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 an axis perpendicular to the plane of the disc and passing through its centre.

Answer 1

For a small elemental ring of width $dx$

$\Rightarrow dI=dM{x}^2$

$=(A+Bx)2\pi xdx{x}^2$

$=(2\pi A{x}^3+2\pi B{x}^4)dx$

$={[\frac{\pi A{x}^4}{2}+\frac{2\pi B{x}^4}{5}]}_0^9$

$=\frac{\pi A{a}^4}{2}+\frac{2\pi B{a}^5}{5}$

Hence, solved.