Question

From a circular disc of radius $R$ and mass $9M$, a small disc of mass $M$ and radius $R/3$ is removed concentrically. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through its centre is

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

Answer 1

The correct option is D $\frac{40}{9}M{R}^2$


Given:
Mass of the disc $=9M$
Mass of removed portion of disc $=M$

The moment of inertia of the complete disc about an axis passing through its centre $O$ and perpendicular to its plane is,
$I_1=\frac{9}{2}M{R}^2$

Now, the moment of inertia of the removed portion 0 now, the moment of inertia of the removed portion of the disc,
$I_2=\frac{1}{2}M{\left(\frac{R}{3}\right)}^2=\frac{1}{18}M{R}^2$

Therefore, moment of inertia of the remaining portion of disc about $O$ is
$I=I_1-I_2=\frac{9M{R}^2}{2}-\frac{M{R}^2}{18}=\frac{80M{R}^2}{18}$
$I=\frac{40M{R}^2}{9}$
Hence , option $\left(B\right)$ is correct.