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 the disc about a perpendicular axis, passing through the centre ?

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

Answer 1

Answer: (A)

$I_{Totaldisc}=\frac{M{R}^2}{2}$
$M_{Removed}=\frac{M}{4}(Mass\propto area)$
$I_{removed}$ (about same Perpendicular axis)
$=\frac{M}{4}\frac{(R/2{)}^2}{2}+\frac{M}{4}(\frac{R}{2}{)}^2=\frac{3M{R}^2}{32}$
$I_{Remaingdisc}=I_{Total}-I_{Removed}$
$=\frac{M{R}^2}{2}-\frac{3}{32}M{R}^2=\frac{13}{32}M{R}^2$