Question

A circular disc of radius $\frac{R}{3}$ is cut from a circular disc of radius $R$ and mass $9M$ as shown. Then moment of inertia of the remaining disc about O perpendicular to the disc is:

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

Answer 1

The correct option is A $4M{R}^2$

The mass of the disc with radius $R$ is $9M$. As the density remains constant the mass of part of the disc with radius $\frac{R}{3}$ will be $M$.
If $m$ is the mass of the smaller disc we get
$\frac{9M}{\pi R^2}=\frac{m}{\pi (\frac{R}{3}{)}^2}$
Thus mass of smaller disc is $M$.
The distance between the centers of the two discs is $R-\frac{R}{3}=\frac{2R}{3}$
We calculate the moment of inertia of the smaller disc about O as $\frac{1}{2}M(\frac{R}{3}{)}^2+M(\frac{2R}{3}{)}^2=\frac{M{R}^2}{2}$(Using parallel axis theorem)
Subtracting this from the moment of inertia value of of the whole disc we get
$\frac{9M{R}^2}{2}-\frac{M{R}^2}{2}=4M{R}^2$