Question

From a disc of radius $R$ an 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

Answer: (A)

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

$\text{Given:-}$ Radius of the disc $=R$

Mass of the disc $=M$

$\text{Solution:-}$

Mass removed, m $=\frac{M}{4}$

Radius of mass removed $=r=\frac{R}{2}$

Moment of inertia of whole disc about its center $=\frac{M{R}^2}{2}$

Moment of inertia of mass removed about it's center $=\frac{m{r}^2}{2}=\frac{M{R}^2}{32}$

By parallel axis theorem ,

Moment of inertia of remaining mass about center of disc $=\frac{M{R}^2}{2}-\frac{3M{R}^2}{32}$

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

$\text{Hence the correct option is A}$