Question

Derive the expression for the moment of inertia of a uniform disc about an axis passing through the centre and perpendicular to the plane.

Answer 1

Consider a disc of mass M and radius R. This disc is made up of many infinitesimally small rings as shown in the figure. Consider one such ring of mass (dm) and thickness (dr) and radius (r). The moment of inertia (dl) of this small ring is,
dI = (dm)R²
As the mass is uniformly distributed, the mass per unit area (σ) is σ = ${\displaystyle \frac{\text{mass}}{\text{area}}}$ = ${\displaystyle \frac{M}{\pi R^2}}$
The mass of the infinitesimally small ring is,
dm = σ 2πr dr = ${\displaystyle \frac{M}{\pi R^2}}$ 2πr dr
where, the term (2πr dr) is the area of this elemental ring (2πr is the length and dr is the thickness), dm = ${\displaystyle \frac{2M}{R^2}}$ r dr
dI = ${\displaystyle \frac{2M}{R^2}}$ r³ dr

Moment of inertia of a uniform disc

The moment of inertia (I) of the entire disc is,

I = ${\displaystyle \int dI}$

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

I = ${\displaystyle \frac{2M}{R^2}{\left[\frac{r^4}{4}\right]}_0^R=\frac{2M}{R^2}[\frac{R^4}{4}-0]}$

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