Following on from my derivation of the moment of inertia of a disk, in this blog I will derive the moment of inertia of an annulus. By an annulus, I mean a disk which has the inner part missing, as shown below.
To derive its moment of inertia, we return to our definition of the moment of inertia, which for a volume element is given by
where is the mass of the volume element . We are going to initially consider the moment of inertia about the zaxis, and so for this annulus it will be
where and are the inner radius and outer radius of the annulus respectively. As with the disk, the mass of the volume element is related to its volume and density via
(assuming that the annulus has a uniform density). The volume element can be found as before by considering a ring at a radius of which a width and a thickness . The volume of this will be
and so we can write the mass as
Thus we can write the moment of inertia as
Integrating this between and we get
But, we can rewrite as (remember can be written as ). So, wen can write Eq. (1) as
The total mass of the annulus can be found by considering the total mass of a disk of radius (which we will call ) and then subtracting the mass of the inner part, a disk of radius (which we will call ). The mass of a disk is just its density multiplied by its area multiplied by its thickness.
so the mass of the annulus is
Substituting this expression for into equation (2) above, we can write that the moment of inertia for an annulus, which goes from an inner radius of to an outer radis of , about the zaxis is
Comparison to the moment of inertia of a disk
As we saw in this blog, the moment of inertia of a disk is . It may therefore seem, at first sight, that the moment of inertia of an annulus is more than that of a disk. This would be true if they have the same mass, but if they have the same thickness and density the mass of an annulus will be much less.
Let us compare the moment of inertia of a disk and an annulus for the 4 following cases.

The same density and thickness,
The same density and thickness,
The same mass,
The same mass,
The same density and thickness,
We are first going to compare the moment of inertia of a disk of mass with that of an annulus which goes from half the radius of the disk to the radius of the disk (i.e. .
For the disk, its mass will be
The mass of the annulus, , will be this mass less the mass of the missing part , so
The moment of inertia of the disk will be
The moment of inertia of the annulus will be
So, for this case, , i.e. slightly less than the disk.
The same density and thickness,
Let us now consider the second case, with an annulus of the same density and thickness as the disk, and its inner radius being 90% of the outer radius, . Now, the mass of the missing part of the disk, will be
which means that the mass of the annulus, is
The moment of inertia of the annulus will then be
and so in this case
which is much less than the moment of inertia of the disk.
The same mass,
In this third case, the mass of the annulus is the same as the mass of the disk, and its inner radius is 50% of the radius of the disk. This would, of course, require the annulus to either have a greater density than the disk, or to be thicker (or both). So, . The moment of inertia of the annulus will be
The same mass,
The last case we will consider is an annulus with its inner radius being 90% of the outer radius, but its mass the same. So, . The moment of inertia of the annulus will be
Summary
To summarise, we have

The same density and thickness,
The same density and thickness,
The same mass,
The same mass,
So, as these calculations show, if keeping the mass of a flywheel down is important, then a larger moment of inertia will be achieved by concentrating most of that mass in the outer parts of the flywheel, as this photograph below shows.
In the next blogpost in this series I will calculate the moment of inertia of a solid sphere.