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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... ... & 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 ... ...egin{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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Off-Diagonal Terms in Moment of Inertia Tensor
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Simple Example