Question

From a circular disc of radius R and mass 9M, a small disc of mass M and radius $\frac{R}{3}$ 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. $\frac{40}{9}M{R}^2$
• B. $M{R}^2$
• C. $4M{R}^2$
• D. $\frac{4}{9}M{R}^2$

Answer 1

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

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 disc with removed portion
$I_2=\frac{1}{2}M{\left(\frac{R}{3}\right)}^2=\frac{1}{18}M{R}^2$
Therefore, moment of inertia of the remaining portion of disc about O is
$I=I_1-I_2$
$=9\frac{M{R}^2}{2}-\frac{M{R}^2}{18}=\frac{40M{R}^2}{9}$