Conservation of Angular Momentum: Ice Skater Spin and Gyroscope Precession

Watch a figure skater go into a spin with her arms stretched wide, then pull them in tight against her chest. She doesn't push off the ice again, doesn't get a second running start — and yet she visibly speeds up, sometimes dramatically. Nothing pushed her. The rule behind that speed-up is the same one that keeps a spinning bicycle wheel from simply toppling over when you hang it off one end of its axle: conservation of angular momentum, $L = I\omega$, the rotational twin of ordinary momentum conservation.

This page has two interactive simulators built around that one idea. The first hands you a single control — hold a button down and watch a skater's arms sweep inward while her spin rate climbs, with a second display right beside her keeping an honest scorecard of what conservation does and doesn't guarantee. The second spins up a gyroscope wheel and traces precession — the slow, steady drift of a fast-spinning object's axis under gravity — which turns out to be angular momentum conservation playing out in a completely different geometry. If you've already read our Torque, Rotational Motion and Moment of Inertia post, this is the natural next question: that page asked what starts something spinning, this one asks what happens once nothing is trying to change the spin at all.

What Is Angular Momentum? The Formula L = Iω Explained

Angular momentum ($L$) is the rotational equivalent of ordinary, linear momentum. Where linear momentum is $p = mv$ — how much "oomph" a moving mass carries in a straight line — angular momentum is:

$L = I\omega$

Here $I$ is the moment of inertia — how the object's mass is distributed relative to its spin axis, the same $I = kMR^2$ quantity from our torque and rotational motion post — and $\omega$ is the angular velocity, in radians per second. A quick sanity check: a 2 kg, 0.5 m solid disk ($I = \tfrac12 MR^2 = 0.25$ kg·m²) spinning at 10 rad/s carries $L = 0.25 \times 10 = 2.5$ kg·m²/s of angular momentum — small numbers for a small object, exactly as you'd expect.

How Is Angular Momentum Different From Linear Momentum?

Linear momentum answers "how hard is it to stop this thing from moving in a straight line?" Angular momentum answers the same question for spinning: how hard is it to stop, or speed up, or slow down, this thing's rotation? Both obey a version of the same law. Absent an outside push, momentum doesn't change; absent an outside twist — a torque — angular momentum doesn't change either. Formally, torque is to angular momentum what force is to linear momentum: $\tau = dL/dt$, the rotational version of Newton's second law. No net torque means $dL/dt = 0$, and $L$ is locked in place.

Why Does a Figure Skater Spin Faster When She Pulls Her Arms In?

Once a skater is spinning on the ice, the vertical axis she's spinning around is nearly torque-free — friction with the ice is small, nowhere near strong enough over the few seconds of a spin to meaningfully change her total angular momentum. So $L$ stays essentially fixed. What she can change, just by moving her own arms, is $I$ — and since $L = I\omega$ has to hold, shrinking $I$ forces $\omega$ to grow to compensate. She isn't summoning energy from nowhere, either: pulling her arms in against the outward pull she feels in her own spinning frame takes real muscular work, and that work is exactly where the extra rotational kinetic energy comes from.

Interactive Ice Skater Spin Simulator

Press and hold the button below — HOLD TO TUCK — and watch the skater's arms sweep in over about a second while her spin rate climbs on the readout above her. Underneath, two rectangles share one vertical axis: both stand exactly as tall as the current spin rate ω. The left one is as wide as the moment of inertia $I$, so its area is $L = I\omega$ — and as the arms come in, that rectangle grows taller and narrower while its area holds dead still, because $L$ is the thing conservation actually promises. The right one keeps a fixed width of $L/2$, so its area is $KE = \tfrac12 L\omega$ — and with no room to shrink sideways, that area is forced to grow as ω climbs. Same height, same starting instant, two different fates: that's conservation of angular momentum and the cost of achieving it, in one picture. The Resting arms slider sets where she settles once you let go, and Starting spin fixes how fast she's turning arms-out — which pins $L$ for everything else on screen, including the worked example below: a fixed core plus two 3 kg arms sweeping between 0.15 m tucked and 0.85 m extended.

Is Kinetic Energy Conserved When a Skater Pulls Her Arms In?

No — and this is the detail that trips up almost everyone meeting conservation of angular momentum for the first time. $L$ stays fixed by definition in this problem; $KE$ does not, and there's no contradiction in that at all. Write kinetic energy in terms of the conserved quantity instead of expanding it the usual way:

$KE = \frac{1}{2}I\omega^2 = \frac{1}{2}(I\omega)\omega = \frac{1}{2}L\omega$

With $L$ pinned, $KE$ is directly proportional to $\omega$ — and $\omega$ itself is inversely proportional to $I$, since $\omega = L/I$. So $KE$ scales as $1/I$ exactly the way $\omega$ does: shrink the moment of inertia by some factor and both the spin rate and the kinetic energy grow by that same factor, never by its square and never by some unrelated amount. In the model on this page that factor is $5.135/0.935 \approx 5.49$, so tucking from fully extended to fully tucked doesn't just spin the skater up 5.49× — it hands her 5.49× the rotational kinetic energy too, from 10.27 J to 56.40 J.

That extra 46.13 J has to come from somewhere, and it isn't free. Pulling a mass inward while it's spinning means working against the outward pull the skater feels in her own rotating frame — the same pull that flings a loose object off a merry-go-round. Her muscles supply that work, and in this idealised model it lands entirely as rotational kinetic energy, since nothing else is there to absorb it. Go back to the simulator above and hold the button down: the right-hand rectangle is drawing exactly this quantity. Its width can't move — $L/2$ is fixed the moment you set a starting spin — so growth is the only thing left for it to do, and every bit of height it gains above the dashed line is a joule her muscles just spent.

What Is the Law of Conservation of Angular Momentum?

The law of conservation of angular momentum states that the total angular momentum of a system stays constant unless an external torque acts on it. It's one of physics' deeper results, tied through Noether's theorem to the fact that the laws of physics don't care which direction you happen to be facing — but you don't need any of that machinery to use it day to day. In practice it means:

$I_1 \omega_1 = I_2 \omega_2 \quad \text{(no external torque acts between state 1 and state 2)}$

That one line solves a surprising range of problems: a skater pulling in her arms, a diver curling into a tuck mid-somersault, a collapsing star spinning up as it shrinks, even a planet's rotation as its shape slowly settles. Every one of them is the same equation wearing different labels.

When Is Angular Momentum Conserved — and When Isn't It?

Angular momentum is conserved for a system with zero net external torque. Internal forces — a skater's own muscles, a diver's own joints — can redistribute mass and change $I$ all they like without breaking the law, because internal forces come in equal-and-opposite pairs that cancel each other's torque within the system. What does break conservation is a torque from outside: ice friction, given long enough, eventually spins a skater down to a stop, and that's exactly an external torque draining $L$ away.

Why Doesn't a Spinning Gyroscope Fall Over? What Causes Precession?

Hang a stationary bicycle wheel off one end of its axle and it does exactly what you'd guess: gravity's torque about the support point tips it over and it falls. Spin that same wheel up fast first, then hang it the same way, and something strange happens — it doesn't fall. Instead, the whole axle sweeps slowly around in a horizontal circle, staying roughly level the entire time. That slow sweep is precession.

Here's the resolution. Gravity's torque never stops trying to tip the wheel — that part hasn't changed. But once the wheel carries substantial spin angular momentum $L = I\omega$ along its axle, that torque acts perpendicular to $L$ rather than trying to shrink it directly. A torque perpendicular to a vector doesn't change that vector's length, only its direction — which is the rotational cousin of the argument that closes our centripetal force calculator post: a force perpendicular to velocity bends a straight path into a circle without ever changing speed. Here, a torque perpendicular to angular momentum bends the spin axis into a circle without ever changing the spin rate. That circling of the axis is precession — instead of toppling, the wheel traces out a slow, steady cone, or, seen from above, a circle.

How Do You Derive the Precession Rate Formula Ω = τ/(Iω)?

Newton's second law for rotation says $d\vec{L}/dt = \vec\tau$. When $\vec\tau$ is perpendicular to $\vec L$ and roughly constant in magnitude — true for a fast-spinning gyroscope, where spin angular momentum dominates completely — $\vec L$'s tip traces a circle instead of growing or shrinking. In a short time $dt$, the vector sweeps through a small angle $d\phi \approx \dfrac{|\vec\tau|\,dt}{|\vec L|}$, arc length over radius, exactly like ordinary circular motion. Divide through by $dt$ and out comes the precession rate:

$\Omega = \frac{\tau}{L} = \frac{\tau}{I\omega}$

with $\tau = Mgd$ for a wheel of mass $M$ whose centre of mass sits a distance $d$ from the pivot. Look at the shape of the result: precession rate is directly proportional to torque and inversely proportional to spin angular momentum. Spin the wheel faster with nothing else changed and it precesses slower — a fast gyroscope is a stubborn, torque-resistant one. This approximation (the "fast top," or steady-precession, approximation) needs $\omega$ comfortably larger than $\Omega$, which is true across most of the simulator's range below. At its slowest-spin, largest-offset corner, though, the ratio falls to about 12, and the simulator itself flags that corner with a caution: a real gyroscope there would visibly nutate.

Interactive Gyroscope Precession Simulator

This simulator draws the physics as vectors instead of asking you to infer them. L points outward along the axle, τ is gravity's torque acting at the wheel, and dL sits at the tip of L, showing which way that torque is about to push it next — sideways, never down. Flip the Wheel state toggle to Wheel at rest and that same τ arrow, unchanged in length or direction, has nothing left to steer: with no spin there's no L, so gravity just tips the wheel over and it falls, the way any unspinning weight on an arm would. Flip back to Wheel spinning and watch that identical torque get redirected into a slow sweep instead — the tip of L traces a level circle marked by a dashed guide, while a tick on the rim shows the wheel's own spin. That tick is deliberately slowed for the eye: the default spin rate is close to 24 revolutions a second, far too fast for a screen refreshing 60 times a second to draw honestly, so only the rim animation is scaled back — the drift around the circle itself always plays at its true, unscaled rate. Drag Spin rate ω and the L arrow visibly lengthens or shortens, showing directly why a faster wheel precesses slower: the same sideways push turns a longer arrow through a smaller angle. Drag Pivot arm d and it's the τ arrow that changes length instead, since moving the wheel's weight further from the pivot is exactly what increases gravity's torque.

Worked Examples: Angular Momentum Conservation Problems