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 $\frac{MR^{2}}{2}$
By parallel axis theorem
$\frac{\mathrm{MR}^2}{2}=\mathrm{I}_{\mathrm{cm}}+\mathrm{M}\left(\frac{4 \mathrm{R}}{3 \pi} \sqrt{2}\right)^2$
$I_{c m}=\frac{M^2}{2}-M\left(\sqrt{2} \frac{4 R}{3 \pi}\right)^2$