Solve the following :


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


Surface density of disc is varying radially so considering an element of thickness 'dr' at a distance ' $r$ ' from center of disc.

$\sigma=\mathrm{A}+\mathrm{Br}=\frac{d m}{d A}$

$d m=(\mathrm{A}+\mathrm{Br})(2 \pi r d r)$

$\int d I_{z z^{r}}=\int d m(r)^{2}$

$I_{Z z^{r}}=\int_{0}^{a}(A+B r)(2 \pi r d r)(r)^{2}$

$=2^{\pi}\left[A^{a} r^{3}+B \int_{0}^{a} r^{4}\right]$

$I_{Z z^{r}}=2 \pi\left[\frac{A a^{4}}{4}+\frac{B a^{5}}{5}\right]$

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