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. $4 \,MR^2$
• B. $ \frac{40}{9} MR^2$
• C. $ 10MR^{2} $
• D. $ \frac{37}{9}MR^{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. $I = I_2 - I_1$
$=\frac{1}{2} 9 M(R)^{2}-\left[\frac{1}{2} M\left(\frac{R}{3}\right)^{2}+\frac{1}{2} M\left(\frac{2 R}{3}\right)^{2}\right]$
$I=4\, M R^{2}$