AP Physics C: Mechanics · Topic 6.4

Topic 6.4: Conservation of Angular Momentum

Unit 6: Energy and Momentum of Rotating Systems10-15% of the multiple-choice section

A system's total angular momentum about a chosen axis is constant when the net external torque about that axis is zero. In AP Physics C the change is the time integral of the net torque, so it is a vector statement about one axis, and it survives even when the rotational inertia is changing.

AP Physics: Unit 6 (topics 6.4 Conservation of Angular Momentum). Topic 6.4 of the current AP Physics C: Mechanics course and exam description, inside Unit 6, weighted 10 to 15% of the multiple-choice section. Two learning objectives, 6.4.A and 6.4.B, and nine essential-knowledge statements counting sub-statements: 6.4.A.1 on the total being a sum about one axis, 6.4.A.2 with sub-statements on Newton's third law for angular impulse, system selection, the nonrigid shape-change case and the angular impulse equalling the change, then 6.4.B.1 that angular momentum is conserved in all interactions, 6.4.B.2 the zero-net-external-torque condition for constancy, and 6.4.B.3 on transfer when it is nonzero. Suggested skills 1.C, 2.B, 3.B and 3.C, although the framework's own sample question on this objective is tagged 2.D, and its exam-weighting page states that required content can be assessed with any skill. No boundary statement: Unit 6 prints exactly one, under Topic 6.5. Calculus differentiators against the identically titled AP Physics 1 Topic 6.4: conservation as a vector statement following from L as a cross product, the change as a time integral of net external torque so that a reversing torque can leave the endpoints matched, and the reason tau = I alpha fails for a shape-changing system while tau = dL/dt holds. Not printed on either course's sheet: any conservation-of-angular-momentum equation, K_rot = L squared over 2I, and tau_net = dL/dt. The framework's Vocabulary appendix section headed Constant or Conserved is the source for the constant against conserved distinction. Sample multiple-choice question 4, answer D, pairs skill 2.D with 6.4.A and 6.4.A.2.

What Topic 6.4 requires

Topic 6.4 carries two learning objectives and nine essential-knowledge statements.

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.

Statement	What it says
6.4.A.1	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 rotational axis
6.4.A.2	Any change to a system's angular momentum must be due to an interaction between the system and its surroundings
6.4.A.2.i	The angular impulse exerted by one object or system on a second is equal and opposite to the angular impulse exerted by the second on the first. This is a direct result of Newton's third law
6.4.A.2.ii	A system may be selected so that the total angular momentum of that system is constant
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 farther from the rotational axis
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
6.4.B.1	Angular momentum is conserved in all interactions
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

Suggested skills: 1.C, 2.B (calculate or estimate an unknown quantity with units from known quantities, by selecting and following a logical computational pathway), 3.B and 3.C. Topic 6.4 prints no boundary statement. Unit 6 prints exactly one, under Topic 6.5.

Worth noting that the suggested-skill list is a suggestion. The framework's sample multiple-choice question that aligns to 6.4.A and 6.4.A.2 is tagged with skill 2.D, which is not in the list above, and the exam-weighting page states plainly that required course content can be assessed with any skill.

Conserved and constant are different words, and the CED defines the difference

Two statements sit next to each other and look like they contradict:

6.4.B.1: "Angular momentum is conserved in all interactions."
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."

If angular momentum is conserved in all interactions, why does 6.4.B.2 attach a condition? Because the two sentences use two different words, and the framework's own appendix, "Vocabulary and Definitions of Important Ideas in AP Physics", has a section headed "Constant or Conserved?" that separates them.

That appendix works the distinction through a spilling box of cereal. The cereal is conserved in every choice of system: it does not cease to exist because it was not selected as part of the system. Whether the amount is constant depends on where you drew the boundary. Include the spilled pieces and it is constant; exclude them and it is not.

Angular momentum works the same way. Conserved is a statement about the universe: it is never created or destroyed, only transferred, which is 6.4.B.1. Constant is a statement about your chosen system over your chosen interval, and 6.4.B.2 and 6.4.B.3 give the test. Zero net external torque about the axis means constant; nonzero means transferred in or out.

On a free-response question the right word matters. "Angular momentum is conserved" on its own is not a justification, and the unit's Preparing for the AP Exam note says as much in general terms: simply referencing an equation, law, or physical principle is not sufficient. Name the system, name the axis, argue that the net external torque about that axis is zero, and only then say the angular momentum is constant.

It is a vector statement, and it is about one axis

Because this course defines angular momentum as $\vec{L} = \vec{r} \times \vec{p}$ , conservation here is a vector statement. Three consequences the algebra-based course cannot reach:

Components conserve separately. If the net external torque has no component along a particular direction, the component of $\vec{L}$ along that direction is constant even if the other components are not. A spinning top has a vertical torque component of zero from gravity, so $L_z$ holds while the horizontal components precess.

Cancellation is vector cancellation. Two identical wheels spinning in opposite senses on the same axle have total $\vec{L} = 0$ . Statement 6.4.A.1 says the total is the sum of the parts' angular momenta about that axis, and a sum of vectors can be zero when neither part is zero. Their kinetic energies, being scalars, still add.

A radial force exerts no torque. This is the cross-product reason behind most Topic 6.4 problems. If $\vec{F}$ is parallel or antiparallel to $\vec{r}$ , then $\vec{r} \times \vec{F} = 0$ . A skater pulling her arms straight in along the radius, a string shortening through a hole in a table, and gravity from a central body all fall into this category, and in every case the angular momentum about that centre is constant no matter how large the force is. The algebra-based course states the outcome; this one can show it in one line.

Statement 6.4.A.1 also fixes a bookkeeping requirement that costs marks when it is missed: the sum has to be taken about a single rotational axis. You cannot add a disk's angular momentum about its own centre to a second disk's angular momentum about a different centre and call the result the system total. Pick one axis first, express every part's contribution about that axis, then add.

Zero net torque, or a time integral that vanishes

Statement 6.4.A.2.iv gives the exact relationship: if the total angular momentum of a system changes, that change is equivalent to the angular impulse exerted on the system. Combined with 6.3.C.2.i, the calculus-based version of the whole topic is one line:

\Delta \vec{L} = \int_{t_1}^{t_2} \vec{\tau}_{\text{net,ext}}\, dt

Read as an integral, that says something the algebra-based $\Delta L = \tau\Delta t$ cannot. The angular momentum at the end of an interval equals the angular momentum at the start whenever the integral is zero, which is a weaker condition than the torque being zero throughout. A torque that pushes one way and then the other, with equal areas under the torque against time graph, leaves a system exactly where it started even though it was never torque free. Worked example 2 is a wheel driven by a sinusoidal torque that ends at rest after turning 3.82 rad.

Be careful with the wording, because the distinction is exam relevant. In that example the angular momentum is not constant during the interval: it rises and falls. What is true is that its net change over the full cycle is zero. Statement 6.4.B.2's implication runs one way only: zero net external torque gives constant angular momentum. Matching endpoints do not, by themselves, give you a torque-free interval.

The practical procedure for an exam question:

Choose the system and say what is in it.
Choose the axis and say where it is.
List the external forces and ask, for each, whether $\vec{r} \times \vec{F}$ about that axis is zero. Radial forces and forces acting at the axis both give zero.
If every external torque is zero, set $L_i = L_f$ . If not, evaluate $\int \tau\, dt$ and use 6.4.A.2.iv.

Step 3 is where the marks are. A force existing is not a torque existing, and this is the step the cross product makes rigorous.

The nonrigid case, which is where the calculus earns its place

Statement 6.4.A.2.iii is the figure-skater sentence: "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 farther from the rotational axis."

A system that changes shape has a rotational inertia that is a function of time. That single fact breaks one equation and leaves another standing, and knowing which is which is the calculus-based content of this topic. From 6.3.C.2.ii:

\tau_{\text{net}} = \frac{dL}{dt} = I\frac{d\omega}{dt} = I\alpha

with the framework's own qualifier attached: the theorem is a direct result of Newton's second law of motion for cases in which rotational inertia is constant. So for a shape-changing system:

$\tau_{\text{net}} = dL/dt$ still holds. It is the general statement.
$\tau_{\text{net}} = I\alpha$ does not. Pulling $I$ out of the derivative is what fails.
Written out properly, $\dfrac{dL}{dt} = \dfrac{d(I\omega)}{dt} = I\dfrac{d\omega}{dt} + \omega\dfrac{dI}{dt}$ , and the second term is exactly the one a constant- $I$ treatment throws away.

With zero external torque the left side is zero, so $I\frac{d\omega}{dt} = -\omega\frac{dI}{dt}$ . Rotational inertia falling means angular speed rising, in the exact proportion that keeps $I\omega$ fixed. That is the skater, and it is also the collapsing star in the framework's sample multiple-choice question 4.

The algebra-based course states the same outcome and reaches it with $I_1\omega_1 = I_2\omega_2$ , which is correct and which you should still use for the arithmetic. The difference is that this course can say why $\tau = I\alpha$ is not the tool for the job, and free-response questions in this course ask why.

Angular momentum constant, kinetic energy not

The standard follow-up question, and the one that separates a memorised answer from an understood one. Start from the two printed lines $K_{\text{rot}} = \frac{1}{2}I\omega^2$ and $L = I\omega$ and eliminate $\omega$ :

K_{\text{rot}} = \frac{1}{2}I\omega^2 = \frac{(I\omega)^2}{2I} = \frac{L^2}{2I}

That relation is not printed on the equation sheet, and it is two lines to derive, so derive it. It settles the whole question at a glance. Hold $L$ fixed and halve $I$ : the angular speed doubles and the kinetic energy doubles. Energy went up while angular momentum did not move.

Where did the energy come from? From work done inside the system. The skater's muscles pull her arms inward against the outward push those arms need to keep circling, and that force acts through a radial displacement, so it does positive work. This is not a violation of anything: angular momentum is protected by the absence of an external torque, and kinetic energy is protected by nothing at all. Nothing in Unit 6 says rotational kinetic energy is conserved.

A collision or a merge goes the other way. Two objects that end up rotating together have lost kinetic energy to internal deformation, exactly as in a perfectly inelastic linear collision, while their total angular momentum about the axis is unchanged. Worked example 3 loses 97% of the kinetic energy and none of the angular momentum.

Choosing the system, and the pivot that conserves L but not p

Statement 6.4.A.2.ii says a system may be selected so that its total angular momentum is constant. That is a licence, and using it well is most of the skill in this topic.

The most useful case in practice: a pivot exerts a force but no torque about itself. A rod hinged at one end and struck by something has a hinge force on it, sometimes a large one, so the linear momentum of the rod-plus-projectile system is not conserved. But the hinge force acts at the hinge, so $\vec{r} = 0$ for it about that point, and $\vec{r} \times \vec{F} = 0$ . Choose the axis at the hinge and angular momentum is constant even though linear momentum is not.

Worked example 3 puts numbers on that: the linear momentum of the system goes from $5.0$ to $5.61\ \mathrm{kg \cdot m/s}$ while the angular momentum about the hinge stays at exactly $3.0\ \mathrm{kg \cdot m^2/s}$ . Both statements are true at once and neither is a mistake.

Statement 6.4.B.3 covers the case you cannot engineer away: if the net external torque is nonzero, angular momentum is transferred between the system and the environment. The transfer is real and quantitative, and 6.4.A.2.iv gives its size. When a merry-go-round is braked, its angular momentum goes into the Earth, whose own angular momentum changes by an amount too small to measure but not zero. That is the sense in which 6.4.B.1 says angular momentum is conserved in all interactions.

One more consequence of Newton's third law, from 6.4.A.2.i: the angular impulse one object exerts on a second is equal and opposite to the angular impulse the second exerts on the first. Two objects interacting therefore have angular momentum changes that are equal and opposite, which is 6.4.A.1 applied to a two-part system and the reason the total holds.

What the equation sheet prints for Topic 6.4

The striking thing about this topic on the sheet is how little of it is there. Checked against the printed Table of Information:

Equation	On the C: Mechanics sheet
$\vec{L} = \vec{r} \times \vec{p} = I\vec{\omega}$	yes
$\Delta L = \int \tau\, dt$	yes
$I_{\text{tot}} = \sum I_i = \sum m_i r_i^2$	yes
$I = \int r^2\, dm$	yes
$I' = I_{\text{cm}} + Md^2$	yes
$K_{\text{rot}} = \frac{1}{2}I\omega^2$	yes
$L_i = L_f$ or $I_1\omega_1 = I_2\omega_2$	no
$K_{\text{rot}} = \frac{L^2}{2I}$	no
$\tau_{\text{net}} = \frac{dL}{dt}$	no
$\sum L_{\text{before}} = \sum L_{\text{after}}$	no

There is no conservation-of-angular-momentum equation printed on either course's sheet. The AP Physics 1 sheet does not print one either. That is not an oversight: a conservation law is a statement about a system and an interval, not a formula, and writing $I_1\omega_1 = I_2\omega_2$ commits you to a rigid body on a fixed axis, which is the special case rather than the rule.

What you write instead is $L_i = L_f$ for the system and axis you named, having justified it from 6.4.B.2. On a free-response question that justification is a scoring point in its own right.

The three rotational-inertia lines are on this list because most Topic 6.4 problems are rotational-inertia problems in disguise. Getting $I_1$ and $I_2$ right, including the parallel-axis theorem when a mass lands off centre, is where the arithmetic usually goes wrong, not in the conservation step.

How Topic 6.4 is tested, and if you are in AP Physics 1

Sample multiple-choice question 4 in the framework aligns to 6.4.A and essential knowledge 6.4.A.2, paired with skill 2.D, predicting new values or factors of change using functional dependence. A spherical star spinning at some initial angular velocity suddenly collapses to half its original radius with no loss of mass, uniform density before and after, and the question asks for the new angular velocity. The published answer is choice D. The route is $I \propto MR^2$ at fixed shape and mass, so quartering $R^2$ quarters $I$ and the angular speed goes up by four.

Skill 2.B carries 20 to 25% of the multiple-choice section and skill 3.B carries 15 to 25%, so a Topic 6.4 question is as likely to ask you to justify a claim as to compute a number. Skill 3.C, justifying with evidence, carries 5 to 10%.

The unit's third sample instructional activity belongs to 6.4 and describes a Create a Plan task: research the rotational inertia of a human body in different configurations, such as arms outstretched against arms pulled in, then obtain footage of an ice skater spinning and pulling in their arms and analyse it to see whether angular momentum is conserved. That is 6.4.A.2.iii turned into a measurement.

AP Physics 1 has a Topic 6.4 with the same title, and the conservation statement itself is word for word the same idea in both courses. The [AP Physics 1 Topic 6.4 page](/ap-physics-1/unit-6-energy-and-momentum-of-rotating-systems/6-4-conservation-of-angular-momentum) is for students in the algebra-based course, who need the system-and-axis reasoning and $I_1\omega_1 = I_2\omega_2$ . This page is for students in AP Physics C: Mechanics, who additionally need conservation as a vector statement, the vanishing time integral, and the reason $\tau = I\alpha$ fails for a shape-changing system while $\tau = dL/dt$ does not. The conceptual core is genuinely shared; the extra machinery here is what the free-response questions ask you to produce.

Where it leads: Topic 6.3 supplies the definitions this topic conserves, and Topic 6.6 is the unit's largest application of it, since gravity from a central body is radial and therefore exerts no torque about that body. The conservation of momentum guide owns the linear routine, and the momentum and collisions practice set has problems to work. The Unit 6 hub lists all six topics.

A skater pulling in: where the extra kinetic energy comes from

A skater spins on frictionless ice with arms out, with rotational inertia $I_1 = 4.5\ \mathrm{kg \cdot m^2}$ at $\omega_1 = 1.6$ rad/s. She pulls her arms in, reducing her rotational inertia to $I_2 = 1.8\ \mathrm{kg \cdot m^2}$ . Find (a) her new angular speed, (b) her rotational kinetic energy before and after, (c) the same answer from $K = L^2/2I$ , and (d) justify why the angular momentum did not change even though a large force was involved.

System: the skater alone. Axis: her vertical spin axis. Convention: her sense of rotation is positive.
(a) The ice is frictionless and the only other external force, gravity, acts along the axis, so the net external torque about the axis is zero and 6.4.B.2 gives constant angular momentum.
$L = I_1\omega_1 = (4.5)(1.6) = 7.2\ \mathrm{kg \cdot m^2/s}$ , so $\omega_2 = L/I_2 = 7.2/1.8 = 4.0\ \mathrm{rad/s}$ .
(b) $K_1 = \frac{1}{2}I_1\omega_1^2 = \frac{1}{2}(4.5)(2.56) = 5.76\ \mathrm{J}$ and $K_2 = \frac{1}{2}I_2\omega_2^2 = \frac{1}{2}(1.8)(16) = 14.4\ \mathrm{J}$ .
The kinetic energy rose by $8.64$ J, a factor of $14.4/5.76 = 2.5$ , which is exactly $I_1/I_2$ .
(c) Derive $K = L^2/2I$ by eliminating $\omega$ : $\frac{1}{2}I\omega^2 = \frac{(I\omega)^2}{2I} = \frac{L^2}{2I}$ . With $L^2 = 51.84$ , $K_1 = 51.84/9.0 = 5.76$ J and $K_2 = 51.84/3.6 = 14.4$ J. The digits match part (b).
That form also gives the general result at a glance: with $L$ fixed, $K$ scales as $1/I$ , so the energy ratio must equal the inverse rotational-inertia ratio, which is the 2.5 above.
(d) The force she uses to pull her arms in points along the radius, from her arms toward the axis. For a radial force, $\vec{r}$ and $\vec{F}$ are antiparallel, so $\vec{\tau} = \vec{r} \times \vec{F} = \vec{0}$ however large the force is. That is the cross-product justification, and it is stronger than saying the force is internal, because it holds even if you split her into two systems.
The $8.64$ J came from her muscles. Her arms are moving in circles, so keeping them there requires an inward force; pulling them inward means that force acts through an inward displacement, and that is positive work done on the rotating system. Angular momentum was protected by the absence of a torque; kinetic energy had nothing protecting it.

(a) $\omega_2 = 4.0$ rad/s. (b) $K$ rises from 5.76 J to 14.4 J. (c) $K = L^2/2I$ gives the same two numbers and shows the ratio must be $I_1/I_2 = 2.5$ . (d) The pulling force is radial, so its cross product with the position vector is exactly zero and it exerts no torque about the spin axis.

A torque that reverses: back where it started, and never torque free

A wheel with $I = 0.50\ \mathrm{kg \cdot m^2}$ is at rest at $t = 0$ . It is driven by $\tau(t) = \tau_0\sin\left(\frac{2\pi t}{T}\right)$ with $\tau_0 = 3.0\ \mathrm{N \cdot m}$ and $T = 2.0$ s, and nothing else acts on it. Find (a) the angular momentum as a function of time, (b) the greatest angular speed reached and when, (c) the angular momentum at $t = 2.0$ s, and (d) the total angle turned over that cycle. Then say precisely what is and is not constant.

Convention: positive torque acts in the sense the wheel first turns. Write $a = 2\pi/T = \pi\ \mathrm{rad/s}$ .
(a) Statement 6.4.A.2.iv with 6.3.C.2.i gives $\Delta L = \int_0^t \tau\, dt'$ , and the sheet's Calculus table prints $\int \sin(ax)\, dx = -\frac{1}{a}\cos(ax)$ .
$L(t) = \int_0^t \tau_0 \sin(a t')\, dt' = \frac{\tau_0}{a}\left[1 - \cos(at)\right] = \frac{3.0}{\pi}\left[1 - \cos(\pi t)\right]\ \mathrm{kg \cdot m^2/s}$ .
(b) The bracket is largest when $\cos(\pi t) = -1$ , at $t = 1.0$ s, giving $L_{\max} = \frac{2(3.0)}{\pi} = 1.910\ \mathrm{kg \cdot m^2/s}$ .
With $I$ constant, $\omega_{\max} = L_{\max}/I = 1.910/0.50 = 3.82\ \mathrm{rad/s}$ , reached at $t = 1.0$ s, which is when the torque passes back through zero.
(c) At $t = 2.0$ s, $\cos(2\pi) = 1$ , so $L = 0$ and the wheel is at rest again. The integral over the full cycle vanishes because the positive and negative areas under the torque against time graph are equal.
(d) $\omega(t) = L(t)/I = \frac{6.0}{\pi}\left[1 - \cos(\pi t)\right]$ , so $\theta = \int_0^{2.0}\omega\, dt = \frac{6.0}{\pi}\left[t - \frac{\sin(\pi t)}{\pi}\right]_0^{2.0} = \frac{6.0}{\pi}(2.0) = \frac{12}{\pi} = 3.82\ \mathrm{rad}$ .
The wheel turned 3.82 rad, or 219 degrees, and finished at rest.
Now the precise statement. Over the interval from 0 to 2.0 s, $\Delta L = 0$ . The angular momentum was not constant during that interval: it rose to 1.91 and came back. And the net external torque was nonzero at every instant except three.
So a matching pair of endpoints is not evidence of a torque-free interval. Statement 6.4.B.2 runs one way: zero net external torque gives constant angular momentum. The converse needs the whole time history, and this is the counterexample. The algebra-based $\Delta L = \tau\Delta t$ cannot even express this problem, because there is no single $\tau$ to put in it.

(a) $L(t) = \frac{\tau_0 T}{2\pi}\left[1 - \cos(2\pi t/T)\right] = \frac{3.0}{\pi}[1 - \cos(\pi t)]\ \mathrm{kg \cdot m^2/s}$ . (b) $\omega_{\max} = 3.82$ rad/s at $t = 1.0$ s. (c) $L = 0$ at $t = 2.0$ s, so the net change over the cycle is zero. (d) 3.82 rad. The angular momentum returned to its starting value without ever having been constant, and without the torque ever being zero for a finite stretch.

A bullet into a hinged rod: angular momentum holds, linear momentum does not

A uniform rod of mass $M = 1.2$ kg and length $L = 0.80$ m hangs at rest from a frictionless hinge at its upper end. A bullet of mass $m = 0.020$ kg travelling horizontally at $v = 250$ m/s strikes the rod perpendicular to it at $d = 0.60$ m below the hinge and embeds itself. Find (a) the angular momentum of the system about the hinge just before impact, (b) the angular speed just after, (c) the fraction of the kinetic energy that survives, and (d) check whether the system's linear momentum is conserved.

System: rod plus bullet. Axis: the hinge. Convention: the sense the bullet drives the rod is positive.
(a) The bullet travels in a straight line, so use 6.3.A.2 with the printed cross-product magnitude: $L_i = \lvert \vec{r} \times \vec{p} \rvert = d\,mv\sin 90^\circ = (0.60)(0.020)(250) = 3.0\ \mathrm{kg \cdot m^2/s}$ . The rod contributes nothing, being at rest.
(b) During the impact the external forces are gravity and the hinge force. Gravity acts through the centre of mass but the impact is brief, so its angular impulse is negligible; the hinge force acts at the axis, where $\vec{r} = \vec{0}$ , so its torque about the hinge is exactly zero. By 6.4.B.2 the angular momentum about the hinge is constant through the collision.
Afterwards the rod and bullet rotate together, so use 6.4.A.1 to add the parts about that one axis: $I_{\text{tot}} = \frac{1}{3}ML^2 + md^2 = \frac{1}{3}(1.2)(0.64) + (0.020)(0.36) = 0.2560 + 0.0072 = 0.2632\ \mathrm{kg \cdot m^2}$ .
$\omega = L_i / I_{\text{tot}} = 3.0/0.2632 = 11.398 \approx 11.4\ \mathrm{rad/s}$ .
(c) $K_i = \frac{1}{2}mv^2 = \frac{1}{2}(0.020)(62500) = 625\ \mathrm{J}$ .
$K_f = \frac{1}{2}I_{\text{tot}}\omega^2 = \frac{1}{2}(0.2632)(129.92) = 17.10\ \mathrm{J}$ .
The surviving fraction is $17.10/625 = 0.0274$ , so 2.7% remains and 608 J went into deformation and heating. Angular momentum was constant to every digit while 97% of the kinetic energy disappeared, which is the rotational version of a perfectly inelastic collision.
(d) Before the impact the system's linear momentum is $mv = (0.020)(250) = 5.0\ \mathrm{kg \cdot m/s}$ . Afterwards the rod's centre of mass, at $0.40$ m from the hinge, moves at $\omega(0.40) = 4.559$ m/s, carrying $(1.2)(4.559) = 5.471\ \mathrm{kg \cdot m/s}$ , and the bullet at $0.60$ m moves at $\omega(0.60) = 6.839$ m/s, carrying $(0.020)(6.839) = 0.137\ \mathrm{kg \cdot m/s}$ .
Total after: $5.61\ \mathrm{kg \cdot m/s}$ , against $5.0$ before. Linear momentum is not conserved, and the difference of $0.61\ \mathrm{N \cdot s}$ is the impulse the hinge delivered.
Both results are correct at once, and that is the point of 6.4.A.2.ii. The hinge force ruins linear momentum conservation and leaves angular momentum about the hinge untouched, because a force applied at the axis has zero lever arm about that axis. Choosing the hinge as the axis is what makes the problem solvable in one line.

(a) $L_i = 3.0\ \mathrm{kg \cdot m^2/s}$ about the hinge. (b) $\omega = 11.4$ rad/s. (c) 2.7% of the kinetic energy survives; 625 J becomes 17.1 J. (d) No. Linear momentum goes from 5.0 to $5.61\ \mathrm{kg \cdot m/s}$ , the extra $0.61\ \mathrm{N \cdot s}$ coming from the hinge, which exerts a force but no torque about itself.

Frequently asked questions

When is angular momentum conserved in AP Physics C?

Essential knowledge 6.4.B.2 gives the test: if the net external torque exerted on the selected object or rigid system is zero, the total angular momentum of that system is constant. Both qualifiers matter. The torque has to be external to the system you chose, and it has to be reckoned about the axis you chose. In AP Physics C the change in angular momentum is the time integral of the net external torque, so the angular momentum at the end of an interval also matches the start whenever that integral vanishes, even if the torque was not zero throughout.

What is the difference between angular momentum being conserved and being constant?

The AP framework's vocabulary appendix separates the two words in a section headed Constant or Conserved. Conserved means never created or destroyed, only transferred, which is why statement 6.4.B.1 can say angular momentum is conserved in all interactions with no conditions attached. Constant means unchanging for the particular system you selected over the interval you are looking at, and that does have a condition: zero net external torque about your axis, per 6.4.B.2. Choose a different system boundary and the same physical situation can have angular momentum that is constant or not.

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

Essential knowledge 6.4.A.2.iii states it: 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 farther from the rotational axis. Pulling her arms in moves mass toward the axis, which lowers her rotational inertia, and with no external torque about the axis the product of rotational inertia and angular speed has to stay the same, so the angular speed rises. The force she pulls with points along the radius, so its cross product with the position vector is zero and it exerts no torque however large it is.

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

No, it increases. Eliminating angular speed between the two printed equations for rotational kinetic energy and angular momentum gives kinetic energy equal to the angular momentum squared divided by twice the rotational inertia. With angular momentum held fixed, halving the rotational inertia doubles the kinetic energy. The extra energy comes from work her muscles do pulling her arms inward against the outward force those arms need to keep circling. Angular momentum is protected by the absence of an external torque; rotational kinetic energy is protected by nothing, and no statement in Unit 6 claims otherwise.

Why does the equation for angular acceleration fail when rotational inertia changes?

Because pulling the rotational inertia outside a time derivative assumes it is a constant. Essential knowledge 6.3.C.2.ii writes net torque as the time derivative of angular momentum, then as rotational inertia times angular acceleration, and attaches the qualifier that the second form is a direct result of Newton's second law for cases in which rotational inertia is constant. Differentiating the product properly gives rotational inertia times angular acceleration plus angular speed times the rate of change of rotational inertia, and it is that second term the constant-inertia version discards. For a system changing shape, only the derivative form is valid.

Is angular momentum conserved when a bullet hits a rod on a hinge?

About the hinge, yes, and this is the standard reason to pick the hinge as the axis. The hinge exerts a force on the rod, sometimes a large one, but that force acts at the axis, so its position vector relative to the axis is zero and its torque about the axis is exactly zero. Gravity's angular impulse over a brief impact is negligible. Linear momentum, by contrast, is not conserved, because the hinge force is external and does change the total. Both statements are true at once, and neither is a mistake.

Is there a conservation of angular momentum equation on the AP Physics C equation sheet?

No, and there is not one on the AP Physics 1 sheet either. What the AP Physics C: Mechanics sheet prints is the definition of angular momentum as a cross product, the change in angular momentum as the time integral of the torque, and three ways of finding a rotational inertia. Setting the initial angular momentum equal to the final one is a step you take after arguing that the net external torque about your chosen axis is zero, and on a free-response question that argument is itself worth credit. Writing the product of rotational inertia and angular speed as equal before and after also quietly commits you to a rigid body on a fixed axis.