AP Physics C: Mechanics · Unit 6 of 7

Unit 6: Energy and Momentum of Rotating Systems

10-15% of the multiple-choice section6 topics

Topics in this unit

Energy and Momentum of Rotating Systems is Unit 6 of AP Physics C: Mechanics, worth 10 to 15 percent of the multiple-choice section over about 13 to 19 class periods. Six topics, eleven learning objectives. Angular momentum is a cross product and work by a torque is an integral.

AP Physics: Unit 6 (topics 6.1 Rotational Kinetic Energy, 6.2 Torque and Work, 6.3 Angular Momentum and Angular Impulse, 6.4 Conservation of Angular Momentum, 6.5 Rolling, 6.6 Motion of Orbiting Satellites). Unit 6 of the current AP Physics C: Mechanics course and exam description, weighted 10 to 15% of the multiple-choice section at about 13 to 19 class periods, against 5 to 8% for the identically titled AP Physics 1 Unit 6. Six topics and eleven learning objectives, with three each under Topics 6.3 and 6.5. The unit prints exactly one boundary statement, under Topic 6.5: rolling friction is beyond the scope of AP Physics C: Mechanics. Topics 6.1, 6.2, 6.3, 6.4 and 6.6 print none. All three of the unit's Derived Equations, which the Required Equations page defines as final results of derivations expected of students on the exam, sit in Topic 6.6: the circular-orbit relations K = -U/2 and E_total = U/2 = -GMm/2r in statement 6.6.A.3, and the escape velocity in 6.6.A.4.ii. None of the three is on the equation sheet. Also absent from the sheet despite appearing in required content: the definition of angular impulse on its own, the change in angular momentum as a difference, the net torque as the time derivative of angular momentum, and the total kinetic energy of a rolling body as a sum. Calculus differentiators against the algebra-based sibling: angular momentum as the cross product of position and linear momentum, work by a torque as an integral over angular position, and angular impulse as an integral over time. Suggested skills by topic: 6.1 uses 1.C, 2.C, 3.B, 3.C; 6.2 uses 1.A, 2.A, 2.C, 3.B; 6.3 uses 1.C, 2.A, 2.C, 3.C; 6.4 uses 1.C, 2.B, 3.B, 3.C; 6.5 uses 1.B, 2.A, 2.D, 3.A, 3.B; 6.6 uses 1.C, 2.A, 3.C.

What the CED requires across Unit 6

Unit 6 of AP Physics C: Mechanics is Energy and Momentum of Rotating Systems. The course and exam description weights it at 10 to 15% of the multiple-choice section and suggests about 13 to 19 class periods. That is the same band as Units 1, 5 and 7. Unit 2 is heaviest at 20 to 25%, Unit 3 is 15 to 25%, and Unit 4 is 10 to 20%.

Six topics carry eleven learning objectives. Topics 6.3 and 6.5 have three each, Topic 6.4 has two, and the remaining three have one apiece.

Topic	Learning objectives	Suggested skills
6.1 Rotational Kinetic Energy	6.1.A	1.C, 2.C, 3.B, 3.C
6.2 Torque and Work	6.2.A	1.A, 2.A, 2.C, 3.B
6.3 Angular Momentum and Angular Impulse	6.3.A, 6.3.B, 6.3.C	1.C, 2.A, 2.C, 3.C
6.4 Conservation of Angular Momentum	6.4.A, 6.4.B	1.C, 2.B, 3.B, 3.C
6.5 Rolling	6.5.A, 6.5.B, 6.5.C	1.B, 2.A, 2.D, 3.A, 3.B
6.6 Motion of Orbiting Satellites	6.6.A	1.C, 2.A, 3.C

Every one of the eleven opens with either "describe" or "relate". The CED says the task verb describe, used in nearly all learning objectives, "encompasses the range of possible graphical, mathematical, or verbal skill applications", so describing angular momentum can mean drawing it, integrating for it, or saying in words what it is about which axis.

The CED's framing is that in Unit 6 students apply their knowledge of energy and momentum to rotating systems, and that similar to the approach used for translational energy and momentum in Units 3 and 4, it matters that students conceptually understand how angular momentum and rotational energy change due to external torques on a system. It adds that articulating the conditions under which the rotational energy or angular momentum of a system remains constant is foundational to working through more complex scenarios, and that students will use the content and skills of both Units 5 and 6 to study orbiting satellites and rolling without slipping.

The essential questions on the unit opener are what keeps a bicycle balanced, why planets move faster when they travel closer to the sun, what satellites and projectiles have in common, and how figure skating would be different if angular momentum were not conserved.

The three places calculus changes the answer, not just the notation

AP Physics 1 has a Unit 6 with the same six topic titles. The two units diverge in three specific equations, and each divergence changes what a question can ask.

Angular momentum becomes a cross product. Statement 6.3.A.1 gives the magnitude about a specific axis as $L = I\omega$ , and then 6.3.A.2 gives the angular momentum of an object about a given point as

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

The AP Physics C: Mechanics formula sheet prints it as $\vec{L} = \vec{r} \times \vec{p} = I\vec{\omega}$ , with arrows on all three. The AP Physics 1 sheet prints two scalars instead, $L = I\omega$ and $L = rmv\sin\theta$ . Same magnitudes, no direction. So this course can ask which way an angular momentum points, and can add two angular momenta that are not parallel; the algebra-based course cannot.

Work by a torque becomes an integral. Statement 6.2.A.2 gives

W = \int_{\theta_1}^{\theta_2} \tau\, d\theta

against the AP Physics 1 sheet's $W = \tau\Delta\theta$ . That is the difference between a torque that stays constant and a torque that does not. Statement 6.2.A.3 then says the work can be found from the area under a graph of torque against angular position, which is the same statement drawn rather than written.

Angular impulse becomes an integral too. Statement 6.3.B.1 defines angular impulse as $\int \tau\, dt$ , and 6.3.C.2.i writes the rotational impulse-momentum theorem as $\Delta L = \int_{t_1}^{t_2} \tau\, dt$ . The AP Physics 1 sheet prints $\Delta L = \tau\Delta t$ . Statement 6.3.C.2.ii then supplies the differential form that no algebra-based course has:

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

with the important qualifier attached: the rotational impulse-momentum theorem is a direct result of Newton's second law for cases in which rotational inertia is constant. Statement 6.3.C.3 adds that the net torque equals the slope of a graph of angular momentum against time.

Read together, those three make a whole class of question available. A torque that varies with angle, a torque that varies with time, and an angular momentum whose direction is part of the answer are all fair game here and out of reach in the algebra-based course.

Unit 6 prints exactly one boundary statement

Across all six topics there is one, and it sits under Topic 6.5. Quoted whole, because it is one sentence:

"Rolling friction is beyond the scope of AP Physics C: Mechanics."

Topics 6.1, 6.2, 6.3, 6.4 and 6.6 print no boundary statement at all. That is worth knowing before you assume the orbital-mechanics results at the end of the unit are fenced off. They are not fenced by a boundary statement. What bounds them is the Derived Equation label, and the CED's Required Equations page defines it precisely: not all equations in the framework appear on the equation sheet, many are provided for reference and guidance "or to demonstrate the final results of derivations expected of students on the exam", and those are denoted Derived Equations.

All three of Unit 6's Derived Equations sit in Topic 6.6. Statement 6.6.A.3 labels

K = -\frac{1}{2}U \qquad E_{\text{total}} = \frac{1}{2}U = -\frac{GMm}{2r}

as derived, for a satellite in a circular orbit. Statement 6.6.A.4.ii labels

v_{\text{esc}} = \sqrt{\frac{2GM}{r}}

as derived, and says in the same breath that the escape velocity of a satellite from a central body of mass $M$ can be derived using conservation of energy laws. So none of the three is printed on the equation sheet, and all three are derivations you are expected to be able to produce. What is printed, and is all you need, is $U_G = -\frac{Gm_1m_2}{r}$ , $\lvert\vec{F}_g\rvert = \frac{Gm_1m_2}{r^2}$ , $K = \frac{1}{2}mv^2$ and $a_c = \frac{v^2}{r} = r\omega^2$ .

The one thing the boundary statement does remove is worth taking literally. Rolling friction is out, which is why statement 6.5.B.2 can say that for ideal cases, rolling without slipping implies that the frictional force does not dissipate any energy from the rolling system.

How the six topics build

6.1 Rotational Kinetic Energy gives $K_{\text{rot}} = \frac{1}{2}I\omega^2$ and then spends its remaining statements on bookkeeping that trips people up. Statement 6.1.A.1.ii says the total kinetic energy of a rigid system is the sum of its rotational kinetic energy about its center of mass and the translational kinetic energy of that center of mass. Statement 6.1.A.2 says a rigid system can have rotational kinetic energy while its center of mass is at rest, because the individual points still have linear speed. And 6.1.A.3 states flatly that rotational kinetic energy is a scalar.

6.2 Torque and Work is the integral above, plus 6.2.A.1: a torque can transfer energy into or out of a rigid system if it is exerted over an angular displacement. No angular displacement, no work, however large the torque.

6.3 Angular Momentum and Angular Impulse carries three objectives: describe the angular momentum (6.3.A), describe the angular impulse delivered by a torque (6.3.B), and relate the change in angular momentum to that angular impulse (6.3.C). Statement 6.3.A.2.i says the selection of the axis about which an object is considered to rotate influences the determination of its angular momentum, and 6.3.A.2.ii says the measured angular momentum of an object travelling 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. Statement 6.3.B.2 adds that angular impulse has the same direction as the torque imparting it.

6.4 Conservation of Angular Momentum is the unit's conceptual centre. Statement 6.4.B.1 says angular momentum is conserved in all interactions. Statement 6.4.A.2.i says the angular impulse exerted by one system on a second is equal and opposite to the reverse, as a direct result of Newton's third law. Statement 6.4.A.2.iii is the figure-skater sentence: the angular speed of a nonrigid system may change without the angular momentum changing if the system changes shape by moving mass closer to or farther from the rotational axis.

6.5 [Rolling](/glossary/rolling-without-slipping) splits into total kinetic energy (6.5.A), rolling without slipping (6.5.B) and rolling while slipping (6.5.C). The rolling constraint is $\Delta x_{\text{cm}} = r\Delta\theta$ , $v_{\text{cm}} = r\omega$ , $a_{\text{cm}} = r\alpha$ . Statement 6.5.C.1 says that when slipping, the motion of the centre of mass and the rotational motion cannot be directly related, and 6.5.C.2 explains why kinetic friction then dissipates energy: the point of application of the friction force moves with respect to the surface.

6.6 Motion of Orbiting Satellites closes the course's mechanics with conservation laws. Statement 6.6.A.2.i says that in circular orbits the system's total mechanical energy, its gravitational potential energy, and the satellite's angular momentum and kinetic energy are all constant. Statement 6.6.A.2.ii says that in elliptical orbits the total mechanical energy and the angular momentum are constant, but the gravitational potential energy and the kinetic energy can each change. That single contrast answers the unit's own essential question about planets moving faster near the sun.

Which Unit 6 equations are printed

Sixteen equations appear in Unit 6's required content. Knowing which are absent from the equation sheet is worth more than knowing which are present.

Equation	Where the CED puts it	Printed on the sheet
$K_{\text{rot}} = \frac{1}{2}I\omega^2$	6.1.A.1	yes
$W = \int \tau\, d\theta$	6.2.A.2	yes
$L = I\omega$	6.3.A.1	yes, as the second half of the vector line
$\vec{L} = \vec{r} \times \vec{p}$	6.3.A.2	yes
angular impulse $= \int \tau\, dt$	6.3.B.1	not by that name
$\Delta L = L - L_0$	6.3.C.1	no
$\Delta L = \int \tau\, dt$	6.3.C.2.i	yes
$\tau_{\text{net}} = \frac{dL}{dt} = I\frac{d\omega}{dt} = I\alpha$	6.3.C.2.ii	no
$K_{\text{tot}} = K_{\text{trans}} + K_{\text{rot}}$	6.5.A.1	no
$\Delta x_{\text{cm}} = r\Delta\theta$	6.5.B.1	yes
$v_{\text{cm}} = r\omega$	6.5.B.1	as $v = r\omega$ , without the subscript
$a_{\text{cm}} = r\alpha$	6.5.B.1	as $a_T = r\alpha$ , tangential subscript
$U_g = -G\frac{m_1m_2}{r}$	6.6.A.2.iii	yes, as $U_G$
$K = -\frac{1}{2}U$	6.6.A.3, derived	no
$E_{\text{total}} = \frac{1}{2}U = -\frac{GMm}{2r}$	6.6.A.3, derived	no
$v_{\text{esc}} = \sqrt{\frac{2GM}{r}}$	6.6.A.4.ii, derived	no

Four observations to carry into the exam.

The angular impulse integral is printed only in its theorem form. The sheet gives $\Delta L = \int \tau\, dt$ , which is 6.3.C.2.i. It does not print $\int \tau\, dt$ separately as the definition of angular impulse, and it does not print $\Delta L = L - L_0$ . Neither absence costs you anything, but do not go looking.

$\tau_{\text{net}} = dL/dt$ is not on the sheet. The $I\alpha$ end of that chain rearranges from the printed rotational second law, $\alpha_{\text{sys}} = \tau_{\text{net}}/I_{\text{sys}}$ , but the derivative form is yours to write, and 6.3.C.2.ii tells you when the two agree: when rotational inertia is constant. A skater pulling her arms in is precisely the case where it does not.

The total kinetic energy of a rolling body is not printed as one equation. The sheet prints $K = \frac{1}{2}mv^2$ and $K_{\text{rot}} = \frac{1}{2}I\omega^2$ separately, and adding them is a step you take, per 6.5.A.1.

All three orbital energy results are derivations. That is the single most useful thing on this page for Topic 6.6. You will not find $v_{\text{esc}}$ or $E_{\text{total}} = -GMm/2r$ on the sheet, and the CED says outright that the escape velocity is derived using conservation of energy laws.

Traps that span more than one topic

Angular momentum is not conserved because nothing is happening. Statement 6.4.B.2 says the total angular momentum of a selected system is constant if the net external torque on it is zero, and 6.4.B.3 says angular momentum is transferred between the system and the environment when that torque is nonzero. The question to ask is always about the net external torque about the chosen axis, not about whether forces exist.

Angular momentum can be conserved while kinetic energy is not. A skater pulling her arms in keeps $L$ and raises $K_{\text{rot}}$ , because $K_{\text{rot}} = L^2/2I$ and $I$ dropped. Her muscles did the work. The reverse pairing, conserving energy while angular momentum changes, is the one to be suspicious of.

A straight-line motion has angular momentum. Statement 6.3.A.2.ii says so, and the cross product shows why: $\vec{r} \times \vec{p}$ is nonzero for any reference point not on the line of travel, and it is constant as the object moves. Choosing a point on the line gives exactly zero. Both answers are correct about their own axes, which is what 6.3.A.2.i is warning you about.

A large torque with no angular displacement does no work. Statement 6.2.A.1 makes the angular displacement the condition. Holding a wrench against a seized bolt transfers no energy to it.

In an elliptical orbit, the potential energy is not constant. Statement 6.6.A.2.ii is explicit: for elliptical orbits, total mechanical energy and the satellite's angular momentum are constant while gravitational potential energy and kinetic energy each change. Reaching for the circular-orbit result $K = -\frac{1}{2}U$ on an elliptical orbit is a defect, because that relation is stated for a circular path in 6.6.A.3.

Rolling without slipping does not mean frictionless. It means, per 6.5.B.2, that in ideal cases the frictional force dissipates no energy. There is still a friction force, and on a ramp it is the force exerting the torque that spins the object up.

The reference point for gravitational potential energy is at infinity. Statement 6.6.A.2.iii defines $U_g$ to be zero when the satellite is an infinite distance from the central object, which is why the printed $U_G$ carries a minus sign and why $\Delta U_g = mg\Delta y$ is a different, local approximation.

How Unit 6 is assessed

The AP Physics C: Mechanics exam is 3 hours long: 42 multiple-choice questions in 85 minutes for half the score, then 4 free-response questions in 95 minutes for the other half, in a fixed order of Mathematical Routines (10 points), Translation Between Representations (12 points), Experimental Design and Analysis (10 points), and Qualitative/Quantitative Translation (8 points). A four-function, scientific, or graphing calculator is allowed on both sections.

On the multiple-choice section, skill 2.A carries 25 to 30%, skill 2.B 20 to 25%, and skills 2.C and 2.D 10 to 15% each; skill 3.B carries 15 to 25% and 3.C carries 5 to 10%. Practice 1 is not assessed there at all. On the free-response section, Practice 1 carries 20 to 35%, Practice 2 carries 40 to 45%, and Practice 3 carries 30 to 35%.

The unit's Building the Science Practices page flags 2.C, 2.D, 3.B and 3.C, and gives its own example: students could describe conceptually what happens to the rotational inertia of a system when the pivot point is moved, and then justify what impact that change will have on the angular acceleration. The Preparing for the AP Exam note for this unit contains the sharpest sentence about grading anywhere in the framework. It says that when writing justifications for claims, simply referencing an equation, law, or physical principle is not sufficient, and gives the example that saying one disk is rolling faster than another because of conservation of energy is not a complete enough answer to earn credit on the free-response section. Students must clearly and concisely explain the steps in their reasoning that lead from the equation, law, or principle to the justification of their claim.

Three of the CED's fifteen sample multiple-choice questions align to Unit 6. Question 4 pairs skill 2.D with objective 6.4.A and essential knowledge 6.4.A.2: a spherical star of uniform density spinning at some initial angular velocity collapses to half its original radius with no loss of mass, and you are asked for the new angular velocity. Question 14 pairs skill 2.A with 6.3.A and 6.3.A.2: two uniform disks of the same mass $M$ and radii $2R$ and $R$ , one spinning and one dropped onto it, spinning together afterwards. Conservation of angular momentum gives $\frac{4}{5}\omega_0$ , since a disk of radius $2R$ has four times the rotational inertia of one of the same mass and radius $R$ . Note that this question supplies the rotational inertia of a uniform disk in its own stem. Question 13 pairs skill 2.A with 6.6.A and 6.6.A.4: a rock launched at twice the escape speed, where energy conservation gives $\sqrt{3}\,v_{\text{esc}}$ far from the planet. All three are derivation or scaling questions, which matches skill 2.A's 25 to 30% weight.

The sample free-response set's Question 4, a Qualitative/Quantitative Translation question worth 8 points, aligns to objectives 1.3.A, 3.4.B, 5.4.A, 5.6.A, 5.4.B and 6.1.A. It races a hollow sphere, a solid sphere and a hoop of equal mass and radius down a ramp, and its published example response argues from rotational inertia to how much gravitational potential energy becomes rotational rather than translational kinetic energy. Its scoring guidelines credit both a Newton's-second-law-in-rotational-form route and a conservation-of-energy route.

Skill 3.A, creating experimental procedures, is listed for Topic 6.5 alone, and both of the unit's rolling sample activities are experiments: a yo-yo released down a ramp whose rotational inertia students find from its final velocity and release height, and a hoop and disk of equal mass and radius rolled down identical ramps with the result explained using energy bar charts and to-scale free-body diagrams. The Progress Check for Unit 6 runs about 18 multiple-choice questions and 4 free-response questions, one of each type.

If you are in AP Physics 1, this is not your page

AP Physics C: Mechanics is a calculus-based introductory college-level course, equivalent to the first course in an introductory college sequence in calculus-based physics. Its only stated prerequisite is that students should have taken, or be concurrently taking, calculus.

AP Physics 1 has a Unit 6 with the same title and the same six topic titles, and it is a different unit in two measurable ways. It is weighted 5 to 8% of that exam's multiple-choice section against 10 to 15% here, so the same six topics carry roughly twice the weight in this course. And its equation sheet prints work by a torque as a product, angular impulse as a product, and angular momentum as two scalars, where this one prints two integrals and a cross product.

If you are in the algebra-based course, the page you want is the AP Physics 1 Unit 6 hub. That page is for AP Physics 1 students; this one is for AP Physics C: Mechanics students. Neither is a reading-level variant of the other.

For the procedures rather than the framework, the conservation of energy guide and the conservation of momentum guide own the step-by-step routines, and the work and power calculator checks arithmetic. The AP Physics C: Mechanics course hub lists all seven units.

Angular momentum of an object moving in a straight line

A 0.40 kg puck slides in a straight line at a constant $6.0\ \mathrm{m/s}$ in the $+x$ direction along the line $y = 1.5$ m. Find its angular momentum about the origin (a) when it is at $x = 2.0$ m and (b) when it is at $x = 5.0$ m, (c) its angular momentum about the point $(0, 1.5\ \mathrm{m})$ , and (d) check the magnitude against the algebra-based form $L = rmv\sin\theta$ .

Declare the convention: $\hat{k}$ points out of the page, so a positive $L_z$ is counterclockwise angular momentum about the chosen point.
(a) Statement 6.3.A.2 gives $\vec{L} = \vec{r} \times \vec{p}$ . With $\vec{p} = (2.4\ \mathrm{kg \cdot m/s})\hat{i}$ from $p = mv = (0.40)(6.0)$ , and both vectors in the plane, $L_z = xp_y - yp_x = (2.0)(0) - (1.5)(2.4) = -3.6\ \mathrm{kg \cdot m^2/s}$ .
So $\vec{L} = -(3.6\ \mathrm{kg \cdot m^2/s})\hat{k}$ , into the page. Nothing is spinning, and the angular momentum is still nonzero. That is what 6.3.A.2.ii means when it says the angular momentum of an object travelling 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.
(b) Repeat at $x = 5.0$ m: $L_z = (5.0)(0) - (1.5)(2.4) = -3.6\ \mathrm{kg \cdot m^2/s}$ . Identical. The $x$ coordinate never enters, because $p_y = 0$ , so the angular momentum is constant during the straight-line motion, exactly as 6.4.B.2 requires when the net external torque about the origin is zero.
(c) About the point $(0, 1.5\ \mathrm{m})$ , the position vector is $\vec{r} = (x)\hat{i}$ with no $y$ component, so $L_z = (x)(0) - (0)(2.4) = 0$ .
Both answers are right about their own reference points, which is 6.3.A.2.i: the selection of the axis influences the determination of the angular momentum. An angular momentum without a stated point is not a number.
(d) Check the magnitude. At $x = 2.0$ m, $r = \sqrt{(2.0)^2 + (1.5)^2} = 2.5$ m and the angle between $\vec{r}$ and $\vec{v}$ has $\sin\theta = 1.5/2.5 = 0.60$ . So $rmv\sin\theta = (2.5)(0.40)(6.0)(0.60) = 3.6\ \mathrm{kg \cdot m^2/s}$ , matching part (a) in magnitude.
The two agree on size and only one of them tells you the sign. That is the whole difference between the two courses' sheets on this line.

(a) $\vec{L} = -(3.6\ \mathrm{kg \cdot m^2/s})\hat{k}$ , into the page. (b) The same $-3.6\ \mathrm{kg \cdot m^2/s}$ , constant as the puck moves. (c) Zero about a point on the line of travel. (d) $rmv\sin\theta = 3.6\ \mathrm{kg \cdot m^2/s}$ , the same magnitude with no direction.

Work done by a torque that changes with angle

A wheel with rotational inertia $I = 0.25\ \mathrm{kg \cdot m^2}$ starts from rest and is driven by a torque that depends on angular position as $\tau(\theta) = 12\ \mathrm{N \cdot m} - (3.0\ \mathrm{N \cdot m/rad})\theta$ . Find (a) the work done on the wheel over the first 4.0 rad, (b) the angular speed at that point, (c) why 4.0 rad is where the angular speed peaks, and (d) what the algebra-based formula $W = \tau\Delta\theta$ would need in order to give the same answer.

Declare the convention: positive $\theta$ is the direction the wheel turns, so the driving torque is positive while it lasts.
(a) Statement 6.2.A.2 gives $W = \int_{\theta_1}^{\theta_2} \tau\, d\theta$ . Here $W = \int_0^{4.0} (12 - 3.0\theta)\, d\theta = \left[12\theta - 1.5\theta^2\right]_0^{4.0}$ .
$W = 12(4.0) - 1.5(4.0)^2 = 48 - 24 = 24\ \mathrm{J}$ .
(b) The wheel starts at rest, so all of that work becomes rotational kinetic energy: $\frac{1}{2}I\omega^2 = 24$ J, giving $\omega^2 = \frac{2(24)}{0.25} = 192\ \mathrm{rad^2/s^2}$ and $\omega = 13.856 \approx 13.9\ \mathrm{rad/s}$ .
(c) Set $\tau = 0$ : $12 - 3.0\theta = 0$ gives $\theta = 4.0$ rad. Past that angle the torque reverses sign, so the integrand goes negative and the wheel starts slowing. The maximum of $\omega$ is where $\tau$ crosses zero, not where $\tau$ is largest.
The graph reading in 6.2.A.3 says the same thing: the work is the area under the torque against angular position curve, and that area stops growing where the curve crosses the axis.
(d) The average torque over $0$ to $4.0$ rad is $\frac{12 + 0}{2} = 6.0\ \mathrm{N \cdot m}$ , and $(6.0)(4.0) = 24$ J, the same number. The product form works only if you already know the average, which for a torque linear in $\theta$ is the midpoint value and in general is not anything simple.
Sanity check the units: $\mathrm{N \cdot m}$ times radians is joules, since the radian is dimensionless. And $\mathrm{kg \cdot m^2}$ times $\mathrm{rad^2/s^2}$ is also joules, so part (b) is consistent.

(a) $W = 24$ J. (b) $\omega = 13.9\ \mathrm{rad/s}$ . (c) The angular speed peaks at $\theta = 4.0$ rad because that is where the torque changes sign, not where it is largest. (d) $W = \tau\Delta\theta$ reproduces 24 J only with the average torque of $6.0\ \mathrm{N \cdot m}$ , which the integral hands you and the product form does not.

The three orbital results, all derived rather than looked up

A 1200 kg satellite is in a circular orbit of radius $8.00 \times 10^6$ m about a planet of mass $M = 5.97 \times 10^{24}$ kg. Using $G = 6.67 \times 10^{-11}\ \mathrm{N \cdot m^2/kg^2}$ , find (a) the orbital speed, (b) the kinetic, potential and total energies, (c) the escape speed from that radius, and (d) the energy that must be added to let the satellite escape.

None of $K = -\frac{1}{2}U$ , $E_{\text{total}} = -\frac{GMm}{2r}$ or $v_{\text{esc}} = \sqrt{2GM/r}$ is on the equation sheet. All three carry the Derived Equation label, so derive them. Start from $GM = (6.67 \times 10^{-11})(5.97 \times 10^{24}) = 3.9820 \times 10^{14}\ \mathrm{m^3/s^2}$ .
(a) A circular orbit needs the gravitational force to supply the centripetal acceleration, and both pieces are printed: $\frac{GMm}{r^2} = \frac{mv^2}{r}$ , so $v = \sqrt{GM/r}$ .
$v = \sqrt{\dfrac{3.9820 \times 10^{14}}{8.00 \times 10^6}} = \sqrt{4.9775 \times 10^{7}} = 7.055 \times 10^3\ \mathrm{m/s}$ , about 7.06 km/s.
(b) $U = -\dfrac{GMm}{r} = -\dfrac{(3.9820 \times 10^{14})(1200)}{8.00 \times 10^6} = -5.973 \times 10^{10}\ \mathrm{J}$ , using the printed $U_G$ .
$K = \frac{1}{2}mv^2 = \frac{1}{2}(1200)(4.9775 \times 10^7) = 2.986 \times 10^{10}\ \mathrm{J}$ . Compare that with $-\frac{1}{2}U = 2.986 \times 10^{10}\ \mathrm{J}$ : the derived relation of 6.6.A.3 falls out of the two printed equations, and the digits agree.
$E_{\text{total}} = K + U = 2.986 \times 10^{10} - 5.973 \times 10^{10} = -2.986 \times 10^{10}\ \mathrm{J}$ , which is $\frac{1}{2}U$ , the other half of 6.6.A.3. The total energy is negative, which is what bound means.
(c) Statement 6.6.A.4 defines escape velocity as the speed at which the mechanical energy of the satellite and central object is zero, so set $\frac{1}{2}mv_{\text{esc}}^2 - \frac{GMm}{r} = 0$ and get $v_{\text{esc}} = \sqrt{2GM/r}$ .
$v_{\text{esc}} = \sqrt{9.9550 \times 10^7} = 9.977 \times 10^3\ \mathrm{m/s}$ , about 9.98 km/s. That is $\sqrt{2}$ times the orbital speed, and $(1.41421)(7.055) = 9.977$ , so the ratio checks.
(d) To reach zero total energy from $-2.986 \times 10^{10}$ J you must add $2.986 \times 10^{10}$ J, which happens to equal the satellite's current kinetic energy. Statement 6.6.A.4.i describes the outcome: the satellite moves away until its speed reaches zero at an infinite distance from the central body.

(a) $v = 7.06\ \mathrm{km/s}$ . (b) $K = 2.986 \times 10^{10}$ J, $U = -5.973 \times 10^{10}$ J, $E_{\text{total}} = -2.986 \times 10^{10}$ J, confirming $K = -\frac{1}{2}U$ and $E_{\text{total}} = \frac{1}{2}U$ numerically. (c) $v_{\text{esc}} = 9.98\ \mathrm{km/s}$ , which is $\sqrt{2}$ times the orbital speed. (d) $2.99 \times 10^{10}$ J, equal in size to the total energy and to the current kinetic energy.

Frequently asked questions

How much of the AP Physics C Mechanics exam is Unit 6?

Unit 6, Energy and Momentum of Rotating Systems, is weighted at 10 to 15% of the multiple-choice section of the AP Physics C: Mechanics exam, and the course description suggests about 13 to 19 class periods for it. Units 1, 5 and 7 carry the same band; Unit 2 is heaviest at 20 to 25%, Unit 3 is 15 to 25%, and Unit 4 is 10 to 20%. This is one of the places the two rotation courses part company: AP Physics 1 weights its Unit 6, with the same six topic titles, at only 5 to 8%.

Does AP Physics C Unit 6 have any boundary statements?

One, and it is a single sentence under Topic 6.5: rolling friction is beyond the scope of AP Physics C: Mechanics. Topics 6.1, 6.2, 6.3, 6.4 and 6.6 print no boundary statement at all. What limits the orbital results at the end of the unit is not a boundary statement but the Derived Equation label, which the framework's Required Equations page defines as marking the final results of derivations expected of students on the exam. All three of the unit's derived equations sit in Topic 6.6.

Is the escape velocity formula on the AP Physics C Mechanics equation sheet?

No. Essential knowledge 6.6.A.4.ii gives the escape velocity as the square root of two times the gravitational constant times the central mass, divided by the distance, and labels it a Derived Equation, adding that it can be derived using conservation of energy laws. Neither it nor the circular-orbit results relating kinetic energy and total energy to the potential energy appear on the sheet. What the sheet does print is everything you need to derive them: the gravitational force, the gravitational potential energy with its minus sign, kinetic energy, and the centripetal acceleration. The route is to set the total mechanical energy to zero and solve for the speed.

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

It has a direction. Essential knowledge 6.3.A.2 of AP Physics C: Mechanics defines the angular momentum of an object about a point as the cross product of the position vector with the linear momentum, and the course's equation sheet prints it as a vector equation that also equals the rotational inertia times the angular velocity vector. The AP Physics 1 sheet prints two scalar forms instead, the rotational inertia times angular speed and the product of distance, mass, speed and the sine of the enclosed angle. The magnitudes agree; only the calculus-based course can ask which way the angular momentum points or add two that are not parallel.

Can angular momentum be conserved while kinetic energy is not?

Yes, and it is the standard Unit 6 question. Essential knowledge 6.4.A.2.iii says the angular speed of a nonrigid system may change without its angular momentum changing if the system changes shape by moving mass closer to or farther from the rotational axis. A skater who pulls her arms in reduces her rotational inertia, so her angular speed rises, and because rotational kinetic energy can be written as the angular momentum squared divided by twice the rotational inertia, that energy rises too. The extra energy comes from the work her muscles do pulling the arms inward. Angular momentum is set by the net external torque, which is zero here; kinetic energy is not protected by anything.

Why is work done by a torque an integral in AP Physics C?

Because the torque does not have to be constant. Essential knowledge 6.2.A.2 gives the work done on a rigid system by a torque as the integral of the torque with respect to angular position, and 6.2.A.3 adds that it can be found from the area under a graph of torque against angular position. The algebra-based sheet prints the product of torque and angular displacement, which is only the special case of a constant torque, or of using an average value you have some way of knowing. Statement 6.2.A.1 also fixes the condition for any work at all: the torque has to be exerted over an angular displacement, so a torque applied to something that does not turn transfers no energy.

Do satellites in elliptical orbits conserve angular momentum?

Yes. Essential knowledge 6.6.A.2.ii says that in elliptical orbits the system's total mechanical energy and the satellite's angular momentum are both constant, while the system's gravitational potential energy and the satellite's kinetic energy can each change. That is why a planet speeds up as it moves closer to the sun: with the angular momentum fixed, a smaller distance requires a larger speed. In circular orbits, statement 6.6.A.2.i says more is constant, namely the total mechanical energy, the gravitational potential energy, the angular momentum and the kinetic energy all at once. The circular-orbit relation between kinetic energy and potential energy is stated only for the circular case, so do not carry it onto an ellipse.