# Moment of inertia of a disc with a hole

I think i have this right, but would like to confirm i did the right thing:

## Homework Statement

I have a lamina disc radius R with a disc of radius r cut out of it from its centre. Total mass of the actual object is M, Find moment of inertia when lamina rotates about axis perpendicular to the lamina through its centre

## Homework Equations

dm = density x area = density x circumference x width = p 2 pi r dr

Inertia = mr2 = Integration sign ( dIr)

## The Attempt at a Solution

I chose to integrate over thin concentric rings with width dr, so that mass of each was

dm = density x area = density x circumference x width = p 2 pi r dr

So that I for each ring about the axis is

dIr = dm r2 = 2 p pi r3 dr

We integrate over all rings that make up this disk ie from radius r to R

So, I = (integral) dIr
= (integral)R to r ( 2 p pi r3 dr)
= 2 p pi (integral)R to r (r3 dr)
= 2 p pi (r4/4)]R to r
= p pi /2 (R4 - r4)

Since the area of this disk with the cutout is pi R2 - pi r2, its density is

p = Mass / area = mass / (pi R2 - pi r2)

Substitute this in the equation for I giving:

I = M pi (R4 - r4) / 2 (pi R2 - pi r2)

since (R4-r4) = (R2+r2)(R2-r2) we can cancel so

I = M (R2 + r2)/2

Thanks!