Angular momentum of a point particle

Consider a particle of mass $m$, position vector ${\bf r}$, and instantaneous velocity ${\bf v}$, which rotates about an axis passing through the origin of our coordinate system. We know that the particle's linear momentum is written

$${\bf p} = m {\bf v},$$ (414)

and satisfies

$$\frac{d{\bf p}}{dt} = {\bf f},$$ (415)

where ${\bf f}$ is the force acting on the particle. Let us search for the rotational equivalent of ${\bf p}$.

Consider the quantity

$${\bf l} = {\bf r}\times {\bf p}.$$ (416)

This quantity--which is known as angular momentum--is a vector of magnitude

$$l = r p \sin\theta,$$ (417)

where $\theta$ is the angle subtended between the directions of ${\bf r}$ and ${\bf p}$. The direction of ${\bf l}$ is defined to be mutually perpendicular to the directions of ${\bf r}$ and ${\bf p}$, in the sense given by the right-hand grip rule. In other words, if vector ${\bf r}$ rotates onto vector ${\bf p}$ (through an angle less than $180^\circ$), and the fingers of the right-hand are aligned with this rotation, then the thumb of the right-hand indicates the direction of ${\bf l}$. See Fig. 85.

Figure 85: Angular momentum of a point particle about the origin.

Let us differentiate Eq. (416) with respect to time. We obtain

$$\frac{d{\bf l}}{dt} = \dot{\bf r} \times {\bf p} + {\bf r} \times \dot{\bf p}.$$ (418)

Note that the derivative of a vector product is formed in much the same manner as the derivative of an ordinary product, except that the order of the various terms is preserved. Now, we know that $\dot{\bf r} = {\bf v} = {\bf p}/m$ and $\dot{\bf p} = {\bf f}$. Hence, we obtain

$$\frac{d{\bf l}}{dt} = \frac{{\bf p} \times {\bf p}}{m} + {\bf r} \times {\bf f}.$$ (419)

However, ${\bf p} \times {\bf p} = {\bf0}$, since the vector product of two parallel vectors is zero. Also,

$${\bf r} \times {\bf f} = \mbox{\boldmath$\tau$},$$ (420)

where $\mbox{\boldmath$\tau$}$ is the torque acting on the particle about an axis passing through the origin. We conclude that

$$\frac{d{\bf l}}{dt} = \mbox{\boldmath$\tau$}.$$ (421)

Of course, this equation is analogous to Eq. (415), which suggests that angular momentum, ${\bf l}$, plays the role of linear momentum, ${\bf p}$, in rotational dynamics.

For the special case of a particle of mass $m$ executing a circular orbit of radius $r$, with instantaneous velocity $v$ and instantaneous angular velocity $\omega$, the magnitude of the particle's angular momentum is simply

$$l = m v r = m \omega r^2.$$ (422)