IIT JEE Physics (1978 to 2018: 41 Years) Topic-wise Complete Solutions

# 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,

1. $4MR^2$
2. $\frac{40}{9}MR^2$
3. $10MR^2$
4. $\frac{37}{9}MR^2$

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 $I_\text{remaining}=4MR^2$.