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Solve the following :

Question:

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.

Solution:

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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