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Relation of Angular to Linear Momentum

Recall (§B.3) that the momentum of a mass $ m$ traveling with velocity $ v$ in a straight line is given by

$\displaystyle p = m v,
$

while the angular momentum of a point-mass $ m$ rotating along a circle of radius $ R$ at $ \omega $ rad/s is given by

$\displaystyle L \eqsp I\omega,
$

where $ I=mR^2$ . The tangential speed of the mass along the circle of radius $ R$ is given by

$\displaystyle v \eqsp R\omega.
$

Expressing the angular momentum $ I$ in terms of $ v$ gives

$\displaystyle L \isdefs I\omega \eqsp I\frac{v}{R} \isdefs mR^2\frac{v}{R} \eqsp Rmv \eqsp Rp. \protect$ (B.18)

Thus, the angular momentum $ L$ is $ R$ times the linear momentum $ p=mv$ .

Linear momentum can be viewed as a renormalized special case of angular momentum in which the radius of rotation goes to infinity.


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``Physical Audio Signal Processing'', by Julius O. Smith III, W3K Publishing, 2010, ISBN 978-0-9745607-2-4.
Copyright © 2014-03-23 by Julius O. Smith III
Center for Computer Research in Music and Acoustics (CCRMA),   Stanford University
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