at
![$ \underline{x}=[x,0,0]^T$](img2894.png){kind=small}
. Then the mass moment of inertia
tensor is
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Consider a mass
at
. Then the mass moment of inertia
tensor is
![$\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].
$](img2895.png)
For the angular-velocity vector
![$ \underline{\omega}=[\omega,0,0]^T$](img2896.png){kind=small}
, we obtain the moment of
inertia
![\begin{displaymath}
I \eqsp \underline{\tilde{\omega}}^T\mathbf{I}\,\underline{\tilde{\omega}}\eqsp
\begin{array}{r}\left[\begin{array}{ccc} 1 & 0 & 0\end{array}\right]\\ [2pt]{}\\ [2pt]{}\end{array}\left[\begin{array}{ccc}
0 & 0 & 0\\ [2pt]
0 & x^2 & 0\\ [2pt]
0 & 0 & x^2
\end{array}\right]\left[\begin{array}{c} 1 \\ [2pt] 0 \\ [2pt] 0\end{array}\right]m \eqsp 0.
\end{displaymath}](img2897.png)
This makes sense because the axis of rotation passes through the point mass, so the moment of inertia should be zero about that axis. On the other hand, if we look at
![$ \underline{\omega}=[0,1,0]^T$](img2898.png){kind=small}
, we get
![$\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
$](img2899.png)
which is what we expected.
