AP Physics 1 · Topic 6.4

Topic 6.4: Conservation of Angular Momentum

Unit 6: Energy and Momentum of Rotating Systems5-8% of the multiple-choice section

A system's total angular momentum stays constant whenever the net external torque on it is zero. Internal torques come in equal and opposite pairs, so they cancel in the total. A skater who pulls her arms in shrinks her rotational inertia, so her angular speed rises to keep the product the same.

AP Physics: Unit 6 (topics 6.4 Conservation of Angular Momentum). AP Physics 1 Unit 6, Topic 6.4. Two learning objectives: 6.4.A, describe the behavior of a system using conservation of angular momentum, and 6.4.B, describe how the selection of a system determines whether the angular momentum of that system changes. Five essential knowledge statements support them: 6.4.A.1 (the total angular momentum of a system about a rotational axis is the sum of the angular momenta of its constituent parts about that axis), 6.4.A.2 (any change must be due to an interaction with the surroundings) with its four sub-statements 6.4.A.2.i (angular impulses between two systems are equal and opposite, a direct result of Newton's third law), 6.4.A.2.ii (a system may be selected so that its total angular momentum is constant), 6.4.A.2.iii (the angular speed of a nonrigid system may change without its angular momentum changing if the system moves mass closer to or further from the axis) and 6.4.A.2.iv (any change equals the angular impulse exerted on the system), plus 6.4.B.1 (angular momentum is conserved in all interactions), 6.4.B.2 (if the net external torque on a selected object or rigid system is zero, the total angular momentum of that system is constant) and 6.4.B.3 (if it is nonzero, angular momentum is transferred between the system and the environment). The CED prints no boundary statement for this topic; the Topic 6.3 boundary statement, which puts the direction of angular momentum and angular impulse beyond the scope of the course, still governs the quantities used. The CED's suggested skills for this topic are 1.B, 2.D, 3.A, 3.B and 3.C. Unit 6 carries 5 to 8 percent of the multiple-choice section and about 8 to 14 class periods.

What Topic 6.4 requires

Topic 6.4 carries two learning objectives, and the second is why this topic rewards slowing down.

6.4.A Describe the behavior of a system using conservation of angular momentum.
6.4.B Describe how the selection of a system determines whether the angular momentum of that system changes.

Those two sentences are the Topic 4.3 objectives rewritten for rotation, and the match is close enough to be worth checking against the source. Objective 4.3.A reads describe the behavior of a system using conservation of linear momentum, which is 6.4.A word for word with linear swapped for angular. Objective 4.3.B reads describe how the selection of a system determines whether the momentum of that system changes, and 6.4.B is that sentence with angular inserted before momentum. The essential knowledge underneath follows the same pattern, so if you already know conservation of linear momentum, much of the work here is translation.

Five essential knowledge statements sit under the objectives: two under 6.4.A, where 6.4.A.2 carries four sub-statements numbered i to iv, and three under 6.4.B. The CED lists five suggested skills for the topic: 1.B, create quantitative graphs with appropriate scales and units, including plotting data; 2.D, predict new values or factors of change of physical quantities using functional dependence between variables; 3.A, create experimental procedures that are appropriate for a given scientific question; 3.B, apply an appropriate law, definition, theoretical relationship, or model to make a claim; and 3.C, justify or support a claim using evidence from experimental data, physical representations, or physical principles or laws. Three of the five are science practice 3 skills, which is the band where experimental design, claims and justifications live, and that tells you what the free-response version of this topic looks like.

No boundary statement is printed for Topic 6.4. Topic 6.3 has one, and it applies to the quantities you carry into this topic: the direction of angular momentum and angular impulse is beyond the scope of the course, so everything here runs on magnitudes and signs along one chosen rotational sense.

Unit 6 is weighted at 5 to 8 percent of the multiple-choice section, and the CED allots the unit about 8 to 14 class periods.

Conserved and constant are two different words (6.4.B)

Essential knowledge 6.4.B.1 is one sentence long: angular momentum is conserved in all interactions. Taken alone that seems to settle everything, so be exact about its scope. The sentence is about what happens in an interaction, not about the boundary you chose to draw around part of one.

The next two statements say what it means for your system.

6.4.B.2 If the net external torque exerted on a selected object or rigid system is zero, the total angular momentum of that system is constant.
6.4.B.3 If the net external torque exerted on a selected object or rigid system is nonzero, angular momentum is transferred between the system and the environment.

The CED is using two words on purpose, and they are not interchangeable. Conserved describes the universe: angular momentum is not created or destroyed, only handed from one thing to another. Constant describes one system over one interval, and it holds only once you have checked the condition in 6.4.B.2. A wheel slowing under a brake loses angular momentum, and nothing about that contradicts 6.4.B.1, because the wheel handed what it lost to the brake.

Everyday speech collapses the two, and you will hear angular momentum is not conserved here used to mean this system's angular momentum is not constant. Translate it when you read it, and keep the words apart when you write, because free-response justifications are marked on wording. The question to ask about a scenario is not whether angular momentum is conserved. It is whether anything outside your system exerts a torque about your axis.

One wording detail is worth holding on to. Statements 6.4.B.2 and 6.4.B.3 are both phrased for a selected object or rigid system. Statement 6.4.A.2.ii, which says a system may be selected so that the total angular momentum of that system is constant, is phrased for a system, with no rigidity attached. That difference matters, because the headline example of this topic, a skater changing shape, is not a rigid system. Statement 6.4.A.2.iii is the one written for her.

The total is a sum about one axis (6.4.A.1)

Statement 6.4.A.1 defines what you are conserving: the total angular momentum of a system about a rotational axis is the sum of the angular momenta of the system's constituent parts about that axis. Three words in that sentence do the work.

Sum. Add the parts. Nothing else.
About a rotational axis. One axis, named before you start.
About that axis. Every part's angular momentum has to be computed about the same axis as every other part's. Mixing axes inside one sum produces a number that means nothing.

Because the boundary statement for Topic 6.3 restricts you to one-dimensional vector conventions, the sum is a sum of signed numbers. Choose a positive rotational sense, usually counterclockwise, and give every contribution a sign. A part turning the other way subtracts. A dropped minus sign here is the cheapest way to lose an angular momentum question.

Parts that are not spinning still count. A child running along a tangent line toward the rim of a playground roundabout carries angular momentum about the roundabout's axis, given by the other printed form, [ $L = rmv\sin\theta$ ](/ap-physics-1/unit-6-energy-and-momentum-of-rotating-systems/6-3-angular-momentum-and-angular-impulse). Leave that term out of the before line and the numbers come out as though the child arrived from nowhere.

Statement 4.3.A.2 says the same thing for linear momentum. The only phrase 6.4.A.1 adds is about that axis, because angular momentum means nothing until a reference is named.

Only an outside interaction can change it (6.4.A.2)

Statement 6.4.A.2 reads that any change to a system's angular momentum must be due to an interaction between the system and its surroundings. Its four sub-statements then supply the reason, the permission, the loophole, and the bookkeeping.

The reason (6.4.A.2.i). The angular impulse exerted by one object or system on a second object or system is equal and opposite to the angular impulse exerted by the second object or system on the first. The CED calls this a direct result of Newton's third law. Two objects that interact start interacting at the same instant and stop at the same instant, so they share the interval $\Delta t$ , and equal and opposite torques over an identical interval give equal and opposite angular impulses, $\Delta L_2 = -\Delta L_1$ . Add the two and you get zero. Every internal torque pairs off this way, which is why no internal torque, however violent, can change a system's total angular momentum.

The permission (6.4.A.2.ii). A system may be selected so that the total angular momentum of that system is constant. Read may be selected as a verb aimed at you: the CED is telling you to build the system deliberately rather than accept whichever one the diagram suggests, and objective 6.4.B is the assessment of whether you can.

The loophole (6.4.A.2.iii). The angular speed of a nonrigid system may change without the angular momentum of the system changing if the system changes shape by moving mass closer to or further from the rotational axis. That is the skater, and it gets its own section below.

The bookkeeping (6.4.A.2.iv). If the total angular momentum of a system changes, that change will be equivalent to the angular impulse exerted on the system. This is Topic 6.3's [ $\Delta L = \tau \Delta t$ ](/ap-physics-1/unit-6-energy-and-momentum-of-rotating-systems/6-3-angular-momentum-and-angular-impulse) doing double duty: when the net external torque is zero it gives you a conservation statement, and when it is not it gives you the size of the change.

Choosing the system, and choosing the axis

The procedure is the one from linear momentum with an extra step, and the extra step is the one people skip.

Name the axis first. Angular momentum has no value until you do, by 6.3.A.2.i, and the same axis has to serve every term on both sides of your equation.
Put everything that takes part inside the system. The roundabout and the child. The turntable and the student on it. The disk and the putty landing on it.
Everything they do to each other is now internal, and by 6.4.A.2.i it cancels in pairs.
Ask which external forces still exert a torque about that axis. Not which forces act: which ones produce a torque.
If none of them delivers a meaningful angular impulse over your interval, the total is constant. Write it before and after, then solve for the unknown.

Step 4 is where the marks are, and it is where rotation differs from translation. A force can be large and still exert no torque about your axis. Three cases cover most exam setups.

Forces applied at the axis have zero lever arm. A roundabout's axle pushes hard on it and exerts no torque about itself.
Forces parallel to the axis exert no torque about it. For a turntable spinning about a vertical axis, gravity on every part points straight down, along the axis, and contributes nothing to the torque about it. That is the reason so many exam scenarios put the rotation in a horizontal plane: it is the geometry that empties step 4.
A force always directed at your chosen point exerts no torque about that point. The sun's pull on a planet points at the sun the whole way round, so $\sin\theta = 0$ in $\tau = rF\sin\theta$ and the planet's angular momentum about the sun is constant. Topic 6.6 develops the orbital case.

Friction is the usual survivor. A roundabout on a rusty bearing feels an external friction torque, so its angular momentum is not constant over a long interval. The fix is the one Topic 4.3 uses for collisions: compare the instant before the interaction with the instant after, over which that torque has had almost no time to deliver an angular impulse.

The skater: angular speed changes, angular momentum does not (6.4.A.2.iii)

The CED lists What do ice skaters do with their arms when they want to spin faster? Why? among its essential questions for Unit 6, so the example is not folklore. Here is the exact answer.

Take the skater and everything she is holding as the system, and take her vertical spin axis. On ice the external torque about that axis is negligible, so by 6.4.B.2 her total angular momentum is constant:

I_i \omega_i = I_f \omega_f

Pulling her arms in moves mass closer to the axis. Rotational inertia depends on where the mass sits, through $I = \sum m_i r_i^2$ from essential knowledge 5.4.A.3, so [ $I$ ](/ap-physics-1/unit-5-torque-and-rotational-dynamics/5-4-rotational-inertia) falls. With the product locked, $\omega$ has to rise by the same factor. Extending her arms runs the whole thing backwards.

Statement 6.4.A.2.iii is careful about the wording, and so should you be: the angular speed of a nonrigid system may change without the angular momentum of the system changing if the system changes shape by moving mass closer to or further from the rotational axis. A rigid system, by essential knowledge 5.1.A.1.i, is one that holds its shape. A skater who moves her arms is not one, which is exactly why she can do this and a flywheel cannot.

The muscles that pull her arms in are internal to the system. By 6.4.A.2.i the torques they exert come in equal and opposite pairs, so they cannot change $L$ . They can and do change other things, which is the subject of the next section.

Two traps live here.

**A larger $\omega$ is not more angular momentum.** $L$ is the product, and the product did not move. Answering that she gained angular momentum contradicts the premise of the question.
The rotational form of Newton's second law is not available. As Topic 6.3 notes, [ $\tau_{\text{net}} = I\alpha$ ](/ap-physics-1/unit-6-energy-and-momentum-of-rotating-systems/6-3-angular-momentum-and-angular-impulse) is derived for cases in which rotational inertia is constant, and here $I$ is precisely what is changing. The skater's angular speed changes with no net external torque at all, which looks like a contradiction only if you reach for the wrong equation.

Constant angular momentum does not mean constant kinetic energy

Rotational kinetic energy is $K = \frac{1}{2}I\omega^2$ . Substitute $\omega = L/I$ and it turns into a statement about $L$ and $I$ alone:

K = \frac{1}{2}I\left(\frac{L}{I}\right)^2 = \frac{L^2}{2I}

Hold $L$ fixed and the kinetic energy is inversely proportional to the rotational inertia. Halve $I$ and $K$ doubles. That single line answers a question examiners ask often, and it settles the sign of the answer without any arithmetic: when the skater pulls her arms in, her rotational kinetic energy increases.

Where does that energy come from? From the skater. To pull her arms inward she exerts an inward force on them while they move inward, so that force does positive work, drawing on chemical energy stored in her body. Nothing external is involved, which is the point: internal forces cannot change a system's angular momentum, but they can change its kinetic energy. Letting her arms back out reverses the exchange.

The linear version of this is on the Topic 4.3 page, where two carts pushed apart by a compressed spring end with more kinetic energy than they started with while the total momentum sits at zero throughout. Same logic, same reason.

Energy can also go the other way. When a lump of putty lands on a spinning disk, or a child steps onto a moving roundabout, friction has to drag the newcomer up to speed and drags the disk down, and that sliding dissipates energy. Angular momentum is still constant; kinetic energy falls. Worked example 1 puts numbers on it. This is the rotational counterpart of the perfectly inelastic collision from Topic 4.4, and the two share a useful check. With $L$ constant,

\frac{K_f}{K_i} = \frac{I_i}{I_f}

so kinetic energy falls whenever the rotational inertia grows and rises whenever it shrinks, by exactly the same factor.

When the net external torque is not zero (6.4.B.3)

Statement 6.4.B.3 covers the other half: if the net external torque exerted on a selected object or rigid system is nonzero, angular momentum is transferred between the system and the environment. Combined with 6.4.A.2.iv, the size of the transfer is the angular impulse, $\Delta L = \tau \Delta t$ .

Three everyday cases where the angular momentum of the obvious system is not constant:

A wheel with a brake pad on it, taking the wheel alone as the system. The friction torque is external and it removes angular momentum steadily.
A rod pivoted at one end and released from horizontal, taking the rod alone. Gravity acts at the center of mass, a lever arm away from the pivot, so the rod's angular momentum grows as it falls.
A yo-yo unwinding on its string. The string tension acts a radius away from the yo-yo's axis.

None of that angular momentum vanished. It ended up in the brake mounting, in the pivot, in Earth. Redraw the boundary so that whatever applies the torque sits inside it and the transfer becomes an internal pair with nowhere to go. One event, two boundaries, two true descriptions, and objective 6.4.B is the CED asking whether you can hold both at once. Worked example 3 runs a single event both ways.

Skill 1.B lands here. A graph of a system's angular momentum against time is flat while the net external torque is zero and sloped while it is not, with the slope equal to the net torque by essential knowledge 6.3.C.3. Being handed such a graph and asked during which interval the system was isolated is a fair translation-between-representations question.

How Topic 6.4 is tested

Three of the five suggested skills for this topic, 3.A, 3.B and 3.C, are about designing, claiming and justifying, and the CED's Unit 6 overview says exactly what it wants from a justification: simply referencing an equation, law, or physical principle is not sufficient. Its own example is that stating one disk is rolling faster than another because of conservation of energy is not a complete enough answer to earn credit on the free-response section. Students must clearly and concisely explain the steps in their reasoning that lead from the equation, law, or physical principle to the justification of their claim in FRQ 1, the Mathematical Routines question. That last clause belongs to the CED's sentence, so the demand is aimed at the first free-response question in particular, but it is the standard worth meeting on any written answer.

Applied here, a full answer to why does the skater speed up has four steps, not one: the net external torque about her spin axis is negligible, so her total angular momentum is constant; angular momentum is the product of rotational inertia and angular speed; pulling her arms in moves mass closer to the axis and so reduces her rotational inertia; therefore her angular speed must rise to hold the product fixed. Naming conservation of angular momentum and stopping is the answer the CED says will not earn the credit.

The other two skills fill in the rest of the question set. Skill 2.D asks for factors of change: a student on a turntable halves her rotational inertia, so her angular speed doubles and her kinetic energy doubles, from $L = I\omega$ and $K = L^2/2I$ . Skill 3.A asks for a procedure: describe an experiment to test whether the angular momentum of a rotating platform plus a dropped ring is constant, and expect follow-up questions on what to measure, how to plot it, and what a slope would mean.

The unit continues into Topic 6.5, rolling, and Topic 6.6, the motion of orbiting satellites, which is the conservation statement on this page applied to a planet under a force that always points at the same spot.

Clay onto a spinning platform: L constant, kinetic energy not

A horizontal platform turns freely about a fixed vertical axle with rotational inertia 24 kg m^2 about that axle, at 3.5 rad/s. A 4.0 kg lump of clay is dropped straight down and lands 1.0 m from the axle, where it sticks. Treat the clay as a point object and ignore friction in the axle. Find (a) the platform's angular velocity afterwards, (b) the rotational kinetic energy before and after, and (c) what happened to the difference.

Choose the system as platform plus clay, take the axle as the axis, and take the platform's sense of rotation as positive. Check step 4 of the procedure: gravity on every part points straight down, parallel to the axis, so it exerts no torque about the axis; the axle's forces act at the axis, so their lever arm is zero; the friction between clay and platform is internal. The net external torque about this axis is zero, so by 6.4.B.2 the total angular momentum is constant.
Before: the platform contributes $L = I\omega = (24\ \text{kg} \cdot \text{m}^2)(3.5\ \text{rad/s}) = 84\ \text{kg} \cdot \text{m}^2/\text{s}$ . The clay contributes nothing, because its velocity is straight down, parallel to the axis, so its angular momentum about that axis is zero. By 6.4.A.1 the total is $84\ \text{kg} \cdot \text{m}^2/\text{s}$ .
(a) After: the clay is stuck at $r = 1.0\ \text{m}$ , so from 5.4.A.2 it adds $I = mr^2 = (4.0\ \text{kg})(1.0\ \text{m})^2 = 4.0\ \text{kg} \cdot \text{m}^2$ , giving a combined $I_f = 24 + 4.0 = 28\ \text{kg} \cdot \text{m}^2$ . Setting the totals equal: $84 = (28)\omega_f$ , so $\omega_f = 3.0\ \text{rad/s}$ .
Sanity check on the direction of the change: adding mass away from the axis raises $I$ , so $\omega$ must fall. It fell from 3.5 to 3.0 rad/s. Had it risen, something would be wrong.
(b) $K_i = \frac{1}{2}I_i\omega_i^2 = \frac{1}{2}(24)(3.5)^2 = \frac{1}{2}(24)(12.25) = 147\ \text{J}$ . And $K_f = \frac{1}{2}I_f\omega_f^2 = \frac{1}{2}(28)(3.0)^2 = \frac{1}{2}(28)(9.0) = 126\ \text{J}$ .
Check that against the shortcut $K = L^2/2I$ : $K_i = (84)^2/(2 \times 24) = 7056/48 = 147\ \text{J}$ and $K_f = 7056/56 = 126\ \text{J}$ . The ratio $K_f/K_i = 126/147 = 0.857$ equals $I_i/I_f = 24/28 = 0.857$ , as it must when $L$ is constant.
(c) The system lost $147 - 126 = 21\ \text{J}$ . While the clay was sliding on the platform, kinetic friction acted through a relative sliding distance and dissipated that energy, mostly as thermal energy in the two surfaces. Nothing about that touches the angular momentum, because the friction torques between clay and platform are an internal pair and cancel by 6.4.A.2.i.

(a) $\omega_f = 3.0\ \text{rad/s}$ . (b) $K_i = 147\ \text{J}$ and $K_f = 126\ \text{J}$ . (c) 21 J was dissipated by friction between the clay and the platform as they slid against each other. This is the rotational counterpart of a perfectly inelastic collision: angular momentum constant, kinetic energy not, and the two questions answered separately.

A student on a turntable: where the extra kinetic energy comes from

A student sits on a freely rotating stool holding a mass in each outstretched hand. Student, stool and masses together have a rotational inertia of 3.6 kg m^2 about the stool's vertical axis, and the system turns at 1.5 rad/s. The student pulls both masses in close to her chest, reducing the system's rotational inertia to 1.2 kg m^2. Friction in the stool is negligible. Find (a) the new angular velocity, (b) the rotational kinetic energy before and after, and (c) the work the student did.

System: student, stool and both masses. Axis: the stool's vertical axis. Gravity is parallel to that axis and the bearing forces act at it, so no external torque acts about the axis and by 6.4.B.2 the total angular momentum is constant. Take the direction of rotation as positive.
(a) $L = I_i\omega_i = (3.6\ \text{kg} \cdot \text{m}^2)(1.5\ \text{rad/s}) = 5.4\ \text{kg} \cdot \text{m}^2/\text{s}$ , and this is the same before and after. So $\omega_f = L/I_f = 5.4/1.2 = 4.5\ \text{rad/s}$ .
Read that as skill 2.D asks you to. The rotational inertia fell by a factor of 3, so the angular speed rose by a factor of 3. Nothing about the specific numbers matters to that conclusion.
(b) $K_i = \frac{1}{2}(3.6)(1.5)^2 = \frac{1}{2}(3.6)(2.25) = 4.05\ \text{J}$ and $K_f = \frac{1}{2}(1.2)(4.5)^2 = \frac{1}{2}(1.2)(20.25) = 12.15\ \text{J}$ .
Cross-check with $K = L^2/2I$ : $K_i = (5.4)^2/(2 \times 3.6) = 29.16/7.2 = 4.05\ \text{J}$ and $K_f = 29.16/2.4 = 12.15\ \text{J}$ . The kinetic energy tripled while the rotational inertia was divided by three, which is $K \propto 1/I$ at constant $L$ .
(c) The kinetic energy rose by $12.15 - 4.05 = 8.10\ \text{J}$ , and no external torque did any work about this axis, so the student supplied it. To haul the masses inward she pulled inward on them while they moved inward, so her force did positive work on them, drawing on chemical energy stored in her body.
If she now lets the masses back out to the original 3.6 kg m^2, the angular momentum stays at $5.4\ \text{kg} \cdot \text{m}^2/\text{s}$ , the angular speed returns to 1.5 rad/s, and the kinetic energy returns to 4.05 J. Her arms do $-8.10\ \text{J}$ of work on the way out.

(a) $\omega_f = 4.5\ \text{rad/s}$ , three times the original. (b) $K_i = 4.05\ \text{J}$ and $K_f = 12.15\ \text{J}$ , also three times. (c) The student did +8.10 J of work, from chemical energy in her muscles, by pulling inward on the masses while they moved inward. Her angular momentum never changed; her kinetic energy tripled. Internal forces are barred from changing the first and free to change the second.

One braked wheel, two different systems

A bicycle wheel of rotational inertia 0.20 kg m^2 spins on a low-friction axle at 30 rad/s. A brake pad is pressed against the rim and exerts a constant friction torque of 0.80 N m on the wheel for 5.0 s. (a) Taking the wheel alone as the system, find its change in angular momentum and its final angular velocity. (b) Say what happened to the missing angular momentum, and name the essential knowledge statements that cover each answer.

Take the wheel's sense of rotation as positive, so the friction torque is $-0.80\ \text{N} \cdot \text{m}$ .
(a) $L_0 = I\omega_0 = (0.20\ \text{kg} \cdot \text{m}^2)(30\ \text{rad/s}) = 6.0\ \text{kg} \cdot \text{m}^2/\text{s}$ . The brake is outside this system, so the net external torque on it is nonzero and 6.4.B.3 applies: the wheel's angular momentum is not constant.
The angular impulse over the interval is $\tau \Delta t = (-0.80\ \text{N} \cdot \text{m})(5.0\ \text{s}) = -4.0\ \text{N} \cdot \text{m} \cdot \text{s}$ , so by 6.4.A.2.iv the change in angular momentum is $\Delta L = -4.0\ \text{kg} \cdot \text{m}^2/\text{s}$ .
So $L_f = 6.0 - 4.0 = 2.0\ \text{kg} \cdot \text{m}^2/\text{s}$ and $\omega_f = L_f/I = 2.0/0.20 = 10\ \text{rad/s}$ .
Cross-check with the rotational form of Newton's second law, available here because the wheel's rotational inertia is constant: $\alpha = \tau/I = -0.80/0.20 = -4.0\ \text{rad/s}^2$ , so $\omega_f = 30 + (-4.0)(5.0) = 10\ \text{rad/s}$ . The two routes agree.
(b) Enlarge the system to wheel plus brake plus frame plus Earth. The friction torque is now internal, and by 6.4.A.2.i the brake's angular impulse on the wheel is matched by an equal and opposite angular impulse of the wheel on the brake. The $4.0\ \text{kg} \cdot \text{m}^2/\text{s}$ was transferred to the frame and through it to Earth, whose rotational inertia is so large that the resulting change in its angular speed is unmeasurable.
Note which quantity did what. The angular momentum fell by two thirds while the kinetic energy fell from $\frac{1}{2}(0.20)(30)^2 = 90\ \text{J}$ to $\frac{1}{2}(0.20)(10)^2 = 10\ \text{J}$ , a factor of nine, because $K$ goes as $\omega^2$ . The angular momentum was transferred; most of that energy was dissipated as heat in the pad and rim.

(a) $\Delta L = -4.0\ \text{kg} \cdot \text{m}^2/\text{s}$ and $\omega_f = 10\ \text{rad/s}$ . (b) The angular momentum was transferred to the brake, the frame and Earth, which is 6.4.B.3 for the small system and 6.4.B.1 for the interaction as a whole. The wheel alone loses angular momentum; nothing anywhere destroys any. The same event is a change or a transfer depending only on where you drew the boundary.

Frequently asked questions

When is angular momentum conserved?

Essential knowledge 6.4.B.1 puts it broadly: angular momentum is conserved in all interactions. What you need for a calculation is the narrower statement in 6.4.B.2, that if the net external torque exerted on a selected object or rigid system is zero, the total angular momentum of that system is constant. So test your own system rather than the universe: ask whether anything outside it exerts a torque about the axis you named. Forces applied at that axis and forces pointing along it do not.

Why does an ice skater spin faster when she pulls her arms in?

Because the external torque about her spin axis is negligible, so her total angular momentum, the product of rotational inertia and angular speed, is constant. Pulling her arms in moves mass closer to the axis, and rotational inertia depends on mass and on how far it sits from the axis, so her rotational inertia falls. With the product fixed, her angular speed has to rise by the same factor. The CED describes this in essential knowledge 6.4.A.2.iii as a nonrigid system changing shape.

Does the skater's kinetic energy stay the same when she pulls her arms in?

No, it increases. Rotational kinetic energy can be written as K = L squared divided by 2I, so with angular momentum constant, kinetic energy is inversely proportional to rotational inertia: halve the rotational inertia and the kinetic energy doubles. The extra energy comes from the skater. She exerts an inward force on her arms while they move inward, so that force does positive work, drawing on chemical energy stored in her body.

Is angular momentum conserved when a child jumps onto a merry-go-round?

The total angular momentum of the child plus merry-go-round about the vertical axle is constant, because gravity is parallel to that axis and the axle's forces have no lever arm about it. Include the child's angular momentum before the landing, using L = rmv sin theta, if the child was running. Kinetic energy is not constant: friction between child and platform dissipates some of it, exactly as in a perfectly inelastic collision.

What is the difference between conservation of angular momentum and conservation of linear momentum?

They are the same statement with rotational quantities substituted. Linear momentum is constant when the net external force on a system is zero; angular momentum is constant when the net external torque about a chosen axis is zero. The AP Physics 1 CED even uses matching wording for the two learning objectives, 4.3.A and 6.4.A. The practical difference is that angular momentum has no value until you name an axis, and that a large force can exert no torque about that axis.

Can a system's angular speed change if no external torque acts on it?

Yes, if the system is not rigid. Essential knowledge 6.4.A.2.iii says the angular speed of a nonrigid system may change without the angular momentum of the system changing if the system changes shape by moving mass closer to or further from the rotational axis. A rigid system cannot do this, because its rotational inertia is fixed. This is also why the rotational form of Newton's second law, which assumes constant rotational inertia, does not apply to a shape-changing system.

Why do planets move faster when they are closer to the sun?

Because the sun's gravitational pull on a planet always points at the sun, its torque about the sun is zero, so the planet's angular momentum about the sun is constant. Written as L = rmv sin theta, a smaller distance r must be paid for by a larger value of v sin theta, so the planet speeds up as it swings in and slows down as it moves out. The CED lists this among its Unit 6 essential questions, and Topic 6.6 develops the orbital case.