AP Physics C: Mechanics Unit 6 Review: Energy and Momentum of Rotating Systems

Q: What's on the AP Physics Mech Unit 6 progress check (MCQ and FRQ)?

The AP Physics C: Mechanics Unit 6 progress check includes both MCQ and FRQ parts drawn from all six unit topics: Rotational Kinetic Energy, Torque and Work, Angular Momentum and Angular Impulse, Conservation of Angular Momentum, Rolling, and Motion of Orbiting Satellites. MCQ questions test conceptual understanding and calculation, while the FRQ section asks you to set up and solve multi-part problems involving angular momentum or rolling motion. For matched practice aligned to the progress check, visit /ap-physics-c-mechanics/unit-6 .

Q: How do I practice AP Physics Mech Unit 6 FRQs?

To practice Unit 6 FRQs, focus on the topics that appear most often in free-response questions: Conservation of Angular Momentum, Rolling without slipping, and Torque and Work. Typical FRQ formats ask you to derive an expression for angular momentum before and after a collision, calculate rotational kinetic energy, or analyze a rolling object on an incline. Work through each part by writing out your setup, applying the correct conservation law, and checking units. Find Unit 6 FRQ practice at /ap-physics-c-mechanics/unit-6 .

Q: Where can I find AP Physics Mech Unit 6 practice questions?

You can find AP Physics C: Mechanics Unit 6 practice questions, including multiple-choice and practice test problems, at /ap-physics-c-mechanics/unit-6 . That page has MCQ sets and FRQs covering all six topics: Rotational Kinetic Energy, Torque and Work, Angular Momentum, Conservation of Angular Momentum, Rolling, and Motion of Orbiting Satellites.

Q: How should I study AP Physics Mech Unit 6?

Start Unit 6 by building a solid feel for rotational kinetic energy and how it connects to the linear kinetic energy formulas you already know. Then work through Torque and Work, Angular Momentum and Angular Impulse, and Conservation of Angular Momentum in order, since each concept builds on the last. Rolling without slipping trips up a lot of students, so practice breaking it into translational and rotational parts separately before combining them. Finish with Motion of Orbiting Satellites, which ties angular momentum to gravity. For topic guides and practice sets, go to /ap-physics-c-mechanics/unit-6 .

10–15% of AP multiple choice
6 topics
AP exam review verified for 2027

Unit 6 covers rotational kinetic energy, work done by torques, angular momentum and angular impulse, conservation of angular momentum, rolling motion combining translation and rotation, and the orbital mechanics of satellites governed by gravitational conservation laws.

Fiveable's AP Physics C: Mechanics Unit 6 review covers 6 topics, with guided practice questions for the unit.

start with 6.1 start unit practiceReview with a tutorFree, 24/7, real people, from UPchieve

AP Physics C: Mechanics unit 6 topics

6.1

Rotational Kinetic Energy

Defines K_rot = 1/2 I omega^2 and establishes that total kinetic energy is the sum of translational and rotational contributions. Introduces the scalar nature of rotational kinetic energy and common moment-of-inertia values.

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6.2

Torque and Work

Develops W = integral tau d theta as the rotational analog of linear work, including the graphical interpretation as area under a tau-versus-theta curve and the rotational work-energy theorem.

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6.3

Angular Momentum and Angular Impulse

Introduces L = I omega and L = r x p, defines angular impulse as the integral of tau dt, and connects them through the rotational impulse-momentum theorem delta L = integral tau dt.

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6.4

Conservation of Angular Momentum

Establishes that total angular momentum is conserved when net external torque is zero, explains how nonrigid systems change omega by redistributing mass, and emphasizes system boundary selection.

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6.5

Rolling

Applies the no-slip constraint v_cm = r omega to combine translational and rotational kinetic energy, distinguishes pure rolling (energy conserved) from slipping (kinetic friction dissipates energy).

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6.6

Motion of Orbiting Satellites

Uses conservation of energy and angular momentum to analyze circular and elliptical orbits, deriving orbital speed, total mechanical energy E = -GMm/(2r), and escape velocity v_esc = sqrt(2GM/r).

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Practice AP Physics C: Mechanics unit 6 questions

Try AP-style questions on unit 6, then use the review notes below for anything you miss.

Example AP-style MCQs

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MCQ

AP-style practice question

Question

Torque $\tau$ decreases linearly from $\tau_{max}$ at $\theta=0$ to zero at $\theta=\Theta$ . Does the total work equal $\tau_{max}\Theta$ ?

No. Work is torque integrated over angle, not a function of $\omega^2$ .

Yes. Treating torque constant yields $W=\tau_{max}\Theta$ , which overestimates actual work.

No. Work equals the triangular area under $\tau$ – $\theta$ , so $W=\tfrac{1}{2}\tau_{max}\Theta$ .

Yes. Assuming initial torque alone determines work ignores torque variation with angle.

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MCQ

AP-style practice question

Question

A bowling ball is released with a forward linear velocity $v$ and a very large forward angular velocity $\omega$ such that $R\omega > v$ . A student claims the friction force acts in the direction of motion. Which justification supports this?

The contact point slips forward relative to the floor, so kinetic friction acts backward.

The contact point is stationary relative to the floor, so static friction acts forward.

The contact point slips backward relative to the floor, so kinetic friction acts forward.

The contact point is stationary relative to the floor, so static friction acts backward.

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Example FRQs

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FRQ

Rotational inertia and angular momentum conservation

4. A uniform horizontal disk of mass $M_d = 4 (see Figure 2).0 \text{ kg}$ and radius $R = 0.50 \text{ m}$ is mounted on a frictionless vertical axle through its center, as shown in Figure 1. The disk has rotational inertia $I_d = \frac{1}{2}M_d R^2$ about the axle. A small block of mass $m = 1.0 \text{ kg}$ is placed on the disk at a distance $r_0 = 0.30 \text{ m}$ from the center. The disk is initially rotating with angular speed $\omega_0 = 8.0 \text{ rad/s}$ , and the block rotates with the disk without slipping due to static friction.

Figure 1. Initial top view: block at r0 = 0.30 m on a rotating disk of radius R = 0.50 m, rotating counterclockwise at ω0 = 8.0 rad/s.

Figure 2. Final top view: block at the rim (r = R = 0.50 m) after sliding outward; disk+block rotate counterclockwise at ωf.

As the block slides outward from $r_0$ to $R$ , the angular speed of the disk changes from $\omega_0$ to $\omega_f$ .

Indicate whether $\omega_f$ is greater than, less than, or equal to $\omega_0$ by writing one of the following.

$\omega_f > \omega_0$
$\omega_f < \omega_0$
$\omega_f = \omega_0$

Justify your answer using qualitative reasoning beyond referencing equations.

Derive an expression for the angular speed $\omega_f$ of the disk after the block reaches the edge at radius $R$ . Express your answer in terms of $M_d$ , $m$ , $R$ , $r_0$ , $\omega_0$ , and physical constants, as appropriate. Begin your derivation by writing a fundamental physics principle or an equation from the reference information.

Indicate whether the rotational kinetic energy of the system increases by an amount greater than, less than, or equal to the work done by the torque. In a different scenario, the block is instead connected to the center of the disk by a massless string of length $r_0 = 0.30 \text{ m}$ . The disk and block again start with angular speed $\omega_0 = 8.0 \text{ rad/s}$ . A motor applies a constant torque $\tau = 5.0 \text{ N}\cdot\text{m}$ to the disk for a time interval $\Delta t = 2.0 \text{ s}$ .

Briefly justify your answer.

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FRQ

Inelastic collision momentum and pendulum energy conservation

3. A puck of measured mass m = 0.0500 kg is launched horizontally with the same unknown speed u on every trial and embeds in a stationary pendulum catcher. Known masses can be secured to the catcher to change the measured total post-collision mass M_T, which includes the puck. After capture, the combined system swings upward, and a motion sensor determines the vertical rise h of its center of mass. Figure 1 shows the setup. External impulse during the collision is negligible, pivot friction and air resistance during the swing are negligible, and use g = 9.80 m/s squared.

Figure 1. Ballistic-pendulum setup and rise measurement.

Describe a procedure for collecting data that can determine u from a linear graph. State what is varied and measured and give one way to reduce uncertainty.

Derive a linear relationship for the collected data. Identify the quantities on both axes and explain how the slope determines u. Use momentum during the collision and mechanical energy during the subsequent swing.

Figure 2. Grid for student graph.

Synthetic idealized processed measurements. Total mass and vertical rise were measured before the displayed quantities were calculated.

Reciprocal total mass 1/M_T (1/kg)	Square root of rise sqrt(h) (sqrt(m))
1.500	0.150
2.000	0.200
2.500	0.250
3.000	0.300
3.500	0.350

The processed measurements are shown in Table 1.

Indicate the quantities that should be graphed.

Vertical axis: Horizontal axis:

ii.

On the blank grid in Figure 2, create the graph using Table 1. Label both axes with units and plot all five points.

iii.

Draw a single best-fit line for the plotted data.

Calculate the experimental value of u. Use the slope of the best-fit line from part C.

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FRQ

Rotational kinetic energy and angular momentum of disk

2. A uniform disk of mass M = 2.0 kg and radius R = 0.40 m is mounted on a fixed horizontal axle through its center. The disk is initially at rest. A motor applies a time-varying torque to the disk, causing it to rotate. The rotational inertia of the disk about its center is $\frac{1}{2}MR^2$ . Friction in the axle is negligible.

Figure 2: Energy bar chart of rotational kinetic energy K at two times, t = 0.60 s and t = 1.2 s.

An energy bar chart can be used to represent the rotational kinetic energy K of the disk at different times. On the energy bar chart in Figure 2, draw shaded bars to represent the rotational kinetic energy of the disk at t = 0.60 s and at t = 1.2 s.

The heights of the shaded bars should be proportional to the rotational kinetic energy at each time.
If the energy is zero at any time, represent this with a distinct line on the zero-energy line.

Derive an expression for the work W done by the motor torque on the disk during the time interval from t = 0 to t = 1.2 s. Express your answer in terms of M, R, and the angular velocities at t = 0 and t = 1.2 s, and physical constants, as appropriate. Begin your derivation by writing a fundamental physics principle or an equation from the reference information. Then calculate the numerical value of W. The disk rotates through an angular displacement of $\theta = 9.6$ rad during the time interval from t = 0 to t = 1.2 s.

Figure 1: Angular velocity ω (rad/s) versus time t (s) for a disk driven by a motor until t = 1.2 s, then braked to rest by t = 2.0 s.

Figure 3: Axes for angular momentum L (kg·m²/s) versus time t (s), to be sketched consistent with Figure 1.

On the axes in Figure 3, sketch a graph of the angular momentum L of the disk as a function of time t for the entire time interval from t = 0 to t = 2.0 s. The graph should be consistent with the angular velocity shown in Figure 1. The angular momentum of the disk changes as torques are applied to it.

Describe whether the student's claim is correct or incorrect. A student claims that the magnitude of the braking torque applied during the time interval from t = 1.2 s to t = 2.0 s is greater than the magnitude of the motor torque applied during the time interval from t = 0 to t = 1.2 s.

Justify your answer using physics principles and the information provided in Figure 1.

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AP Physics C: Mechanics unit 6 practice test

A timed, AP-style test built only from unit 6 questions — multiple choice plus free response, scored with AI feedback. Use the standard format or build your own.

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Unit 6 review notes

6.1

Rotational Kinetic Energy

A rigid body spinning about any axis has rotational kinetic energy K_rot = 1/2 I omega^2. This is a scalar, so it adds directly to translational kinetic energy. When a rigid body both translates and rotates, K_tot = 1/2 m v_cm^2 + 1/2 I omega^2. A system can have rotational kinetic energy even when its center of mass is stationary, because individual mass elements still have linear speed.

K_rot = 1/2 I omega^2: Rotational kinetic energy depends on the moment of inertia I about the rotation axis and the angular velocity omega. It is a scalar.
Total kinetic energy: K_tot = K_trans + K_rot = 1/2 m v_cm^2 + 1/2 I omega^2 for any rigid body that both translates and rotates.
Stationary center of mass: A spinning object whose center of mass is at rest still has rotational kinetic energy because its mass elements move in circles.
Moment of inertia I: Quantifies how mass is distributed relative to the rotation axis; determines how much rotational kinetic energy a body stores at a given omega.

A solid disk (I = 1/2 m r^2) and a hoop (I = m r^2) of equal mass and radius spin at the same omega. Which has more rotational kinetic energy, and by what factor?

Object	Moment of inertia I	K_rot at same omega
Solid disk	1/2 m r^2	1/4 m r^2 omega^2
Hoop (thin ring)	m r^2	1/2 m r^2 omega^2
Solid sphere	2/5 m r^2	1/5 m r^2 omega^2
Thin rod (center)	1/12 m L^2	1/24 m L^2 omega^2

6.2

Torque and Work

A torque transfers energy into or out of a rotating system when it acts through an angular displacement. The work done is W = integral from theta_1 to theta_2 of tau d theta. For constant torque, W = tau * delta theta. On a torque-versus-angular-position graph, work equals the area under the curve. The rotational work-energy theorem states that the net work done by all torques equals the change in rotational kinetic energy.

W = integral tau d theta: General expression for work done by a torque over an angular displacement; reduces to W = tau * delta theta when torque is constant.
Area under tau vs theta graph: Graphical method to find rotational work without integration; positive area means energy added, negative area means energy removed.
Rotational work-energy theorem: Net rotational work equals the change in rotational kinetic energy: W_net = delta K_rot.
Sign of rotational work: Work is positive when torque and angular displacement are in the same direction, negative when they are opposite.

A torque that varies as tau = 3 theta (in N*m, theta in radians) acts on a disk from theta = 0 to theta = 2 rad. What is the total work done on the disk?

Quantity	Linear analog	Rotational version
Work	W = integral F dx	W = integral tau d theta
Constant-force/torque work	W = F * delta x	W = tau * delta theta
Work-energy theorem	W_net = delta K_trans	W_net = delta K_rot

6.3

Angular Momentum and Angular Impulse

Angular momentum has two useful forms: L = I omega for a rigid body about a fixed axis, and L = r x p (vector cross product) for any object about a chosen point. The choice of reference point matters because it changes the value of L. Even an object moving in a straight line has nonzero angular momentum about any point not on its line of motion. Angular impulse equals the integral of tau dt and equals the change in angular momentum: delta L = integral tau dt. On a torque-versus-time graph, angular impulse is the area under the curve, and the slope of an L-versus-time graph equals the net torque.

L = I omega: Angular momentum of a rigid body rotating about a fixed axis; units are kg*m^2/s.
L = r x p: Vector angular momentum of any object about a reference point; magnitude is m v r sin(theta) where theta is the angle between r and v.
Angular impulse = integral tau dt: The rotational analog of linear impulse; equals the change in angular momentum of the system.
tau_net = dL/dt: Net external torque equals the time rate of change of angular momentum; for constant I this reduces to tau_net = I alpha.
Torque-time graph area: The area under a tau-versus-t graph gives the angular impulse delivered to the system over that time interval.

A point mass moves in a straight line at constant velocity. Explain why its angular momentum about a point off its path is nonzero and constant.

Concept	Linear version	Rotational version
Momentum	p = m v	L = I omega or r x p
Impulse	J = integral F dt	Angular impulse = integral tau dt
Impulse-momentum theorem	J = delta p	Angular impulse = delta L
Newton's second law	F_net = dp/dt	tau_net = dL/dt

6.4

Conservation of Angular Momentum

The total angular momentum of a system is the sum of the angular momenta of all its parts. If the net external torque on a system is zero, total angular momentum is conserved. If the net external torque is nonzero, angular momentum is transferred between the system and its environment by an amount equal to the angular impulse. A nonrigid system can change its angular speed without any external torque if it redistributes mass, changing I while L stays constant (the classic spinning-skater scenario). System selection is critical: choose a system boundary where all relevant torques are internal to apply conservation.

Conservation condition: Total angular momentum is constant when the net external torque about the chosen axis is zero.
L_total = sum of L_i: Angular momenta of all parts of a system add as vectors about the same axis.
Nonrigid system redistribution: When a skater pulls arms inward, I decreases and omega increases so that L = I omega remains constant.
Newton's third law for torques: Internal torque pairs are equal and opposite, so they cancel in the total angular momentum of the system.
System boundary selection: Choosing the system to include all objects exerting torques on each other makes those torques internal and allows conservation to apply.

A student sits on a frictionless rotating stool holding a spinning bicycle wheel with its axis vertical. The student flips the wheel upside down. Describe what happens to the student's rotation and explain why using angular momentum conservation.

Condition	Net external torque	Angular momentum
Isolated system	Zero	Conserved
External torque applied	Nonzero	Changes by angular impulse
Nonrigid, no external torque	Zero	Conserved; omega changes as I changes

6.5

Rolling Motion

When an object rolls without slipping, its center-of-mass translation and its rotation are linked by v_cm = r omega and a_cm = r alpha. Total kinetic energy is K_tot = 1/2 m v_cm^2 + 1/2 I omega^2. Because static friction does no work in pure rolling, mechanical energy is conserved and you can use energy methods directly. When an object slips, kinetic friction acts at the contact point, the no-slip constraint breaks down, and kinetic friction dissipates energy from the system.

No-slip condition: v_cm = r omega and a_cm = r alpha; links translational and rotational motion so the contact point has zero velocity relative to the surface.
K_tot = K_trans + K_rot: For a rolling object: K_tot = 1/2 m v_cm^2 + 1/2 I omega^2; substitute v_cm = r omega to express everything in terms of v_cm.
Static friction in pure rolling: Provides the torque needed to maintain rolling but does no work because the contact point has zero velocity; mechanical energy is conserved.
Slipping condition: v_cm is not equal to r omega; kinetic friction acts and dissipates energy, so mechanical energy is not conserved.
Rolling on an incline: Use energy conservation: mgh = 1/2 m v_cm^2 + 1/2 I omega^2. Objects with larger I/mr^2 ratios reach the bottom more slowly.

A solid sphere (I = 2/5 m r^2) and a hollow sphere (I = 2/3 m r^2) of equal mass and radius start from rest at the top of the same incline. Which reaches the bottom first, and why?

Object	I / (m r^2)	Fraction of K_tot that is rotational
Hoop	1	1/2
Hollow sphere	2/3	2/5
Solid disk	1/2	1/3
Solid sphere	2/5	2/7

6.6

Motion of Orbiting Satellites

For a satellite of negligible mass orbiting a massive central body M, gravity provides centripetal force and two conservation laws govern the motion. In circular orbits, total mechanical energy E = -GMm/(2r), kinetic energy K = GMm/(2r), and gravitational potential energy U = -GMm/r are all constant, as is angular momentum L = m v r. In elliptical orbits, total mechanical energy and angular momentum are still conserved, but K and U each vary as the satellite moves closer to or farther from the central body. Escape velocity from radius r is v_esc = sqrt(2GM/r), which corresponds to total mechanical energy equal to zero.

Circular orbital speed: v_circ = sqrt(GM/r); derived by setting gravitational force equal to centripetal force.
Total mechanical energy (circular orbit): E = -GMm/(2r); negative value indicates a bound orbit. K = -E and U = 2E.
Gravitational potential energy: U = -GMm/r; defined as zero at infinite separation, so U is always negative for bound systems.
Escape velocity: v_esc = sqrt(2GM/r); the minimum speed needed for a satellite to escape to infinity with zero kinetic energy remaining.
Elliptical orbit conservation: Total mechanical energy and angular momentum L = m v r sin(theta) are constant throughout an elliptical orbit; K and U individually vary.

A satellite in a circular orbit of radius r is given a brief forward thrust that increases its speed. Describe qualitatively what happens to its orbit shape, total energy, and angular momentum immediately after the thrust.

Orbit type	Total energy E	Angular momentum L	K and U individually
Circular	Constant, -GMm/(2r)	Constant	Both constant
Elliptical	Constant (negative)	Constant	Each varies
Escape (parabolic)	Zero	Constant	Each varies

Key terms

Term	Definition
angular displacement	The angle in radians through which a rigid body rotates about a specified axis, calculated as delta theta = theta minus theta_0. It appears in the rotational work integral W = integral tau d theta.
impulse-momentum theorem	In rotational form: the angular impulse delivered to a system equals its change in angular momentum, delta L = integral tau dt. This is the rotational analog of J = delta p.
scalar	A physical quantity described by magnitude only, without direction. Rotational kinetic energy and work are scalars, which is why K_rot adds directly to K_trans.
two-body gravitational system	A system of a massive central object and an orbiting satellite interacting only via gravity. When the satellite mass is negligible, the central object is treated as stationary and conservation laws govern the satellite's orbit.

How to review unit 6

Topics students miss most

Miss rates come from Fiveable multiple-choice practice.

6.6

Motion of Orbiting Satellites

43%246 tries

6.3

Angular Momentum and Angular Impulse

42%498 tries

6.4

Conservation of Angular Momentum

34%271 tries

Common unit 6 mistakes

Forgetting to include both K_trans and K_rot

For a rolling object, total kinetic energy is K_tot = 1/2 m v_cm^2 + 1/2 I omega^2. Counting only translational kinetic energy underestimates the total and gives wrong speeds at the bottom of inclines.

Applying angular momentum conservation when external torques exist

Conservation of angular momentum requires zero net external torque about the chosen axis. Friction from a surface or a pivot exerting a torque means angular momentum is not conserved for that system.

Confusing the two forms of angular momentum

L = I omega applies to rigid bodies rotating about a fixed axis. L = r x p applies to any object about any reference point, including a point mass moving in a straight line. Using the wrong form leads to incorrect magnitudes and directions.

Treating gravitational potential energy as mgh in orbital problems

For satellites, gravitational potential energy is U = -GMm/r, not mgh. Using mgh is only valid near Earth's surface. Orbital energy calculations require U = -GMm/r and E = -GMm/(2r).

Assuming static friction does work during rolling

In pure rolling without slipping, the contact point has zero instantaneous velocity, so static friction does no work and mechanical energy is conserved. Only kinetic friction during slipping dissipates energy.

How this unit shows up on the AP exam

Multi-step energy and momentum accounting

Free-response questions in this unit frequently require you to apply both energy conservation and angular momentum conservation in sequence within the same problem. A common structure is: find the angular speed after a collision or mass redistribution using angular momentum conservation, then use that result to find kinetic energy or height using energy conservation. Keeping track of which quantities are conserved at each stage is the central skill.

Graphical interpretation of rotational quantities

Multiple-choice and free-response questions often present torque-versus-time or torque-versus-angle graphs and ask you to extract angular impulse or work as areas under the curve, or to identify net torque as the slope of an angular momentum graph. Practicing these graphical readings directly from the equations W = integral tau d theta and delta L = integral tau dt is essential.

Justifying conservation law applicability

A recurring task is explaining in words why a conservation law does or does not apply to a given system. For angular momentum, you must identify whether net external torque is zero and justify your system boundary choice. For energy in rolling problems, you must state whether the no-slip condition holds and whether friction does work. Clear, equation-linked justifications are expected in free-response scoring.

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Frequently Asked Questions

What's on the AP Physics Mech Unit 6 progress check (MCQ and FRQ)?

The AP Physics C: Mechanics Unit 6 progress check includes both MCQ and FRQ parts drawn from all six unit topics: Rotational Kinetic Energy, Torque and Work, Angular Momentum and Angular Impulse, Conservation of Angular Momentum, Rolling, and Motion of Orbiting Satellites. MCQ questions test conceptual understanding and calculation, while the FRQ section asks you to set up and solve multi-part problems involving angular momentum or rolling motion. For matched practice aligned to the progress check, visit /ap-physics-c-mechanics/unit-6.

How do I practice AP Physics Mech Unit 6 FRQs?

To practice Unit 6 FRQs, focus on the topics that appear most often in free-response questions: Conservation of Angular Momentum, Rolling without slipping, and Torque and Work. Typical FRQ formats ask you to derive an expression for angular momentum before and after a collision, calculate rotational kinetic energy, or analyze a rolling object on an incline. Work through each part by writing out your setup, applying the correct conservation law, and checking units. Find Unit 6 FRQ practice at /ap-physics-c-mechanics/unit-6.

Where can I find AP Physics Mech Unit 6 practice questions?

You can find AP Physics C: Mechanics Unit 6 practice questions, including multiple-choice and practice test problems, at /ap-physics-c-mechanics/unit-6. That page has MCQ sets and FRQs covering all six topics: Rotational Kinetic Energy, Torque and Work, Angular Momentum, Conservation of Angular Momentum, Rolling, and Motion of Orbiting Satellites.

How should I study AP Physics Mech Unit 6?

Start Unit 6 by building a solid feel for rotational kinetic energy and how it connects to the linear kinetic energy formulas you already know. Then work through Torque and Work, Angular Momentum and Angular Impulse, and Conservation of Angular Momentum in order, since each concept builds on the last. Rolling without slipping trips up a lot of students, so practice breaking it into translational and rotational parts separately before combining them. Finish with Motion of Orbiting Satellites, which ties angular momentum to gravity. For topic guides and practice sets, go to /ap-physics-c-mechanics/unit-6.

Ready to review Unit 6?Start with the notes, check the topic cards, and use the practice or resource links when they are available for this course.

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