PlanetPhysics/Inertia Tensor

The inertia tensor is straight forward to calculate in theory from equation (1). However, in practice finding the inertia tensor of an object is quite involved and high precision is needed in applications such as spacecraft design.

$I=\left[{\begin{matrix}\sum _{i}m_{i}(r_{i}^{2}-x_{i}^{2})&-\sum _{i}m_{i}x_{i}y_{i}&-\sum _{i}m_{i}x_{i}z_{i}\\-\sum _{i}m_{i}y_{i}x_{i}&\sum _{i}m_{i}(r_{i}^{2}-y_{i}^{2})&-\sum _{i}m_{i}y_{i}z_{i}\\-\sum _{i}m_{i}z_{i}x_{i}&-\sum _{i}m_{i}z_{i}y_{i}&\sum _{i}m_{i}(r_{i}^{2}-z_{i}^{2})\end{matrix}}\right]$

Units for the inertia tensor are $[kg/m^{2}]$. Terminology to reference the elements of the inertia tensor are the moments of inertia, which are the diagonal elements, $I_{11}$, $I_{22}$, and $I_{33}$.

and the products of inertia, which are the off-diagonal elements. A quick inspection of (1) reveals that the inertia tensor is symmetric, so the the products of inertia are

$I_{12}=I_{21}$ $I_{13}=I_{31}$ $I_{23}=I_{32}$

To help build intuition on working with the inertia tensor, a few simple cases are looked at. For the example of a solid sphere, we found that the moment of inertia of a Solid Sphere was

$I={\frac {2}{5}}MR^{2}$

This value was just the rotational inertia about a single axis that was a diameter of the sphere. It is easy to see that the general case about any of the sphere's body axes, x,y or z is taken care of by the inertia tensor with the moments of inertia being

$I_{11}=I_{22}=I_{33}=I$

and the products of inertia equal to 0. For this case of the body axes being diameters of the sphere, then the body axes are also the principle axes and this says a lot about the symmetry of the chosen axes. For an axis of rotation that is not a diameter of the sphere, then there will be non-zero products of inertia and no more symmetry, so this axis is not a principle axis. If you visualize moving the origin of the body axes anywhere other than the orgin, then the set of coordinate body axes are not principle axes of rotation.

Another simple case is for the moment of inertia of a solid cylinder. Here we find that the diagonal elements are different based on their symmetry. For the moment of inertia about the z axis, we found

$I_{33}={\frac {1}{2}}MR^{2}$

and for the symmetric axes of x and y

$I_{11}=I_{22}={\frac {1}{4}}MR^{2}+{\frac {1}{12}}ML^{2}$

If you imagine a very dense pencil, where R is small and L is large, you might have an inertia tensor that looks like

$I=\left[{\begin{matrix}100&0&0\\0&100&0\\0&0&10\end{matrix}}\right]$

Notice how this gives us a feel for how the mass of the body is distributed. For a stubby cylinder, like a disk the inertia tensor would be something like

$I=\left[{\begin{matrix}10&0&0\\0&10&0\\0&0&100\end{matrix}}\right]$