Example with Coupled Rotations
Now let the mass be located at
so that
![\begin{eqnarray*}
\mathbf{I}&=& m \left(\left\Vert\,\underline{x}\,\right\Vert^2...
... & 0\\ [2pt]
-1 & 1 & 0\\ [2pt]
0 & 0 & 2
\end{array}\right].
\end{eqnarray*}](http://www.dsprelated.com/josimages_new/pasp/img2897.png)
We expect
to yield zero for the moment of inertia, and
sure enough
. Similarly, the vector angular
momentum is zero, since
.
For
, the result is
![\begin{displaymath}
\mathbf{I}\eqsp
\begin{array}{r}\left[\begin{array}{ccc} 1 ...
...egin{array}{c} 1 \\ [2pt] 0 \\ [2pt] 0\end{array}\right]m = m,
\end{displaymath}](http://www.dsprelated.com/josimages_new/pasp/img2902.png)





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Off-Diagonal Terms in Moment of Inertia Tensor
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Simple Example