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Angular Motion in the Space-Fixed Frame

Let's now consider angular motion in the presence of linear motion of the center of mass. In general, we have [270]

$\displaystyle \underline{L}\eqsp \sum \underline{x}\times \underline{p}
$

where the sum is over all mass particles in the rigid body, and $ \underline{p}$ denotes the vector linear momentum for each particle. That is, the angular momentum is given by the tangential component of the linear momentum times the associated moment arm. Using the chain rule for differentiation, we find

$\displaystyle \underline{\tau}\eqsp \dot{\underline{L}} \eqsp \frac{d}{dt}\sum ...
...m (\underline{v}\times\underline{p}+ \underline{x}\times \dot{\underline{p}}).
$

However, $ \underline{v}\times \underline{p}=\underline{v}\times m\underline{v}=\underline{0}$, so that

$\displaystyle \underline{\tau}\eqsp \dot{\underline{L}} \eqsp \sum \underline{x}\times \dot{\underline{p}}
\eqsp \sum \underline{x}\times \underline{f}
$

which is the sum of moments of all external forces.


Previous: Body-Fixed and Space-Fixed Frames of Reference
Next: Euler's Equations for Rotations in the Body-Fixed Frame

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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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