## Introduction

A while ago I wanted to know the moment of inertia of a tetrahedron. I'd forgotten some of the basic stuff, and the calculation was a bit fiddly so I thought I'd write it up on here. There are three related articles:

2. A toy problem: the sphere.2
3. The final calculation.3

## General principles

When things are rotating, the obvious thing to measure is the angular velocity4 $$\boldsymbol{\omega}$$. You can imagine taking two photographs of the rotating object, a fraction of a second apart, then working out the line which is (instantaneously) at rest, and the rate at which the body is rotating about that line.

Another important quantity is the angular momentum5 $$\textbf{L}$$, which is defined as

$\textbf{L} = m\, \textbf{r} \times \left( \boldsymbol{\omega} \times \textbf{r} \right).$

It's important because the angular momentum doesn't change unless a torque is applied.

## The moment of inertia

The angular momentum is linear in $$\boldsymbol{\omega}$$, so it makes sense to define the moment of inertia6 tensor $$\textbf{I}$$ which satisfies

$\textbf{L} = \textbf{I} \, \boldsymbol{\omega}.$

Obviously $$\textbf{I}$$ is some quadratic function of $$\textbf{r}$$, but which one ?

Recall this identity for the vector triple product:7

$\textbf{a} \times (\textbf{b} \times \textbf{c}) = (\textbf{c} . \textbf{a}) \textbf{b} - (\textbf{b} . \textbf{a}) \textbf{c},$

and apply it:

\begin{align} \frac{1}{m} \textbf{L} &= \textbf{r} \times \left( \boldsymbol{\omega} \times \textbf{r} \right), \\ &= (\textbf{r} . \textbf{r}) \; \boldsymbol{\omega} - (\textbf{r} . \boldsymbol{\omega}) \textbf{r}, \\ &= (\textbf{r}^T \textbf{r}) \; \boldsymbol{\mathbb{I}} \; \boldsymbol{\omega} - \textbf{r} \textbf{r}^T \boldsymbol{\omega},\\ &= \left( (\textbf{r}^T \textbf{r}) \; \boldsymbol{\mathbb{I}} - \textbf{r} \textbf{r}^T \right) \boldsymbol{\omega}, \\ \frac{1}{m}\textbf{I} &= (\textbf{r}^T \textbf{r}) \; \boldsymbol{\mathbb{I}} - \textbf{r} \textbf{r}^T. \end{align}

where $$\mathbb{I}$$ is the identity matrix.

## Extended bodies

Although the discussion above relates to a point mass, it's simple to extend to general bodies: just sum (or integrate) the contribution from each elemental mass.

If the body has lots of symmetry this might be easy, but the general case is often fiddly to get right.

## In component form

When it comes to calculating a particular $$\textbf{I}$$, we'll probably need the components:

\begin{align} \frac{1}{m}\textbf{I} &= (\textbf{r}^T \textbf{r}) \; \boldsymbol{\mathbb{I}} - \textbf{r} \textbf{r}^T, \\ &= \left(\begin{array}{c} x & y & z \end{array}\right) \left(\begin{array}{ccc} x \\ y \\ z \end{array}\right) \left(\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right) - \left(\begin{array}{ccc} x \\ y \\ z \end{array}\right) \left(\begin{array}{c} x & y & z \end{array}\right), \\ &= (x^2+y^2+z^2) \left(\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right) - \left(\begin{array}{ccc} x^2 & xy & zx \\ xy & y^2 & yz \\ zx & yz & z^2 \end{array}\right), \\ &= \left(\begin{array}{ccc} y^2 + z^2 & -xy & -zx \\ -xy & z^2 + x^2 & -yz \\ -zx & -yz & x^2 + y^2 \end{array} \right). \end{align}

## A computational trick

It's often easier to calculate first the second moment of $$m$$:

\begin{align} \textbf{C} &= m \, \textbf{r}\; \textbf{r}^T, \\ &= m\, \left(\begin{array}{ccc} x^2 & xy & zx \\ xy & y^2 & yz \\ zx & yz & z^2 \end{array} \right), \end{align}

then evaluate its trace,

$\textrm{Tr} \, \textbf{C} = \sum_i C_{ii} = x^2 + y^2 + z^2,$

and so calculate $$\textbf{I}$$:

$\textbf{I} = \left( \textrm{Tr} \, \textbf{C} \right) \; \mathbb{I} - \textbf{C}.$

## The scalar simplification

In general $$\textbf{I}$$ will be a positive semi-definite matrix (otherwise the kinetic energy might be negative!).

Accordingly we can always find some frame in which $$\textbf{I}$$ is diagonal. If the body is rotating about one of the axes of that frame, then the the angular velocity and momentum will be colinear, and we can write a simple scalar equation:

$L = I \omega.$

Usually this will only be useful if the object has been carefully spun about a particular direction, or if the object has lots of symmetry. For example, the sphere's moment of inertia (about its centre) is a multiple of the identity matrix and so diagonal in all frames.

If we consider the $$x$$-component equation then:

\begin{align} L_x &= I_{xx} \omega_x, \\ &= m (y^2 + z^2) \omega_x, \\ &= m r^2 \omega_x, \end{align}

where $$r^2$$ is just the radius of rotation about the $$x$$-axis.

For an extended body:

$L_x = \left( \int r^2 dm \right) \omega_x.$