Moment of inertia of circular flywheel

1. Dec 5, 2013

KiNGGeexD

Hi I'm new to this forum so still getting used to it!

I have to find the moment of inertia of a circular flywheel which has radius, a and mass m! But also had area density (mass per unit area) of ρ0,

Where ρ0=ρa(1+r/a)

I know moment of inertia is the integral of a^2 dm and in order to get mass m I just multiply the above expression by area of the flywheel which I think is pi*a^2

I end up with a rather foolish answer when I have competed the intergratiom!
Any help would be great

Thanks guys

2. Dec 6, 2013

Vineeth T

Is the answer (9∏$\rho$a5 /10) ?
If its' correct pls. reply so that i would post my solution.

3. Dec 6, 2013

4. Dec 6, 2013

haruspex

That's not how this forum works. We don't just post solutions. The idea is to provides hints and guide the thread originator to the solution.

5. Dec 6, 2013

Vineeth T

Oh! Sorry then.
The idea is to take a differential strip(a ring of thickness dr) at a general distance r, find its' M.I and then integrate it to get M.I of the whole disc(a flywheel).

6. Dec 7, 2013

KiNGGeexD

Yea I know sorry!
Well what I have done is used polar coordinates and defined my moment of inertia as of course

I= integral of r^2 dm

And my element of mass is just ρ0+[1+r/α] *2pi a^2 but I am having trouble with integrating the expression! Are my boundaries a -0 or r-0?

7. Dec 7, 2013

haruspex

Seems to me there's some confusion between variables and constants here.
In the OP you had r as a variable from 0 to a (constant); ρ0 as a variable from ρ a to 2 ρ a, ρ being a constant. That's very unusual - I would have expected ρ and ρ0 to have reversed roles. Assuming you had them crossed over, I'll write ρ(r) = ρ0a(1+r/a). (Maybe you meant ρa, not ρ a, in the OP, but that's just a change to the constant ρ0.)
You want to integrate wrt r. At radius r the density is ρ(r). To get the element of mass you need to multiply that by the element of area, i.e. the area of an annulus between radius r and radius r+dr. In your above post, you seem to have that as 2pi a^2, which is obviously wrong. Also, is "ρ0+[1+r/α]" a typo? Should it be ρ0[1+r/α], or ρ0a[1+r/α]?

8. Dec 7, 2013

KiNGGeexD

Why is 2pi a^2 obviously wrong? It is the area of the flywheel is it not?

9. Dec 7, 2013

haruspex

Well, no, the area of the whole flywheel would be pi a2, not 2pi a2. But for the purposes of the integrand we want the area of an element between radius r and radius r+dr, not the whole flywheel, right? That is independent of a.