Question

Find out the moment of inertia of a semi-circular disc about an axis passing through its center of mass and perpendicular to the plane? (mass =M and radius=R)

Answer 1

Mass per unit area $=\frac{M}{\frac{\pi R^2}{2}}=\frac{2M}{\pi R^2}$

Considered a half ring of thickness $dr$ and radius $r$. we can consider that the half circular disc is compose of such half circular ring.

The mass of the half ring

$=$Area $\times$ mass per unit areas

$=2\pi rdr\times \frac{2M}{\pi R^2}$

$=\frac{4M}{R^2}r.dr$

Moment of inertia of this about a perpendicular axis passing through centre is

$=\frac{4M}{r^2}r.dr.r^2$

$=\frac{4M}{R^2}{r}^{3}dr$

so, total moment of inertia

$\frac{4M}{R^2}{\int }_0^R{r}^{3}dr$

$=\frac{4M}{R^2}\times \frac{R6}{4}$

$=M{R}^2$