Question

From a circular disc of radius R and mass 9M, a small disc of radius $\frac{R}{3}$ is removed as shown in figure. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc passing through O is

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

Answer 1

The correct option is A $4{MR}^2$

Solution:- (A) $4M{R}^2$

As we know that the moment of inertia of complete disc about a perpendicular axis passing through centre $O$ is given as-

$I=\frac{1}{2}M{R}^2$

Whereas $M$ and $R$ are the mass and radius of the disc respectively.

Given that the mass of disc is $9M$ and radius is $R$.

Therefore,

$I_1=\frac{1}{2}\times 9M\times R^2=\frac{9}{2}M{R}^2$

Now, mass of cut out disc of radius $\frac{R}{3}$-

$m=\frac{9M}{\pi R^2}\times \pi {\left(\frac{R}{3}\right)}^2$

$\Rightarrow m=M$

Now using the theorem of parallel axis, the moment of inertia of cut out disc about the perpendicular axis passing through cenre $O$ is,

$I_2=\frac{1}{2}\times M\times {\left(\frac{R}{3}\right)}^{2}M{\left(\frac{2R}{3}\right)}^2$

$\Rightarrow I_2=\frac{1}{2}M{R}^2$

Therefore, the moment of inertia of residue disc is-

$I=I_1-I_2$

$\Rightarrow I=\frac{9}{2}M{R}^2-\frac{1}{2}M{R}^2$

$\Rightarrow I=4M{R}^2$

Hence the moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc passing through $O$ is $4M{R}^2$.