Question

A thin uniform disc of mass 9$M$ and of radius R. A disc of radius $\frac{R}{3}$ is cut as shown in figure. Find the moment of inertia of the remaining disc about an axis passing through 0 and perpendicular to the plane of disc.

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

Answer 1

The correct option is C 4${MR}^2$

Moment of Inertia of complete disc

about perpendicular axis passing

through centre O.

$I_1=\frac{1}{2}\times \left(9M\right)R^2=\frac{9}{2}M{R}^2$

The mass of cut out disc, if radius is $\left(\frac{R}{3}\right)$

$m=\frac{9M}{\pi R^2}\pi (\frac{R}{3}{)}^2=M$

Now, by parallel axis Theorem,

$MOI$ of cut out disc about perpendicular

axis passing Through center O is,

$I_2=\frac{1}{2}M(\frac{R}{3}{)}^2+M(\frac{2R}{3}{)}^2$

$=\frac{M{R}^2}{2}$

MOI of residue disc

$I=I_1-I_2$

$=\frac{9M{R}^2}{2}-\frac{M{R}^2}{2}=4M{R}^2$