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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.

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## Derivation of the moment of inertia of a disk

In physics, the rotational equivalent of mass is something called the moment of inertia. The definition of the moment of inertia of a volume element $dV$ which has a mass $dm$ is given by

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

where $r_{\perp}$ is the perpendicular distance from the axis of rotation to the volume element. To find the total moment of inertia of an object, we need to sum the moment of inertia of all the volume elements in the object over all values of distance from the axis of rotation. Normally we consider the moment of inertia about the vertical (z-axis), and we tend to denote this by $I_{zz}$. We can write

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

The moment of inertia about the other two cardinal axes are denoted by $I_{xx}$ and $I_{yy}$, but we can consider the moment of inertia about any convenient axis.

## Derivation of the moment of inertia of a disk

In this blog, I will derive the moment of inertia of a disk. In upcoming blogs I will derive other moments of inertia, e.g. for an annulus, a solid sphere, a spherical shell and a hollow sphere with a very thin shell.

For our purposes, a disk is a solid circle with a small thickness $t$ ($t \ll r$, small in comparison to the radius of the disk). If it has a thickness which is comparable to its radius, it becomes a cylinder, which we will discuss in a future blog. So, our disk looks something like this.

A disk of small thickness $t$, with a radius of $r$

To calculate the moment of inertia of this disk about the z-axis, we sum the moment of inertia of a volume element $dV$ from the centre (where $r=0$) to the outer radius $r$.

$I_{zz} = \int_{r=0} ^{r=r} r_{\perp} ^{2} dm \text{ (Equ. 1)}$

The mass element $dm$ is related to the volume element $dV$ via the equation
$dm = \rho dV$ (where $\rho$ is the density of the volume element). We will assume in this example that the density $\rho(r)$ of the disk is uniform; but in principle if we know its dependence on $r, \; \rho (r) = f(r)$, this would not be a problem.

The volume element $dV$ can be calculated by considering a ring at a radius $r$ with a width $dr$ and a thickness $t$. The volume of this ring is just this rings circumference multiplied by its width multiplied by its thickness.

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

so we can write

$dm = \rho (2 \pi r dr) t$

and hence we can write equation (1) as

$I_{zz} = \int_{r=0} ^{r=r} r_{\perp} ^{2} \rho (2 \pi r dr) t = 2 \pi \rho t \int_{r=0} ^{r=r} r_{\perp} ^{3} dr$

Integrating between a radius of $r=0$ and $r$, we get

$I_{zz} = 2 \pi \rho t [ \frac{ r^{4} }{ 4 } -0 ] = \frac{1}{2} \pi \rho t r^{4} \text{ (Equ. 2)}$

If we now define the total mass of the disk as $M$, where

$M = \rho V$

and $V$ is the total volume of the disk. The total volume of the disk is just its area multiplied by its thickness,

$V = \pi r^{2} t$

and so the total mass is

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

Using this, we can re-write equation (2) as

$\boxed{ I_{zz} = \frac{1}{2} \pi \rho t r^{4} = \frac{1}{2} Mr^{2} }$

## What are the moments of inertia about the x and y-axes?

To find the moment of inertia about the x or the y-axis we use the perpendicular axis theorem. This states that, for objects which lie within a plane, the moment of inertia about the axis parallel to this plane is given by

$I_{zz} = I_{xx} + I_{yy}$

where $I_{xx}$ and $I_{yy}$ are the two moments of inertia in the plane and perpendicular to each other.

We can see from the symmetry of the disk that the moment of inertia about the x and y-axes will be the same, so $I_{zz} = 2I_{xx}$. Therefore we can write

$\boxed{ I_{xx} = I_{yy} = \frac{1}{2}I_{zz} = \frac{1}{4} Mr^{2} }$

## Flywheels

Flywheels are used to store rotational energy. This is useful when the source of energy is not continuous, as they can help provide a continuous source of energy. They are used in many types of motors including modern cars.

It is because of an disk’s moment of inertia that it can store rotational energy in this way. Just as with mass in the linear case, it requires a force to change the rotational speed (angular velocity) of an object. The larger the moment of inertia, the larger the force required to change its angular velocity. As we can see above from the equation for the moment of inertia of a disk, for two flywheels of the same mass a thinner larger one will store more energy than a thicker smaller one because its moment of inertia increases as the square of the radius of the disk.

Sometimes mass is a critical factor, and next time I will consider the case of an annulus, where the inner part of the disk is removed.

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