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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\mathbf{E}-\underline{x}\underline{x}^T\right) \eqsp
m\left[\begin{array}{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{\tilde{\omega}}\eqsp
\begin{array}{r}\left[\begin{array}{ccc} 1 & 0 & 0\end{array}\right]\\ [2pt]{}\\ [2pt]{}\end{array}\left[\begin{array}{ccc}
0 & 0 & 0\\ [2pt]
0 & x^2 & 0\\ [2pt]
0 & 0 & x^2
\end{array}\right]\left[\begin{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{\omega}}\eqsp \begin{array}{r}\left[\begin{array}{ccc} 0 & 1 & 0\end{array}\right]\\ [2pt]{}\\ [2pt]{}\end{array}\left[\begin{array}{ccc}
0 & 0 & 0\\ [2pt]
0 & x^2 & 0\\ [2pt]
0 & 0 & x^2
\end{array}\right]
\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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``Physical Audio Signal Processing'', by Julius O. Smith III, W3K Publishing, 2010, ISBN 978-0-9745607-2-4.
Copyright © 2014-03-23 by Julius O. Smith III
Center for Computer Research in Music and Acoustics (CCRMA),   Stanford University
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