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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})$).


Previous: Cross-Product Magnitude
Next: Tangential Velocity as a Cross Product

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About the Author: Julius Orion Smith III
Julius Smith's background is in electrical engineering (BS Rice 1975, PhD Stanford 1983). He is presently Professor of Music and Associate Professor (by courtesy) of Electrical Engineering at Stanford's Center for Computer Research in Music and Acoustics (CCRMA), teaching courses and pursuing research related to signal processing applied to music and audio systems. See http://ccrma.stanford.edu/~jos/ for details.


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