Question

Moment of inertia of a thin uniform semicircular wire having mass $M$, radius $R$ and centre $C$ about an axis passing through point $O$ and perpendicular to its plane is

• A. $2M{R}^2$
• B. $2M{R}^2(1-\frac{1}{\pi })$
• C. $M{R}^2(1-\frac{2}{\pi })$
• D. $2M{R}^2(1-\frac{2}{\pi })$

Answer 1

The correct option is D $2M{R}^2(1-\frac{2}{\pi })$

$I_1=Icm+m{\left(\frac{2R}{\pi }\right)}^2$

$I_2=Icm+m{\left(R-\frac{2R}{\pi }\right)}^2$

$I_1-I_2=m\left(\frac{4R}{{\pi }^2}\right)-m\left[R^2+\frac{4R^2}{{\pi }^2}-\frac{4R^2}{\pi }\right]$

$m{R}^2-I_2=m{R}^2+\frac{4R^2}{\pi }$

$I_2=2m{R}^2-\frac{4R^2}{\pi }$

$=2m{R}^2\left(1-\frac{2}{\pi }\right)$

$\therefore$ Option $D$ is correct.