Question

From a disc of radius $R$ , a concentric circular portion of the radius $r$ is cut out so as to leave an annular disc of mass $M$ . The moment of inertia of this annular disc about the axis perpendicular to its plane and passing through its centre of gravity is

• A. $\frac{1}{2} \, M \, \left(R^{2 \, } + r^{2}\right)$
• B. $\frac{1}{2}M\left(R^{2 \, } - r^{2}\right)$
• C. $\frac{1}{2}M\left(R^{4 \, } + r^{4}\right)$
• D. $\frac{1}{2}M'\left(R^{4 \, } - r^{4}\right)$

Answer 1

Answer: (A)

Mass per unit area $=\frac{M}{\pi\left(R^{2}-r^{2}\right)}$
$\therefore $ mass of the whole $\text{disc}=\frac{M \cdot \pi R^{2}}{\pi\left(R^{2}-r^{2}\right)}=\frac{M R^{2}}{\left(R^{2}-r^{2}\right)}$
Moment of inertia of the disc of radius $r$
$=\frac{1}{2} \frac{M \cdot \pi R^{2}}{2 \pi\left(R^{2}-r^{2}\right)} \cdot r^{2}=\frac{1}{2} \frac{M r^{4}}{\left(R^{2}-r^{2}\right)}$
Moment of inertia of the whole disc $=\frac{1}{2} \frac{M r^{2} \cdot r^{2}}{\left(R^{2}-r^{2}\right)}=\frac{M r^{4}}{2\left(R^{2}-r^{2}\right)}$
$\therefore $ Moment of inertia of the annular disc $=\frac{M}{2}\left[\frac{R^{4}-r^{4}}{R^{2}-r^{2}}\right]=\frac{M\left(R^{2}+r^{2}\right)}{2}$