AP Physics 1 · Topic 6.3

Topic 6.3: Angular Momentum and Angular Impulse

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

Angular momentum measures how much rotation something has. The AP Physics 1 sheet prints two forms: L equals I omega for a rigid system about an axis, and L equals rmv sin theta for any object about a chosen point. Angular impulse is torque times time, and it equals the change in angular momentum.

AP Physics: Unit 6 (topics 6.3 Angular Momentum and Angular Impulse). AP Physics 1 Unit 6, Topic 6.3. Three learning objectives: 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; and 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. Nine essential knowledge statements support them, including 6.3.A.1 (L = I omega for a rigid system about a specific axis), 6.3.A.2 (L = rmv sin theta for an object about a given point), 6.3.A.2.i (the choice of axis influences the value), 6.3.B.1 (angular impulse is tau times delta t), 6.3.B.3 (angular impulse is the area under a torque vs time graph), 6.3.C.1 (delta L = L minus L naught), 6.3.C.2.i (angular impulse equals the change in angular momentum, delta L = tau times delta t), 6.3.C.2.ii (tau net = delta L over delta t = I times delta omega over delta t = I alpha, for constant rotational inertia), 6.3.C.3 (net torque is the slope of an angular momentum vs time graph) and 6.3.C.4 (angular impulse is the area under a net external torque vs time graph). The boundary statement says that while AP Physics 1 expects students to mathematically manipulate the magnitude of angular momentum using one-dimensional vector conventions, the direction of angular momentum and angular impulse is beyond the scope of the course. The CED's suggested skills for this topic are 1.B, 2.A, 2.D and 3.B. Unit 6 carries 5 to 8 percent of the multiple-choice section and about 8 to 14 class periods. The AP Physics 1 equation sheet prints L = I omega, L = rmv sin theta, and delta L = tau times delta t.

What Topic 6.3 requires

Topic 6.3 carries three learning objectives, and each owns a different piece of one idea.

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.

Nine essential knowledge statements sit under them: two under 6.3.A, three under 6.3.B, and four under 6.3.C. Four further sub-statements hang off those, namely 6.3.A.2.i, 6.3.A.2.ii, 6.3.C.2.i and 6.3.C.2.ii. The CED lists four suggested skills for the topic: 1.B, create quantitative graphs with appropriate scales and units, including plotting data; 2.A, derive a symbolic expression from known quantities by selecting and following a logical mathematical pathway; 2.D, predict new values or factors of change of physical quantities using functional dependence between variables; and 3.B, apply an appropriate law, definition, theoretical relationship, or model to make a claim.

One boundary statement is printed for this topic, and it is worth reading before you start practicing. While AP Physics 1 expects that students can mathematically manipulate the magnitude of angular momentum using one-dimensional vector conventions, the direction of angular momentum and angular impulse is beyond the scope of the course. Magnitudes and signs along one chosen rotational sense are in. Vector directions in three dimensions are out.

Unit 6 is weighted at 5 to 8 percent of the multiple-choice section, and the CED allots it about 8 to 14 class periods. Topic 6.3 is the rotational twin of Topic 4.2, and reading the two side by side is the quickest route through this material.

Both forms of angular momentum are printed on the sheet

Essential knowledge 6.3.A.1 and 6.3.A.2 give two equations, and they describe two different situations.

L = I\omega

That is 6.3.A.1: the magnitude of the angular momentum of a rigid system about a specific axis. Use it when something is spinning, and read $I$ as the rotational inertia about that same axis.

L = rmv\sin\theta

That is 6.3.A.2: the magnitude of the angular momentum of an object about a given point. Nothing here has to be spinning. Use it when a single object is moving and you want its angular momentum about a point you have chosen.

Both lines are printed in the rotational column of the AP Physics 1 equation sheet, one directly beneath the other, and so is $\Delta L = \tau \Delta t$ . Those three lines are the whole of what the sheet says about angular momentum. Arriving at the exam knowing only $L = I\omega$ is a real way to lose marks here, because a question about a puck sliding past a pivot needs the other one.

Two further details about the sheet, because the gap between what it prints and what the CED says is where careless answers come from.

The sheet gives no symbol for angular impulse. Its symbol key lists $J$ for impulse in the translational column and $L$ for angular momentum in the rotational column, and angular impulse appears only as the right-hand side of $\Delta L = \tau \Delta t$ . The phrase angular impulse with its own equation, $\tau \Delta t$ , comes from essential knowledge 6.3.B.1, not from the sheet.
The sheet prints rotational Newton's second law solved for angular acceleration, as $\alpha_{\text{sys}} = \frac{\sum \tau}{I_{\text{sys}}} = \frac{\tau_{\text{net}}}{I_{\text{sys}}}$ . The $\tau_{\text{net}} = \Delta L / \Delta t$ form that essential knowledge 6.3.C.2.ii uses is not printed anywhere. Both are examinable; only one is given to you.

An object moving in a straight line has angular momentum about a point

$L = rmv\sin\theta$ works for an object that is not rotating at all, and essential knowledge 6.3.A.2.ii spells out what the answer 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. Four inputs. Three are properties of the object; the first is a consequence of the point you picked.

Draw the geometry once and it stays drawn. Mark the point, mark the object, draw the line $r$ from the point to the object, and draw the velocity arrow. The angle $\theta$ sits between those last two arrows. Then

r_{\perp} = r\sin\theta

is the perpendicular distance from the point to the line the object is travelling along, so the equation reads equally well as $L = m v r_{\perp}$ . It is the same trigonometric move the sheet makes for torque, where $\tau = r_{\perp}F = rF\sin\theta$ , and essential knowledge 5.3.A.2 names that perpendicular distance the lever arm. Topic 5.3 works the geometry through for forces.

Two consequences follow at once.

Angular momentum is not a property an object carries by itself. Statement 6.3.A.2.i says the selection of the axis about which an object is considered to rotate influences the determination of the angular momentum of that object. Change the point, change the number. A question that asks for angular momentum without naming a point or an axis has not finished asking.
An object heading straight at your chosen point has zero angular momentum about it. There $r$ and $v$ lie along the same line, $\theta$ is $0$ or $180$ degrees, $\sin\theta = 0$ , and $r_{\perp} = 0$ . The same object, at the same instant, has plenty of angular momentum about a point off to the side.

For an object moving at constant velocity, the angular momentum about any fixed point is the same all the way along, even though $r$ and $\theta$ both change continuously. Worked example 2 checks that at two instants. It has to come out that way: with no net force there is no torque about the point, so $\Delta L = \tau \Delta t$ gives zero change.

Angular impulse is torque multiplied by time (6.3.B)

Essential knowledge 6.3.B.1 defines angular impulse as the product of the torque exerted on an object or rigid system and the time interval during which the torque is exerted, and gives the relevant equation as

\text{angular impulse} = \tau \Delta t

Set that beside the linear definition in essential knowledge 4.2.A.2, $\vec{J} = \vec{F}_{\text{avg}} \Delta t$ , and the correspondence is exact: swap force for torque and you are finished. Topic 4.2 is the page to read alongside this one, and the impulse-momentum theorem guide carries the linear procedure in full.

The unit is the newton meter second, and expanding the newton shows it to be a kilogram meter squared per second, so angular impulse and angular momentum share a unit exactly as impulse and momentum do. Checking that the two sides of your working carry the same unit catches an algebra slip in seconds.

Statement 6.3.B.2 fixes the direction: angular impulse has the same direction as the torque exerted on the object or system. Read it together with the topic's boundary statement, which puts the direction of angular momentum and angular impulse beyond the scope of the course. What is left for the exam is the one-dimensional version. Pick a rotational sense as positive, write it down, and let signs carry direction from there. A torque in the negative sense delivers a negative angular impulse and reduces $L$ .

One caution transfers straight from the linear case. Any single torque delivers its own angular impulse, but the angular impulse that equals the system's change in angular momentum is the one from the net torque. If three torques act and the problem hands you one of them, multiplying that one by $\Delta t$ gives that torque's contribution and not the answer.

The rotational impulse-momentum theorem (6.3.C)

Change in angular momentum is defined the way every change is defined, final minus initial. That is essential knowledge 6.3.C.1:

\Delta L = L - L_0

Statement 6.3.C.2 then names the relationship, and 6.3.C.2.i states it: the angular impulse exerted on an object or rigid system is equal to the change in angular momentum of that object or rigid system. This is the line the equation sheet prints:

\Delta L = \tau \Delta t

Read left to right it says a torque acting for a while changes how much rotation something has. Read right to left it says every change in angular momentum has a torque and a duration behind it. The method is the one from linear momentum: whichever side of the equation the question gives you, cross to the other.

Statement 6.3.C.2.ii supplies the derivation, and the CED's own equation carries three equals signs:

\tau_{\text{net}} = \frac{\Delta L}{\Delta t} = I \frac{\Delta \omega}{\Delta t} = I\alpha

The CED says the rotational form of the impulse-momentum theorem is a direct result of the rotational form of Newton's second law of motion for cases in which rotational inertia is constant. That final clause is doing real work. Pulling $I$ out in front of the $\Delta$ in the middle step is legal only when $I$ does not change, and Topic 6.4 is built entirely on systems where $I$ deliberately does change. There $\tau_{\text{net}} = I\alpha$ stops being available while $\Delta L = \tau \Delta t$ keeps working, which is the reason the CED bothers to state the caveat.

The direction of the CED's claim is the reverse of the linear one, and it is worth noticing. In Topic 4.2, essential knowledge 4.2.B.3 says Newton's second law of motion is a direct result of the impulse-momentum theorem applied to systems with constant mass. In 6.3.C.2.ii the rotational impulse-momentum theorem is the direct result of Newton's second law in rotational form. Either way the two statements are the same physics, but if a free-response question asks you to derive one from the other, the CED's wording tells you which way it expects you to travel.

Reading the graphs (6.3.B.3, 6.3.C.3 and 6.3.C.4)

Skill 1.B, creating quantitative graphs, is listed for this topic, and the CED attaches three separate graph statements to it.

6.3.B.3 The angular impulse delivered to an object or rigid system by a torque can be found from the area under the curve of a graph of the torque as a function of time.
6.3.C.4 The angular impulse delivered to an object is equal to the area under the curve of a graph of the net external torque exerted on an object as a function of time.
6.3.C.3 The net torque exerted on an object is equal to the slope of the graph of the angular momentum of an object as a function of time.

The first two are not the same statement, and the difference is the word net. A graph of one torque gives that torque's angular impulse. A graph of the net external torque gives the angular impulse that equals $\Delta L$ . Plot only the motor's torque on a wheel that is also being braked and the area you measure belongs to the motor, not to the wheel's angular momentum.

Graph	Its slope gives	The area under it gives
Net external torque vs time	no standard quantity in AP Physics 1	angular impulse, equal to $\Delta L$ (6.3.C.4)
Angular momentum vs time	net torque (6.3.C.3)	no standard quantity in AP Physics 1
Torque vs angular position	no standard quantity in AP Physics 1	work done by that torque (6.2.A.3)
Angular velocity vs time	angular acceleration	angular displacement

Rows three and four belong to Topic 6.2 and to Topic 5.1. Reading a torque against angle graph as though it were a torque against time graph turns an angular impulse into a quantity of work, and the two carry different units, so check the horizontal axis label first.

To read an area, split straight-sided graphs into rectangles and triangles, add the pieces, and count any area below the time axis as negative. To read a slope, remember that a horizontal angular momentum line means zero net torque, a straight sloped line means a constant net torque equal to that slope, and a curve means the net torque is itself varying from instant to instant.

Radians, units, and the linear-to-rotational dictionary

Angular quantities in this unit are in radians. Essential knowledge 5.1.A.1 defines angular displacement as the measurement of the angle, in radians, through which a point on a rigid system rotates about a specified axis, and $\omega$ , $\alpha$ and everything downstream inherit that. Set your calculator to radians and convert any angle handed to you in degrees or revolutions before it goes anywhere near $v = r\omega$ or $W = \tau \Delta\theta$ .

There is one exception, and it lives on this page. The $\theta$ in $L = rmv\sin\theta$ is not an angular position. It is the geometric angle between two arrows, the same kind of $\theta$ as in $\tau = rF\sin\theta$ , and you can take its sine in degrees without a second thought. Two different quantities are wearing the same letter on the same equation sheet.

Linear quantity	Rotational counterpart	On the AP Physics 1 sheet
Mass $m$	Rotational inertia $I$	$I = \sum m_i r_i^2$
Velocity $v$	Angular velocity $\omega$	$v = r\omega$
Force $F$	Torque $\tau$	$\tau = r_{\perp}F = rF\sin\theta$
Momentum $\vec{p} = m\vec{v}$	Angular momentum $L = I\omega$	both printed
Impulse $\vec{J} = \vec{F}_{\text{avg}} \Delta t = \Delta \vec{p}$	Angular impulse $\tau \Delta t$	printed as $\Delta L = \tau \Delta t$
Kinetic energy $\frac{1}{2}mv^2$	Rotational kinetic energy $\frac{1}{2}I\omega^2$	both printed

Angular momentum is measured in $\text{kg} \cdot \text{m}^2/\text{s}$ and angular impulse in $\text{N} \cdot \text{m} \cdot \text{s}$ , and those are the same unit written two ways. Radians are dimensionless, which is why they never appear in either unit even though $\omega$ is quoted in radians per second. The rotational kinematics guide sets out the rest of the dictionary.

How Topic 6.3 is tested

The four suggested skills point at four question shapes.

Read an angular impulse off a torque against time graph, then find an angular velocity (skill 1.B). Worked example 1 below is that question end to end.
Derive a symbolic expression (skill 2.A). Show that a constant torque $\tau$ applied for a time $t$ to a system of rotational inertia $I$ starting from rest leaves it turning at $\omega = \tau t / I$ , working in symbols throughout.
Compare two scenarios (skill 2.D). The same angular impulse is delivered to a system whose rotational inertia is three times larger. What happens to the final angular speed? Report the ratio, one third, rather than recomputing anything.
Apply a law or definition to make a claim (skill 3.B). A puck slides past a pivot in a straight line. Does it have angular momentum about the pivot? Justify the claim with 6.3.A.2 rather than with intuition.

Section I of the exam is 42 multiple-choice questions in 85 minutes and Section II is four free-response questions in 95 minutes, each worth half the score, and a four-function, scientific or graphing calculator is allowed on both. The CED's Progress Check for Unit 6 lists about 18 multiple-choice questions and four free-response questions.

The CED also lists an optional sample instructional activity for this topic: 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 rotating. Treat that as a demonstration rather than exam preparation. The topic's boundary statement puts the direction of angular momentum beyond the scope of the course, and the Topic 5.5 boundary statement adds that AP Physics 1 does not expect students to simultaneously analyze rotation in multiple planes.

From here, Topic 6.4 asks the question this topic sets up: what happens to a system when the net external torque on it is zero?

Angular impulse from a torque vs time graph

A flywheel of rotational inertia 1.2 kg m^2 about its axle is already turning at 5.0 rad/s in the counterclockwise sense. A sensor records the net external torque on it: the torque rises linearly from 0 to 6.0 N m between t = 0 and t = 2.0 s, holds at 6.0 N m until t = 5.0 s, then switches to a constant -3.0 N m from t = 5.0 s to t = 7.0 s. Find (a) the angular impulse delivered in each stage, (b) the angular velocity at the two stage boundaries t = 5.0 s and t = 7.0 s, and (c) the average net torque across the full 7.0 s.

Take counterclockwise as positive and keep that convention to the last line. Essential knowledge 6.3.C.4: the angular impulse delivered is the signed area under the net external torque curve. The starting angular momentum is $L_0 = I\omega_0 = (1.2\ \text{kg} \cdot \text{m}^2)(5.0\ \text{rad/s}) = 6.0\ \text{kg} \cdot \text{m}^2/\text{s}$ .
(a) Stage 1 is a triangle of base 2.0 s and height 6.0 N m: $\frac{1}{2}(2.0\ \text{s})(6.0\ \text{N} \cdot \text{m}) = 6.0\ \text{N} \cdot \text{m} \cdot \text{s}$ .
Stage 2 is a rectangle of width $5.0 - 2.0 = 3.0\ \text{s}$ and height 6.0 N m: $(6.0)(3.0) = 18\ \text{N} \cdot \text{m} \cdot \text{s}$ .
Stage 3 lies below the time axis, so its area is negative: $(-3.0\ \text{N} \cdot \text{m})(2.0\ \text{s}) = -6.0\ \text{N} \cdot \text{m} \cdot \text{s}$ .
(b) Through $t = 5.0\ \text{s}$ the angular impulse totals $6.0 + 18 = 24\ \text{N} \cdot \text{m} \cdot \text{s}$ , so by 6.3.C.2.i $L = L_0 + \Delta L = 6.0 + 24 = 30\ \text{kg} \cdot \text{m}^2/\text{s}$ and $\omega = L/I = 30/1.2 = 25\ \text{rad/s}$ .
Over the full 7.0 s the angular impulse is $24 - 6.0 = 18\ \text{N} \cdot \text{m} \cdot \text{s}$ , so $L = 6.0 + 18 = 24\ \text{kg} \cdot \text{m}^2/\text{s}$ and $\omega = 24/1.2 = 20\ \text{rad/s}$ . The third stage took angular momentum back out without ever turning the flywheel around, since the running total stayed positive throughout.
Cross-check stage 2 against $\tau_{\text{net}} = I\alpha$ , which is available here because the flywheel's rotational inertia is constant. At $t = 2.0\ \text{s}$ , $L = 6.0 + 6.0 = 12\ \text{kg} \cdot \text{m}^2/\text{s}$ , so $\omega = 10\ \text{rad/s}$ . Then $\alpha = \tau/I = 6.0/1.2 = 5.0\ \text{rad/s}^2$ , and after 3.0 more seconds $\omega = 10 + (5.0)(3.0) = 25\ \text{rad/s}$ , which matches.
(c) $\tau_{\text{avg}} = \Delta L / \Delta t = (18\ \text{N} \cdot \text{m} \cdot \text{s})/(7.0\ \text{s}) = 2.5714... = 2.6\ \text{N} \cdot \text{m}$ to two significant figures. Draw a rectangle of that height across the whole 7.0 s and it encloses the same signed area as the three-stage shape, which is what the average is defined to be.

(a) $+6.0$ , $+18$ and $-6.0\ \text{N} \cdot \text{m} \cdot \text{s}$ , for a total of $+18\ \text{N} \cdot \text{m} \cdot \text{s}$ . (b) 25 rad/s at $t = 5.0\ \text{s}$ and 20 rad/s at $t = 7.0\ \text{s}$ . (c) $\tau_{\text{avg}} = 2.6\ \text{N} \cdot \text{m}$ , well below the 6.0 N m peak. Between $t = 2.0\ \text{s}$ and $t = 5.0\ \text{s}$ a graph of $L$ against time would be a straight line of slope $(30 - 12)/(5.0 - 2.0) = 6.0\ \text{N} \cdot \text{m}$ , which is essential knowledge 6.3.C.3 read off the other graph.

The angular momentum of a puck sliding in a straight line

A 0.40 kg puck slides in a straight line across frictionless ice at a constant 6.0 m/s. Point P is a fixed point on the ice, off to one side of the puck's path. At one instant the puck is 3.0 m from P, and the angle between the line from P to the puck and the puck's velocity is 30 degrees. (a) Find the puck's angular momentum about P at that instant. (b) Find it again at the puck's closest approach to P. (c) Find its angular momentum about a point Q that lies on the puck's own line of travel.

The puck is not spinning, so $L = I\omega$ is no use. Essential knowledge 6.3.A.2 is the statement that applies: the magnitude of the angular momentum of an object about a given point is $L = rmv \sin\theta$ .
(a) Substitute directly: $L = (3.0\ \text{m})(0.40\ \text{kg})(6.0\ \text{m/s})\sin 30^\circ = (7.2)(0.500) = 3.6\ \text{kg} \cdot \text{m}^2/\text{s}$ .
Check it the other way round. The perpendicular distance from P to the puck's path is $r_{\perp} = r\sin\theta = (3.0)(0.500) = 1.5\ \text{m}$ , and $L = m v r_{\perp} = (0.40)(6.0)(1.5) = 3.6\ \text{kg} \cdot \text{m}^2/\text{s}$ . Same number, and $r_{\perp}$ is the piece of geometry that does not change as the puck slides.
(b) At closest approach the line from P to the puck is perpendicular to the velocity, so $\theta = 90^\circ$ and $r$ has shrunk to $r_{\perp} = 1.5\ \text{m}$ : $L = (1.5)(0.40)(6.0)\sin 90^\circ = 3.6\ \text{kg} \cdot \text{m}^2/\text{s}$ .
The two factors moved in opposite directions and cancelled exactly: $r$ fell from 3.0 m to 1.5 m while $\sin\theta$ rose from 0.500 to 1.000. That is not a coincidence. The ice is frictionless and the puck travels at constant velocity, so the net force on it is zero, the net torque about P is zero, and $\Delta L = \tau \Delta t$ gives $\Delta L = 0$ .
For a sense of scale, the puck covers $\sqrt{(3.0)^2 - (1.5)^2} = \sqrt{6.75} = 2.6\ \text{m}$ between the two instants, which at 6.0 m/s takes about 0.43 s.
(c) A point Q on the puck's own line of travel has $r_{\perp} = 0$ , because the perpendicular distance from a point to a line running through it is zero. Equivalently $\theta$ is $0^\circ$ or $180^\circ$ , so $\sin\theta = 0$ . The puck's angular momentum about Q is zero at every instant.

(a) and (b) both give $L = 3.6\ \text{kg} \cdot \text{m}^2/\text{s}$ about P, unchanged as the puck slides past. (c) $L = 0$ about any point on the puck's own line of travel. The same puck at the same instant has two different angular momenta because two different reference points were chosen, which is exactly what essential knowledge 6.3.A.2.i is warning you about.

One rigid system, two equations, one answer

Two 0.75 kg balls are fixed to the ends of a light rod and spin about an axis through the rod's center, each ball 0.30 m from the axis, at a constant 8.0 rad/s. Find the system's angular momentum about that axis twice: once from L = I omega, and once by adding L = rmv sin theta for the two balls separately.

From essential knowledge 5.4.A.2 and 5.4.A.3, the rotational inertia of the pair about the central axis is $I = \sum m_i r_i^2 = 2(0.75\ \text{kg})(0.30\ \text{m})^2 = 2(0.75)(0.090) = 0.135\ \text{kg} \cdot \text{m}^2$ . The rod is light, so it contributes nothing.
Essential knowledge 6.3.A.1: $L = I\omega = (0.135\ \text{kg} \cdot \text{m}^2)(8.0\ \text{rad/s}) = 1.08\ \text{kg} \cdot \text{m}^2/\text{s}$ .
Now the other route. Each ball travels a circle of radius 0.30 m, so its speed is $v = r\omega = (0.30\ \text{m})(8.0\ \text{rad/s}) = 2.4\ \text{m/s}$ , and in circular motion the velocity is always perpendicular to the radius, so $\theta = 90^\circ$ and $\sin\theta = 1$ .
Essential knowledge 6.3.A.2, applied to one ball: $L = rmv\sin\theta = (0.30)(0.75)(2.4)(1) = 0.54\ \text{kg} \cdot \text{m}^2/\text{s}$ .
Both balls circle the same way about the same axis, so their angular momenta add rather than cancel: $L_{\text{total}} = 2(0.54) = 1.08\ \text{kg} \cdot \text{m}^2/\text{s}$ .

$L = 1.08\ \text{kg} \cdot \text{m}^2/\text{s}$ by either route. $L = I\omega$ is not a separate law from $L = rmv\sin\theta$ : it is what you get by applying the point-object form to every piece of a rigid system and adding the results, which is why the CED can print both without contradiction. Adding up the parts is also the statement Topic 6.4 opens with, in essential knowledge 6.4.A.1.

Frequently asked questions

What is angular momentum in AP Physics 1?

Angular momentum is the rotational counterpart of linear momentum, and the AP Physics 1 equation sheet prints two expressions for its magnitude. For a rigid system spinning about a specific axis it is L = I omega, the rotational inertia times the angular velocity (essential knowledge 6.3.A.1). For an object about a given point it is L = rmv sin theta (essential knowledge 6.3.A.2). Its unit is the kilogram meter squared per second.

What are the two angular momentum equations on the AP Physics 1 formula sheet?

L = I omega and L = rmv sin theta, printed one directly beneath the other in the rotational column, along with a third line, delta L = tau times delta t. Those three lines are the whole of what the sheet says about angular momentum. The first applies to a rigid system about a specific axis, the second to any object about a given point, and the third relates a change in angular momentum to the angular impulse that caused it.

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

Yes, about any point that does not lie on its line of travel. Essential knowledge 6.3.A.2 gives the magnitude as L = rmv sin theta, where theta is the angle between the line from the point to the object and the object's velocity. Because r sin theta is the perpendicular distance from the point to the object's path, the same object has different angular momenta about different points, and zero angular momentum about any point on its own line of travel.

What is angular impulse?

Essential knowledge 6.3.B.1 defines angular impulse as the product of the torque exerted on an object or rigid system and the time interval during which that torque is exerted, written as tau times delta t. It is the rotational counterpart of impulse, which is force times time. Its unit is the newton meter second, which is the same unit as angular momentum, and 6.3.C.2.i says the angular impulse exerted on a system equals that system's change in angular momentum.

What does the area under a torque vs time graph represent?

The angular impulse delivered over that interval. Essential knowledge 6.3.B.3 states it for a single torque, and 6.3.C.4 states it for the net external torque, in which case the area also equals the change in angular momentum. Split straight-sided graphs into rectangles and triangles and count area below the time axis as negative. Do not confuse this graph with torque against angular position, whose area is work.

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

The net torque on the object, from essential knowledge 6.3.C.3. A horizontal line means the net torque is zero and the angular momentum is constant, a straight sloped line means a constant net torque equal to that slope, and a curve means the net torque is itself varying. The area under an angular momentum against time graph has no standard meaning in AP Physics 1.

Does AP Physics 1 test the direction of angular momentum?

Not as a vector direction. The boundary statement for Topic 6.3 says that while AP Physics 1 expects students to mathematically manipulate the magnitude of angular momentum using one-dimensional vector conventions, the direction of angular momentum and angular impulse is beyond the scope of the course. In practice you choose one rotational sense as positive and carry direction with plus and minus signs. Right-hand rules and gyroscopic precession are not required.