Question

One quarter section is cut from a uniform circular disc of radius R. This section 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. $\frac{1}{2}M{R}^2$
• B. $\frac{1}{4}M{R}^2$
• C. $\frac{1}{8}M{R}^2$
• D. $\sqrt{2}M{R}^2$

Answer 1

The correct option is A $\frac{1}{2}M{R}^2$

The $MI$ of $\frac{3}{4}$th of the disc can be considered as difference of $MI$ of the full disc and $MI$ of $\frac{1}{4}$th of the disc.
Given, Mass of $\frac{3}{4}$th disc $=M$
Hence, mass of the full circular disc is $M^{\prime }=\frac{4}{3}M$
Let $MI$ of the full disc be $I_{full}$
Let $MI$ of one quarter of disc be $I_{quarter}$
Thus, $MI$ of $\frac{3}{4}$th of disc: $I=I_{full}-I_{quarter}=I_{full}-\frac{1}{4}{I}_{full}=\frac{3}{4}{I}_{full}=\frac{3}{4}\times (\frac{1}{2}\times \left(\frac{4}{3}M\right)R^2)=\frac{1}{2}M{R}^2$