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 $ I = mR^2 = m\cdot
\vert\vert\,\underline{x}-{\cal P}_{\underline{\omega}}(\und...
...erline{\tilde{\omega}}^T\underline{x})\underline{\tilde{\omega}}\,\vert\vert ^2$, where $ \underline{\tilde{\omega}}\isdeftext \underline{\omega}/ \vert\vert\,\underline{\omega}\,\vert\vert $. 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
$

where $ R= \vert\vert\,\underline{x}\,\vert\vert \sin(\theta_{\underline{\tilde{\omega}}\underline{x}})$ is the distance from the rotation axis out to the point $ \underline{x}$ (which equals the length of the vector $ \underline{x}-{\cal P}_{\underline{\omega}}(\underline{x})$).


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Tangential Velocity as a Cross Product
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Cross-Product Magnitude