Question

From a circular disc of radius $R$ and mass $9M$, a small disc of radius $R/3$ is removed from the disc, the moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through $O$ is

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

Answer 1

Answer: (A)

As the mass is uniformly distributed on the disc. so mass density (per unit area) $=\frac{9M}{\pi R^2}$
Mass of removed portion $=\frac{9M}{\pi R^2}\times {\left(\frac{R}{3}\right)}^2$
$=M$
So moment of inertia of the removed portion about the stated axis by theorem of parallel axis.
$I_1=\frac{1}{2}M{\left(\frac{R}{3}\right)}^2+M{\left(\frac{2R}{3}\right)}^2$
If the disc would not have been removed, then the moment of inertia of complete disc about the started axis.
$I_2=\frac{1}{2}9M(R^2)$
So, the moment of inertia of the disc about required axis. $I=I_2-I_1$
$=\frac{1}{2}9M(R{)}^2-\left[\frac{1}{2}M{\left(\frac{R}{3}\right)}^2+\frac{1}{2}M{\left(\frac{2R}{3}\right)}^2\right]$
$I=4M{R}^2$