inertia about given axis $=\frac{\left(4M\right)R^2}{2}=2M{R}^2$

Hence, by symmetry for the given quater of disc, moment of inertia about the axis passing through the center of the original disc.
$I=\frac{2M{R}^2}{4}=\frac{M{R}^2}{2}$
$I=\frac{M{R}^2}{2}$