Inertia Tensor Derivation
In this post, I will derive the value of the inertia tensor from the definition of angular momentum. I’ve included this for completeness, but it’s not required to understand the rest of the series.
Caution: In this post, the variable $\b{L}$ is referring to angular momentum. In all future posts it will refer to length, since angular momentum will never come up again. Please do not conflate these variables.
Point Mass
As always, we start with a point mass $m$ at position $\b{r}$ rotating around a fixed axis of rotation $\b{\omega}$. Its angular momentum is described by the equation.
\[\b{L} = \b{r} \times \b{p} = \b{r} \times (m \b{v}) = m \left [ \b{r} \times (\b{\omega} \times \b{r}) \right ]\]Here, $\times$ is the cross product \(\b{u} \times \b{v} = \left \lvert \begin{array}{ccc} \u{x} & \u{y} & \u{z} \\ u_x & u_y & u_z \\ v_x & v_y & v_z \end{array} \right \rvert = (u_y v_z - u_z v_y) \; \u{x} + (u_z v_x - u_x v_z) \; \u{y} + (u_x v_y - u_y v_x) \; \u{z}\) . Unlike our analysis in the previous post, we cannot in general assume that $\b{r}$ is perpendicular to $\b{v}$. Thus, we use the triple cross product identity.
\[\b{a} \times (\b{b} \times \b{c}) = (\b{a} \cdot \b{c}) \b{b} - (\b{a} \cdot \b{b}) \b{c}\]Substituting in for $\b{a} = \b{c} = \b{r}$ and $\b{b} = \b{\omega}$ we get
\[\b{L} = m \left [ r^2 \b{\omega} - (\b{r} \cdot \b{\omega}) \b{r} \right ]\]Let $\b{r} = x \; \u{x} + y \; \u{y} + z \; \u{z}$ and $\b{\omega} = \omega_x \; \u{x} + \omega_y \; \u{y} + \omega_z \; \u{z}$. Expanding the vector notation we get.
\[\begin{align} L_x &= m \left [ r^2 \omega_x - (\b{r} \cdot \b{\omega}) x \right ] \\[10pt] &= m \left [ r^2 \omega_x - (x \omega_x + y \omega_y + z \omega_z) x \right ] \\[10pt] &= m \left [ (r^2 - x^2)\omega_x - xy \omega_y - xz \omega_z \right ] \\[10pt] &= m \left [ (y^2 + z^2)\omega_x - xy \omega_y - xz \omega_z \right ] \\[10pt] \end{align}\]Of course, we can do exactly the same thing for the $y$ and $z$ components to get
\[\b{L} = m \left [ ((y^2 + z^2)\omega_x - xy \omega_y - xz \omega_z) \; \u{x} + (- yx \omega_x + (z^2 + x^2) \omega_y - yz \omega_z) \; \u{y} + (- zx \omega_x - zy \omega_y + (x^2 + y^2)\omega_z) \; \u{z} \right ]\]There’s a lot of repetition here. We can write this much more compactly as follow
\[\b{L} = m \begin{bmatrix} y^2 + z^2 & - xy & - xz \\ - yx & z^2 + x^2 & - yz \\ - zx & - zy & x^2 + y^2 \end{bmatrix} \b{\omega} = \m{I} \; \b{\omega}\]This matrix in the middle (along with the mass) is called the inertia tensor, denoted in this series by $\m{I}$.
A Collection of Point Masses
Suppose we have a collection of $n$ point masses $m_i$ at position $\b{r}_i$ all rotating around a fixed axis $\b{\omega}$. We define the angular momentum of the system as follows.
\[\b{L}_{\text{system}} = \sum_{i=1}^n \b{L}_i = \sum_{i=1}^n m_i \begin{bmatrix} y_i^2 + z_i^2 & - x_iy_i & - x_iz_i \\ - y_ix_i & z_i^2 + x_i^2 & - y_iz_i \\ - z_ix_i & - z_iy_i & x_i^2 + y_i^2 \end{bmatrix} \b{\omega}\]Notice that $\b{\omega}$ is the same for all point masses, so we factor it out of the sum. This means we can sum all inertia tensors first in order to get an effective inertia tensor for the system.
\[\b{L}_{\text{system}} = \left ( \sum_{i=1}^n \m{I_i} \right ) \b{\omega} = \m{ I_{\text{system}} } \; \b{\omega}\]Exactly the same as in the simplified case.
Rigid Body
Now suppose we have some rigid body $\mathcal{G}$ rotating around a fixed axis $\b{\omega}$. We think of this object as an continuous collection of point masses. In other words, we consider each infinitesmal mass and integrate along the geometry.
\[\b{L} = \int_{\mathcal{G}} \begin{bmatrix} y^2 + z^2 & - xy & - xz \\ - yx & z^2 + x^2 & - yz \\ - zx & - zy & x^2 + y^2 \end{bmatrix} \b{\omega} \; dm = \left ( \int_{\mathcal{G}} \begin{bmatrix} y^2 + z^2 & - xy & - xz \\ - yx & z^2 + x^2 & - yz \\ - zx & - zy & x^2 + y^2 \end{bmatrix} \; dm \right ) \b{\omega}\]Just as in the system of particles, $\b{\omega}$ is constant and therefore can be factored out of the integral. Thus, we get the effective inertia tensor of the rigid body.
\[\m{ I_{\mathcal{G}} } = \int_{\mathcal{G}} \begin{bmatrix} y^2 + z^2 & - xy & - xz \\ - yx & z^2 + x^2 & - yz \\ - zx & - zy & x^2 + y^2 \end{bmatrix} \; dm\]This gives us the definition from the previous post.
A Compact Formula
Sometimes you will see the elements of the inertia written as follows. Let $(x_1, x_2, x_3) = (x, y, z)$ and let $\delta_{ij}$ be the Kronecker delta \(\delta_{ij} = \begin{cases} 1 &\text{if } i = j \\ 0 &\text{if } i \neq j \end{cases}\) .
\[I_{ij} = \int_{\mathcal{G}} ( r^2 \delta_{ij} - x_i x_j ) \; dm\]If we let $\otimes$ denote the outer product \(\boldsymbol{u} \otimes \boldsymbol{v} = \begin{bmatrix} u_1 v_1 & u_1 v_2 & \cdots & u_1 v_n \\ u_2 v_1 & u_2 v_2 & \cdots & u_2 v_n \\ \vdots & \vdots & \ddots & \vdots \\ u_m v_1 & u_m v_2 & \cdots & u_m v_n \end{bmatrix}\) and $\m{ \mathbb{I} }$ be the identity matrix \(\left [ \ \mathbb{I} \ \right ] = \begin{bmatrix} 1 & 0 & \cdots & 0 \\ 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 1 \end{bmatrix}\) , then we can write the following.
\[\m{I_{\mathcal{G}}} = \int_{\mathcal{G}} ( r^2 \m{ \mathbb{I} } - \b{r} \otimes \b{r} ) \; dm\]Calculating the Moment of Inertia for an Arbitrary Axis
Given an arbitrary axis of rotation $\b{\omega}$, the objects angular momentum is $\b{L} = \m{I} \; \b{\omega}$. This is a vector quantity. What if we are only interested in the angular momentum in the direction of the axis of rotation? We can compute this component by simply taking the dot product of the angular momentum with the unit vector in the axis of rotation.
\[L_{\b{\omega}} = \b{L} \cdot \u{\omega} = \u{\omega}^T \b{L}\]Now, we substitute in the expression for the angular momentum in terms of the rotational inertial.
\[L_{\b{\omega}} = \u{\omega}^T \; \left ( \m{I} \; \b{\omega} \right ) = \left ( \u{\omega}^T \; \m{I} \; \u{\omega} \right ) \omega\]Therefore, given any arbitrary axis of rotation and the inertia tensor, the moment of inertia about that axis is computed as follows
\[I_{\b{\omega}} = \u{\omega}^T \; \m{I} \; \u{\omega}\]Products of Inertia Revisited
Let $(x_1, x_2, x_3) = (x, y, z)$. Then
\[I_{ij} = \u{x}_i^T \; \m{I} \; \u{x}_j\]Applying this to what we derived above, suppose we have two axes of rotation $\b{\omega}$ and $\b{\psi}$, then
\[\u{\psi}^T \m{I} \; \u{\omega}\]vaguely measures the angular momentum in the $\b{\phi}$ direction when the object is rotating about the $\b{\omega}$ axis. If we consider the shared plane occupied by $\b{\omega}$ and $\b{\psi}$, then the product of inertia is a measure of the mass distribution in that plane.