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 through its centre.

Answer 1

Given in question

Radius $=r$

width $=2r$

Surface density depends upon distance fro center

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

Area of ring $=2\pi v.dv$

$\therefore$ Moment of inertia of ring perpendicular axis from center.

$=(A+Br)\times 2\pi vdv\times r^2=(2\pi r^3+2\pi r+B).drr=0tor=a={[2\pi r^4\frac{A}{4}+2\pi r^5\frac{B}{5}]}_0^a=2\pi \frac{A{a}^4}{4}+\frac{B{a}^5}{5}$