Question

From a disc of radius R and mass M, a circular hole of diameter R, whose rim passes through the centre is cut. What is the moment of inertia of the remaining part of disc about a perpendicular axis, passing through the centre?

• A. $9M{R}^2/32$
• B. $15M{R}^2/32$
• C. $13M{R}^2/32$
• D. $11M{R}^2/32$

Answer 1

The correct option is C $13M{R}^2/32$

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

Mass of removed part $=\frac{M}{4}(\because Mass\propto Area)$
Moment of inertia of removed part about same perpendicular axis is $I_{removed}=\frac{M}{4}\frac{(R/2{)}^2}{2}+\frac{M}{4}{\left(\frac{R}{2}\right)}^2=\frac{3M{R}^2}{32}$
Moment of inertia of remaining part of disc about perpendicular axis passing through centre is $I_{remainingdisc}=I_{Total}-I_{removed}$
$=\frac{M{R}^2}{2}-\frac{3M{R}^2}{32}=\frac{13}{32}M{R}^2$