Question

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

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

Answer 1

The correct option is A $\frac{13M{R}^2}{32}$


so the moment of inertia of the remaining disk=
$I_{remain}=I_{disk}-I_{outside}$

$I_{remain}=\frac{M{R}^2}{2}-\frac{\frac{M}{4}(\frac{R}{2}{)}^2}{2}+\frac{M}{4}(\frac{R}{2}{)}^2$
$I_{remain}=\frac{13M{R}^2}{32}$