11.5 Conservation of Angular Momentum
The Most Extreme Spinners in the Universe
A massive star collapses at the end of its life. Its core --- once the size of the Sun --- crushes down to a neutron star roughly 10 km across. The radius decreases by a factor of about 1000. The star was rotating before the collapse, perhaps once every few weeks. Afterward, the neutron star spins hundreds of times per second. Its angular velocity has increased by a factor of roughly 1,000,000.
No external torques caused this speed-up. Nothing pushed the star to spin faster. The collapse was entirely internal --- gravity pulling the star's own mass inward. And yet the result is the most extreme spinning object in the known universe: a pulsar, rotating so fast that if you stood on its surface, you would be flung off at a significant fraction of the speed of light.
How can an object spin a million times faster without any external torque acting on it?
The answer is a conservation law --- the rotational counterpart of the momentum conservation you learned in Section 8.3. It is the centerpiece of this chapter, and it will explain figure skaters, collapsing stars, and turntable collisions with a single principle.
Prediction
Before you read on: A turntable is spinning freely on a frictionless bearing. You hold a heavy disk directly above the turntable and drop it straight down (no sideways push, no spin of your own). The disk lands on the turntable and sticks to it.
Does the turntable:
(a) Speed up
(b) Slow down
(c) Stay at the same angular velocity
Commit to your answer and your reasoning before continuing.
[Interactive: Predict-Then-Reveal. The student selects one of the three options. After committing, the student sees: "The answer is (b): the turntable slows down. The dropped disk had zero angular momentum (it was not spinning). The turntable had angular momentum $L = I_{\text{turntable}}\omega_i$. After the disk lands, the combined system has a larger moment of inertia ($I_{\text{turntable}} + I_{\text{disk}}$) but the same total angular momentum. Since $L$ is fixed and $I$ increased, $\omega$ must decrease. No external torque acted --- the forces between the disk and turntable during the landing are internal to the system."]
Notice what happened. The turntable visibly changed its motion --- it slowed down --- and yet nothing external acted on it. The change was caused by a redistribution of mass within the system. Angular momentum did not change. Angular velocity did.
This distinction between the conserved quantity ($L$) and the observable motion ($\omega$) is the heart of this section.
The Guiding Question
What remains fixed when there is no external torque, even when the motion visibly changes?
In Section 8.3, you learned that the total linear momentum of an isolated system is constant, even when the individual objects within it speed up, slow down, or reverse direction. The internal forces cancel in pairs (Newton's third law), and only external forces can change the total momentum.
The same logic applies to rotation. Internal torques --- torques that objects within the system exert on each other --- cancel in pairs. Only external torques can change the total angular momentum. When there are no external torques, the total angular momentum is locked in place, no matter how dramatically the motion rearranges itself.
Exploration: The Turntable Lab
[Interactive: Angular Momentum Conservation Explorer. A top-down view shows a turntable spinning on a frictionless bearing. A small mass (displayed as a dot) sits on the turntable at some distance $r$ from the center. The student can drag the mass radially --- toward the center or toward the rim. The display shows:
- The moment of inertia $I$ of the system (turntable + mass), updating in real time as the mass moves.
- The angular velocity $\omega$, updating in real time.
- The angular momentum $L = I\omega$, displayed as a fixed bar that does not change.
- Numerical readouts of $I$, $\omega$, and $L$.
Phase 1 --- Drag the mass inward. The student drags the mass from the rim toward the center.
Guided prompt: "What happens to $\omega$ as you move the mass inward? What happens to $I$? What happens to $L$?"
Phase 2 --- Drag the mass outward. The student drags the mass from the center toward the rim.
Guided prompt: "Now what happens? Can you make $\omega$ as small as you want by moving the mass far enough out? What about $L$?"
Phase 3 --- Drop a second disk. A button labeled "Drop disk" appears. Clicking it drops a non-spinning disk onto the turntable. The combined moment of inertia increases, and $\omega$ decreases. The $L$ bar remains unchanged.
Guided prompt: "The moment of inertia just increased. What adjusted to keep $L$ constant? Where did the 'lost' angular velocity go?"]
Before you read on: After working through the exploration, state the pattern you observed in one sentence. What quantity stayed constant? What quantities changed? What determined the direction of the change?
If you did the exploration carefully, you found:
- Moving the mass inward decreases $I$ and increases $\omega$. Moving it outward increases $I$ and decreases $\omega$.
- Dropping a second disk increases $I$ and decreases $\omega$.
- In every case, the product $L = I\omega$ remained exactly constant.
The system's angular momentum was set when the turntable started spinning. With no external torques (frictionless bearing), nothing can change it. When you rearrange the mass distribution ($I$ changes), the angular velocity $\omega$ must adjust so that the product $I\omega$ stays the same.
Concept Reveal: Conservation of Angular Momentum
Let us make this precise, following the same structure as the linear momentum derivation in Section 8.3.
In Section 11.4, you saw that the angular impulse-momentum theorem gives:
$$\sum \tau_{\text{ext}} = \frac{dL_{\text{total}}}{dt}$$
The rate of change of the total angular momentum of a system depends only on the external torques. Internal torques --- torques between objects within the system --- cancel in pairs, just as internal forces cancel for linear momentum.
The conservation law follows immediately. If the net external torque is zero, then:
$$\frac{dL_{\text{total}}}{dt} = 0$$
which means:
$$\boxed{L_{\text{total}} = \text{constant} \quad \text{(when } \sum \tau_{\text{ext}} = 0\text{)}}$$
For a system where the moment of inertia can change (like a skater pulling in her arms, or a mass sliding on a turntable), this takes the form:
$$\boxed{I_i \omega_i = I_f \omega_f}$$
This is the conservation of angular momentum: when no net external torque acts on a system, the total angular momentum does not change.
The Structural Parallel
This is worth pausing on. Look at the two conservation laws side by side:
| Translational (Section 8.3) | Rotational (this section) | |
|---|---|---|
| Quantity | $\vec{p} = m\vec{v}$ | $L = I\omega$ |
| Rate of change | $\sum \vec{F}{\text{ext}} = d\vec{p}/dt$} | $\sum \tau_{\text{ext}} = dL_{\text{total}}/dt$ |
| Conservation condition | $\sum \vec{F}_{\text{ext}} = 0$ | $\sum \tau_{\text{ext}} = 0$ |
| Conserved quantity | $\vec{p}_{\text{total}} = \text{constant}$ | $L_{\text{total}} = \text{constant}$ |
| Why internal interactions cancel | Newton's third law: $\vec{F}{A\to B} = -\vec{F}$ | Newton's third law: $\tau_{A\to B} = -\tau_{B\to A}$ |
The architecture is identical. The same reasoning --- internal interactions cancel, so only external influences can change the total --- produces the same kind of conservation law. If you understood why linear momentum is conserved, you already understand why angular momentum is conserved. The physics is the same; only the vocabulary has changed.
Two Types of Problems
Conservation of angular momentum appears in two distinct flavors, and recognizing which type you are facing saves time.
Type 1: Reconfiguration
The system's moment of inertia changes, but no objects enter or leave and no external torques act. The same angular momentum is redistributed.
Example: The figure skater. A skater spins with arms extended ($I_i = 4.0 \text{ kg}\cdot\text{m}^2$, $\omega_i = 2.0 \text{ rad/s}$). She pulls her arms in, reducing her moment of inertia to $I_f = 1.6 \text{ kg}\cdot\text{m}^2$. Find her new angular velocity.
No external torques act (the ice is nearly frictionless, and the vertical axis of rotation means gravity produces no torque about that axis). Angular momentum is conserved:
$$I_i \omega_i = I_f \omega_f$$
$$\omega_f = \frac{I_i}{I_f}\omega_i = \frac{4.0}{1.6}(2.0) = 5.0 \text{ rad/s}$$
Her angular velocity increases by a factor of 2.5. She spins much faster --- visibly and dramatically --- without any external push.
Type 2: Rotational collision
An object joins (or separates from) a rotating system. The total angular momentum before equals the total angular momentum after.
Example: The dropped disk. A turntable ($I_t = 0.040 \text{ kg}\cdot\text{m}^2$) spins at $\omega_i = 10 \text{ rad/s}$. A non-spinning disk ($I_d = 0.020 \text{ kg}\cdot\text{m}^2$) is dropped onto it. Find the final angular velocity.
Before: only the turntable spins. $L_i = I_t \omega_i = (0.040)(10) = 0.40 \text{ kg}\cdot\text{m}^2\text{/s}$.
After: the turntable and disk spin together. $L_f = (I_t + I_d)\omega_f$.
Conservation:
$$I_t \omega_i = (I_t + I_d)\omega_f$$
$$\omega_f = \frac{I_t}{I_t + I_d}\omega_i = \frac{0.040}{0.060}(10) = 6.7 \text{ rad/s}$$
The turntable slows down. The disk, initially not spinning, absorbs some of the angular momentum. The total $L$ is unchanged --- it is now shared between the two objects.
Spaced Retrieval
Before moving to practice, test your recall of earlier material.
Recall prompt 1: What is the angular impulse-momentum theorem, and how does it relate torque to angular momentum? (Section 11.4)
Recall prompt 2: What is the definition of angular momentum for a rigid body rotating about a fixed axis? (Section 11.3)
Recall prompt 3: In Section 8.3, what condition must hold for linear momentum to be conserved? What is the analogous condition here?
Practice
Layer 1: Concrete
Problem 1. A merry-go-round (uniform disk, $M = 200 \text{ kg}$, $R = 2.0 \text{ m}$) spins freely at $\omega_i = 1.5 \text{ rad/s}$. A child ($m = 40 \text{ kg}$) standing at the rim jumps off tangentially (leaves the merry-go-round). After the child jumps off, the merry-go-round spins at $\omega_f = 2.0 \text{ rad/s}$. What is the child's tangential speed as she leaves?
Check your answer
**System:** Merry-go-round + child. The bearing is frictionless, so no external torques act about the vertical axis. Angular momentum is conserved. **Before:** The child is at the rim, rotating with the merry-go-round. $$I_{\text{mgr}} = \tfrac{1}{2}MR^2 = \tfrac{1}{2}(200)(2.0)^2 = 400 \text{ kg}\cdot\text{m}^2$$ The child at the rim contributes $I_{\text{child}} = mR^2 = (40)(2.0)^2 = 160 \text{ kg}\cdot\text{m}^2$. $$L_i = (I_{\text{mgr}} + I_{\text{child}})\omega_i = (400 + 160)(1.5) = 840 \text{ kg}\cdot\text{m}^2\text{/s}$$ **After:** The merry-go-round spins alone at $\omega_f = 2.0 \text{ rad/s}$. The child moves tangentially at speed $v$. $$L_f = I_{\text{mgr}}\omega_f + m v R = (400)(2.0) + (40)(v)(2.0)$$ Setting $L_i = L_f$: $$840 = 800 + 80v$$ $$v = \frac{40}{80} = 0.50 \text{ m/s}$$ The child leaves at 0.50 m/s tangentially. Notice: she moves slower than the rim speed before the jump ($v_{\text{rim}} = R\omega_i = 3.0$ m/s). By leaving slowly, she transfers angular momentum back to the merry-go-round, which speeds up.Problem 2. A neutron star has moment of inertia $I_i = 1.0 \times 10^{38} \text{ kg}\cdot\text{m}^2$ and rotates once every 20 days. After further gravitational collapse, its moment of inertia decreases to $I_f = 1.0 \times 10^{32} \text{ kg}\cdot\text{m}^2$. Find the new rotation period.
Check your answer
No external torques act during the collapse (the gravitational forces are internal). Angular momentum is conserved. $$I_i \omega_i = I_f \omega_f$$ Since $\omega = 2\pi / T$: $$I_i \cdot \frac{2\pi}{T_i} = I_f \cdot \frac{2\pi}{T_f}$$ $$T_f = T_i \cdot \frac{I_f}{I_i} = 20 \text{ days} \times \frac{1.0 \times 10^{32}}{1.0 \times 10^{38}} = 20 \times 10^{-6} \text{ days}$$ Converting: $T_f = 20 \times 10^{-6} \times 86{,}400 \text{ s} \approx 1.7 \text{ s}$. The star now completes a full rotation roughly every 1.7 seconds. The moment of inertia decreased by a factor of $10^6$, so the angular velocity increased by a factor of $10^6$, and the period decreased by a factor of $10^6$. This is how pulsars form --- and why they spin so extraordinarily fast.Layer 2: Pattern
Problem 3. For each scenario below, decide whether the angular momentum of the stated system is conserved about the given axis. Justify your reasoning by identifying any external torques.
(a) A figure skater spins on frictionless ice, pulling her arms in. System: the skater. Axis: the vertical axis through her center.
(b) A ball on a string is whirled in a horizontal circle at constant speed. System: the ball. Axis: the vertical axis through the center of the circle.
(c) A disk spinning on an axle is gradually slowed by friction in the bearing. System: the disk. Axis: the axle.
(d) A potter's wheel is spun up by the potter's foot pressing on a kick bar. System: the wheel. Axis: the wheel's axle.
(e) Two gears mesh and rotate together, with no friction at their axles. System: both gears together. Axis: any fixed point.
Check your answer
**(a)** Conserved. The ice exerts no frictional torque (frictionless). Gravity and the normal force both act through the axis of rotation, so they produce zero torque about that axis. No external torque acts. Angular momentum is conserved --- this is why she spins faster when she pulls her arms in. **(b)** Conserved. The tension in the string points radially inward (toward the center). A radial force produces zero torque about the center ($\tau = rF\sin\theta$ with $\theta = 0$). The ball's angular momentum about the center is constant. **(c)** Not conserved. Friction in the bearing exerts an external torque on the disk (opposing rotation). This torque reduces the disk's angular momentum over time. If you expanded the system to include the bearing and whatever it is mounted to, friction would become internal --- but for the disk alone, it is an external torque. **(d)** Not conserved. The potter's foot applies a tangential force to the kick bar, creating an external torque on the wheel. This torque increases the wheel's angular momentum. **(e)** Conserved. The contact forces between the gears are internal to the system. Newton's third law ensures the torque gear A exerts on gear B is equal and opposite to the torque gear B exerts on gear A. With frictionless axles, no external torques act on the two-gear system. The total angular momentum is constant (even though individual gears speed up or slow down). **The pattern:** The question is always the same one from Section 8.3, translated to rotation: are there net external torques on the system you have defined? If not, angular momentum is conserved. If yes, it is not. Expanding your system boundary can turn external torques into internal ones.Layer 3: Structure
Problem 4. A figure skater pulls her arms in and spins faster. Her angular momentum is conserved, so $I_i \omega_i = I_f \omega_f$.
But consider her kinetic energy. Before: $KE_i = \frac{1}{2}I_i \omega_i^2$. After: $KE_f = \frac{1}{2}I_f \omega_f^2$.
(a) Show that $KE_f > KE_i$. (Hint: express $KE_f$ in terms of $KE_i$ using the conservation equation.)
(b) The kinetic energy increased, but no external work was done (no external torques). Where did the extra energy come from?
Check your answer
**(a)** From conservation: $\omega_f = (I_i / I_f)\omega_i$. Substitute into $KE_f$: $$KE_f = \frac{1}{2}I_f \omega_f^2 = \frac{1}{2}I_f \left(\frac{I_i}{I_f}\right)^2 \omega_i^2 = \frac{1}{2}\frac{I_i^2}{I_f}\omega_i^2 = \frac{I_i}{I_f} \cdot \frac{1}{2}I_i\omega_i^2 = \frac{I_i}{I_f}\, KE_i$$ Since the skater pulled her arms in, $I_f < I_i$, so $I_i / I_f > 1$, and therefore $KE_f > KE_i$. If the skater reduces her moment of inertia by half, her kinetic energy *doubles*. **(b)** The extra kinetic energy comes from **the skater's muscles** --- internal energy stored in her body. Pulling your arms inward against the centrifugal tendency requires muscular effort. The skater does internal work on herself. This work converts chemical energy (from her muscles) into rotational kinetic energy. This is an important lesson: **angular momentum conservation does not imply energy conservation**. The conserved quantity ($L$) and the kinetic energy ($KE = L^2 / 2I$) are different things. When $I$ decreases at constant $L$, kinetic energy increases. The "missing" energy has a physical source --- the internal work done to change the configuration. This parallels the translational case. In Section 8.3, two skaters push off from rest. Momentum is conserved (stays zero), but kinetic energy increases (both skaters start moving). The energy comes from the skaters' muscles. Conservation of momentum does not mean conservation of kinetic energy.Layer 4: Debug
Problem 5. A turntable spins on an axle with a small but nonzero frictional torque from the bearing. A student drops a disk onto it and writes:
"Angular momentum is conserved: $I_t \omega_i = (I_t + I_d)\omega_f$."
Is this valid? Under what conditions, if any, would this calculation give a good approximation?
Check your answer
Strictly, this is **not valid**. The frictional torque from the bearing is an external torque on the system. It acts continuously, so angular momentum is not exactly conserved. However, the student's calculation may still be a **good approximation** if the drop happens quickly. During the brief moment when the disk lands and the two objects reach a common angular velocity, the frictional torque from the bearing delivers a very small angular impulse (because the time interval is short and the frictional torque is small). The internal torques between the disk and turntable during the collision are very large by comparison. So the angular momentum changes negligibly during the collision itself. This is exactly the same reasoning used for linear momentum in collisions on surfaces with friction (Section 8.3): if the collision is fast, the friction impulse during the collision is tiny, and momentum is approximately conserved during the impact. The student should state this assumption explicitly: "The collision is brief, so the angular impulse from bearing friction during the collision is negligible compared to the internal angular impulse between the disk and turntable. Therefore, angular momentum is approximately conserved during the collision." After the collision, friction will gradually slow the combined system. But the conservation equation applies to the instant of the collision, not to the long-term behavior. **The lesson:** Conservation laws require conditions. Always check whether those conditions hold --- and if they hold only approximately, say so.Layer 5: Creation
Problem 6. Design a physical system in which angular momentum is conserved but kinetic energy is not. Describe the system, explain why angular momentum is conserved, and state whether kinetic energy increases or decreases.
Check your answer
There are many valid answers. Here are two: **System 1: Disk dropped onto a turntable (kinetic energy decreases).** A turntable spins freely. A non-spinning disk is dropped onto it. Angular momentum is conserved (no external torques). But kinetic energy decreases: $$KE_i = \frac{1}{2}I_t\omega_i^2, \qquad KE_f = \frac{1}{2}(I_t + I_d)\omega_f^2$$ Since $\omega_f = I_t\omega_i / (I_t + I_d)$, you can verify that $KE_f < KE_i$. The "lost" kinetic energy is converted to thermal energy by friction between the surfaces during the collision (the disk and turntable initially slide against each other until they reach the same angular velocity). This is the rotational analog of a perfectly inelastic collision. **System 2: Skater pulling arms in (kinetic energy increases).** A spinning skater on frictionless ice pulls her arms inward. Angular momentum is conserved (no external torques). Kinetic energy increases (as shown in Problem 4). The extra energy comes from the skater's muscles doing internal work. The key insight: angular momentum conservation constrains one relationship ($I_i\omega_i = I_f\omega_f$), but it says nothing about what happens to energy. Energy can increase (internal work adds it), decrease (friction or deformation removes it), or stay the same (a special case). Conservation of angular momentum and conservation of energy are independent principles.Reflection
What makes conservation of angular momentum surprising in everyday experience?
In daily life, spinning objects usually slow down. Tops wobble and stop. Wheels coast to a halt. We are not accustomed to seeing angular momentum conserved, because friction and air resistance provide external torques that drain angular momentum away.
So when you see a skater pull her arms in and spin breathtakingly fast --- seemingly "from nowhere" --- it looks like magic. But it is not. She did not gain angular momentum. She redistributed it: the same $L$, packed into a smaller $I$, demands a larger $\omega$. The effect is vivid precisely because our everyday intuition is trained by friction to expect things to slow down, not speed up.
The neutron star is the same idea, pushed to an extreme. A lazy stellar rotation becomes hundreds of revolutions per second --- not because anything spun it up, but because the same angular momentum was compressed into a vastly smaller moment of inertia. The conservation law does not care whether the system is a dancer or a dead star. The principle is the same.
This is the power of conservation laws: they cut through complexity. You do not need to know the details of how the star collapsed, or how the skater's muscles worked, or how the disk landed on the turntable. You need only one fact --- no external torques acted --- and the rest follows from algebra.
Looking Ahead
You now have the full conservation toolkit for rotation: rotational kinetic energy (Section 11.1), work by torques (Section 11.2), angular momentum (Section 11.3), angular impulse (Section 11.4), and conservation of angular momentum (this section).
But a question remains. In the figure skater problem, you saw that angular momentum is conserved while kinetic energy increases. In the dropped-disk problem, angular momentum is conserved while kinetic energy decreases. In earlier chapters, you saw similar behavior with linear momentum: it can be conserved while kinetic energy changes (inelastic collisions) or while kinetic energy is also conserved (elastic collisions).
These observations raise a practical challenge: when a real system both translates and rotates --- like a ball rolling down a ramp, or a cylinder unwinding from a string --- how do you combine the translational and rotational tools? Do you use forces and torques? Energy conservation? Angular momentum? All three?
In Section 11.6, you will face exactly this question. You will learn to combine translational and rotational energy, compare the available methods, and develop judgment about which tool to reach for first. That is where the full power of the rotational toolkit comes together.