Question

A circular hole of radius $\frac{R}{4}$ is made in a thin uniform disc having mass $M$ and radius $R$, as shown in figure. The moment of inertia of the remaining portion of the disc about an axis passing through the point $O$ and perpendicular to the plane of the disc is :

• A. $\frac{197M{R}^2}{256}$
• B. $\frac{19M{R}^2}{512}$
• C. $\frac{237M{R}^2}{512}$
• D. $\frac{219M{R}^2}{256}$

Answer 1

The correct option is C $\frac{237M{R}^2}{512}$

Let the mass per unit area of disc be $\sigma$

Now, moment of inertia of removed mass

$\left(M_2\right)=(\frac{\left(\sigma A^{\prime }\right){\left(\frac{R}{4}\right)}^2}{2}+\left(\sigma A^{\prime }\right){\left(\frac{3R}{4}\right)}^2)$

Using parallel axis theorem

$M_2=\left(\sigma A^{\prime }\right)(\frac{R^2}{32}+\frac{9R^2}{16})$

$A^{\prime }=\pi {\left(\frac{R}{4}\right)}^2=\frac{\pi R^2}{16}$

$M_2=\left(\frac{\sigma \pi R^4}{16}\right)\left(\frac{19}{32}\right)$

Also, moment of inertia of complete disc is:

$\Rightarrow M_1=\left(\frac{\left(\sigma A\right)\times R^2}{2}\right)$

Effect moment of inertia = $M_1-M_2$

$M_0=\frac{\sigma }{2}\pi R^4-\frac{\sigma \pi R^4.19}{512}$

$M_0=\left(\sigma \pi R^2\right).R^2.\frac{237}{512}$

$\sigma \pi R^2=M{M}_0=\frac{237}{512}M{R}^2$