Question

How to derive the formula for the moment of inertia of a disc about an axis passing through its center and perpendicular to its plane?

Answer 1

Step 1: Assume an elementary ring

Let us assume a disc of mass $M$ and radius $R$, having infinitesimal small rings whose radius is $dr$ and their mass is $dm$.

Therefore, the moment of inertia of this infinitesimal small ring is given by

$dI=\left(dm\right)R^2$

Density of this small ring is

$\rho =\frac{M}{{\mathrm{\pi R}}^2}$

$2\mathrm{\pi rdr}$ is the area of the infinitesimal small ring.

Step 2: Derive the mass of the elementary ring

Therefore, the mass of the ring can be given as,

$dm=2\mathrm{\pi r}.\mathrm{\rho }.\mathrm{dr}$

$=\frac{M}{{\mathrm{\pi R}}^2}.2\mathrm{\pi r}.\mathrm{dr}$

$=\frac{2M}{R^2}r.dr$

Step 3: Establish the moment of inertia

Therefore, the moment of inertia of a disc about an axis passing through its center and perpendicular to its plane is

$I=\int dm.r^2$

$={\int }_0^R\frac{2M}{R^2}{r}^{3}dr$

$=\frac{2M}{R^2}{\int }_0^R{r}^{3}dr$

$=\frac{2M}{R^2}{\left[\frac{r^4}{4}\right]}_0^R$

$=\frac{2M}{R^2}\left[\frac{R^4}{4}-0\right]$

$=\frac{1}{2}M{R}^2$

Hence, the moment of inertia of a disc about an axis passing through its center and perpendicular to its plane is $\frac{1}{2}M{R}^2$.