Question

A circular disc of radius R/3 is cut from a circular disc of radius R and mass 9 M as shown. Then moment of inertia of remaining disc about O perpendicular to the plane of 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 Moment of Inertia of the complete disc, of radius R and mass 9M, i.e., disc without any cut-out is:

$I_{total}=9M{R}^2/2$ $......\left(1\right)$

Mass of the cut-out portion of Radius $R/3$, $m$ = $9M\frac{\pi (R/3{)}^2}{\pi R^2}=M$

Moment of Inertia of the cut-out portion, a disk of Mass about the axis passing through O is :

$I_{cutout}=I_{o{}^{\prime }}+m(\frac{2R}{3}{)}^2=\frac{m(R/3{)}^2}{2}+m(\frac{2R}{3}{)}^2=M{R}^2/2$ $.......\left(2\right)$

The moment of Inertia of remaining disc is $I_{remaining}=I_{total}-I_{cutout}$

$⟹$ $I_{remaining}=9M{R}^2/2-M{R}^2/2=4M{R}^2$