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

$$p = m v,$$

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

$$L \eqsp I\omega,$$

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

$$v \eqsp R\omega.$$

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

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