Question

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

• A. $4M{R}^2$
• B. $\frac{40}{4}M{R}^2$
• C. $10M{R}^2$
• D. $\frac{37}{9}M{R}^2$

Answer 1

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

Let $\sigma$ be the mass per unit area. The total mass of the disc = $\sigma \times \pi R^2=9M.$

M = Mass of the circular disc cut
= $\sigma \times \pi {\left(\frac{R}{3}\right)}^2$
= $\sigma \times \frac{\pi R^2}{9}=M$
Let us consider the above system as a complete disc of mass 9M and negative mass M superimposed on it.
Moment of inertia $\left(I_1\right)$ of the complete disc = $9M{R}^2/2$ about an axis passing through O and perpendicular to the plane of the disc.
MI of the cutout portion about an axis passing through O' and perpendicular to the plane of disc is $\frac{1}{2}\times M\times {\left(\frac{R}{3}\right)}^2$
Therefore, MI $\left(I_2\right)$ of the cutout portion about an axis passing through O and perpendicular to the plane of disc is $[\frac{1}{2}\times M\times {\left(\frac{R}{3}\right)}^2+M\times {\left(\frac{2R}{3}\right)}^2]$
[Using perpendicular axis theorem]
Therefore, the total MI of the system about an axis passing through O and perpendicular to the plane of the disc is
$I=I_1+I_2$
$=\frac{1}{2}9M{R}^2-{[\frac{1}{2}\times M\times {\left(\frac{R}{3}\right)}^2+M\times {\left(\frac{2R}{3}\right)}^2]}^2$
$=\frac{1}{2}9M{R}^2-M{R}^2[\frac{1}{18}+\frac{4}{9}]$