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. $\frac{1}{2}{M}^{\prime }\left(R^4-r^4\right)$

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 $=\frac{M}{2}\left[\frac{R^4-r^4}{R^2-r^2}\right]=\frac{M\left(R^2+r^2\right)}{2}$