Angular Momentum and Its Conservation

Beginner

Sit on an office chair that turns freely, hold your arms and legs out, and get a friend to give you a gentle spin. Now pull them in tight: suddenly you whirl much faster. Nobody pushed you; you only changed your shape. Here is why, step by step.

Step 1. A spinning object carries an amount of spin, which physicists call angular momentum. It depends on how fast the object turns and on how far its mass lies from the axis: mass far out counts for a lot, mass close in counts for little.

Step 2. Like ordinary momentum, this amount of spin cannot change by itself; only a twist from outside, such as friction, can change it. So when the mass moves closer to the axis, the object must turn faster to keep the same amount of spin, and when the mass moves out again, it slows down.

Step 3. The animation shows a figure skater from above and from the side. Arms out, she turns slowly; as she pulls them in she whirls three times faster, as the spin-speed bar shows, and spreading them again slows her down. The ice is so slippery that it hardly twists her at all.

Step 4. The same rule works far beyond the ice rink. Divers tuck into a ball to somersault quickly and stretch out before they reach the water. A planet moves fastest when it swings closest to the Sun, because the Sun's pull cannot change its amount of spin around the Sun.

Check yourself: a pizza chef spins a ball of dough, and it spreads out into a wide, thin disc. What happens to how fast it turns? Answer: it slows down, because its mass has moved farther from the axis.

Intermediate

What angular momentum measures

A skater spinning on ice turns faster when she pulls her arms in, although the slippery ice hardly twists her at all. Angular momentum $L$ is the amount of rotation a spinning body carries; like ordinary momentum, only something outside can change it. It depends on how fast the body turns and how its mass is spread around the axis. The rate of turning is the angular velocity $\omega$, in rad/s: one turn is $2\pi$ rad, so one turn per second is about 6.28 rad/s. Anticlockwise seen from above, the way the skater turns, counts as positive.

The spread of the mass is measured by the moment of inertia, $I=\sum m{r}^2$, in kg·m²: every mass times the square of its distance $r$ from the axis, so a mass twice as far out counts four times as much. For a body turning about a fixed axis,
$L=I\omega$
in kg·m²/s. For a small mass on a circle, $I=m{r}^2$ and $\omega =v/r$, so $L=mvr$: its momentum times its distance from the axis. Only motion around the axis counts; a mass heading straight for the axis has none.

Torque changes angular momentum

A force changes momentum; a torque changes angular momentum. The lever arm of a force is the perpendicular distance from the axis to the force's line of action. The torque of the force is the force times its lever arm, in N·m. Only torques from outside the system count:
${\tau }_{\mathrm{ext}}=\frac{dL}{dt}$
Here $dL/dt$ is the rate of change of $L$, the change $\Delta L$ divided by a short time $\Delta t$: a steady 2 N·m adds 2 kg·m²/s every second.

Forces between parts of the system come in equal and opposite pairs along one line, so their torques cancel: they can pass angular momentum from one part to another but never create it. With no external torque the total is fixed, whatever the system does to its shape:
${\tau }_{\mathrm{ext}}=0\to I_1{\omega }_1=I_2{\omega }_2$
Unlike mass, a moment of inertia changes whenever the body changes shape, and $\omega$ then changes in inverse proportion.

The working equations

Problems about spinning bodies use a short toolkit:

• $L=I\omega$
• $I=\sum m{r}^2$
• $L=mvr\text{(point mass on a circle)}$
• $I_1{\omega }_1=I_2{\omega }_2$
• $K=\frac{1}{2}I{\omega }^2=\frac{L^2}{2I}$

List every mass and its distance from the axis before and after, and work out $I_1$ and $I_2$. Check that nothing outside exerts a torque about the axis, then set $I_1{\omega }_1=I_2{\omega }_2$. Compute energies last, from $K=L^2/(2I)$: at fixed $L$ the kinetic energy is inversely proportional to $I$.

Worked example: the skater in the animation

The animation models the skater as a body with $I_0=0.72\mathrm{kg}\cdot {\mathrm{m}}^2$ plus two arms, each treated as a 3 kg mass at a distance $r$ from the axis, so $I=I_0+2m{r}^2$. She starts with her arms out, at $r=0.60\mathrm{m}$, turning at 2.5 rad/s, and pulls them in to $r=0.20\mathrm{m}$. The ice exerts no torque about the vertical axis.
$I_1=0.72+2\cdot 3\cdot 0.60^2=2.88\mathrm{kg}\cdot {\mathrm{m}}^2$
$L=I_1{\omega }_1=2.88\cdot 2.5=7.20\mathrm{kg}\cdot {\mathrm{m}}^2/s$
$I_2=0.72+2\cdot 3\cdot 0.20^2=0.96\mathrm{kg}\cdot {\mathrm{m}}^2$
${\omega }_2=L/I_2=7.20/0.96=7.5rad/s$

Her moment of inertia drops to a third, so she spins three times faster, from 0.40 to 1.19 turns per second. The gold $I$, blue $\omega$ and purple $L$ readouts end the pull on these values, and the line $2.88\times 2.5=0.96\times 7.5$ below the conservation law checks the product. Her kinetic energy rises from $\frac{1}{2}\cdot 2.88\cdot 2.5^2=9\mathrm{J}$ to $\frac{1}{2}\cdot 0.96\cdot 7.5^2=27\mathrm{J}$; her muscles supply the 18 J difference.

The green arrows are the velocities of the arm masses, $v=\omega r$. With the arms out, $v=2.5\cdot 0.60=1.5m/s$; with the arms in, $v=7.5\cdot 0.20=1.5m/s$ again, because $\omega$ triples while $r$ falls to a third. Her hands end up no faster; her whole body turns faster.

Worked example: a child jumps onto a merry-go-round

A merry-go-round, a uniform disc of 100 kg and radius 1.5 m on a freely turning axle, is at rest. A 30 kg child runs at 4 m/s along a line that just grazes its rim, jumps on at the edge and holds tight. How fast does it turn?

Running in a straight line, she already has angular momentum about the axis: $L=mvr=30\cdot 4\cdot 1.5=180\mathrm{kg}\cdot {\mathrm{m}}^2/s$. The disc has $I_d=\frac{1}{2}M{R}^2=\frac{1}{2}\cdot 100\cdot 1.5^2=112.5\mathrm{kg}\cdot {\mathrm{m}}^2$ and the child at the rim adds $m{R}^2=67.5\mathrm{kg}\cdot {\mathrm{m}}^2$, so the total is $180\mathrm{kg}\cdot {\mathrm{m}}^2$ and
$\omega =L/I=180/180=1.0rad/s$

The kinetic energy falls from $\frac{1}{2}\cdot 30\cdot 4^2=240\mathrm{J}$ to $\frac{1}{2}\cdot 180\cdot 1.0^2=90\mathrm{J}$: as when two colliding bodies stick together, angular momentum survives the landing while 150 J becomes heat. If she then walks to the centre, her share of $I$ drops to zero (treating her as a point), so $\omega =180/112.5=1.6rad/s$ and $K=\frac{1}{2}\cdot 112.5\cdot 1.6^2=144\mathrm{J}$. The extra 54 J is the work she does pulling herself inward.

Reading the bars and arrows

As the arms come in, the gold $I$ bar shrinks to a third, the blue $\omega$ bar grows threefold and the purple $L$ bar never moves: the two changing quantities trade off so that their product stays at $7.20\mathrm{kg}\cdot {\mathrm{m}}^2/s$. Plotted against each other, $\omega$ and $I$ lie on a hyperbola, $\omega =L/I$.

In the loop's first 1.5 s, arms out, she turns through 3.75 rad, about 0.6 of a turn; in the 2.4 s with her arms in, through 18 rad, almost three turns. While the arms move, the green arrows tilt, inward during the pull and outward during the release, and grow to about 1.8 m/s, because the hands also move along the radius; when the arms stop, the arrows are tangent again at 1.5 m/s.

Common mistakes

• Kinetic energy is not conserved when a body changes shape. If it were, a third of the moment of inertia would give only $\sqrt{3}$ times the spin, 4.33 rad/s instead of 7.5 rad/s; in fact $K$ triples, from 9 J to 27 J.
• Pulling the arms in puts no torque on the skater as a whole. Arms and torso push on each other, but these internal torques only move angular momentum around: the arms' share $2m{r}^2\omega$ falls from 5.40 to 1.80 kg·m²/s as the body's $I_0\omega$ rises from 1.80 to 5.40.
• The moment of inertia goes with the square of the distance. Bringing the arm masses to a third of their distance cuts their share $2m{r}^2$ from 2.16 to 0.24 kg·m², a ninth; the total falls only to a third because the body's part stays.
• Conservation needs zero external torque about the axis. Gravity and the push of the ice are vertical and exert none about the vertical axis; the slight friction of the blade does, so real spins die away.

Where you meet it

Divers and gymnasts take off with a fixed angular momentum, tuck to turn several times faster and open out before landing. A helicopter needs a tail rotor: the engine turns the main rotor one way, and with no outside torque the body would turn the other way. Collapsing stars spin up enormously: an Earth-sized core turning once a day, shrunk to a ball 10 km in radius, would have its $I$ cut by $(6400/10{)}^2\approx 4\times 10^5$ and spin about five times a second, like a typical pulsar.

Next, Gravitation and Orbits turns this law into Kepler's law of equal areas: gravity exerts no torque about the Sun, so a planet moves fastest when closest to it. Atomic Models: the Bohr Atom shows angular momentum coming in whole steps.

Advanced

Angular momentum as a cross product

Momentum measures how much straight-line motion a body carries. Angular momentum does the same job for motion around a point, and it stays constant whenever nothing outside twists the system. To define it, first choose a reference point O. A particle of mass $m$ at position $\vec{r}$ from O, moving with velocity $\vec{v}$, has momentum $\vec{p}=m\vec{v}$ and angular momentum about O
$\vec{L}=\vec{r}\times \vec{p}$

Legend

$O$	Chosen point about which $\vec{L}$ is taken.
$\vec{r}$	Position of the particle, from $O$ to the particle.	$\mathrm{m}$
$\vec{p}$	Momentum $m\vec{v}$ of the particle, along its line of motion.	$\mathrm{kg}\cdot m/s$
$\phi$	Angle between $\vec{r}$ and $\vec{p}$, marked at the particle.	$\mathrm{rad}$
$b$	Lever arm: perpendicular distance from $O$ to the line of motion, $b=r\sin \phi$.	$\mathrm{m}$
$\vec{L}$	Angular momentum about $O$: size $rp\sin \phi =pb$, out of the page by the right-hand rule.	$\mathrm{kg}\cdot {\mathrm{m}}^2/s$

In Figure 1 the vector $\vec{r}$ runs from O to the particle, and $\vec{p}$ points along the dashed line of motion. The angle $\phi$ between them is marked at the particle, from the continuation of $\vec{r}$ round to $\vec{p}$. A cross product of two vectors has the size of the first times the second times the sine of the angle between them, so $L=rp\sin \phi$.

Now look at the segment $b$. It is drawn from O to the line of motion and meets it at a right angle, so $b$ is the shortest distance from O to that line. It closes a right triangle whose longest side is $\vec{r}$. The triangle's angle at the particle lies between the direction back to O and the direction back along the line; these are the reverses of $\vec{r}$ and $\vec{p}$, so that angle is $\phi$ again. The side opposite it is $b=r\sin \phi$, and therefore
$L=rp\sin \phi =pb$
The distance $b$ is the lever arm of the momentum. In words: angular momentum is momentum times its lever arm.

The direction of $\vec{L}$ is at right angles to both $\vec{r}$ and $\vec{p}$, and the right-hand rule picks which way: curl the fingers of your right hand in the sense in which the particle goes round O, and your thumb points along $\vec{L}$. In Figure 1 the particle goes anticlockwise round O, so $\vec{L}$ points out of the page, towards you; the circled dot at O marks this. In the same way the skater, turning anticlockwise seen from above, has $\vec{L}$ pointing up.

Three consequences follow at once. First, only motion around O counts: a particle heading straight towards O or straight away from it has $\phi =0$ or $\phi =\pi$, so $L=0$. Second, $\vec{L}$ depends on the choice of O. A puck sliding freely at speed $v$ has $L=mvb$ about any point a distance $b$ from its line, the same at every instant, and $L=0$ about any point on the line. Third, $\vec{L}$ measures how fast area is swept out. In a short time $dt$ the line from O to the particle sweeps a thin triangle of area $\frac{1}{2}|\vec{r}\times \vec{v}|dt$, so the area swept per second is $|\vec{r}\times \vec{v}|/2=L/(2m)$.

Torque, internal forces and conservation

What changes $\vec{L}$? For one particle, differentiate $\vec{L}=\vec{r}\times \vec{p}$ with the product rule, using $d\vec{r}/dt=\vec{v}$ and Newton's second law, $d\vec{p}/dt=\vec{F}$:
$\frac{d\vec{L}}{dt}=\vec{v}\times m\vec{v}+\vec{r}\times \vec{F}=\vec{\tau }$
The first term is zero because the cross product of two parallel vectors vanishes, and $\vec{v}$ is parallel to $m\vec{v}$. The second term is the torque $\vec{\tau }=\vec{r}\times \vec{F}$ of the force about O. By the same geometry as in Figure 1, its size is the force times its own lever arm, the distance from O to the line along which the force acts.

For a system of particles, add this equation over all of them. Each particle feels external forces and forces from the other particles, and the question is what the internal ones do to the total.

Legend

$O$	Point about which the torques are taken.
${\vec{r}}_1$	Position of particle 1, from $O$.	$\mathrm{m}$
${\vec{r}}_2$	Position of particle 2, from $O$.	$\mathrm{m}$
$\vec{F}$	Force on particle 1 from particle 2, along the line joining them. Particle 2 feels $-\vec{F}$.	$\mathrm{N}$
$b$	Lever arm of both forces: distance from $O$ to the line joining the particles.	$\mathrm{m}$
${\vec{\tau }}_1$	Torque of $\vec{F}$ on particle 1 about $O$: size $Fb$, into the page.	$\mathrm{N}\cdot \mathrm{m}$
${\vec{\tau }}_2$	Torque of $-\vec{F}$ on particle 2 about $O$: size $Fb$, out of the page.	$\mathrm{N}\cdot \mathrm{m}$

Figure 2 shows one internal pair. Particle 1 feels the force $\vec{F}$ from particle 2, and particle 2 feels $-\vec{F}$ from particle 1: equal and opposite, by Newton's third law. Both forces act along the dashed line joining the two particles. This is the strong form of the third law, and gravity and contact forces obey it.

Look at the lever arms. Both forces lie on the same line, and that line is one distance $b$ from O, so both forces have the lever arm $b$, whatever the positions ${\vec{r}}_1$ and ${\vec{r}}_2$ are. Their torques therefore have the same size, $Fb$. Their senses are opposite: $\vec{F}$ turns particle 1 clockwise round O, so ${\vec{\tau }}_1$ points into the page, while $-\vec{F}$ turns particle 2 anticlockwise, so ${\vec{\tau }}_2$ points out of it. The two torques cancel.

The argument works about any point and for every pair, so all internal torques drop out of the sum. What remains is
$\frac{d\vec{L}}{dt}={\vec{\tau }}_{\mathrm{ext}}$
where $\vec{L}$ is now the total angular momentum and ${\vec{\tau }}_{\mathrm{ext}}$ the total torque of the external forces. It is the rotational twin of $d\vec{p}/dt={\vec{F}}_{\mathrm{ext}}$. It holds for any collection of particles, rigid or not, and it also holds about the centre of mass, even when that point accelerates.

With no external torque, $\vec{L}$ stays constant, however the system changes its shape. A diver in flight shows this: gravity acts at her centre of mass, so it has no lever arm about that point and no torque, and she keeps her angular momentum all the way to the water.

A central force: equal areas

The first payoff is in orbits. A central force is one that always points along the line to a fixed point. That line passes through the point, so the force has no lever arm about it and no torque.

Legend

$S$	The Sun, the fixed point the force always points to.
$r$	Distance from $S$ to the planet; this line sweeps out each sector.	$\mathrm{m}$
$\vec{F}$	Force on the planet, always along the line to $S$.	$\mathrm{N}$
$\vec{v}$	Velocity of the planet, tangent to its orbit.	$m/s$
$A_1$	Area swept by the line from $S$ to the planet in a time $\Delta t$ near $S$.	${\mathrm{m}}^2$
$A_2$	Area swept in the same time $\Delta t$ far from $S$; equal to $A_1$.	${\mathrm{m}}^2$

In Figure 3 the force $\vec{F}$ on the planet lies along the line $r$ to the Sun $S$, so $\vec{r}\times \vec{F}=0$ and the planet's angular momentum about $S$ never changes. That has two consequences. Its direction is fixed, and because $\vec{L}$ is at right angles to both $\vec{r}$ and $\vec{v}$, the orbit stays in one plane. Its size is fixed, so the line from $S$ to the planet sweeps area at the steady rate $L/(2m)$ found in the first section.

Now compare the two shaded sectors, each swept in the same time $\Delta t$. Near $S$ the planet is fast but the line $r$ is short, so $A_1$ is a short, wide fan. Far from $S$ the planet is slow but $r$ is long, so $A_2$ is a long, narrow slice. The area swept per second, $\frac{1}{2}vb$, is the same in both places, so $A_1=A_2$, however different the shapes look. This is Kepler's second law, equal areas in equal times, which he published in 1609, nearly eighty years before Newton showed why it holds. The argument never uses the inverse-square law, so equal areas hold for every central force.

Spinning bodies: what the graphs show

Most everyday rotation is a body turning about a fixed axis, and there angular momentum takes a simple form. A small mass $m$ at distance $r$ from the axis moves on a circle at speed $\omega r$, where $\omega$ is the angular velocity in rad/s. Its momentum $m\omega r$ is at right angles to the radius, so its lever arm about the axis is $r$, and its angular momentum along the axis is $m{r}^2\omega$. Adding all the masses,
$L=\sum m{r}^2\omega =I\omega$
where $I=\sum m{r}^2$ is the moment of inertia. Unlike mass, $I$ changes when the body changes shape.

The animation models the skater as a torso with $I_0=0.72\mathrm{kg}\cdot {\mathrm{m}}^2$ plus two arm masses of $m=3\mathrm{kg}$ each at distance $r$ from the axis, so $I=I_0+2m{r}^2$. The ice exerts no torque about the vertical axis, so $L$ is fixed and $\omega =L/I$.

The expert graphs show exactly this. $L(t)$ is a flat line at $7.20\mathrm{kg}\cdot {\mathrm{m}}^2/s$: its slope is the external torque, which is zero. $I(t)$ falls from 2.88 to 0.96 kg·m² as $r$ goes from 0.60 to 0.20 m, and $\omega (t)=L/I(t)$ rises from 2.5 to 7.5 rad/s. It does not rise evenly. It climbs slowly at first and fastest about two thirds of the way through the pull, because the same drop in $I$ raises $L/I$ more when $I$ is already small; at the very end the arms slow to a stop.

Halfway in, at $r=0.40\mathrm{m}$, $I=0.72+2\cdot 3\cdot 0.40^2=1.68\mathrm{kg}\cdot {\mathrm{m}}^2$ and $\omega =7.20/1.68=4.29rad/s$, just over a third of the way from 2.5 to 7.5 rad/s.

The $K(t)$ graph has exactly the shape of $\omega (t)$. The reason is $K=\frac{1}{2}I{\omega }^2=\frac{1}{2}L\omega$: with $L$ fixed, $K$ is simply $\omega$ times $L/2=3.6\mathrm{J}\cdot \mathrm{s}$. That gives 9 J at 2.5 rad/s and 27 J at 7.5 rad/s. The angular momentum stays the same while the kinetic energy triples.

Who pays for the extra energy

The kinetic energy rose by $27-9=18\mathrm{J}$, yet no external torque acted. The energy must come from the skater herself, through the work her arms do as they pull in.

Seen from the rink, each arm mass moves on a circle, so it needs a centripetal pull $m{\omega }^{2}r$ towards the axis, and her arms supply it. While the masses move inward, each pull points along their motion, so it does positive work. To count that work, keep $L$ fixed and write $K=L^2/(2I)$. Because $I=I_0+2m{r}^2$, a small change $dr$ of the radius changes $I$ by $dI=4mrdr$. Then, using $L/I=\omega$,
$dK=-\frac{L^2}{2I^2}dI=-2m{\omega }^{2}rdr$
As the arms come in, $dr$ is negative and $dK$ positive. The right-hand side is exactly the work of the two pulls: $2m{\omega }^{2}r$ times the inward step $|dr|$. Adding up the steps from $I_1$ to $I_2$,
$W=\Delta K=\frac{L^2}{2}\left(\frac{1}{I_2}-\frac{1}{I_1}\right)$
which for the skater is $25.92\cdot (1/0.96-1/2.88)=18\mathrm{J}$. Letting the arms out again takes the same 18 J back out of the spin.

The $W$ readout counts this work as it is done. During the pull it runs slightly ahead of the rise in $K$, by at most 0.60 J, because the work also pays for the arm masses' motion towards the axis: kinetic energy $m{\dot{r}}^2$ for the pair, where $\dot{r}$ is their radial speed. Both reach 18 J when the arms stop.

Seen from her own turning frame, the same story is told with different forces. There her arms do the same work against the centrifugal force. The arm masses moving inward also feel the Coriolis force, which acts sideways and pushes them ahead of the torso, in the direction of the spin. Held back at the shoulders, they press forward on them and spin the torso up. The torso's own angular momentum $I_0\omega$ grows at the rate $I_{0}d\omega /dt$, so that is the internal torque. At mid-pull the slope of the $\omega (t)$ graph is about $5.5rad/{\mathrm{s}}^2$, and the torque is $I_{0}d\omega /dt\approx 0.72\cdot 5.5\approx 4.0\mathrm{N}\cdot \mathrm{m}$.

When the axis turns: precession

So far $\vec{L}$ kept its direction. But $\vec{L}$ is a vector, and a torque at right angles to it changes its direction rather than its size. This is what keeps a spinning gyroscope from toppling.

Legend

$d$	Distance from the pivot to the line of the weight.	$\mathrm{m}$
$mg$	Weight of the wheel, acting at its centre.	$\mathrm{N}$
$\vec{L}$	Spin angular momentum of the wheel, along the axle.	$\mathrm{kg}\cdot {\mathrm{m}}^2/s$
$\vec{\tau }$	Torque of the weight about the pivot, $\tau =mgd$: horizontal and at right angles to $\vec{L}$.	$\mathrm{N}\cdot \mathrm{m}$
$d\vec{L}$	Change of $\vec{L}$ in a short time $dt$: $d\vec{L}=\vec{\tau }dt$.	$\mathrm{kg}\cdot {\mathrm{m}}^2/s$
${\vec{L}}^{\prime }$	Angular momentum after $dt$: $\vec{L}+d\vec{L}$.	$\mathrm{kg}\cdot {\mathrm{m}}^2/s$
$d\beta$	Angle through which $\vec{L}$ turns in $dt$: $d\beta =\tau dt/L$.	$\mathrm{rad}$
$\Omega$	Rate at which the axle swings round, $\Omega =d\beta /dt=\tau /L$.	$rad/s$

Figure 4 shows the standard example: a wheel spinning fast on a horizontal axle that is held at one end only. Its spin angular momentum $\vec{L}$ points along the axle. On the left, seen from the side, the weight $mg$ acts at the wheel's centre, a distance $d$ from the pivot. Its torque about the pivot has the size $\tau =mgd$ and is horizontal, at right angles to the axle. On the right, seen from above, $\vec{\tau }$ stands at right angles to $\vec{L}$.

Now apply $d\vec{L}/dt=\vec{\tau }$. In a short time $dt$ the angular momentum gains $d\vec{L}=\vec{\tau }dt$, parallel to the torque and so sideways to $\vec{L}$. The new vector ${\vec{L}}^{\prime }=\vec{L}+d\vec{L}$ has the same length as $\vec{L}$ to first order in $dt$, but it has swung through the small angle $d\beta =dL/L=\tau dt/L$. (The figure draws a large $dt$ so that the angle can be seen.) The axle follows $\vec{L}$, so instead of falling it swings round horizontally at the rate
$\Omega =\frac{d\beta }{dt}=\frac{\tau }{L}$
In vector form, $\vec{\tau }=\vec{\Omega }\times \vec{L}$, with $\vec{\Omega }$ pointing up for the anticlockwise swing in Figure 4.

The same holds for a top or a gyroscope with spin angular momentum $I\omega$ and its centre of mass a distance $d$ from the pivot along its axis. If the axis is tilted from the vertical, the lever arm of the weight and the horizontal part of $\vec{L}$ both carry the sine of the tilt, and that factor cancels. So at any tilt
$\Omega =\frac{mgd}{I\omega }\text{(gyroscopic precession)}$

A bicycle wheel with $I=0.25\mathrm{kg}\cdot {\mathrm{m}}^2$ and a mass of 2.5 kg is spun at 40 rad/s and supported at one end of its axle, 0.20 m from the wheel's centre. It has $L=0.25\cdot 40=10\mathrm{kg}\cdot {\mathrm{m}}^2/s$, and its weight exerts the torque $\tau =2.5\cdot 9.81\cdot 0.20\approx 4.9\mathrm{N}\cdot \mathrm{m}$. It does not fall: it swings round at $\Omega =4.9/10\approx 0.49rad/s$, once every 13 s. The Earth is a slow top: the Sun and the Moon pull on its equatorial bulge, and its axis precesses once in about 26 000 years.

One more consequence of $\vec{L}$ being a vector: it is parallel to $\vec{\omega }$ only for rotation about a principal axis of the body, such as an axis of symmetry. A wheel mounted slightly tilted on its axle has $\vec{L}$ off the axle. As the wheel turns, $\vec{L}$ sweeps round the axle and so keeps changing, and by $d\vec{L}/dt=\vec{\tau }$ the bearings must keep supplying the torque. That is why car wheels are balanced.

Worked problem: a spin that dies away

Real ice is not frictionless. Suppose the blade exerts a steady friction torque $\tau =0.30\mathrm{N}\cdot \mathrm{m}$ against the spin, starting from the animation's $L_0=7.20\mathrm{kg}\cdot {\mathrm{m}}^2/s$. How long does the spin last, and how many turns does she make with her arms held in ($I=0.96\mathrm{kg}\cdot {\mathrm{m}}^2$), and with them held out ($I=2.88\mathrm{kg}\cdot {\mathrm{m}}^2$)?

How long: now $dL/dt=-\tau$, so $L$ falls at a steady rate, $L(t)=7.20-0.30t$, and reaches zero after $T=L_0/\tau =24\mathrm{s}$. Her shape does not enter: in both cases the flat $L(t)$ graph would become the same falling straight line.

How many turns: the angle does depend on her shape, because $\omega =L(t)/I$. The angle turned is the integral of $\omega$ over the 24 s:
$\theta ={\int }_0^T\frac{L_0-\tau t}{I}dt=\frac{L_0^2}{2\tau I}$
This is the area of a triangle under the falling line, with height $L_0/I$ and base $L_0/\tau$. With her arms in, $\theta =51.84/(2\cdot 0.30\cdot 0.96)=90\mathrm{rad}$, about 14.3 turns. With her arms out, it is only 30 rad, about 4.8 turns.

Energy confirms it. The friction torque does the work $\tau \theta$ against the spin, and $0.30\cdot 90=27\mathrm{J}$ and $0.30\cdot 30=9\mathrm{J}$ are exactly her two starting kinetic energies. Pulling the arms in does not make the spin last longer; it packs three times as many turns into the same 24 s.

Limits of validity

The skater model is idealised: real arms are spread-out masses, not two points, and friction and air drag never quite vanish. The conservation law itself is sturdier than any model. The proof above used the strong form of Newton's third law, with internal forces along the lines joining the particles. Gravity and contact forces obey it; magnetic forces between moving charges need not. Yet nothing is lost, because the electromagnetic field carries angular momentum of its own, and the law holds once the field's share is counted. Circularly polarised light, for example, hands angular momentum to whatever absorbs it.

The precession formula assumes $\Omega \ll \omega$, so that the spin carries almost all of the angular momentum; for the bicycle wheel the ratio is about 1 %. Spin the wheel at only 10 rad/s and $\Omega$ reaches 2.0 rad/s, a fifth of $\omega$. The axle then visibly dips and nods as it goes round, a motion called nutation, instead of gliding smoothly.

At the atomic scale angular momentum is quantised: its component along any axis takes only values spaced by the reduced Planck constant, $\hslash =1.055\times 10^{-34}\mathrm{J}\cdot \mathrm{s}$. The skater's $7.20\mathrm{kg}\cdot {\mathrm{m}}^2/s$ is about $7\times 10^{34}$ such steps, so the graininess is invisible. For an electron in an atom it is everything. The electron also carries an intrinsic angular momentum, its spin, with components $\pm \hslash /2$, which is not the rotation of any little ball. Special relativity keeps $\vec{L}=\vec{r}\times \vec{p}$, with the relativistic momentum, and keeps its conservation.

Where this leads

Why is angular momentum conserved at all? Because space has no preferred direction. By Noether's theorem every continuous symmetry of the laws of physics brings a conserved quantity: the sameness of space from place to place gives momentum, the sameness of time gives energy, and the sameness of all directions gives angular momentum.

The law reaches far. In Gravitation and Orbits it explains equal areas, and the term $L^2/(2m{r}^2)$ in the radial energy acts as a centrifugal barrier that keeps planets from falling into the Sun. In Atomic Models: the Bohr Atom, the rule $L=n\hslash$ fixes the energy levels of hydrogen. Spin, counted in units of $\hslash$, separates the matter particles of Particles and Forces: the Standard Model, with spin 1/2, from the force carriers, with spin 1. And the aligned spins of electrons make iron magnetic, as Magnetic Fields of Magnets and Currents describes.