Mass Moment of Inertia as a Cross Product
In Eq.(B.14) above, the mass moment of inertia was expressed
in terms of orthogonal projection as
, where
. In terms of the vector cross
product, we can now express it as
![$\displaystyle I \eqsp m\cdot(\underline{\tilde{\omega}}\times \underline{x})^2 ...
...cdot\sin(\theta_{\underline{\tilde{\omega}}\underline{x}})\right]^2
\eqsp mR^2
$](http://www.dsprelated.com/josimages_new/pasp/img2845.png)
![$ R= \vert\vert\,\underline{x}\,\vert\vert \sin(\theta_{\underline{\tilde{\omega}}\underline{x}})$](http://www.dsprelated.com/josimages_new/pasp/img2846.png)
![$ \underline{x}$](http://www.dsprelated.com/josimages_new/pasp/img260.png)
![$ \underline{x}-{\cal P}_{\underline{\omega}}(\underline{x})$](http://www.dsprelated.com/josimages_new/pasp/img2847.png)
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Tangential Velocity as a Cross Product
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Cross-Product Magnitude