Question

From a uniform circular disc of radius $R$ and mass $9M$, a small disc of radius $R/3$ is removed as shown in the figure. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through centre of disc is:

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

Answer 1

The correct option is C

$4M{R}^2$

Step 1. Given data

Mass of circular disc is, $9M$

Radius of disc is, $R$

Radius of small disc removed is, $R/3$

Step 2: Calculating mass of small disc

Mass per unit area of disc is, $\frac{9M}{\pi R^2}$

Mass of small disc is

$$\begin{gathered} \frac{9M}{\pi R^2}\times areaofsmalldisc \\ =\frac{9M}{\pi R^2}\times \pi {\left(\frac{R}{3}\right)}^2 \\ =M \end{gathered}$$

Step 3: Finding the moment of inertia

According to parallel axis theorem the moment of inertia about a point at a distance is, $I_h=I+M{h}^2$,

Where $I$ is the moment of inertia about centre and$h$ is the distance between the two axis.

The moment of inertia of a small disc about the centre of original disc is $\frac{M{\left({\displaystyle \raisebox{1ex}{$R$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\right)}^2}{2}+M{\left(\frac{2R}{3}\right)}^2$

Moment of inertia of remaining disc is the difference between moment of inertia of original disc about centre and Moment of inertia of small disc about the centre of original disc is

$$\begin{gathered} I=\frac{\left(9M\right)R^2}{2}-\left(\frac{M{\left({\displaystyle \raisebox{1ex}{$R$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\right)}^2}{2}+M{\left(\frac{2R}{3}\right)}^2\right) \\ I=\frac{\left(9M\right)R^2}{2}-\left(\frac{M{\left({\displaystyle R}\right)}^2}{18}+\frac{4M}{9}{\left(R\right)}^2\right) \\ I=\frac{M{R}^2}{2}\left(9-\frac{1}{9}-\frac{8}{9}\right) \\ I=\frac{M{R}^2}{2}\left(8\right)=4M{R}^2 \end{gathered}$$

Hence, option C is correct.