# Moment of Inertia

## Problems from IIT JEE

**Problem (IIT JEE 2005):**
From a circular disc of radius $R$ and mass $9M$, a small disc of radius $R/3$ is removed. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through $O$ is,

- $4MR^2$
- $\frac{40}{9}MR^2$
- $10MR^2$
- $\frac{37}{9}MR^2$

**Solution:**
The moment of inertia of disc of mass $9M$ and radius $R$ about an axis perpendicular to its plane and passing through its centre O is,
\begin{align}
I_\text{total}=\tfrac{1}{2}(9M)R^2=\tfrac{9}{2}MR^2.\nonumber
\end{align}
The mass of removed disc is $\frac{9M}{\pi R^2}\frac{\pi R^2}{9}=M$. The parallel axis theorem gives moment of inertia of the removed disc about axis passing through $O$ as,
\begin{align}
I_\text{removed}=\tfrac{1}{2}M\left(\tfrac{R}{3}\right)^{\!2}+Md^2=\tfrac{1}{18}MR^2+M\left(\tfrac{2R}{3}\right)^{\!2}=\tfrac{1}{2}MR^2.\nonumber
\end{align}
Using, $I_\text{total}=I_\text{remaining}+I_\text{removed}$, we get $I_\text{remaining}=4MR^2$.