Using Integration?

Question

Using Integration, Calculate the Rotational Inertia of a Solid cylinder (or disk) of mass M, radius R, and length L about central axis
Hint: Equation: I=(1/2)MR^2

Answer 1

The moment of inertia of a point mass m at a radius r is given by:

I = mr²

So the moment of inertia contributed by a mass of uniform surface density σ over a small region dA will be given by σr² dA. So we need merely integrate this over the entire disk. In other words, the moment of inertia over the entire disk is:

I = [disk]∬σ r² dA

We will find it easiest to integrate in polar coordinates. The limits of integration are 0≤θ≤2π and 0≤r≤R. Thus we have:

I = [0, 2π]∫ [0, R]∫ σr² r dr dθ

(don't forget the extra factor of r when switching to polar coordinates)

Now it's just a straightforward iterated integral:

I = [0, 2π]∫ [0, R]∫ σr³ dr dθ
I = [0, 2π]∫ σR⁴/4 |[0, R] dθ
I = [0, 2π]∫ σR⁴/4 dθ
I = 2πσR⁴/4 (since the integrand is constant w.r.t θ)
I = πσR⁴/2

Now we just need to calculate the surface area density. Since the disk is (presumably, it really should have been stated explicitly though) of uniform density, the surface mass density will be the mass of the disk divided by the surface area of the disk, which is πR². So we have:

I = π(M/(πR²))R⁴/2
I = MR²/2

Which is the equation you were given.