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

Answer 1

Answer: (B)

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

As mass is proportional to area, $M_{Removed}=\frac{M}{4}$

Now, about the same perpendicular axis:

$I_{Removed}=\frac{M}{4}\frac{(R/2{)}^2}{2}+\frac{M}{4}{\left(\frac{R}{2}\right)}^2=\frac{3M{R}^2}{32}$

$\Rightarrow I_{RemainingDisc}=I_{Total}-I_{Removed}$

$=\frac{M{R}^2}{2}-\frac{3M{R}^2}{32}$

$=\frac{13M{R}^2}{32}$