inertia tensor
InertiaTensor
2006-03-28 00:25:45
2006-03-28 10:07:30
Definition
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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.
\begin{equation}
I =
\left[ \begin{array}{ccc}
\displaystyle\sum_i m_i (r^2_i - x^2_i) & -\displaystyle\sum_i m_i x_i y_i & -\displaystyle\sum_i m_i x_i z_i \\
-\displaystyle\sum_i m_i y_i x_i & \displaystyle\sum_i m_i (r^2_i - y^2_i) & -\displaystyle\sum_i m_i y_i z_i \\
-\displaystyle\sum_i m_i z_i x_i & -\displaystyle\sum_i m_i z_i y_i & \displaystyle\sum_i m_i (r^2_i - z^2_i) \end{array} \right] \end{equation}
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
$$I = \frac{2}{5} M R^2 $$
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.