AP exam review verified for 2027

AP Physics C: Mechanics Unit 6 Review: Energy & Momentum

Q: What topics are covered in AP Physics Mech Unit 6?

AP Physics C: Mechanics Unit 6 covers 6 topics: Rotational Kinetic Energy, Torque and Work, Angular Momentum and Angular Impulse, Conservation of Angular Momentum, Rolling, and Motion of Orbiting Satellites. Together they apply energy and momentum concepts to rotating systems, including objects rolling without slipping and orbiting satellites. See the full topic list at /ap-physics-c-mechanics/unit-6 .

Q: How much of the AP Physics Mech exam is Unit 6?

Unit 6 makes up 10-15% of the AP Physics C: Mechanics exam. That weight comes from topics like Rotational Kinetic Energy, Conservation of Angular Momentum, Rolling, and Motion of Orbiting Satellites. It's a meaningful chunk of the test, so understanding when angular momentum is conserved versus when it changes is especially important.

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 .

Review AP Physics C: Mechanics Unit 6 to build fluency with rotational kinetic energy, angular momentum, conservation laws, rolling motion, and satellite orbits. This unit carries 10-15% of the exam and pulls together energy and momentum tools from earlier units and applies them to spinning and orbiting systems.

Use the topic guides, practice questions, and FRQ practice available on Fiveable to work through each concept before your exam.

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AP exam review verified for 2027

What is AP Physics C: Mechanics unit 6?

Unit 6 asks you to apply two of the most powerful tools in mechanics, energy conservation and momentum conservation, to systems that rotate. The unit builds from the rotational inertia and torque ideas in Unit 5 and adds the energy and momentum accounting needed to solve complex spinning and orbiting problems.

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.

Energy in rotating systems

Rotational kinetic energy K_rot = 1/2 I omega^2 is a scalar that adds directly to translational kinetic energy. For a rolling object, K_tot = 1/2 m v_cm^2 + 1/2 I omega^2. Work done by a torque equals the integral of tau d theta, and the rotational work-energy theorem connects net rotational work to the change in K_rot.

Angular momentum and its conservation

Angular momentum is L = I omega for a rigid body about a fixed axis, or L = r x p for any object about a point. Angular impulse equals the integral of tau dt and equals the change in L. When net external torque is zero, total angular momentum is conserved, even if the system changes shape and redistributes mass.

Rolling and orbital motion

Rolling without slipping links translation and rotation through v_cm = r omega, a_cm = r alpha. Static friction does no work, so mechanical energy is conserved. For satellites, circular orbits have constant L, K, and total mechanical energy E = -GMm/(2r), while elliptical orbits conserve L and total energy but allow K and gravitational potential energy to vary.

Conservation laws govern rotating and orbiting systems

The central skill in Unit 6 is recognizing when angular momentum is conserved (zero net external torque) versus when it changes (nonzero net external torque equals angular impulse). The same logic applies to energy: identify whether work is done by external torques or whether the no-slip condition allows mechanical energy conservation. These two questions, is angular momentum conserved? is mechanical energy conserved? drive nearly every problem in the unit.

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 snapshot

Hardest AP Physics C: Mechanics unit 6 topics

This snapshot uses Fiveable practice activity to show where students tend to miss questions and which review moves are worth prioritizing first.

61%average MCQ accuracy

Across 2.0k multiple-choice practice attempts for this unit.

2.0kMCQ attempts

Practice activity included in this snapshot.

49%average FRQ score

Across 8 scored free-response attempts for this unit.

Hardest topics in unit 6

MCQ miss rate

6.6

Motion of Orbiting Satellites

Review Motion of Orbiting Satellites with attention to how the concept appears in AP-style source and evidence questions.

43%246 tries

6.3

Angular Momentum and Angular Impulse

Review Angular Momentum and Angular Impulse with attention to how the concept appears in AP-style source and evidence questions.

42%498 tries

6.4

Conservation of Angular Momentum

Review Conservation of Angular Momentum with attention to how the concept appears in AP-style source and evidence questions.

34%271 tries

6.1

Rotational Kinetic Energy

Review Rotational Kinetic Energy with attention to how the concept appears in AP-style source and evidence questions.

33%379 tries

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

Practice AP Physics C: Mechanics unit 6 questions

Try AP-style multiple-choice questions and written prompts after you review the notes.

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 equals the triangular area under $\tau$ – $\theta$ , so $W=\tfrac{1}{2}\tau_{max}\Theta$ .

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.

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 backward relative to the floor, so kinetic friction acts forward.

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 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.

Create a clean, monochrome (black line art on white background) top-view physics setup diagram.

Overall layout:
- The disk is a perfect circle centered in the frame, occupying most of the image width.
- The center of the disk is marked with a small solid dot (axle location) and a short crosshair (thin vertical and horizontal lines crossing at the dot) to emphasize the exact center.

Disk labeling and given values:
- Near the disk (outside the circle, to the upper-left of the disk), place the text label: "Uniform disk" on one line and directly beneath it "Md = 4.0 kg".
- Also show the disk radius value explicitly: place the text "R = 0.50 m" next to the radius label described below.

Radius R measurement (center to rim):
- Draw a straight radius line from the exact center dot to the rim pointing horizontally to the right.
- Put a single arrowhead at the rim end of this radius line (so it reads as a measurement from center outward).
- Place the label "R = 0.50 m" just above this horizontal radius line, centered along the line.

Block placement at r0:
- Place a small rectangular block on the disk along the same horizontal radius line used for R, located between the center and the rim.
- The block’s center is exactly on that rightward horizontal radius line.
- The block must be closer to the center than to the rim, consistent with r0 < R.
- Draw a separate measurement line from the center dot to the near side of the block along the same rightward horizontal direction; end this measurement line at the block location with a small arrowhead pointing toward the block.
- Label this inner measurement clearly as "r0 = 0.30 m" placed just below the measurement line, centered along it.
- Attach a leader arrow from the text "Block, m = 1.0 kg" to the block rectangle, with the text placed inside the disk region in the upper-right quadrant but not overlapping the rotation arrow.

Rotation indication (ω0):
- Draw a bold curved arrow tracing a circular arc around the center, located in the upper-left quadrant of the disk interior (so it does not overlap the block on the right).
- The curved arrow must indicate counterclockwise rotation (arrowhead pointing in the counterclockwise sense).
- Place the label "ω0 = 8.0 rad/s" next to the curved arrow, aligned so it unambiguously refers to the rotation arrow.

Clarity constraints:
- All text must be horizontal and legible.
- Do not include any perspective; it is strictly a top view.
- No extra forces (no friction arrows) are drawn in this figure.
- Ensure both measurement labels "r0 = 0.30 m" and "R = 0.50 m" are present and correspond to the correct center-to-block and center-to-rim distances, respectively.

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

Create a clean, monochrome (black line art on white background) top-view physics setup diagram matching the style of Figure 1.

Overall layout:
- Draw the same circular disk centered in the frame, with the center axle marked by the same small solid dot and crosshair.

Disk labeling and given values:
- Near the disk (outside the circle, to the upper-left of the disk), place the text label: "Uniform disk" and beneath it "Md = 4.0 kg".
- The disk radius must again be explicitly labeled as "R = 0.50 m" as described below.

Radius R measurement (center to rim):
- Draw a straight radius line from the exact center dot to the rim pointing horizontally to the right, identical in direction to Figure 1.
- Put a single arrowhead at the rim end.
- Place the label "R = 0.50 m" just above the radius line, centered along it.

Block at the edge (r = R):
- Place the same small rectangular block on the disk exactly at the rim on the rightmost point of the circle, centered on the rightward horizontal radius line.
- The block should visually touch the circumference so it is clearly at the edge.
- Attach a leader arrow from the text "Block, m = 1.0 kg" to the block rectangle, with the text placed inside the disk region in the upper-right quadrant but not overlapping other labels.
- Add a short text label near the block reading "r = R" to explicitly indicate the block’s radial position equals the disk radius.

Rotation indication (ωf):
- Draw a bold curved arrow tracing a circular arc around the center, located in the upper-left quadrant of the disk interior.
- The curved arrow must indicate counterclockwise rotation.
- Place the label "ωf" next to the curved arrow, with no numeric value attached.
- Do NOT include any text stating a comparison such as "ωf < ω0" inside the diagram; only label the final angular speed as ωf.

Clarity constraints:
- All text is horizontal and legible.
- No perspective; strict top view.
- No force arrows.
- Ensure the block is unambiguously at the rim and the label "R = 0.50 m" is present.

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 $\tau = 5.0 \text{ N·m}$ to the disk for a time interval $\Delta t = 2.0 \text{ s}$ .

Briefly justify your answer.

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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.

A two-bar vertical bar chart on a white background.

Axes (must be exact):
- Vertical axis on the left labeled: "K (J)".
- Vertical axis range: from 0 to 60.
- Vertical tick marks and tick labels: 0, 10, 20, 30, 40, 50, 60.
- A bold horizontal zero-energy baseline is drawn at K = 0.
- Horizontal axis label centered below: "Time" (no units on this axis because time values are in the category labels).
- Two category tick positions on the horizontal axis:
1) left category label printed under the first bar: "t = 0.60 s"
2) right category label printed under the second bar: "t = 1.2 s"
- The origin value "0" is printed at the bottom of the vertical axis where it meets the baseline.
- No arrows on axes (bar-chart style axes).

Bar geometry and style (uniform for both bars):
- Exactly two bars, equal width.
- Bar width: each bar spans a little under half the distance between the two category centers, leaving a visible gap between bars (gap is roughly one bar-width).
- Fill: solid medium-gray shading.
- Outline: solid black outline of consistent thickness.

Exact bar heights (must be numerically accurate to the scenario):
- Compute I for a uniform disk using I = 1/2 M R^2 with M = 2.0 kg and R = 0.40 m, giving I = 0.16 kg·m^2.
- From Figure 1, ω increases linearly from 0 at t = 0 to 16 rad/s at t = 1.2 s, so at t = 0.60 s (half of 1.2 s) the angular velocity is exactly 8 rad/s.
- Rotational kinetic energy is K = 1/2 I ω^2.
- Left bar (t = 0.60 s): height equals K = 1/2 × 0.16 × 8^2 = 5.12 J.
- Right bar (t = 1.2 s): height equals K = 1/2 × 0.16 × 16^2 = 20.48 J.
- The tops of the bars align exactly with these y-values on the K-axis scale (not rounded to the nearest tick).

Error bars (required, non-zero, realistic):
- Each bar includes a vertical black error bar with horizontal caps.
- Cap width: each cap spans roughly half the bar width.
- Left bar mean 5.12 J with ±10% uncertainty: error bar extends from 4.61 J to 5.63 J.
- Right bar mean 20.48 J with ±10% uncertainty: error bar extends from 18.43 J to 22.53 J.

Zero-energy representation rule:
- If any bar would be exactly zero, it would be drawn as a distinct thick line segment lying on the K = 0 baseline at that category position. (In this chart, neither time has zero energy, so both are bars above the baseline.)

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.

A clean 2D Cartesian graph with a white background and no title inside the plot area.

Axes (must be exact):
- Horizontal axis label centered below axis: "t (s)".
- Horizontal axis range: from 0 to 2.0.
- Horizontal tick marks and tick labels: 0, 0.4, 0.8, 1.2, 1.6, 2.0. (All these numbers are printed on the axis.)
- Vertical axis label rotated vertically along the left side: "ω (rad/s)".
- Vertical axis range: from 0 to 20.
- Vertical tick marks and tick labels: 0, 4, 8, 12, 16, 20. (All these numbers are printed on the axis.)
- The origin at the axes intersection is explicitly labeled with the numeral "0".
- Arrows are drawn at the positive (right) end of the t-axis and the positive (up) end of the ω-axis.

Curve shape and behavior (must be exact and piecewise linear):
- The plotted ω(t) curve is a solid black line of medium thickness.
- Segment 1 (motor on): A perfectly straight line beginning at the origin and rising with constant positive slope, passing exactly through the labeled tick intersection where t equals 1.2 and ω equals 16. This line is not curved (no concavity).
- Corner/kink at the motor shutoff time: At the vertical tick labeled 1.2 s, the graph has a sharp corner (slope changes abruptly). There is no rounding of the corner.
- Segment 2 (braking): From t = 1.2 s onward, a perfectly straight line with constant negative slope descending to meet the time axis exactly at the rightmost tick labeled 2.0 s, where ω equals 0. This line is not curved (no concavity).

Endpoints and markers:
- Use closed, filled circular markers at the start point at t = 0 and at the final resting point at t = 2.0 to emphasize that ω equals 0 at both times.
- Use a closed, filled circular marker at the corner point at t = 1.2 to emphasize ω equals 16 at that instant.

Critical numerical constraints to enforce:
- Maximum ω shown is exactly 16 rad/s and it occurs exactly at t = 1.2 s.
- ω equals exactly 0 at t = 0 and again exactly 0 at t = 2.0 s.
- The increasing segment is linear from 0 to 1.2 s, and the decreasing segment is linear from 1.2 to 2.0 s, with no curvature and no additional segments.

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

A clean 2D Cartesian set of axes with a light square grid.

Axes (must be exact):
- Horizontal axis label centered below: "t (s)".
- Horizontal axis range: from 0 to 2.0.
- Horizontal tick marks and tick labels: 0, 0.4, 0.8, 1.2, 1.6, 2.0.
- Vertical axis label rotated along left side: "L (kg·m²/s)".
- Vertical axis range: from 0 to 3.0.
- Vertical tick marks and tick labels: 0, 0.5, 1.0, 1.5, 2.0, 2.5, 3.0.
- The origin at the axes intersection is labeled with the numeral "0".
- Arrows are drawn at the positive (right) end of the t-axis and the positive (up) end of the L-axis.

Grid (must be exact):
- Vertical grid lines at every 0.4 s tick.
- Horizontal grid lines at every 0.5 kg·m²/s tick.
- Grid lines are thin and light gray so the student’s sketch stands out.

No curve is pre-drawn (blank plotting area):
- Do not include any plotted data or line for L(t).
- Only the axes, tick marks, tick labels, and grid are present.

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.

write this FRQ

FRQ

Rotational inertia from falling mass acceleration

3. A solid disk of mass M = 2.4 kg and radius R = 0.30 m is mounted on a horizontal axle through its center, allowing it to rotate freely with negligible friction. A string is wrapped around the outer edge of the disk. The free end of the string is attached to a hanging block of mass m = 0.50 kg, as shown in Figure 1. The system is released from rest, and the block falls a vertical distance while the disk rotates.

Figure 1. Disk-and-string apparatus: a hanging mass unwinds a string from a freely rotating solid disk (M = 2.4 kg, R = 0.30 m) on a low-friction horizontal axle.

Create a clean black-and-white physics apparatus diagram with thin lines and clear labels. No background shading.

Overall layout:
- The solid disk is drawn as a perfect circle occupying the left half of the figure, centered vertically.
- A horizontal axle passes through the exact center of the disk and extends a short distance to both left and right of the disk.
- Two small support brackets (simple U-shaped stands) hold the axle at the left and right sides of the disk; label one bracket "support".

Disk details (must be explicitly labeled):
- Label the disk "solid disk".
- Next to the disk label, include the given values exactly as text: "M = 2.4 kg".
- Draw a radius line from the disk center to the rightmost point on the rim. Label this line "R = 0.30 m" with the label placed just above the radius line.

String path and tangency (unambiguous geometry):
- A thin string is wrapped around the outer rim of the disk and leaves the disk at the rightmost point on the rim (the 3 o’clock position), tangent to the circle.
- From that tangency point, the string extends horizontally to the right for a short distance, then turns smoothly over a small, fixed, frictionless guide hook (drawn as a small open circle or hook) located in the upper-right quadrant of the figure, and then continues straight downward as a vertical segment.
- The vertical string segment is positioned in the right half of the figure, clearly separated from the disk by visible white space.

Hanging block:
- A rectangular block is attached to the bottom end of the vertical string segment.
- Center the block directly below the vertical string segment so the string is vertical.
- Label the block "block" and add the mass label next to it: "m = 0.50 kg".

Fall distance indicator h (must be a measurable vertical distance marker):
- To the immediate right of the block, draw a vertical double-headed arrow labeled "h".
- The top tip of the double-headed arrow is level with the block’s initial position (top surface of the block at release).
- The bottom tip of the double-headed arrow is below the block by a clearly visible amount, marking the lower position after falling.
- Draw two thin horizontal reference tick marks (short lines) extending leftward from the top and bottom tips of the h-arrow toward the block to indicate the initial and final vertical levels.

Rotation direction:
- Draw a curved arrow around the disk rim indicating the positive rotation direction.
- The curved arrow is drawn on the upper half of the disk, following the circular rim.
- The arrowhead points counterclockwise (CCW) so that unwinding at the rightmost tangency causes the string to move downward and the disk to rotate CCW.
- Label this curved arrow "positive rotation".

Axle and sensor label (to match the prompt’s allowed instruments):
- On the right side of the axle (just outside the disk), draw a small box attached to the axle labeled "angular velocity sensor".

Text and labeling rules:
- All labels are horizontal, readable, and do not overlap lines.
- Include only the labels specified above; no extra numeric values besides M, R, and m.

Students are asked to experimentally determine the moment of inertia I of the disk using a linear graph. To determine I, the students are permitted to use measurements from only a meterstick, a stopwatch, and the angular velocity sensor.

Describe an experimental procedure using the described setup to collect data that would allow the students to determine an experimental value of I using a linear graph. Include any steps necessary to reduce experimental uncertainty.

Describe how the data collected in part A could be graphed and how that graph would be analyzed to determine the value of I. You may use the known values M = 2.4 kg, R = 0.30 m, m = 0.50 kg, and g = 10 m/s $^2$ in your analysis.

Figure 2. Same disk on axle, now slowed by a brake pad that provides a constant friction torque τ_f; disk starts with angular speed ω₀ and rotates until it stops.

Create a clean black-and-white physics apparatus diagram, matching the style of Figure 1.

Overall layout:
- Draw the same disk-and-axle assembly as Figure 1, but without the hanging block emphasized; the focus is the braking.
- The disk is a perfect circle placed in the left half of the figure, centered vertically.
- The horizontal axle passes through the exact disk center with support brackets on both sides.

Initial angular velocity indication:
- On the face of the disk, near the top edge, place a bold curved arrow following the rim direction.
- The arrow direction is counterclockwise (CCW).
- Label this arrow "ω₀" (Greek omega with subscript zero), with the label positioned next to the arrow.

Brake pad and contact geometry (constant friction torque):
- Draw a rectangular brake pad on the right side of the disk, positioned so its left face touches the disk’s rim exactly at the 3 o’clock point (rightmost point of the circle).
- The brake pad is mounted on a short horizontal arm that extends to the right and connects to a fixed support (a vertical post) in the right third of the figure.
- Label the rectangular pad "brake pad".
- At the contact point between pad and rim, add a small filled dot to emphasize contact.

Friction torque label:
- Near the axle, draw a small curved arrow opposite the direction of ω₀ (clockwise), indicating the torque exerted by friction on the disk.
- Label this opposing curved arrow "τ_f".

Stopping indication:
- To the right of the disk, include a simple two-stage annotation with text only (no timeline):
- Near the disk with the ω₀ arrow: text "initial".
- Near the disk’s center on a second small sketch-outline of the disk (or a small disk icon) with no ω arrow: text "stops".
- If using the small second disk icon, place it in the lower-right quadrant, smaller than the main disk, and show it without any rotation arrow to indicate ω = 0.

Optional string depiction (kept minimal to avoid clutter):
- If a string is shown, depict only a short segment tangent at the top of the disk rim and label it "string" without a hanging mass.

Text rules:
- All labels are clear, non-overlapping, and horizontal.
- Use the exact symbols "ω₀" and "τ_f" as visible text.

Figure 3. Blank graph grid for plotting experimental data and drawing a best-fit line.

Draw a blank Cartesian graph grid with bold axes and light square gridlines.

Axes and scale (must be exact):
- Horizontal axis label centered below the axis: "x" (no units shown on the printed grid).
- Vertical axis label centered to the left of the axis: "y" (no units shown on the printed grid).
- The horizontal axis runs from 0 at the origin to 10 at the far right end.
- The vertical axis runs from 0 at the origin to 25 at the top end.
- The origin is at the bottom-left corner of the grid area and is labeled with the numeral "0" at the intersection of the axes.
- Place arrowheads on the positive (right) end of the horizontal axis and on the positive (top) end of the vertical axis.

Tick marks and numeric labels (must be exact):
- Horizontal axis: tick marks every 1 unit. Numeric labels shown at 0, 2, 4, 6, 8, 10.
- Vertical axis: tick marks every 5 units. Numeric labels shown at 0, 5, 10, 15, 20, 25.

Gridlines:
- Light gridlines form squares corresponding to 1 unit by 1 unit in the horizontal direction and 1 unit by 1 unit in the vertical direction.
- Emphasize (slightly darker) the horizontal gridlines at y = 5, 10, 15, 20, 25 and the vertical gridlines at x = 2, 4, 6, 8, 10.

No plotted data:
- Do not include any points, lines, or curve on the grid.
- Do not include a title.
- Keep ample blank margins for students to write axis labels and units later if needed.

$\omega_0$ (rad/s)	$t_{stop}$ (s)
4.2	1.8
6.5	2.7
8.1	3.4
10.3	4.3
12.6	5.3

The experiment is modified as shown in Figure 2. The hanging block is removed, and a brake pad is installed that applies a constant frictional torque $\tau_f$ to the disk when activated. The disk is given an initial angular velocity $\omega_0$ and then the brake is applied. Students measure the time $t_{stop}$ it takes for the disk to come to rest from the initial angular velocity.

The moment of inertia of the disk is I = 0.12 kg·m $^2$ . The students' measurements of $\omega_0$ and $t_{stop}$ are shown in Table 1.

Indicate two quantities, either measured quantities from Table 1 or additional calculated quantities, that could be graphed to produce a straight line that could be used to determine $\tau_f$ .

Vertical axis: Horizontal axis:

ii.

On the grid provided in Figure 3, create a graph of the quantities indicated in part C(i).

•

Use Table 2 to record the measured or calculated quantities that you will plot.

•

Clearly label the axes, including units as appropriate.

•

Plot the points you recorded in Table 2.

iii.

Draw a best-fit line to the data graphed in part C(ii).

Using the best-fit line that you drew in part C(iii), calculate an experimental value for $\tau_f$ . Using the best-fit line from part C(iii) and the known value I = 0.12 kg·m $^2$ .

write this FRQ

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.

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.

Final unit 6 review checklist

Unit 6 final review checklistUse this list to confirm you can handle every major skill before your exam.
Calculate rotational and total kinetic energyApply K_rot = 1/2 I omega^2 and K_tot = K_trans + K_rot for rigid bodies with known moments of inertia such as disks, hoops, and spheres.
Find work done by a torqueEvaluate W = integral tau d theta analytically for variable torques and graphically as area under a tau-versus-theta curve; apply the rotational work-energy theorem.
Compute angular momentum in both formsUse L = I omega for rigid bodies about a fixed axis and L = r x p (magnitude m v r sin theta) for point masses about any reference point.
Apply conservation of angular momentumIdentify whether net external torque is zero, select the correct system boundary, and solve for unknown angular speeds when moment of inertia changes.
Solve rolling problems with energy methodsUse v_cm = r omega to link translation and rotation, write K_tot = 1/2 m v_cm^2 + 1/2 I omega^2, and apply energy conservation for rolling without slipping on inclines.
Analyze satellite orbits with conservation lawsDerive circular orbital speed and total energy, use E = -GMm/(2r) and v_esc = sqrt(2GM/r), and explain which quantities are conserved in elliptical versus circular orbits.

How to study unit 6

Step 1: Rotational kinetic energy and torque work (Topics 6.1-6.2)Read the Topic 6.1 and 6.2 guides on Fiveable. Practice writing K_rot = 1/2 I omega^2 for disks, hoops, and spheres, then evaluate W = integral tau d theta for both constant and variable torques. Confirm you can read area off a tau-versus-theta graph.

Step 2: Angular momentum and angular impulse (Topic 6.3)Work through the Topic 6.3 guide. Practice computing L = I omega for rigid bodies and L = r x p for point masses at various reference points. Sketch torque-versus-time graphs and identify angular impulse as the area under the curve.

Step 3: Conservation of angular momentum (Topic 6.4)Use the Topic 6.4 guide to practice identifying zero-external-torque conditions and selecting system boundaries. Solve problems where a skater, diver, or rotating platform changes shape, and verify that L = I omega is conserved while omega changes.

Step 4: Rolling motion (Topic 6.5)Work through the Topic 6.5 guide. Set up K_tot = 1/2 m v_cm^2 + 1/2 I omega^2 with the no-slip substitution v_cm = r omega for incline problems. Compare final speeds for objects with different I values and explain why the hoop is slowest.

Step 5: Satellite orbits and full-unit FRQ practice (Topic 6.6)Read the Topic 6.6 guide and derive E = -GMm/(2r) and v_esc = sqrt(2GM/r) from scratch. Then use the available FRQ practice on Fiveable to work multi-part problems that combine rolling, angular momentum conservation, or orbital energy in a single question.

More ways to review

Topic study guides

Open the individual guides for Unit 6 when you want a closer review of one topic.

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FRQ practice

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Cram archive videos

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Cheatsheets

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Score calculator

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Frequently Asked Questions

What topics are covered in AP Physics Mech Unit 6?

AP Physics C: Mechanics Unit 6 covers 6 topics: Rotational Kinetic Energy, Torque and Work, Angular Momentum and Angular Impulse, Conservation of Angular Momentum, Rolling, and Motion of Orbiting Satellites. Together they apply energy and momentum concepts to rotating systems, including objects rolling without slipping and orbiting satellites. See the full topic list at /ap-physics-c-mechanics/unit-6.

How much of the AP Physics Mech exam is Unit 6?

Unit 6 makes up 10-15% of the AP Physics C: Mechanics exam. That weight comes from topics like Rotational Kinetic Energy, Conservation of Angular Momentum, Rolling, and Motion of Orbiting Satellites. It's a meaningful chunk of the test, so understanding when angular momentum is conserved versus when it changes is especially important.

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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