12.11 Angular Momentum

FIGURE 12.51 shows a particle that, at this instant, is located at position $\vec{r}$ and is moving with momentum $\vec{p} = m \vec{v}$ . Together, $\vec{r}$ and $\vec{p}$ define the plane of motion. We define the particle’s angular momentum $\vec{L}$ relative to the origin to be the vector

\vec{L} \equiv \vec{r} \times \vec{p} = (m r v sin β, direction of right-hand rule)

(12.49)

Because of the cross product, the angular momentum vector is perpendicular to the plane of motion. The units of angular momentum are $kg m^{2} / s$ .

The figure illustrates two vector arrows labeled vector r and vector p on the x y z coordinate space.

The figure illustrates three vector arrows labeled vector r, vector p, and vector L on the x y z coordinate space.

Full Alternative Text

NOTE Angular momentum is the rotational equivalent of linear momentum in much the same way that torque is the rotational equivalent of force. Notice that the vector definitions are parallel: $\vec{τ} \equiv \vec{r} \times \vec{F}$ and $\vec{L} \equiv \vec{r} \times \vec{p}$ .

Angular momentum, like torque, is about the point from which $\vec{r}$ is measured. A different origin would yield a different angular momentum. Angular momentum is especially simple for a particle in circular motion. As FIGURE 12.52 shows, the angle $β$ between $\vec{p}$ (or $\vec{v}$ ) and $\vec{r}$ is always $90 °$ if we make the obvious choice of measuring $\vec{r}$ from the center of the circle. For motion in the xy-plane, the angular momentum vector $\vec{L}$ —which must be perpendicular to the plane of motion—is entirely along the z-axis:

L_{z} = m r v_{t} = m r^{2} ω (particle in circular motion)

(12.50)

where $v_{t}$ is the tangential component of velocity. Our sign convention makes $L_{z}$ , like $ω$ , positive for a ccw rotation, negative for a cw rotation.

The figure illustrates the cross product of two vectors on an x y coordinate plane.

FIGURE 12.52 Full Alternative Text

In Chapter 11, we found that Newton’s second law for a particle can be written $\vec{F}_{net} = d \vec{p} / d t$ . There’s a similar connection between torque and angular momentum. To show this, we take the time derivative of $\vec{L}$ :

\begin{array}{l} \frac{d \vec{L}}{d t} = \frac{d}{d t} (\vec{r} \times \vec{p}) & = \frac{d \vec{r}}{d t} \times \vec{p} + \vec{r} \times \frac{d \vec{p}}{d t} \\ = \vec{v} \times \vec{p} + \vec{r} \times {\vec{F}}_{net} \end{array}

(12.51)

where we used Equation 12.47 for the derivative of a cross product. We also used the definitions $\vec{v} = d \vec{r} / d t$ and $\vec{F}_{net} = d \vec{p} / d t$ .

Vectors $\vec{v}$ and $\vec{p}$ are parallel, and the cross product of two parallel vectors is $\vec{0}$ . Thus the first term in Equation 12.51 vanishes. The second term $\vec{r} \times \vec{F}_{net}$ is the net torque, $\vec{τ}_{net} = \vec{τ}_{1} + \vec{τ}_{2} + \dots$ , so we arrive at

\frac{d \vec{L}}{d t} = \vec{τ}_{net}

(12.52)

Equation 12.52, which says a net torque causes the particle’s angular momentum to change, is the rotational equivalent of $d \vec{p} / d t = \vec{F}_{net}$ .

12.11 Angular Momentum

FIGURE 12.51 The angular momentum vector $\vec{L}$ .

FIGURE 12.52 Angular momentum of circular motion.