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Example with Coupled Rotations

Now let the mass $ m$ be located at $ \underline{x}=[1,1,0]^T$ so that

\begin{eqnarray*}
\mathbf{I}&=& m \left(\left\Vert\,\underline{x}\,\right\Vert^2\mathbf{E}-\underline{x}\underline{x}^T\right) \eqsp
m\left(2\mathbf{E}-\left[\begin{array}{ccc}
1 & 1 & 0\\ [2pt]
1 & 1 & 0\\ [2pt]
0 & 0 & 0
\end{array}\right]\right)\\ [5pt]
&=& m\left[\begin{array}{rrr}
1 & -1 & 0\\ [2pt]
-1 & 1 & 0\\ [2pt]
0 & 0 & 2
\end{array}\right].
\end{eqnarray*}

We expect $ \underline{\omega}=[1,1,0]$ to yield zero for the moment of inertia, and sure enough $ \underline{\tilde{\omega}}^T\mathbf{I}\,\underline{\tilde{\omega}}=0$ . Similarly, the vector angular momentum is zero, since $ \mathbf{I}\,\underline{\omega}=\underline{0}$ .

For $ \underline{\omega}=[1,0,0]^T$ , the result is

\begin{displaymath}
\mathbf{I}\eqsp
\begin{array}{r}\left[\begin{array}{ccc} 1 & 0 & 0\end{array}\right]\\ [2pt]{}\\ [2pt]{}\end{array}\left[\begin{array}{rrr}
1 & -1 & 0\\ [2pt]
-1 & 1 & 0\\ [2pt]
0 & 0 & 2
\end{array}\right]
\left[\begin{array}{c} 1 \\ [2pt] 0 \\ [2pt] 0\end{array}\right]m = m,
\end{displaymath}

which makes sense because the distance from the axis $ \underline{e}_1$ to $ \underline{x}$ is one. The same result is obtained for rotation about $ \underline{\omega}=\underline{e}_2$ . For $ \underline{\omega}=\underline{e}_3$ , however, the result is $ I=2m = m \vert\vert\,\underline{x}\,\vert\vert ^2$ , as expect.


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``Physical Audio Signal Processing'', by Julius O. Smith III, W3K Publishing, 2010, ISBN 978-0-9745607-2-4.
Copyright © 2014-06-11 by Julius O. Smith III
Center for Computer Research in Music and Acoustics (CCRMA),   Stanford University
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