Question

Moment of inertia of a uniform quarter disc of radius $R$ and mass $M$ about an axis through its centre of mass and perpendicular to its plane is

• A. $\frac{{\mathrm{MR}}^2}{2}-\mathrm{M}{\left(\frac{4\mathrm{R}}{3\pi }\right)}^2$
• B. $\frac{{\mathrm{MR}}^2}{2}-\mathrm{M}{\left(\sqrt{2}\frac{4\mathrm{R}}{3\pi }\right)}^2$
• C. $\frac{{\mathrm{MR}}^2}{2}-\mathrm{M}{\left(\frac{4\mathrm{R}}{2\pi }\right)}^2$
• D. $\frac{{\mathrm{MR}}^2}{2}+\mathrm{M}{\left(\sqrt{2}\frac{4\mathrm{R}}{3\pi }\right)}^2\frac{9}{32\pi }$

Answer 1

Answer: (B)

Moment of Inertia of the object about $O$ is $\frac{M{R}^2}{2}$
By parallel axis theorem
$\frac{{\mathrm{MR}}^2}{2}={\mathrm{I}}_{\mathrm{cm}}+\mathrm{M}{\left(\frac{4\mathrm{R}}{3\pi }\sqrt{2}\right)}^2$
$I_{cm}=\frac{M^2}{2}-M{\left(\sqrt{2}\frac{4R}{3\pi }\right)}^2$