## 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:

1. Some basic results.1
3. The final calculation.3

## A rotating solid sphere

Consider a solid uniform sphere rotating about an axis through its centre. Explicitly, assume that is has radius $$a$$ and density $$\mu$$. We'll work with spherical polars $$(\rho, \theta, \phi)$$ centred on the centre of the sphere.

As we discussed in the general notes,4 the sphere has sufficient symmetry that we can treat the problem with scalars.

Basically we need to consider an element of the sphere, work out its distance to the rotation axis and thus its moment of inertia. Then we'll just integrate over the sphere.

Hopefully it's clear that an element at $$(\rho, \theta, \phi)$$ moves in a circle of radius $$\rho \sin \theta$$, and has mass $$\mu \rho^2 \sin \theta \; d\rho \; d\theta \; d\phi$$. So,

\begin{align} I_{sphere} &= \int r^2 dm,\\ &= \int_0^a \int_0^{\pi} \int_0^{2 \pi} (\rho \sin \theta)^2 \mu \rho^2 \sin \theta \; d\phi \; d\theta \; d\rho,\\ &= 2 \pi \mu \left(\int_0^a \rho^4 d\rho \right) \left( \int_0^\pi \sin^3 \theta \; d\theta \right),\\ &= 2 \pi \mu \frac{a^5}{5} \frac{4}{3},\\ &= \frac{8}{15} \pi \mu a^5. \end{align}

It's more usual to express this in terms of the mass $$M$$ rather than the density:

\begin{align} I_{sphere} &= \frac{8}{15} \pi \mu a^5,\\ &= \left( \frac{4}{3} \pi a^3 \mu \right) \frac{2}{5} a^2,\\ &= \frac{2}{5} M a^2. \end{align}

Happily this agrees with Wikipedia!5

If we want the full moment of inertia tensor, the symmetry of the sphere implies that it's diagonal viz.:

$\textbf{I} = \frac{2}{5} M a^2 \mathbb{I},$

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

## A less thoughtful approach

That was nice and easy, but our job was simplified because the sphere has lots of symmetry. Let's redo the calculation in a slightly more formal way. It will be messier but easier to apply when the symmetry deserts us.

We'll use the computation trick6 we met earlier or in other words, begin by integrating $$\mu \; \textbf{r} \textbf{r}^T$$ over the volume.

$\textbf{C} = \int \mu \; \left(\begin{array}{ccc} x^2 & xy & zx \\ xy & y^2 & yz \\ zx & yz & z^2 \end{array} \right) \;dV$

The sphere is highly symmetric, and so we only have to do two integrals. Firstly, the diagonal terms:

\begin{align} C_{xx} = C_{yy} = C_{zz} &= \int \; z^2 \; \mu \; dV,\\ &= \int_0^a \int_0^{\pi} \int_0^{2 \pi} (\rho \cos \theta)^2 \mu \; \rho^2 \sin \theta \; d\phi \; d\theta \; d\rho,\\ &= 2 \pi \mu \left(\int_0^a \rho^4 d\rho \right) \left( \int_0^\pi \cos^2 \theta \sin \theta \; d\theta \right),\\ &= \frac{4}{15} \pi \mu a^5,\\ &= \frac{1}{5} M a^2, \end{align}

And now the off-diagonals—which are all zero because the integrand's odd over the volume:

\begin{align} C_{xy} = C_{yz} = C_{zx} &= \int \; xy \; \mu \; dV, \\ &= 0. \end{align}

Thus,

\begin{align} \textbf{C} &= \frac{1}{5} M a^2 \mathbb{I}, \\ \textrm{Tr} \textbf{C} &= \frac{3}{5} M a^2. \end{align}

and so (as expected),

\begin{align} \textbf{I} &= \left( \textrm{Tr} \, \textbf{C} \right) \; \mathbb{I} - \textbf{C}, \\ &= \frac{2}{5} M a^2 \mathbb{I}. \end{align}