Question

The surface density (mass/area) of a circular disc of radius a depends on the distance from the center as ρ(r) = A+Br, Find its moment of inertia about the line perpendicular to the plane of the disc through its center.

Answer 1

Given in the question -
Assume, Radius = r
Width = dr
The surface density depends upon the distance from the center.
$\rho (r) = A+Br$

Now, Area of the ring = $2\pi r.dr$
Mass = $(A+Br)*2\pi r.dr$

Hence Moment of inertia of ring. Perpendicular axis from the center,
$=(A+Br)*2\pi r.dr * r^2$
$= (2\pi r^3A+2\pi r^4B).dr$

Now moment of inertia(disc) Integrating.
$= \int\limits(2\pi r^3A+2\pi r^4B).dr$

{Taking Limit at r =0 to r = a}
$[2\pi r^4A/4+2\pi r^5B/5]$,
Applying the limit, we get
$\boxed{2\pi{Aa^4/4+Ba^5/5}}$


Hope it helps

Answer 2

I = ∫ r² dm

A ring of thickness dr has mass

dm = ρ(r) * 2πr * dr, so

I = ∫ r² * ρ(r) * 2πr * dr = 2π ∫ r³ * (A + Br) * dr → from 0 to a

I = 2π * (Aa^4/4 + Ba^5/5) = 2π(5A + 4Ba)*a^4 / 20

I = π(5A + 4Ba)*a^4 / 10 ◄

which might be expressed other ways.

Is this plausible? Well, if the disk had constant density, then B = 0

and density = A = mass / area = M / πa².

Plugging this (and B = 0) into my expression for I gives

I = π(5*M / πa²)*a^4 / 10 = ½Ma²