Parallel Axis Theorem
There are a few places in this post that a diagram would be useful. I just didn’t have time to make one. I appologize and will try to find time to make them in the future.
A Fixed Axis of Rotation
Theorem
Suppose we know the moment of inertia of object $\mathcal{G}$ with respect to a fixed axis of rotation passing through its center of mass. Now consider another axis of rotation which is parallel to and a distance $D$ away from the original axis.
Then the parallel axis theorem says the following.
\[I_{\text{parallel axis}} = I_{\text{central axis}} + MD^2\]Proof
Define a Cartesian coordinate system with respect to any fixed origin. Fix some arbitrary point in object $\mathcal{G}$, and let $\b{r}$ be its position vector. Let $r_{\text{central axis}}$ and $r_{\text{parallel axis}}$ be the shortest distance from this fixed point to their respective axes.
Notice that
\[r_{\text{parallel axis}} = r_{\text{central axis}} + D\]Now, let’s find the moment of inertia with respect to the parallel axis of rotation.
\[\begin{align} I_{\text{parallel axis}} &= \int_{\mathcal{G}} r_{\text{parallel axis}}^2 \;dm \\[10pt] &= \int_{\mathcal{G}} (r_{\text{central axis}} + D)^2 \;dm \\[10pt] &= \int_{\mathcal{G}} (r_{\text{central axis}}^2 + D^2 + 2 D r_{\text{central axis}}) \;dm \\[10pt] &= \left ( \int_{\mathcal{G}} r_{\text{central axis}}^2 \;dm \right ) + \left ( \int_{\mathcal{G}} D^2 \;dm \right ) + \left ( \int_{\mathcal{G}} 2 D r_{\text{central axis}} \;dm \right ) \\[10pt] &= \left ( \int_{\mathcal{G}} r_{\text{central axis}}^2 \;dm \right ) + \left ( D^2 \int_{\mathcal{G}} \;dm \right ) + \left ( 2 D \int_{\mathcal{G}} r_{\text{central axis}} \;dm \right ) \\[10pt] &= \left ( I_{\text{central axis}} \right ) + \left ( M D^2 \right ) + \left ( 0 \right ) \\[10pt] &= I_{\text{central axis}} + M D^2 \end{align}\]Lastly, we just have to argue why \(\displaystyle \int_{\mathcal{G}} r_{\text{central axis}} \;dm = 0\). Essentially this is due to symmetry, since a central axis of rotation (by definition) passes through the center of mass of the object. This completes the proof.
Generalized Parallel Axis Theorem
Theorem
We can generalize the parallel axis theorem so that it can be applied to an inertia tensor. Fix any coordinate system. Suppose we know the inertia tensor of a object $\mathcal{G}$ with mass $M$ such that its center of mass is positioned at the origin, denoted $\m{I_{\text{cm}}}$. Now suppose that object $\mathcal{G}$ is translated anywhere in the coordinate system such that its center of mass is now at coordinate $\b{D} = D_x \; \u{x} + D_y \; \u{y} + D_z \; \u{z}$.
Then, the generalized parallel axis theorem tells us the following.
\[\m{ I_{\text{translated}} } = \m{I_{\text{cm}}} + M \begin{bmatrix} D_y^2 + D_z^2 & - D_xD_y & - D_xD_z \\ - D_yD_x & D_z^2 + D_x^2 & - D_yD_z \\ - D_zD_x & - D_zD_y & D_x^2 + D_y^2 \end{bmatrix}\]We can also express this compactly, where $\otimes$ 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 \(\left [ \ \mathbb{I} \ \right ] = \begin{bmatrix} 1 & 0 & \cdots & 0 \\ 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 1 \end{bmatrix}\) .
\[\m{ I_{\text{translated}} } = \m{I_{\text{cm}}} + M (D^2 \m{\mathbb{I}} - \b{D} \otimes \b{D})\]Proof
This is a bit more technical, but it’s the same idea as the first proof. Fix some arbitrary point in object $\mathcal{G}$. Let $\b{r} = x \; \u{x} + y \; \u{y} + z \; \u{z}$ be its original position vector and let $\b{r}’ = x’ \; \u{x} + y’ \; \u{y} + z’ \; \u{z}$ denote its position vector after translated.
Therefore, $\b{r}’ = \b{r} + \b{D}$. Unrolling this element-wise, we have
\[x' = x + D_x \qquad y' = y + D_y \qquad z' = z + D_z\]Now, let’s find the inertia tensor of the translated object.
\[\begin{align} \m{ I_{\text{translated}} } &= \int_{\mathcal{G}} \begin{bmatrix} (y')^2 + (z')^2 & - x'y' & - x'z' \\ - y'x' & (z')^2 + (x')^2 & - y'z' \\ - z'x' & - z'y' & (x')^2 + (y')^2 \end{bmatrix} \;dm \\[10pt] &= \int_{\mathcal{G}} \begin{bmatrix} (y + D_y)^2 + (z + D_z)^2 & - (x + D_x)(y + D_y) & - (x + D_x)(z + D_z) \\ - (y + D_y)(x + D_x) & (z + D_z)^2 + (x + D_x)^2 & - (y + D_y)(z + D_z) \\ - (z + D_z)(x + D_x) & - (z + D_z)(y + D_y) & (x + D_x)^2 + (y + D_y)^2 \end{bmatrix} \;dm \\[10pt] &= \int_{\mathcal{G}} \begin{bmatrix} (y^2 + 2yD_y + D_y^2) + (z^2 + 2zD_z + D_z^2) & - xy - xD_x - yD_y - D_xD_y & - xz - xD_x - zD_z - D_xD_z \\ - yx - yD_y - xD_x - D_yD_x & (z^2 + 2zD_z + D_z^2) + (x^2 + 2xD_x + D_x^2) & - yz - yD_y - zD_z - D_yD_z \\ - zx - zD_z - xD_x - D_zD_x & - zy - zD_z - yD_y - D_zD_y & (x^2 + 2xD_x + D_x^2) + (y^2 + 2yD_y + D_y^2) \end{bmatrix} \;dm \\[10pt] &= \int_{\mathcal{G}} \begin{bmatrix} y^2 + z^2 & - xy & - xz \\ - yx & z^2 + x^2 & - yz \\ - zx & - zy & x^2 + y^2 \end{bmatrix} \;dm + D_x \begin{bmatrix} 0 & - 1 & - 1 \\ - 1 & 2 & 0 \\ - 1 & 0 & 2 \end{bmatrix} \int_{\mathcal{G}} x \; dm + D_y \begin{bmatrix} 2 & - 1 & 0 \\ - 1 & 0 & - 1 \\ 0 & - 1 & 2 \end{bmatrix} \int_{\mathcal{G}} y \; dm + D_z \begin{bmatrix} 2 & 0 & - 1 \\ 0 & 2 & - 1 \\ - 1 & - 1 & 0 \end{bmatrix} \int_{\mathcal{G}} z \; dm + \begin{bmatrix} D_y^2 + D_z^2 & - D_xD_y & - D_xD_z \\ - D_yD_x & D_z^2 + D_x^2 & - D_yD_z \\ - D_zD_x & - D_zD_y & D_x^2 + D_y^2 \end{bmatrix} \int_{\mathcal{G}} \; dm \\[10pt] &= \m{I_{\text{cm}}} + M \begin{bmatrix} D_y^2 + D_z^2 & - D_xD_y & - D_xD_z \\ - D_yD_x & D_z^2 + D_x^2 & - D_yD_z \\ - D_zD_x & - D_zD_y & D_x^2 + D_y^2 \end{bmatrix} \end{align}\]Just as before, we know that
\[\int_{\mathcal{G}} x \; dm = \int_{\mathcal{G}} y \; dm = \int_{\mathcal{G}} z \; dm = 0\]since object $\mathcal{G}$ is assumed to have its center of mass at the origin.
An Alternative Formulation
This is an interesting reformation that I discovered, which I think it more elegant. Consider the translation vector $\b{D}$, and consider the angles it makes with respect to the $x$, $y$, and $z$ axes, which we label $\alpha_x$, $\alpha_y$, and $\alpha_z$, respectively.
Using some geometry, you can prove that.
\[\begin{align} D_y^2 + D_z^2 &= D \sin \alpha_x &\quad D_x &= D \cos \alpha_x \\[10pt] D_z^2 + D_x^2 &= D \sin \alpha_y &\quad D_y &= D \cos \alpha_y \\[10pt] D_x^2 + D_y^2 &= D \sin \alpha_z &\quad D_z &= D \cos \alpha_z \end{align}\]Therefore, we can refactor the inertia tensor as the following.
\[\m{I_{\text{translated}}} = \m{I_{\text{cm}}} + M D^2 \begin{bmatrix} \sin^2 \alpha_x & - \cos \alpha_x \cos \alpha_y & - \cos \alpha_x \cos \alpha_z \\ - \cos \alpha_y \cos \alpha_x & \sin^2 \alpha_y & - \cos \alpha_y \cos \alpha_z \\ - \cos \alpha_z \cos \alpha_x & - \cos \alpha_z \cos \alpha_y & \sin^2 \alpha_z \end{bmatrix}\]This can be written compactly as follows, where $\b{\alpha} = [\alpha_x \ \ \alpha_y \ \ \alpha_z]^T$ and $\cos(\b{\alpha})$ is evaluated element-wise.
\[\m{I_{\text{translated}}} = \m{I_{\text{cm}}} + M D^2 (\m{\mathbb{I}} - \cos(\b{\alpha}) \otimes \cos(\b{\alpha}))\]There are two nice things about this. First, the matrix that is left over is unitary, which means it is a pure rotation matrix. Second, when you start taking special cases usually the angle of the object with respect to the axis of rotation is easier to work with.