Question

For a semicircular ring of mass $M$ and radius $R$, the moment of inertia about C.O.M and perpendicular to plane is

• A. $M{R}^2$
• B. $M{R}^2(1-\frac{4}{{\pi }^2})$
• C. $M{R}^2(4-\frac{1}{{\pi }^2})$
• D. $M{R}^2(4{\pi }^2-1)$

Answer 1

The correct option is B $M{R}^2(1-\frac{4}{{\pi }^2})$


$AC=\frac{2R}{\pi }$ (distance to CoM of a semicircular ring)
$I_{A_{\perp }}=M{R}^2$
$I_{A_{\perp }}=I_{{cm}_{\perp }}+M.d^2$
$I_{{cm}_{\perp }}=I_{A_{\perp }}-M{\left(\frac{2R}{\pi }\right)}^2$
$I_{{cm}_{\perp }}=M{R}^2-M.\frac{4R^2}{{\pi }^2}$
$I_{{cm}_{\perp }}=M{R}^2(1-\frac{4}{{\pi }^2})$