Question

Moment of inertia of a thin semi circular disc ( mass$=M$ & radius$=R$) about an axis through point O and perpendicular to plane of disc, is given by :

• A. $\frac{1}{4}M{R}^2$
• B. $\frac{1}{2}M{R}^2$
• C. $\frac{1}{8}M{R}^2$
• D. $M{R}^2$

Answer 1

The correct option is B $\frac{1}{2}M{R}^2$

We know that the moment of inertia of a whole circle with mass $M$ is $I=\frac{1}{2}M{R}^2$. But in this case the mass of half of the circle is $M$ so $2M$ will be the mass of the whole circle.
So the moment of inertia of the whole of the circle is $2\times I=2\times \frac{1}{2}M{R}^2=M{R}^2$ and then the moment of inertia of half of the circle is half of this value i.e. $I=\frac{1}{2}M{R}^2$.