Conservation of angular momentum

From torque to a conservation law. Angular momentum $\vec L$ and net external torque $\vec\tau_{ext}$ on a system are linked by the rotational analogue of Newton's second law, already met under torque and angular momentum:

where: $\vec L$ is the system's total angular momentum (kg m²/s) about a chosen axis, $\vec\tau_{ext}$ is the net external torque about that same axis, and $t$ is time. If $\vec\tau_{ext}$ is zero at every instant, the rate of change of $\vec L$ is zero, and a quantity whose rate of change is zero simply does not change:

This is the law of conservation of angular momentum: in the absence of a net external torque, the total angular momentum of a system stays constant in both magnitude and direction. It is exactly the rotational counterpart of "linear momentum is conserved when net external force is zero."

The working form. For a rigid body — or any system whose moment of inertia about the rotation axis can change internally, such as a person on a frictionless stool pulling their arms in — the law is used as

where: $I$ is the moment of inertia about the axis (kg m²), and $\omega$ is the angular velocity (rad/s) at that instant. Read backwards, this equation is doing all the physical work: if $I$ drops (mass pulled closer to the axis), $\omega$ must rise in exactly the same proportion so the product stays fixed, and vice versa. That is the entire mechanism behind a faster-spinning skater with arms tucked in, and a slower-spinning skater with arms stretched wide.

Energy is not part of the deal. Angular momentum staying constant does not mean kinetic rotational energy stays constant. When the skater pulls her arms in, her rotational kinetic energy actually increases — the extra energy is supplied by the muscular work she does pulling her arms inward against the effective outward pull needed to keep them moving in a circle of shrinking radius. The law guarantees $L$ is fixed; it says nothing about $KE=\tfrac12 I\omega^2$, which can rise or fall depending on how the system reshapes itself.

A student sits on a frictionless rotating stool with arms outstretched, holding a $2\ \text{kg}$ dumbbell in each hand at $0.8\ \text{m}$ from the rotation axis. The student and stool (excluding dumbbells) have moment of inertia $2.5\ \text{kg m}^2$ and are spinning at $2\ \text{rad/s}$. The student pulls the dumbbells in to $0.2\ \text{m}$ from the axis. Find the new angular velocity, and check what happens to the rotational kinetic energy.

Step 1. Total initial moment of inertia. Each dumbbell contributes $mr^2=(2)(0.8)^2=1.28\ \text{kg m}^2$, so both together give $I_{dumbbells,1}=2(1.28)=2.56\ \text{kg m}^2$. Adding the student's own $2.5\ \text{kg m}^2$: $I_1=2.5+2.56=5.06\ \text{kg m}^2$.

Step 2. Total final moment of inertia. Each dumbbell now contributes $(2)(0.2)^2=0.08\ \text{kg m}^2$, so both give $I_{dumbbells,2}=2(0.08)=0.16\ \text{kg m}^2$. Total: $I_2=2.5+0.16=2.66\ \text{kg m}^2$.

Step 3. Apply conservation of angular momentum, since the only torques are internal (the student's own muscular effort) and the stool axis is frictionless: $I_1\omega_1=I_2\omega_2$, so $(5.06)(2)=(2.66)\,\omega_2$, giving $\omega_2=\dfrac{10.12}{2.66}=3.80\ \text{rad/s}$.

Step 4. Check the energy, since $L$ conservation does not imply $KE$ conservation. Initial $KE_1=\tfrac12 I_1\omega_1^2=\tfrac12(5.06)(2)^2=10.12\ \text{J}$. Final $KE_2=\tfrac12 I_2\omega_2^2=\tfrac12(2.66)(3.80)^2\approx19.21\ \text{J}$. Kinetic energy nearly doubles even though $L$ stayed fixed — the extra energy came from the student's own muscles pulling the dumbbells inward.

The student's spin rate increases from $2\ \text{rad/s}$ to $3.80\ \text{rad/s}$, while rotational kinetic energy rises from about $10.1\ \text{J}$ to about $19.2\ \text{J}$.

Competitive stretch

A uniform disc of moment of inertia $0.4\ \text{kg m}^2$ spins freely at $6\ \text{rad/s}$ about its central axis. An identical stationary disc ($0.4\ \text{kg m}^2$) is gently dropped on top, coaxially, and friction between the two brings them to a common angular velocity — an internal effect, so no external torque acts about the shared axis. Initial angular momentum comes only from the first disc: $L_1=I_1\omega_1=(0.4)(6)=2.4\ \text{kg m}^2/\text{s}$. In the final state both discs share $\omega_f$ with combined moment of inertia $I_1+I_2=0.8\ \text{kg m}^2$, so $2.4=(0.8)\omega_f$, giving $\omega_f=3\ \text{rad/s}$ — exactly half the initial rate, since the total moment of inertia doubled while $L$ stayed fixed.

• Assuming kinetic energy is also conserved whenever angular momentum is conserved — as the worked example shows, $KE$ generally changes even while $L$ stays fixed.
• Applying $I_1\omega_1=I_2\omega_2$ when an external torque is actually present (a motor driving the rotation, or friction at the bearing itself) — the law only holds when net external torque about the chosen axis is genuinely zero.
• Forgetting to include every contribution to $I$ on each side of the equation — both the person's own moment of inertia and each held mass, or all colliding bodies in a merger.
• Mixing up which axis "no external torque" refers to — angular momentum about one axis can be conserved even while angular momentum about a different axis through the same body is changing.

Ask the tutor: want to see how this same law gives Kepler's second law for a planet or comet orbiting the Sun, or work through another stool-and-masses variant with numbers?