Free Books

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.
Next Section:
Angular Momentum Vector in Matrix Form
Previous Section:
Tangential Velocity as a Cross Product