Question

From a circular disc of radius $R$ and mass $9M$, a small disc of mass $M$ and radius $\frac{R}{3}$ is removed concentrically. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through its centre is

• A. $\frac{40}{9} MR^2$
• B. $MR^2$
• C. $4MR^2$
• D. $\frac{4}{9}MR^2$

Answer 1

Answer: (A)

$\frac{40}{9} MR^2$

Answer 2

Mass of the disc = 9M
Mass of removed portion of disc = M

The moment of inertia of the complete disc about an axis passing through its centre O and perpendcular to its plane is
$I_1=\frac{9}{2}MR^2$
Now, the moment of inertia of the disc with removed portion
$I_{2}=\frac{1}{2}M\left(\frac{R}{3}\right)^{2}=\frac{1}{18}MR^{2}$
Therefore, moment of inertia of the remaining portion of disc about O is
$I=I_1-I_2=9\frac{MR^2}{2}-\frac{MR^2}{18}=\frac{40MR^2}{9}$