Energy and Momentum of Rotating Systems

Name: AP Physics 1: Algebra-Based Unit 6: Energy and Momentum of Rotating Systems
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5–8% of the exam7 lessons · 99 min42 terms

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What this unit covers

The topics below follow the published Physics 1 course framework for Unit 6. This unit is worth 5–8% of the exam, so budget your time against that rather than against how long the unit takes to teach.

Rotational KEAngular momentumConservationRolling

Lessons in this unit

Formulas in Unit 6

Rotational and total kinetic energy

KE_rot = ½Iω² · KE_total = ½mv² + ½Iω²

For a rolling object, add both terms. Energy conservation from a height h then reads mgh = ½mv² + ½Iω².

Angular momentum

L = Iω · L_point = mvr

Units kg·m²/s. For a point mass on a circle, r is its distance from the axis. Angular momentum is a vector along the spin axis.

Conservation of angular momentum

I₁ω₁ = I₂ω₂

Valid when net external torque is zero. If I drops to half, ω doubles; if I doubles, ω halves.

Rolling condition

v = rω

Ties center speed to spin rate for rolling without slipping. Also a = rα for the accelerations.

Total kinetic energy of a rolling object

K = ½mv² + ½Iω² and with I = cmR² plus ω = v/R: K = ½(1 + c) mv²

c is the shape coefficient: 2/5 for a solid sphere, ½ for a disk, 1 for a hoop.

Angular impulse–momentum theorem

τ Δt = ΔL = I ω_f − I ω_i

The exact analogue of F Δt = Δp. For a varying torque, the angular impulse is the area under a torque–time graph.

Every term in Unit 6

All 42 terms we publish for Energy and Momentum of Rotating Systems, with definitions. Reading them through is the fastest way to find the ones you cannot define — then drill those in cram mode until you can produce them without the prompt.

Conservation of angular momentum: L is constant when the net external TORQUE on the system is zero. Note the qualifier is about torque, not force — a system can have net force and still conserve angular momentum about the right axis.
Rolling race: Down the same ramp, objects with smaller I/mr² arrive first — a solid sphere beats a solid cylinder, which beats a hoop, regardless of mass or radius.
Why ω rises when I falls: L = Iω is fixed, so halving I doubles ω. Kinetic energy ½Iω² then increases — supplied by the work done pulling the mass inward.
Rotational work and power: W = τθ and P = τω, matching the translational forms with rotational quantities substituted.
Rotational kinetic energy: KE_rot = ½Iω², the rotational twin of ½mv². A spinning object at rest in translation still has kinetic energy.
Total kinetic energy of a rolling object: KE = ½mv² + ½Iω², translation plus rotation. Using only ½mv² for a rolling object undercounts the energy and is the commonest error in this unit.
Why a rolling object is slower down a ramp: Some gravitational potential energy goes into rotation instead of translation, so less is left for linear speed. A sliding frictionless block beats every rolling object.
Racing shapes down a ramp: Order is set by I/MR² alone, not by mass or radius: sphere (0.4) beats disk (0.5) beats hoop (1.0). Two spheres of different mass arrive together.
Work done by a torque: W = τθ with θ in radians, the rotational analogue of W = Fd.
Rotational power: P = τω, the rotational analogue of P = Fv.
Angular momentum: L = Iω for a rotating body, and L = mvr sin θ for a particle about a point. A vector, and conserved under the right condition.
The spinning skater: Pulling arms in reduces I, so ω rises to keep L = Iω constant. Kinetic energy INCREASES, supplied by the work the skater does pulling in — energy is not conserved here even though angular momentum is.
Angular momentum of a particle: L = mvr sin θ about a chosen point. An object moving in a straight line has angular momentum about any point not on that line.
Angular impulse: τΔt = ΔL, the rotational analogue of the impulse-momentum theorem.
Rotational collisions: When objects couple rotationally — a blob landing on a turntable — conserve angular momentum, not kinetic energy. Iₐωₐ = (Iₐ + I_b)ω′.
Why kinetic energy is lost in a rotational collision: The same reason as a linear inelastic collision: the coupling forces dissipate energy. Angular momentum survives; kinetic energy does not.
Comparing linear and rotational analogues: x↔θ, v↔ω, a↔α, m↔I, F↔τ, p↔L, ½mv²↔½Iω², FΔt↔τΔt. Learning the map converts every linear result into a rotational one.
Rolling down a ramp by energy: mgh = ½mv² + ½Iω² with ω = v/r. Substituting gives v = √(2gh / (1 + I/mr²)), which shows why the shape factor sets the finishing order.
Angular momentum in orbits: A planet in an elliptical orbit conserves angular momentum, so it moves faster when closer — which is Kepler's second law derived from a conservation principle.
When to use energy vs angular momentum: Angular momentum for collisions and couplings where energy is lost; energy for smooth rolling and falling where nothing dissipates. Choosing wrong is the main failure mode here.
Torque, angular momentum and the second law: Στ = ΔL/Δt, the rotational form of ΣF = Δp/Δt. Zero net torque therefore means constant angular momentum.
A system that is both rolling and colliding: Handle the collision with angular momentum, then the subsequent motion with energy. Trying to do both with one principle is where these problems break down.
Moment of inertia depends on the axis: The same rod has I = 1/12 ML² about its center and ⅓ML² about its end. Stating the axis is part of stating the moment of inertia.
A rolling object carries two kinetic energies: K = ½mv² + ½Iω². Not double-counting: the center of mass is translating and the body is spinning about it, and both are real motions.
K = ½(1 + c)mv² for a rolling body: Substituting I = cmR² and ω = v/R collapses both terms into one. Mass and radius vanish from the ratio, so only the shape matters.
Why the sphere wins the rolling race: v = √(2gh/(1 + c)), so the smallest shape coefficient gives the largest speed. Sphere (2/5) beats disk (½) beats hoop (1), regardless of mass or size.
A sliding block beats every roller: With no rotation, all of mgh goes to ½mv² and v = √(2gh). Rolling always costs speed because part of the energy is diverted into spin.
Fraction of a rolling sphere's energy that is rotational: 2/7. From (2/5)/(1 + 2/5) — the coefficient c is not itself the fraction.
The condition for angular momentum conservation: Zero net external TORQUE, not zero net force. A pivot force acts through the pivot and exerts no torque about it, which is what makes rod collisions solvable.
K = L²/2I: The rotational twin of K = p²/2m. With L fixed, any reduction in rotational inertia raises the kinetic energy — the skater paradox in one line.
Why the skater gains kinetic energy: Angular momentum is conserved, but she does work with her arms pulling mass inward against its tendency to keep circling. The energy comes from her muscles, which is why it is tiring.
A particle moving in a straight line has angular momentum: L = mvr_⊥ about any point not on its line of motion, and it is constant. Only about a point on that line is it zero.
Rotational inelastic collision: A child jumping onto a merry-go-round, or two disks pressed together. Angular momentum about the axis is conserved; rotational kinetic energy is not.
Angular impulse from a torque–time graph: The area under the curve gives ΔL, exactly as the area under a force–time graph gives Δp. Spreading a given ΔL over a longer time lowers the torque needed.
Ball strikes a pivoted rod: the three moves: The ball arrives with L = mvr_⊥ about the pivot; angular momentum about the pivot is conserved through impact; afterward I = I_rod + mr². Then, and only then, switch to energy for the swing.
Where to strike a pivoted rod for maximum effect: As far from the pivot as possible, since the angular momentum delivered is mvr_⊥. Strike at the pivot and the rod does not turn at all.
Angular momentum units: L is in kg m^2/s, which is also N m s. Matching units is a quick check that a moment of inertia has not been used where a mass belongs.
Rotational kinetic energy is not a separate kind of energy: 1/2 I omega^2 is ordinary kinetic energy, summed over parts of the body moving at different speeds. It appears in the same conservation equation as every other term.
Angular momentum of a rigid body: L = I omega for a body rotating about a fixed axis. For a particle it is L = mvr sin(theta), and the two must agree when the particle is the body.
Why a diver tucks: Tucking cuts the moment of inertia, and with L fixed omega rises, so the rotation speeds up. Opening out before entry raises I again and slows the spin for a clean entry.
Angular momentum is conserved about a chosen axis: The statement is meaningless without naming the axis. A quantity conserved about the pivot need not be conserved about the center of mass.
Torque does work only through an angle: W = tau theta. Holding a heavy object still takes force and effort but does zero work, because nothing rotates through any angle.

What examiners penalize here

In a "race down the ramp" question, the object that puts *less* energy into rotation (smaller I relative to mR²) ends up moving faster. That is a direct consequence of splitting mgh between ½mv² and ½Iω².
Angular momentum is a vector along the rotation axis. On the AP exam its direction matters most in conservation problems, where the total vector — magnitude and direction — must stay constant.
Conservation of angular momentum needs zero *external* torque. Internal changes — pulling in arms, dropping on clay — do not violate it. Check that nothing outside the system is applying a torque before you set L constant.
For ramp-race questions, rank shapes by how much of mgh goes into rotation: sphere (⅖MR²) beats disk (½MR²) beats hoop (MR²). Mass and radius cancel out, so the *shape* alone decides the winner.
A "which arrives first" ranking task is answered entirely by the rotational inertia coefficient. Say that the object with the smaller coefficient devotes a smaller fraction of its energy to rotation and therefore has more translational speed — that reasoning is the rubric point, not the number.
Say "no net external torque acts about the axis, so angular momentum is conserved" in words before writing I₁ω₁ = I₂ω₂. Naming the condition is a rubric point independent of the algebra, and it forces you to check whether the condition actually holds.
For a collision that then leads to a swing, write two clearly separated stages on your paper: "Stage 1, collision — angular momentum" and "Stage 2, swing — energy". Rubrics score the two stages independently, so a correct stage 2 still earns points even after a slip in stage 1.

Practice Physics 1

Our practice bank is drawn from across the whole course rather than filtered to one unit, which is closer to how the exam asks anyway — it will not tell you which unit a question is testing.

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Questions about this unit

How much of the AP Physics 1: Algebra-Based exam is Unit 6?

Unit 6, Energy and Momentum of Rotating Systems, is worth 5–8% of the Physics 1 multiple-choice section according to the published course framework. Across all 8 units that makes it a middling share, roughly what an even split across units would give.

What topics are covered in Physics 1 Unit 6?

Energy and Momentum of Rotating Systems covers Rotational KE, Angular momentum, Conservation and Rolling. We publish 42 terms with definitions for this unit, all of them on this page.

How should I study Physics 1 Unit 6?

Read the 7 lessons below first — about 100 minutes — then drill the 42 terms in cram mode until you can produce each definition from memory rather than just recognize it. Recognition is what makes a unit feel finished when it is not. Finish with practice questions and read the explanation for every one you get right by elimination as well as the ones you miss.

All 8 units of AP Physics 1: Algebra-Based

Unit names, topics and exam weights follow the published College Board course framework for AP Physics 1: Algebra-Based. AP® is a trademark registered by the College Board, which does not endorse this site.