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Simple Example

Consider a mass $ m$ at $ \underline{x}=[x,0,0]^T$. Then the mass moment of inertia tensor is

$\displaystyle \mathbf{I}\eqsp m \left(\left\Vert\,\underline{x}\,\right\Vert^2\... ...ay}{ccc} 0 & 0 & 0\\ [2pt] 0 & 1 & 0\\ [2pt] 0 & 0 & 1 \end{array}\right]. $

For the angular-velocity vector $ \underline{\omega}=[\omega,0,0]^T$, we obtain the moment of inertia

\begin{displaymath} I \eqsp \underline{\tilde{\omega}}^T\mathbf{I}\,\underline{\... ...{array}{c} 1 \\ [2pt] 0 \\ [2pt] 0\end{array}\right]m \eqsp 0. \end{displaymath}

This makes sense because the axis of rotation passes through the point mass, so the moment of inertia should be zero about that axis. On the other hand, if we look at $ \underline{\omega}=[0,1,0]^T$, we get

$\displaystyle I \eqsp \underline{\tilde{\omega}}^T\mathbf{I}\,\underline{\tilde... ...] \left[\begin{array}{c} 0 \\ [2pt] 1 \\ [2pt] 0\end{array}\right] \eqsp m x^2 $

which is what we expected.

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Example with Coupled Rotations
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Angular Momentum Vector in Matrix Form