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## Derivation of the moment of inertia of an annulus

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.

An annulus is a disk of small thickness $t$ with the inner part missing. The annulus goes from some inner radius $r_{1}$ to an outer radius $r$.

To derive its moment of inertia, we return to our definition of the moment of inertia, which for a volume element $dV$ is given by

$dI = r_{\perp}^{2} dm$

where $dm$ is the mass of the volume element $dV$. We are going to initially consider the moment of inertia about the z-axis, and so for this annulus it will be

$I_{zz} = \int _{r_{1}} ^{r} r_{\perp}^{2} dm$

where $r_{1}$ and $r$ are the inner radius and outer radius of the annulus respectively. As with the disk, the mass $dm$ of the volume element $dV$ is related to its volume and density via

$dm = \rho dV$

(assuming that the annulus has a uniform density). The volume element $dV$ can be found as before by considering a ring at a radius of $r$ which a width $dr$ and a thickness $t$. The volume of this will be

$dV = (2 \pi r dr) t$

and so we can write the mass $dm$ as

$dm = (2 \pi \rho t)rdr$

Thus we can write the moment of inertia $I_{zz}$ as

$I_{zz} = \int _{r_{1}} ^{r} r_{\perp}^{2} dm = 2 \pi \rho t \int _{r_{1}} ^{r} r_{\perp}^{3} dr$

Integrating this between $r_{1}$ and $r$ we get

$I_{zz} = 2 \pi \rho t [ \frac{ r^{4} - r_{1}^{4} }{4} ] = \frac{1}{2} \pi \rho t (r^4 - r_{1}^{4}) \text{ (Equ. 1)}$

But, we can re-write $(r^{4} - r_{1}^{4})$ as $(r^{2} + r_{1}^{2})(r^{2} - r_{1}^{2})$ (remember $x^{2} - y^{2}$ can be written as $(x+y)(x-y)$). So, wen can write Eq. (1) as

$I_{zz} = \frac{1}{2} \pi \rho t (r^{2} + r_{1}^{2})(r^{2} - r_{1}^{2}) \text{ (Equ. 2)}$

The total mass $M_{a}$ of the annulus can be found by considering the total mass of a disk of radius $r$ (which we will call $M_{2}$) and then subtracting the mass of the inner part, a disk of radius $r_{1}$ (which we will call $M_{1}$). The mass of a disk is just its density multiplied by its area multiplied by its thickness.

$M_{2} = \pi \rho t r^{2} \text{ and } M_{1} = \pi \rho t r_{1}^{2}$

so the mass $M_{a}$ of the annulus is

$M_{a} = M_{2} - M_{1} = \pi \rho t r^{2}- \pi \rho t r_{1}^{2} = \pi \rho t (r^{2} - r_{1}^{2})$

Substituting this expression for $M_{a}$ into equation (2) above, we can write that the moment of inertia for an annulus, which goes from an inner radius of $r_{1}$ to an outer radis of $r$, about the z-axis is

$\boxed{ I_{zz} = \frac{1}{2} M_{a} (r^{2} + r_{1}^{2}) }$

## Comparison to the moment of inertia of a disk

As we saw in this blog, the moment of inertia of a disk is $I_{zz} = \frac{1}{2} Mr^{2}$. 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, $r_{1} = 0.5 r$
The same density and thickness, $r_{1} = 0.9 r$
The same mass, $r_{1} = 0.5 r$
The same mass, $r_{1} = 0.9 r$

## The same density and thickness, $r_{1}=0.5r$

We are first going to compare the moment of inertia of a disk of mass $M$ with that of an annulus which goes from half the radius of the disk to the radius of the disk (i.e. $r_{1} \text{ the inner radius of the annulus, is } = 0.5 r$.

For the disk, its mass will be

$M = \rho t (\pi r^{2}) = \pi \rho t r^{2}$

The mass of the annulus, $M_{a}$, will be this mass less the mass of the missing part $M_{1}$, so

$M_{a} = M - M_{1} = M - \pi \rho t (r_{1})^{2} = \pi \rho t (r^{2} - (0.5r)^{2})= \pi \rho t (1-0.25)r^{2}$

$M_{a} = \pi \rho t (0.75)r^{2} = 0.75 M$

The moment of inertia of the disk will be

$I_{d} = \frac{1}{2} M r^{2}$
The moment of inertia of the annulus will be

$I_{a} = \frac{1}{2} M_{a} (r^{2} + r_{1}^{2}) = \frac{1}{2} (0.75M)(r^{2} + (0.5r)^{2}) = \frac{1}{2} (0.75M)(1.25r^{2}) = \frac{1}{2} (0.9375) M r^{2}$

So, for this case, $I_{a} = 0.9375 I_{d}$, i.e. slightly less than the disk.

## The same density and thickness, $r_{1}=0.9r$

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, $r_{1} = 0.9r$. Now, the mass of the missing part of the disk, $M_{1}$ will be

$M_{1} = \rho t (\pi r_{1}^{2}) = \rho t \pi (0.9r)^{2} = 0.81 \rho t \pi r^{2} = 0.81M$

which means that the mass of the annulus, $M_{a}$ is

$M_{a} = M - M_{1} = M-0.81M=0.19M$

The moment of inertia of the annulus will then be

$I_{a} = \frac{1}{2}M_{a}(r^{2}+r_{1}^{2}) = \frac{1}{2}(0.19M)(r^{2}+(0.9r)^{2})=\frac{1}{2}(0.19M)((1.81)r^{2} = \frac{1}{2}(0.1539)Mr^{2}$

and so in this case

$I_{a} = 0.1539 I_{d}$

which is much less than the moment of inertia of the disk.

## The same mass, $r_{1}=0.5r$

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, $M_{a} = M$. The moment of inertia of the annulus will be

$I_{a} = \frac{1}{2} M(r^{2} + r_{1}^{2}) = \frac{1}{2} M(r^{2} + (0.5r)^{2}) = \frac{1}{2} M(r^{2} + 0.25r^{2}) = \frac{1}{2} M(1.25)r^{2}$

$I_{a}= 1.25 I_{d}$

## The same mass, $r_{1}=0.9r$

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, $M_{a} = M$. The moment of inertia of the annulus will be

$I_{a} = \frac{1}{2} M(r^{2} + r_{1}^{2}) = \frac{1}{2} M(r^{2} + (0.9r)^{2}) = \frac{1}{2} M(r^{2} + 0.81r^{2}) = \frac{1}{2} M(1.81)r^{2}$

$I_{a}= 1.81 I_{d}$

## Summary

To summarise, we have

The same density and thickness, $r_{1} = 0.5 r, \; \; I_{a}=0.9375 I_{d}$
The same density and thickness, $r_{1} = 0.9 r, \; \; I_{a}=0.1539 I_{d}$
The same mass, $r_{1} = 0.5 r, \; \; I_{a}=1.25 I_{d}$
The same mass, $r_{1} = 0.9 r, \; \; I_{a}=1.81 I_{d}$

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.

If keeping mass down is important, a flywheel’s moment of inertia can be increased by concentrating most of the mass in its outer parts

In the next blogpost in this series I will calculate the moment of inertia of a solid sphere.