Question

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

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

Answer 1

Answer: (C)

$4M{R}^2$

Given:

Mass of dics $9M$

Mass of removed part $m=\frac{9M}{\pi R^2}\times \pi (R/3{)}^3=M$

Moment of inertia of Disc without hole about axis perpendicular to the plane passing through center is

$I_1=\frac{9M{R}^2}{2}$

Moment if inertia of hole about axis passing through center of disc

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

The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through centre of disc

$I=I_1-I_2$

$I=\frac{9M{R}^2}{2}-[\frac{M(\frac{R}{3}{)}^2}{2}+M(\frac{2R}{3}{)}^2]=4M{R}^2$