Question

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

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

Answer 1

The correct option is B $\frac{1}{2}M(R^2+r^2)$

Suppose, $M$ be tha mass of the annular disc of outer radius $R$ and inner radius $r$.
The surface mass density $=\sigma =\frac{\text{mass}}{\text{area}}$
$=\frac{M}{\pi (R^2-r^2)}$
Mass of elementary ring of radius $x$ and thickness $dx$
$=\frac{M}{\pi (R^2-r^2)}\times 2\pi xdx=\frac{2Mxdx}{(R^2-r^2)}$
MI of ring about an axis passing through the centre of mass and perpendicular to its plane is
$dI=\frac{2Mxdx}{(R^2-r^2)}{x}^2=\frac{2M{x}^{3}dx}{(R^2-r^2)}$
$\therefore I=\underset{r}{\overset{R}{\int }}\frac{2M{x}^{3}dx}{(R^2-r^2)}=\frac{2M}{(R^2-r^2)}\left[\frac{R^4-r^4}{4r}\right]$
$=\frac{1}{2}M(R^2+r^2)$
Hence, the correct answer is option (b).