Principal Axis Theorem
In the previous post on the parallel axis theorem, we say how the inertia tensor updates when the object $\mathcal{G}$ is translated in our fixed coordiante system. In this post, we are going to see how the inertia tensor is updated when the object $\mathcal{G}$ is rotated.
Principal Axes
If we are analyzing the moment of inertia of an object about a fixed axis of rotation, then clearly rotating about this axis does not change its inertia. In the figure below, each of the point masses have the same rotational inertia.
However, if we have an inertia tensor, then it gets a bit more complicated. The inertia tensor does not remain constant when the object is rotated. For example, consider a point mass on the $xy$ plane a distance $R$ from the origin.
The inertia tensor to describe this is the following. As an exercise, verify this using the general formula for the inertia tensor of a point pass given a previous post.
\[\m{I} = mR^2 \begin{bmatrix} \sin^2 \phi & - \sin \phi \cos \phi & 0 \\ - \sin \phi \cos \phi & \cos^2 \phi & 0 \\ 0 & 0 & 1 \end{bmatrix}\]Notice that if $\phi = \pi/2$, then the inertia tensor becomes
\[\m{I} = mR^2 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}\]This is a much nicer inertia tensor to work with, and we really haven’t changed anything about the object; all we have done is move our coordinate system.
A principal axis is a coordiante system such that the inertia tensor is diagonal. In otherwords, all product of inertia values are $0$. It is analogous to finding the eigenvector decomposition of a linear transformation.
The Principal Axis Theorem
Theorem
A rigid body $\mathcal{G}$ can always be rotated such that its inertia tensor is diagonal.
Stated more mathematically rigorous, given the inertia tensor $\m{I_{\mathcal{G}}}$ of any rigid body $\mathcal{G}$, we can always find an orthonormal basis.
Proof
Lets generalize a bit and consider any real-valued $n \times n$ matrix $\m{A} \in \mathbb{R}^{n \times n}$ which is symmetric. Therefore, we can apply the Spectral Theorem for symmetric matrices, which says that there must exist $n$ distinct eigenvectors $\b{v}_1, \b{v}_2, \ldots, \b{v}_n$ (not necessarily normalized) and cooresponding eigenvalues $\lambda_1, \lambda_2, \ldots, \lambda_n$. Let
\[\m{V} = \begin{bmatrix} \vert & & \vert \\ \b{v}_1 & \cdots & \b{v}_n \\ \vert & & \vert \end{bmatrix} \qquad\qquad \m{\Lambda} = \begin{bmatrix} \lambda_1 & \cdots & 0 \\ \vdots & \ddots & \\ 0 & & \lambda_n \end{bmatrix}\]Therefore, the eigenvector decomposition of $\m{A}$ is the following.
\[\m{A} = \m{V} \ \m{\Lambda} \ \m{V}^{-1}\]Therefore, $\u{v}_1, \u{v}_2, \ldots, \u{v}_n$ forms an orthonormal basis of $\m{A}$, which completes the proof.
Specializing back down to $\m{I_{\mathcal{G}}}$. By definition, it is real-valued and symmetric, so we can apply the results above. Therefore, we can write
\[\m{I_{\mathcal{G}}} = \m{V} \ \m{\Lambda} \ \m{V}^{-1}\]where $\u{v}_1, \u{v}_2, \ldots, \u{v}_n$ are the principle axes, and $\m{\Lambda}$ is the diagonalized inertia tensor with respect to the principle axes.