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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.

### 14 Responses

1. […] 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 […]

2. on 27/10/2016 at 10:05 | Reply Deepak Suwalka

Thanks, dear. It’s a nice post and really helpful. I like the way you have described it. It’s really appreciable. But if you provide some examples of Moment of Inertia of a disk. Then it’ll be a great post.

• I don’t understand your comment. I thought that I had given examples of the moment of inertia of a disk in this post. Can you explain better what it is that you are asking me to do? Thank you.

• on 22/11/2016 at 09:43 Deepak Suwalka

I’m asking for more examples and problems based on it.

• Look them up! My blog shows how to calculate the moment of inertia for a disk from first principles. There are plenty of examples on the web of what the moment of inertia is for different types of objects.

3. on 29/11/2016 at 08:00 | Reply Deepak Suwalka

Ohh sorry I didn’t check.

4. This really helped a lot .
THANKS RhEvans.

• Thank you, you’re very welcome

NICE

6. great work

7. Thanks a lot for the great work.

8. Very clearly explained, thank you!

9. Great post on Moment of Inertia. Reading an article like this for moment of inertia can be extremely useful.

10. How is the volume of the disk = 2*pi*r*dr*t ?
Where did the 2 come from?
It’s just = pi*r*dr*t (pi*r^2*h).
help?!