Conservation of angular momentum

A spinning body carries angular momentum $L=I\omega$, the rotational version of $p=mv$. The big idea is a conservation law: when the net external torque on a system is zero, its total angular momentum cannot change.

The classic case is a figure skater. She pulls her arms in, her moment of inertia drops, and she spins faster, with no outside torque. The product $I\omega$ is the same before and after, so a smaller $I$ forces a larger $\omega$: $I_1\omega_1=I_2\omega_2$. The same logic runs collapsing stars, spinning stools, and rotational collisions.

Four habits carry the topic: conserve the product $I\omega$ (not $\omega$ by itself), remember that kinetic energy is not conserved along with it, check that the net external torque is really zero, and keep the sign of the angular momentum vector.

Torque changes angular momentum: $\sum\tau_{\text{ext}}=\dfrac{dL}{dt}$. So if the net external torque is zero, $\dfrac{dL}{dt}=0$ and $L$ is constant. That is the only condition.

Forces inside the system do not count. When the skater pulls her arms in, her muscles pull inward along the radius; those internal forces exert no net torque about the spin axis, so $L$ holds. But a brake on a rim, a foot dragging on the ground, axle friction, or air drag all act from outside at a lever arm. Each is an external torque, and each changes $L$.

• $L=I\omega$ is conserved when $\sum\tau_{\text{ext}}=0$.
• Then $I_1\omega_1=I_2\omega_2$ across any change in the mass distribution.

Angular momentum is conserved only when the net external torque is zero. Internal forces never change it; friction, brakes, and outside pushes do.

The most common slip is to read "angular momentum is conserved" as "the spin rate is conserved." It is the product $I\omega$ that holds, not $\omega$. When $I$ goes down, $\omega$ must go up by the same factor.

Take a skater with $I_1=6$ kg$\cdot$m$^2$ spinning at $\omega_1=2$ rad/s. She pulls in to $I_2=2$ kg$\cdot$m$^2$. With no external torque, $I_1\omega_1=I_2\omega_2$, so

• $\omega_2=\dfrac{I_1\omega_1}{I_2}=\dfrac{(6)(2)}{2}=6$ rad/s.

Cutting the inertia to a third tripled the spin. Two related slips: leaving $\omega$ unchanged, and inverting the ratio (using $I_2/I_1$ instead of $I_1/I_2$). When two bodies couple, the same rule applies, but the final inertia is the sum of both, so divide $L$ by the total.

Conservation fixes the product $I\omega$, so a smaller moment of inertia means a faster spin: $\omega_2=I_1\omega_1/I_2$. Do not hold $\omega$ fixed or invert the ratio.

Conserving $L$ does not conserve the rotational kinetic energy $K=\tfrac{1}{2}I\omega^2$. For a single body spinning about a fixed axis, $L$ and $\omega$ point the same way, so you can write $K=\tfrac{1}{2}L\omega$ (always positive): $L$ is fixed but $\omega$ changes, so $K$ changes too.

For the skater above, $K_1=\tfrac{1}{2}(6)(2)^2=12$ J and $K_2=\tfrac{1}{2}(2)(6)^2=36$ J. The energy tripled. That energy is real work: she pulls inward against the outward pull of the spinning masses. Angular momentum is not an energy bank that supplies it.

$K=\tfrac{1}{2}L\omega$ changes when $\omega$ does, so kinetic energy is not conserved when $L$ is. In a pull-in it rises (work is done); in an inelastic collision it falls (energy is lost).

Rotational collisions use the same law. When a lump strikes a disk and sticks, or two disks couple, there is no external torque during the brief contact, so angular momentum is conserved: $L_{\text{before}}=L_{\text{after}}$. Kinetic energy is not, because the bodies deform and stick. Never solve a sticking collision with energy conservation.

A particle hitting a rim has angular momentum $L=mvR$ about the axis. After it sticks, the moment of inertia is the disk's plus $mR^2$, and $\omega_f=L/I_{\text{tot}}$.

Angular momentum is also a vector along the axis, so direction matters. Counterclockwise and clockwise spins carry opposite signs. When opposite spins couple, add the signed angular momenta and divide by the total inertia; equal and opposite spins cancel to zero. Adding magnitudes, or dropping the sign, is the trap.

Collisions conserve $L$ but not $K$: use $L=mvR$ and divide by the total inertia, never energy conservation. Keep the sign of $L$, since opposite spins can cancel.

When $L$ is conserved, the wrong turns are predictable: freezing $\omega$ instead of the product $I\omega$, assuming energy rides along, overlooking an external torque, or ignoring sign.

• Holding $\omega$ fixedReading conservation of $L$ as conservation of the spin rate, or inverting the inertia ratio. Fix: $\omega_2=I_1\omega_1/I_2$.
• K conserved tooAssuming kinetic energy is conserved along with $L$. Fix: $K=\tfrac{1}{2}L\omega$ rises in a pull-in and falls in a collision.
• External torqueApplying conservation when a brake, friction, or outside push acts. Fix: $L$ is conserved only if $\sum\tau_{\text{ext}}=0$.
• Dropped signTreating $L$ as a scalar, so opposite spins add instead of cancel. Fix: sum the signed $L$, divide by the total inertia.

Conserve the product $I\omega$, remember that the energy is not conserved with it, check for an external torque, and keep the sign. The whole topic comes down to those four habits.