Question

The surface density (mass/area) of a circular disc of radius a depends on the distance from the centre as $\rho \left(r\right)=A+Br$. Find its moment of inertia about the line perpendicular to the plane of the disc thorough its centre.

Answer 1

Consider a ring of thickness dx at a distance r from the centre of the disc.
Mass of the ring , $dm=\left(A+\mathrm{B}r\right)\times 2\pi r\times dr$



Therefore, moment of inertia about the centre,
$I={\int }_0^a{r}^{2}dm$
$$\begin{gathered} ={\int }_o^{a}d\left(\mathrm{A}+\mathrm{B}r\right)2\pi r.dr\times r^2 \\ ={\int }_o^{a}2\pi \mathrm{A}{r}^{3}dr+{\int }_o^{a}2\pi \mathrm{B}{r}^{4}dr \\ ={\left[2\pi \mathrm{A}\left(\frac{r^4}{4}\right)+2\pi \mathrm{B}\left(\frac{r^5}{5}\right)\right]}_0^a \\ =2\pi \left(\frac{\mathrm{A}{a}^4}{4}+\frac{\mathrm{B}{a}^5}{5}\right) \end{gathered}$$