Question

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

Answer 1

Surface density,

$\rho \left(r\right)=A+Br$

Mass of the ring

$dm=\left[\pi \right(r+dr{)}^2-{\pi r}^2]\rho =2\pi rdr\rho$

[$dr$ is very small, so neglected]

$M.I$ of the disc abot $XY,$

${\int }_0^{a}dI={\int }_0^{a}dm.r^2$

$\therefore I={\int }_{r=0}^{r=a}dm.r^2={\int }_0^{a}2\pi r\rho dr.r^2$

$=2\pi {\int }_0^a{r}^3(A+Br)dr$

$=2\pi \left[A{\int }_0^a{r}^{3}dr+B{\int }_0^a{r}^{4}dr\right]=2\pi [A.\frac{a^4}{4}+B.\frac{a^5}{5}]I=2\pi a^4(\frac{A}{4}+B.\frac{a}{5})$