Question

A disc has mass $9\text{m}$. A hole of radius $\frac{R}{3}$ is cut from it as shown in the figure. The moment of inertia of remaining part about an axis passing through the centre ‘$O$’ of the disc and perpendicular to the plane of the disc is:

• A. $8m{R}^2$
• B. $4m{R}^2$
• C. $\frac{40}{9}m{R}^2$
• D. $\frac{37}{9}m{R}^2$

Answer 1

The correct option is B $4m{R}^2$

The initial moment of inertia of the disc of mass $9m$ is
$I_0=\frac{\left(9m\right)R^2}{2}$
Mass of removed disc is $\frac{9m}{9}=m$

From the parallel axis theorem
Moment of inertia of $\frac{R}{3}$ disc about point $O$ is
$I=I_{\text{com}}+m{\left(\frac{2R}{3}\right)}^2=\frac{m{\left(\frac{R}{3}\right)}^2}{2}+\frac{4R^2}{9}m$
$I=\frac{m{R}^2}{2}$
Moment of inertia of the remaining disc $=\frac{9m{R}^2}{2}-\frac{m{R}^2}{2}=4m{R}^2$