= Angular momentum
{title2=$\mathbf L=\mathbf r\times\mathbf p$}
{wiki}
= Angular momenta
{synonym}
<Angular momentum> measures how a <point mass>'s <momentum> passes around a chosen origin. Motion straight toward or away from that origin contributes zero; transverse <momentum> contributes more when it passes farther away. The picture shows the sign convention before the calculation.
\Image[/angular-momentum-lever-arm.png]
{title=Positive, negative and zero angular momentum about the same origin}
Fix an origin $O$ and nonrotating axes in an <inertial frame>. For a nonrelativistic <point mass> of constant <mass> $m>0$, let $\mathbf r$ be its <position> from $O$ and $\mathbf p=m\mathbf v$ its <mechanical momentum> in that same <reference frame>. Its <angular momentum> is the <cross product>
$$
\mathbf L_O=\mathbf r\times\mathbf p,\qquad
|\mathbf L_O|=rp\sin\phi.
$$
Here $r=|\mathbf r|$, $p=|\mathbf p|$, and $\phi\in[0,\pi]$ is the <angle> between them. The <right-hand rule> gives the direction: curl from $\mathbf r$ toward $\mathbf p$, and the thumb points along $\mathbf L_O$. Thus <angular momentum> is an <axial vector>, perpendicular to both <position> and <momentum>. Its <International System of Units> unit is $\mathrm{kg\,m^2/s}$. A <point mass> need not follow a <circle> to have <angular momentum>.<University Physics Volume 1 by William Moebs and others/11.2 Angular Momentum>[section 11.2: Angular Momentum]{ref}
With $x$ rightward, $y$ upward, and $+z$ out of the page, the <cross product> gives $L_z=xp_y-yp_x$. Take
$$
\mathbf r=(2,0,0)\,\mathrm m,\qquad
\mathbf p=(0,3,0)\,\mathrm{kg\,m/s}.
$$
Then $\mathbf L_O=(0,0,6)\,\mathrm{kg\,m^2/s}$. Reversing the <momentum> makes $L_z=-6\,\mathrm{kg\,m^2/s}$; radial <momentum> $\mathbf p=(3,0,0)\,\mathrm{kg\,m/s}$ gives $\mathbf L_O=\mathbf0$. The factor $r\sin\phi$ is the <lever arm> to the line through the particle parallel to its <momentum>.
For motion in the $xy$ plane, <plane polar coordinates> give $\mathbf v=\dot r\widehat{\mathbf r}+r\dot\theta\widehat{\boldsymbol\theta}$, with $\widehat{\mathbf r}\times\widehat{\boldsymbol\theta}=\widehat{\mathbf z}$. Consequently
$$
L_z=mr^2\dot\theta,\qquad r^2\dot\theta=\frac{L_z}{m}.
$$
The signed <angular velocity> $\dot\theta$ is positive counterclockwise viewed from $+z$ toward the plane, with $\theta$ measured in <radians>. Radial <velocity> does not enter the <cross product>. Under <conservation of angular momentum>, this relation predicts faster angular motion at smaller $r$ and constant <areal velocity> $\dot A=L_z/(2m)$ for consistently oriented swept <area>. That is the geometric content of <Kepler's second law>.
For several <point masses>, add their <angular momenta> about the same origin: $\mathbf L_O=\sum_i\mathbf r_i\times\mathbf p_i$. The origin matters: <angular momentum under an origin shift> explains why the same moving particle can have different values about two fixed points. The <central axis of a momentum system> and <angular momentum decomposition about the centre of mass> make this dependence useful.
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