AP Physics C: Mechanics · Topic 6.3

Topic 6.3: Angular Momentum and Angular Impulse

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

In AP Physics C: Mechanics the angular momentum of an object about a point is the cross product of its position vector with its linear momentum, so it has a direction. Angular impulse is the integral of torque over time, and it equals the change in angular momentum.

AP Physics: Unit 6 (topics 6.3 Angular Momentum and Angular Impulse). Topic 6.3 of the current AP Physics C: Mechanics course and exam description, inside Unit 6, weighted 10 to 15% of the multiple-choice section. Three learning objectives, 6.3.A, 6.3.B and 6.3.C, tying Topic 6.5 for the most in the unit, and thirteen essential-knowledge statements counting sub-statements, more than any other topic in Unit 6. Key statements: 6.3.A.1 gives L = I omega as a magnitude about a specific axis; 6.3.A.2 defines the angular momentum about a point as the cross product of position with linear momentum; 6.3.A.2.i and .ii cover the axis dependence and straight-line motion; 6.3.B.1 defines angular impulse with the relevant equation integral of tau dt; 6.3.B.2 gives its direction; 6.3.B.3, 6.3.C.3 and 6.3.C.4 are the three graph statements; 6.3.C.1 gives delta L as a difference; 6.3.C.2.i is the theorem and 6.3.C.2.ii the differential form with its constant-rotational-inertia qualifier. Suggested skills 1.C, 2.A, 2.C and 3.C. No boundary statement: Unit 6 prints exactly one, under Topic 6.5. Calculus differentiator against the identically titled AP Physics 1 Topic 6.3: angular momentum is a genuine cross product with a direction against two Physics 1 scalars, and angular impulse is a time integral against tau delta t. Sheet detail verified at 300 dpi against the printed Table of Information: the vector line carries arrows on L, r, p and omega but not on I, while the impulse line delta L equals the integral of tau dt carries no arrows and no limits. Printed in the Vectors table of the same appendix and commonly missing from transcriptions: the cross-product magnitude AB sin theta. Not printed: tau_net = dL/dt, delta L = L minus L naught, angular impulse as its own named line, and L = rmv sin theta. Sample multiple-choice question 14, answer C, pairs skill 2.A with 6.3.A and 6.3.A.2.

What Topic 6.3 requires

Topic 6.3 carries three learning objectives, tying Topic 6.5 for the most in Unit 6, and thirteen essential-knowledge statements, which is more than any other topic in the unit. Counted individually: five under 6.1, three under 6.2, thirteen here, nine under 6.4, five under 6.5 and nine under 6.6.

Objective	What it asks
6.3.A	Describe the angular momentum of an object or rigid system
6.3.B	Describe the angular impulse delivered to an object or rigid system by a torque
6.3.C	Relate the change in angular momentum of an object or rigid system to the angular impulse given to that object or rigid system

Statement	What it says
6.3.A.1	The magnitude of the angular momentum of a rigid system about a specific axis can be described with $L = I\omega$
6.3.A.2	The angular momentum of an object about a given point is $\vec{L} = \vec{r} \times \vec{p}$
6.3.A.2.i	The selection of the axis about which an object is considered to rotate influences the determination of the angular momentum of that object
6.3.A.2.ii	The measured angular momentum of an object traveling in a straight line depends on the distance between the reference point and the object, the mass, the speed, and the angle between the radial distance and the velocity
6.3.B.1	Angular impulse is defined as the product of the torque and the time interval during which it is exerted. Relevant equation: $\int \tau\, dt$
6.3.B.2	Angular impulse has the same direction as the torque imparting it
6.3.B.3	It can be found from the area under a graph of the torque as a function of time
6.3.C.1	$\Delta L = L - L_0$
6.3.C.2	A rotational form of the impulse-momentum theorem relates the angular impulse delivered and the change in angular momentum
6.3.C.2.i	The angular impulse equals the change in angular momentum. Relevant equation: $\Delta L = \int_{t_1}^{t_2} \tau\, dt$
6.3.C.2.ii	The theorem is a direct result of Newton's second law for cases in which rotational inertia is constant: $\tau_{\text{net}} = \frac{dL}{dt} = I\frac{d\omega}{dt} = I\alpha$
6.3.C.3	The net torque equals the slope of a graph of angular momentum against time
6.3.C.4	The angular impulse equals the area under a graph of the net external torque against time

Suggested skills: 1.C, 2.A, 2.C and 3.C. Topic 6.3 prints no boundary statement. Unit 6 prints exactly one, under Topic 6.5.

The cross product is the whole difference

AP Physics 1's equation sheet prints two scalar angular momentum lines, $L = I\omega$ and $L = rmv\sin\theta$ . The AP Physics C: Mechanics sheet prints one vector line:

\vec{L} = \vec{r} \times \vec{p} = I\vec{\omega}

Rendered at high resolution off the printed Table of Information, the arrows sit on $\vec{L}$ , $\vec{r}$ , $\vec{p}$ and $\vec{\omega}$ , and not on $I$ , which is a scalar. The magnitudes the two courses compute agree. Only one of the two sheets tells you which way the answer points.

That single change opens a class of question the algebra-based course cannot ask:

Which way does it point? Right-hand rule: fingers along $\vec{r}$ , curl toward $\vec{p}$ , thumb gives $\vec{L}$ . For a body spinning in a plane, $\vec{L}$ is along the rotation axis, and the sense is the one your fingers curl with.
Add two angular momenta that are not parallel. Two flywheels on perpendicular axles have a total angular momentum of magnitude $\sqrt{L_1^2 + L_2^2}$ , pointing along neither axle. Scalar forms cannot express that.
Two counter-rotating wheels. Their angular momenta cancel to zero while their rotational kinetic energies add, because energy is a scalar and 6.1.A.3 says so.
Full three-dimensional geometry. With three components in each input, the cross product has three components out. Worked example 1 evaluates one.

Notice what is not on the C: Mechanics sheet: the scalar $L = rmv\sin\theta$ that AP Physics 1 prints. You are not missing it, because the same appendix prints, in its Vectors table,

\lvert \vec{A} \times \vec{B} \rvert = AB\sin\theta

Apply that to $\vec{r} \times \vec{p}$ and the Physics 1 scalar drops out immediately as $\lvert \vec{L} \rvert = rp\sin\theta = rmv\sin\theta$ . This is worth knowing because equation-sheet transcriptions and cheat sheets often omit the Vectors, Calculus, Geometry and Trigonometry tables, so it is easy to read a partial copy of the sheet and conclude the cross-product magnitude rule is absent. It is printed.

Two forms, and when the scalar one applies

Statements 6.3.A.1 and 6.3.A.2 are not two ways of saying one thing.

6.3.A.1 is careful in its wording: "The magnitude of the angular momentum of a rigid system about a specific axis can be described with the equation $L = I\omega$ ." Note magnitude, note about a specific axis, and note rigid system.

6.3.A.2 is the general definition: the angular momentum of an object about a given point is $\vec{L} = \vec{r} \times \vec{p}$ . It applies to anything with momentum, spinning or not, rigid or not.

The relationship: for a rigid body rotating about a fixed symmetry axis, summing $\vec{r} \times \vec{p}$ over all the mass elements gives a resultant along the axis with magnitude $I\omega$ , which is why the sheet writes both on one line. Away from that special case they are not interchangeable, and $I\omega$ gives only the component of $\vec{L}$ along the axis you chose. Worked example 3 builds a case where the axial component is $1.2\ \mathrm{kg \cdot m^2/s}$ while the full vector has magnitude $1.5\ \mathrm{kg \cdot m^2/s}$ .

Practical rule for the exam. Use $L = I\omega$ when a rigid body turns about a stated axis and you have its rotational inertia about that axis. Use $\vec{L} = \vec{r} \times \vec{p}$ for a particle, for a point mass on the end of something, when the reference point is off the rotation axis, or when a direction is asked for.

The axis is part of the answer (6.3.A.2.i and 6.3.A.2.ii)

Statement 6.3.A.2.i is short and consequential: the selection of the axis about which an object is considered to rotate influences the determination of the angular momentum of that object.

An angular momentum with no stated reference point is not a number. Change the point and you change $\vec{r}$ , so you change $\vec{r} \times \vec{p}$ . Two students can compute different angular momenta for the same object at the same instant and both be right.

Statement 6.3.A.2.ii applies that to the case most people find counterintuitive, and it is worth reading in full: "The measured angular momentum of an object traveling in a straight line depends on the distance between the reference point and the object, the mass of the object, the speed of the object, and the angle between the radial distance and the velocity of the object."

So a puck sliding in a straight line at constant speed has angular momentum, and it is constant. Nothing is spinning. The cross product explains it: about any reference point off the line of travel, $\lvert \vec{r} \times \vec{p} \rvert = r p \sin\theta$ , and the combination $r\sin\theta$ is just the perpendicular distance from the point to the line, which does not change as the puck slides. About a point on the line, $\vec{r}$ and $\vec{p}$ are parallel, $\sin\theta = 0$ , and the angular momentum is exactly zero.

Both answers are constant, which they must be: 6.4.B.2 says the total angular momentum of a selected system is constant when the net external torque on it is zero, and a puck sliding freely has no torque about any point.

The practical habit: write down your reference point before you write down any angular momentum. Then use the same point for the initial and final states of the problem. Switching points midway is the defect that a sign error usually turns out to be.

Angular impulse is an integral, whatever the sentence says

Statement 6.3.B.1 is worth quoting exactly, because its prose and its equation do not agree in generality. The sentence reads: "Angular impulse is defined as the product of the torque exerted on an object or rigid system and the time interval during which the torque is exerted." The relevant equation printed directly beneath it reads:

\text{angular impulse} = \int \tau\, dt

Take the equation as the operative statement, as the AP Physics 1 sheet's $\Delta L = \tau\Delta t$ shows what a genuine product form looks like. The word "product" in that sentence describes the constant-torque case; the integral is what this course actually gives you, and a Physics C question is free to specify a torque that varies with time.

What the integral buys, exactly as in Topic 6.2: a torque that is not constant. Worked example 2 uses $\tau(t) = \tau_0 e^{-t/T}$ , which is tractable because the Calculus table in the same appendix prints $\int e^{ax}\, dx = \frac{1}{a}e^{ax}$ .

Statement 6.3.B.2 supplies the direction: angular impulse has the same direction as the torque imparting it. That matters when the torque changes direction during the interval, because then the vector integral is not the integral of the magnitudes. For the one-axis problems this course usually poses, the direction reduces to a sign, and the sign convention you declared at the start does the work.

The units are $\mathrm{N \cdot m \cdot s}$ , which is $\mathrm{kg \cdot m^2/s}$ , the units of angular momentum. That is not a coincidence; it is the theorem in the next section.

The rotational impulse-momentum theorem, and its one qualifier

Statement 6.3.C.2 says a rotational form of the impulse-momentum theorem relates the angular impulse delivered and the change in angular momentum. Its two sub-statements are where the useful detail is.

6.3.C.2.i gives the theorem itself: the angular impulse exerted on an object or rigid system is equal to the change in angular momentum of that object or rigid system, with the relevant equation

\Delta L = \int_{t_1}^{t_2} \tau\, dt

That exact line is printed on the equation sheet, without limits. Worth noting precisely: on the printed sheet this line carries no vector arrows, unlike the $\vec{L} = \vec{r} \times \vec{p} = I\vec{\omega}$ line above it. The framework writes it the same way. So the sheet asserts the vector nature in the definition and then states the theorem in scalar form, which is the form you will use on a fixed-axis problem.

6.3.C.2.ii supplies the differential form, and the qualifier attached to it is the sentence to memorise: "The rotational form of the impulse-momentum theorem is a direct result of Newton's second law of motion for cases in which rotational inertia is constant."

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

Read that chain from the left. The first equality, $\tau_{\text{net}} = dL/dt$ , always holds. The second one needs $I$ to be constant so that it can be pulled outside the derivative. A skater pulling her arms in is exactly the case where it cannot: $I$ is a function of time, so $\tau_{\text{net}} = I\alpha$ fails while $\tau_{\text{net}} = dL/dt$ still holds. That distinction is the calculus-based version of Unit 6's central conceptual point, and the algebra-based course has no equation that can state it.

Note also that $\tau_{\text{net}} = dL/dt$ is not printed on the equation sheet. What is printed is $\alpha_{\text{sys}} = \frac{\sum \tau}{I_{\text{sys}}} = \frac{\tau_{\text{net}}}{I_{\text{sys}}}$ , which is the $I\alpha$ end of the chain rearranged, and which therefore already assumes constant $I$ . If a question involves a changing rotational inertia, you write the derivative form yourself, and 6.3.C.2.ii is your citation for it.

Three graph statements, and which is which

Topic 6.3 contains three separate graph-reading statements plus one inherited from Topic 6.2. They are close enough together to be confused, so here they are side by side.

Graph	Read	Gives	Statement
torque against angular position	area	work, in joules	6.2.A.3
torque against time	area	angular impulse	6.3.B.3
net external torque against time	area	angular impulse	6.3.C.4
angular momentum against time	slope	net torque	6.3.C.3

The distinction between 6.3.B.3 and 6.3.C.4 is which torque is plotted. Under a graph of one torque you get the angular impulse that torque delivered. Under a graph of the net external torque you get the total change in angular momentum. If a wheel has a drive torque and a friction torque and the graph shows only the drive torque, its area is not $\Delta L$ .

The slope statement, 6.3.C.3, is the differential form drawn: differentiate the $L$ against $t$ curve and you have a torque that never appears in the problem text. A flat stretch means zero net torque and constant angular momentum, which is 6.4.B.2 read off a picture.

Skill 1.C, creating qualitative sketches of graphs, is listed for this topic and belongs to Science Practice 1, which is assessed on the free-response section rather than the multiple-choice section and carries 20 to 35% there. Sketching the $L$ against $t$ curve that matches a described torque, with the right curvature and the right flat regions, is what 1.C asks for here.

What the equation sheet prints for Topic 6.3

Checked line by line against the printed Table of Information in the AP Physics C: Mechanics course and exam description, including its third page, which carries the Geometry, Trigonometry, Vectors, Calculus and Identities tables:

Equation	On the C: Mechanics sheet
$\vec{L} = \vec{r} \times \vec{p} = I\vec{\omega}$	yes, one line, arrows on all four vectors
$\Delta L = \int \tau\, dt$	yes, without limits and without arrows
$\lvert \vec{A} \times \vec{B} \rvert = AB\sin\theta$	yes, in the Vectors table
$\vec{\tau} = \vec{r} \times \vec{F}$	yes
$\vec{p} = m\vec{v}$	yes
$\vec{F}_{\text{net}} = \frac{d\vec{p}}{dt}$	yes, the translational counterpart
$\alpha_{\text{sys}} = \frac{\sum \tau}{I_{\text{sys}}} = \frac{\tau_{\text{net}}}{I_{\text{sys}}}$	yes
$\vec{r} = A\hat{i} + B\hat{j} + C\hat{k}$	yes, in the Vectors table
angular impulse $= \int \tau\, dt$ as its own named line	no, only in the theorem form above
$\Delta L = L - L_0$	no
$\tau_{\text{net}} = \frac{dL}{dt}$	no
$L = rmv\sin\theta$	no, that is the AP Physics 1 sheet
$L = I\omega$ as a standalone scalar line	no, only as the right-hand end of the vector line

Three things follow. $\tau_{\text{net}} = dL/dt$ is the one absence that matters, because it is the only equation that survives when rotational inertia changes; write it from 6.3.C.2.ii. The Physics 1 scalar is derivable in one step from the cross-product line and the printed cross-product magnitude, so a question phrased in Physics 1 language is still answerable from this sheet. The Calculus table is on the sheet, with the power rule, the exponential and logarithmic integrals and the sine and cosine integrals, which is a strong hint about the shape of the $\tau(t)$ functions the exam uses.

How Topic 6.3 is tested, and if you are in AP Physics 1

Sample multiple-choice question 14 in the framework aligns to 6.3.A and essential knowledge 6.3.A.2, paired with skill 2.A. Two uniform disks, A and B, each of mass $M$ , have radii $2R$ and $R$ . Disk A spins in a horizontal plane about its center at $\omega_0$ ; disk B is held at rest, then released and falls onto the center of disk A so that both spin together. The question asks for the final angular velocity. Its stem supplies the rotational inertia of a uniform disk, and the published answer is choice C. The route is conservation of angular momentum with $I_A = \frac{1}{2}M(2R)^2 = 2MR^2$ and $I_B = \frac{1}{2}MR^2$ , giving $\omega_f = \frac{2MR^2}{2.5MR^2}\omega_0 = \frac{4}{5}\omega_0$ . A disk of twice the radius has four times the rotational inertia of one with the same mass, which is the scaling the question is really testing.

Skill 2.A carries 25 to 30% of the multiple-choice section, the largest single share, and 2.C carries 10 to 15%. That describes Topic 6.3 questions well: derive a symbolic expression, or compare one quantity at two times or about two points.

The unit's second sample instructional activity is a 6.3 activity: hand students a set of fidget spinners and ask them to explain why it is difficult to change the plane of rotation of a spinner while it is spinning. That question has no answer in scalar language. It is the direction of $\vec{L}$ , and the torque needed to change it, which is why the activity sits under this topic rather than under Topic 6.1.

AP Physics 1 has a Topic 6.3 with the same title. The [AP Physics 1 Topic 6.3 page](/ap-physics-1/unit-6-energy-and-momentum-of-rotating-systems/6-3-angular-momentum-and-angular-impulse) is for students in the algebra-based course, who need two scalar forms and the area under a torque against time graph. This page is for students in AP Physics C: Mechanics, who need the cross product, the direction, the time integral, and the qualifier on $\tau_{\text{net}} = I\alpha$ . If you are in Physics 1, the cross-product geometry on this page is not on your exam.

Where Topic 6.3 leads: Topic 6.4 turns the theorem into a conservation law, and Topic 6.2 is the other integral of the same torque. The impulse-momentum theorem guide owns the translational routine, and linear compared with angular velocity is the two-minute version of the dictionary between them. The Unit 6 hub lists all six topics.

A full three-dimensional cross product, checked two ways

A particle of mass $m = 0.50$ kg is at position $\vec{r} = (3.0\hat{i} - 2.0\hat{j} + 1.0\hat{k})$ m relative to the origin and moves with velocity $\vec{v} = (4.0\hat{i} + 5.0\hat{j} + 2.0\hat{k})$ m/s. Find (a) its angular momentum about the origin as a vector, (b) the magnitude of that angular momentum, and (c) confirm the magnitude using the printed cross-product magnitude rule and the angle between the two vectors.

First the linear momentum, from the printed $\vec{p} = m\vec{v}$ : $\vec{p} = 0.50(4.0\hat{i} + 5.0\hat{j} + 2.0\hat{k}) = (2.0\hat{i} + 2.5\hat{j} + 1.0\hat{k})\ \mathrm{kg \cdot m/s}$ .
(a) Statement 6.3.A.2 gives $\vec{L} = \vec{r} \times \vec{p}$ . Component by component: $L_x = r_y p_z - r_z p_y = (-2.0)(1.0) - (1.0)(2.5) = -4.5$ .
$L_y = r_z p_x - r_x p_z = (1.0)(2.0) - (3.0)(1.0) = -1.0$ .
$L_z = r_x p_y - r_y p_x = (3.0)(2.5) - (-2.0)(2.0) = 7.5 + 4.0 = 11.5$ .
$\vec{L} = (-4.5\hat{i} - 1.0\hat{j} + 11.5\hat{k})\ \mathrm{kg \cdot m^2/s}$ . Nothing here is spinning, and the angular momentum is nonzero in all three components.
(b) $\lvert \vec{L} \rvert = \sqrt{(-4.5)^2 + (-1.0)^2 + (11.5)^2} = \sqrt{20.25 + 1.00 + 132.25} = \sqrt{153.5} = 12.39\ \mathrm{kg \cdot m^2/s}$ .
(c) The Vectors table on the sheet prints $\lvert \vec{A} \times \vec{B} \rvert = AB\sin\theta$ , so find $\theta$ from the dot product on the line above it, $\vec{A} \cdot \vec{B} = AB\cos\theta$ .
$\vec{r} \cdot \vec{p} = (3.0)(2.0) + (-2.0)(2.5) + (1.0)(1.0) = 6.0 - 5.0 + 1.0 = 2.0$ . With $r = \sqrt{14} = 3.7417$ m and $p = \sqrt{11.25} = 3.3541\ \mathrm{kg \cdot m/s}$ , $\cos\theta = 2.0/12.550 = 0.15936$ , so $\theta = 80.83^\circ$ and $\sin\theta = 0.98722$ .
$rp\sin\theta = (12.550)(0.98722) = 12.39\ \mathrm{kg \cdot m^2/s}$ , matching part (b).
Sanity check the geometry: $\vec{L}$ should be perpendicular to both $\vec{r}$ and $\vec{p}$ . $\vec{L} \cdot \vec{r} = (-4.5)(3.0) + (-1.0)(-2.0) + (11.5)(1.0) = -13.5 + 2.0 + 11.5 = 0$ , as required.
This calculation is the one that has no counterpart in the algebra-based course. Its sheet's $L = rmv\sin\theta$ would give the 12.39, and nothing at all about the direction.

(a) $\vec{L} = (-4.5\hat{i} - 1.0\hat{j} + 11.5\hat{k})\ \mathrm{kg \cdot m^2/s}$ . (b) $12.39\ \mathrm{kg \cdot m^2/s}$ . (c) $rp\sin\theta$ with $\theta = 80.83^\circ$ gives the same 12.39, and $\vec{L} \cdot \vec{r} = 0$ confirms the direction is perpendicular to both inputs.

A torque that dies away, integrated over time

A flywheel with rotational inertia $I = 2.5\ \mathrm{kg \cdot m^2}$ is at rest at $t = 0$ . A motor applies a torque $\tau(t) = \tau_0 e^{-t/T}$ with $\tau_0 = 8.0\ \mathrm{N \cdot m}$ and $T = 4.0$ s, and there is no other torque. Find (a) the angular impulse delivered by $t = 4.0$ s and the angular speed then, (b) the angular speed the flywheel approaches as time goes on, (c) the time at which it reaches half of that final angular speed, and (d) verify statement 6.3.C.3 by differentiating your result.

Convention: the torque and the resulting rotation are both positive, and $I$ is constant, so 6.3.C.2.ii applies in full.
(a) Statement 6.3.B.1 gives the angular impulse as $\int \tau\, dt$ , and the sheet's Calculus table prints $\int e^{ax}\, dx = \frac{1}{a}e^{ax}$ .
$\Delta L(t) = \int_0^t \tau_0 e^{-t'/T}\, dt' = \tau_0 T\left(1 - e^{-t/T}\right)$ .
At $t = 4.0$ s, which is one time constant, $\Delta L = (8.0)(4.0)(1 - e^{-1}) = 32(1 - 0.36788) = 32(0.63212) = 20.228\ \mathrm{kg \cdot m^2/s}$ .
By 6.3.C.2.i the angular impulse is the change in angular momentum, and the wheel started at rest, so $L = 20.228\ \mathrm{kg \cdot m^2/s}$ . With $I$ constant, $\omega = L/I = 20.228/2.5 = 8.091 \approx 8.09\ \mathrm{rad/s}$ .
(b) As $t$ grows, $e^{-t/T} \to 0$ and $\Delta L \to \tau_0 T = (8.0)(4.0) = 32\ \mathrm{kg \cdot m^2/s}$ , so $\omega \to 32/2.5 = 12.8\ \mathrm{rad/s}$ .
The torque never reaches exactly zero and the wheel never quite reaches 12.8 rad/s. A finite total angular impulse from a torque that lasts forever is the shape of result the product form $\tau\Delta t$ cannot produce at all: it would give an unbounded answer.
(c) Set $\tau_0 T(1 - e^{-t/T}) = \frac{1}{2}\tau_0 T$ , so $e^{-t/T} = 0.5$ and $t = T\ln 2 = 4.0(0.6931) = 2.77$ s.
Note that the half-way time does not depend on $\tau_0$ or on $I$ , only on $T$ . Skill 2.D, predicting new values using functional dependence, is exactly this kind of reading.
(d) Statement 6.3.C.3 says the net torque is the slope of the angular momentum against time graph. Differentiate the answer to part (a): $\frac{dL}{dt} = \tau_0 T \cdot \frac{1}{T}e^{-t/T} = \tau_0 e^{-t/T}$ , which is $\tau(t)$ exactly.
Numerically at $t = 4.0$ s the slope is $8.0e^{-1} = 2.94\ \mathrm{N \cdot m}$ , and the graph of $L$ against $t$ is a curve that flattens toward the horizontal line at 32, never reaching it.

(a) $\Delta L = 20.2\ \mathrm{kg \cdot m^2/s}$ and $\omega = 8.09$ rad/s at $t = 4.0$ s. (b) The angular speed approaches $\tau_0 T / I = 12.8$ rad/s. (c) At $t = T\ln 2 = 2.77$ s, independent of both the peak torque and the rotational inertia. (d) Differentiating $L(t)$ returns $\tau_0 e^{-t/T}$ , which is the applied torque, confirming 6.3.C.3.

Where the scalar form loses information

A 0.30 kg ball on a light string moves at a constant 5.0 m/s in a horizontal circle of radius 0.80 m. Find its angular momentum (a) about the center of the circle, using the cross product, (b) about that same center using $L = I\omega$ , and (c) about a point $O$ on the rotation axis but 0.60 m below the plane of the circle. Then say what parts (a) and (c) tell you about statement 6.3.A.2.i.

Set axes at the instant in question: let $\hat{x}$ point from the axis out to the ball, $\hat{y}$ point along the ball's velocity, and $\hat{z}$ point up along the rotation axis. These three are mutually perpendicular at that instant.
$p = mv = (0.30)(5.0) = 1.5\ \mathrm{kg \cdot m/s}$ , directed along $+\hat{y}$ .
(a) About the center of the circle, $\vec{r} = (0.80\ \mathrm{m})\hat{x}$ , perpendicular to $\vec{p}$ , so $\lvert \vec{L} \rvert = rp\sin 90^\circ = (0.80)(1.5) = 1.2\ \mathrm{kg \cdot m^2/s}$ .
Direction by the right-hand rule: $\hat{x} \times \hat{y} = \hat{z}$ , so $\vec{L} = (1.2\ \mathrm{kg \cdot m^2/s})\hat{z}$ , straight up the rotation axis.
(b) Statement 6.3.A.1 with the ball treated as a point mass on a circle: $I = mr^2 = (0.30)(0.80)^2 = 0.192\ \mathrm{kg \cdot m^2}$ and $\omega = v/r = 5.0/0.80 = 6.25\ \mathrm{rad/s}$ , so $L = I\omega = (0.192)(6.25) = 1.2\ \mathrm{kg \cdot m^2/s}$ . The two forms agree, as the single sheet line says they should for this case.
(c) About $O$ , the position vector gains a vertical component: $\vec{r}_O = (0.80\hat{x} + 0.60\hat{z})$ m, with $\lvert \vec{r}_O \rvert = \sqrt{0.64 + 0.36} = 1.00$ m. The momentum is unchanged.
$\vec{L}_O = \vec{r}_O \times \vec{p} = (0.80\hat{x} + 0.60\hat{z}) \times (1.5\hat{y})$ . Using $\hat{x} \times \hat{y} = \hat{z}$ and $\hat{z} \times \hat{y} = -\hat{x}$ : $\vec{L}_O = (1.2)\hat{z} - (0.90)\hat{x}$ , in $\mathrm{kg \cdot m^2/s}$ .
$\lvert \vec{L}_O \rvert = \sqrt{0.81 + 1.44} = \sqrt{2.25} = 1.5\ \mathrm{kg \cdot m^2/s}$ , which is also $\lvert \vec{r}_O \rvert p = (1.00)(1.5)$ , since $\vec{r}_O$ and $\vec{p}$ are still perpendicular.
Now read the two answers together. About $O$ , the angular momentum is larger, 1.5 rather than 1.2, and it does not point along the rotation axis: it tilts, with a horizontal component of 0.90 pointing back toward the axis at that instant.
Its axial component is exactly 1.2, the same number $L = I\omega$ gave. That is the general relationship. $I\omega$ delivers the component of $\vec{L}$ along the chosen axis and nothing else; 6.3.A.2 delivers the whole vector.
One instant later the ball has moved around the circle, so the horizontal component of $\vec{L}_O$ has swung with it. About $O$ the angular momentum has constant magnitude and changing direction, so it is not constant, and by 6.4.B.2 that means the net external torque about $O$ is not zero. About the center of the circle, $\vec{L}$ is genuinely constant. Statement 6.3.A.2.i is warning you about exactly this: the point you choose changes the answer, including whether the answer is constant.

(a) $1.2\ \mathrm{kg \cdot m^2/s}$ along the rotation axis. (b) $I\omega = (0.192)(6.25) = 1.2\ \mathrm{kg \cdot m^2/s}$ , the same. (c) About $O$ , $\vec{L}_O = -0.90\hat{x} + 1.2\hat{z}$ with magnitude $1.5\ \mathrm{kg \cdot m^2/s}$ , tilted off the axis. The scalar form returns only the axial component, which is the information 6.3.A.2.i says depends on your choice of axis.

Frequently asked questions

What is angular momentum in AP Physics C Mechanics?

Essential knowledge 6.3.A.2 defines the angular momentum of an object about a given point as the cross product of the position vector from that point to the object with the object's linear momentum. The equation sheet prints it as a vector equation that also equals the rotational inertia times the angular velocity vector. Its units are kilogram metres squared per second. Because it is a cross product, it has a direction, given by the right-hand rule, and that direction is part of the answer in this course in a way it is not in AP Physics 1.

Is the cross product magnitude formula on the AP Physics C equation sheet?

Yes. The magnitude of a cross product, written as the product of the two magnitudes and the sine of the angle between them, is printed in the Vectors table on the third page of the AP Physics C Table of Information, alongside the dot product, the unit-vector form of a vector, and component addition. That page is often left out of condensed transcriptions of the sheet, which is why students sometimes believe it is absent. Applying it to the angular momentum definition immediately gives the scalar form the AP Physics 1 sheet prints, namely distance times mass times speed times the sine of the enclosed angle.

How is angular momentum different in AP Physics C than in AP Physics 1?

It has a direction. AP Physics C: Mechanics defines it as a cross product of position with linear momentum, and its equation sheet prints a single vector line with arrows on the angular momentum, the position, the momentum and the angular velocity. AP Physics 1 prints two scalar forms instead, rotational inertia times angular speed and distance times mass times speed times the sine of the enclosed angle. The magnitudes agree. Only the calculus-based course can ask which way an angular momentum points, add two that are not parallel, or work in three dimensions.

Can an object moving in a straight line have angular momentum?

Yes, and essential knowledge 6.3.A.2.ii states that its measured value depends on the distance between the reference point and the object, the object's mass and speed, and the angle between the radial distance and the velocity. The cross product shows why: about any point off the line of travel, the perpendicular distance from the point to the line is fixed, so the angular momentum is nonzero and constant even though nothing is spinning. About a point on the line itself, the position vector and the momentum are parallel, the sine is zero, and the angular momentum is exactly zero. Both answers are correct about their own reference points.

What is angular impulse in AP Physics C?

It is the integral of the torque with respect to time, printed as the relevant equation under essential knowledge 6.3.B.1. Statement 6.3.C.2.i then says it equals the change in angular momentum of the object or rigid system, and the equation sheet prints that theorem form directly. Statement 6.3.B.2 adds that angular impulse points in the same direction as the torque that delivered it. The units are newton metre seconds, which is the same combination as kilogram metres squared per second, the units of angular momentum, as the theorem requires.

When does net torque equal rotational inertia times angular acceleration?

When the rotational inertia is constant. Essential knowledge 6.3.C.2.ii attaches exactly that condition, saying the rotational form of the impulse-momentum theorem is a direct result of Newton's second law of motion for cases in which rotational inertia is constant, and it writes the chain from net torque to the time derivative of angular momentum to rotational inertia times angular acceleration. The first link, net torque equals the time derivative of angular momentum, holds always. The later links require pulling the rotational inertia out of the derivative, which fails for a system that changes shape while it turns, such as a skater drawing her arms in.

What does the slope of an angular momentum versus time graph represent?

The net torque exerted on the object or rigid system, per essential knowledge 6.3.C.3. A flat section means zero net torque and therefore constant angular momentum. A steepening curve means a growing net torque. This is the graphical counterpart of the area statements: the area under a graph of torque against time gives the angular impulse, per 6.3.B.3 for a single torque and 6.3.C.4 for the net external torque, while the slope of the angular momentum graph runs the relationship the other way.