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UNIT 6About 13 min + practice

Energy and Momentum of Rotating Systems

Rotation adds energy stores and a new conserved quantity.

What you’ll learn

  • Account for rotational kinetic energy.
  • Use angular momentum conservation with stated external torques.
  • Analyze rolling and rotational collisions.
01

Before you begin

A rolling object can have both center-of-mass translation and rotation. Angular momentum is measured about a specified axis. Conservation of angular momentum requires negligible net external torque about that axis over the relevant interval.

Explain these starting ideas in your own words. Revisit them whenever a later step feels unclear.

02

Rotational energy is not translational energy

A rigid object rotating about a fixed axis has kinetic energy ½Iω2. An object that both translates and rotates can have center-of-mass translational kinetic energy plus rotational kinetic energy about its center of mass. Omitting either term can predict an impossible speed.

Work done by a torque through angular displacement changes rotational kinetic energy under the appropriate fixed-axis model. For constant torque, work is τΔθ, and rotational power is τω when torque and angular velocity are aligned. Use radians.

A modern three-bladed wind turbine beside traditional windmills.
Moving air can do mechanical work

Wind transfers energy to rotating blades. A turbine can convert that mechanical energy into electrical energy; older windmills can use it directly for mechanical tasks.

Photo: Matthew T Rader · Source · CC BY-SA 4.0 · Unmodified.
K = Mvcm2 + Icmω2
A torque changes angular momentumFor constant net torque 3 N·m and initial angular momentum zero, L=3t. The slope of L versus time equals net external torque.
A torque changes angular momentum05101501.252.53.755 Time (s)Angular momentum (kg·m²/s)L=3t
Read figure values as text

L=3t: 0: 0; 5: 15

PAUSE & TRY IT

Why must rolling energy include two terms?

Reveal answer

The center of mass translates while the body rotates about it.

03

Rolling without slipping imposes a constraint

For rolling on a stationary surface without slipping, vcm = Rω. The contact point is instantaneously at rest relative to the surface, while other points have different velocities. That condition does not imply every point has zero acceleration.

In ideal rolling down a fixed incline, static friction can supply a torque without dissipating energy at the instantaneously stationary contact. Mechanical energy can therefore be conserved under suitable assumptions. Rolling with slipping requires different energy accounting because kinetic friction can generate thermal energy.

04

Angular momentum depends on an axis

For a rigid body about a fixed principal axis, angular momentum can be written Iω. A moving particle has angular momentum about a point determined by its position and momentum; the perpendicular distance to its line of motion matters. A particle moving in a straight line can have nonzero angular momentum about a point off that line.

Net external torque changes total angular momentum. If its impulse over the interval is negligible, angular momentum is conserved about the chosen axis. Internal interactions can redistribute angular momentum among parts of a system.

L = Iω
ΔL = τavgΔt for the relevant net torque

PAUSE & TRY IT

Can a straight-moving particle have angular momentum?

Reveal answer

Yes, about a point not on its line of motion.

05

Changing inertia can change speed without conserving energy

If a rotating person pulls masses inward while external torque is negligible, moment of inertia decreases and angular speed increases to preserve angular momentum. Rotational kinetic energy increases because the person does internal work. Angular-momentum conservation does not imply constant rotational kinetic energy.

A rotational collision, such as a lump sticking to a rotating disk, can conserve angular momentum about the axle while losing mechanical energy. External axle forces may have zero torque about that axle even when they produce an impulse, making angular rather than linear momentum the useful conservation law.

Conserved angular momentum does not imply constant energyIllustrative model, not collected experimental data. For L=6 kg·m²/s, angular speed is ω=L/I. Reducing rotational inertia increases angular speed when external torque is negligible.
Conserved angular momentum does not imply constant energy024601.534.56 Rotational inertia (kg·m²)Angular speed (rad/s)ω=6/I
Read figure values as text

ω=6/I: 1: 6; 1.1041666666666667: 5.433962264150943; 1.2083333333333333: 4.9655172413793105; 1.3125: 4.571428571428571; 1.4166666666666667: 4.235294117647059; 1.5208333333333335: 3.9452054794520546; 1.625: 3.6923076923076925; 1.7291666666666665: 3.4698795180722897; 1.8333333333333335: 3.2727272727272725; 1.9375: 3.096774193548387; 2.041666666666667: 2.9387755102040813; 2.145833333333333: 2.7961165048543695; 2.25: 2.6666666666666665; 2.354166666666667: 2.548672566371681; 2.458333333333333: 2.440677966101695; 2.5625: 2.341463414634146; 2.666666666666667: 2.2499999999999996; 2.770833333333333: 2.165413533834587; 2.875: 2.0869565217391304; 2.979166666666667: 2.0139860139860137; 3.0833333333333335: 1.9459459459459458; 3.1875: 1.8823529411764706; 3.2916666666666665: 1.8227848101265824; 3.3958333333333335: 1.7668711656441718; 3.5: 1.7142857142857142; 3.6041666666666665: 1.6647398843930636; 3.7083333333333335: 1.6179775280898876; 3.8125: 1.5737704918032787; 3.9166666666666665: 1.5319148936170213; 4.020833333333334: 1.4922279792746111; 4.125: 1.4545454545454546; 4.229166666666666: 1.4187192118226604; 4.333333333333334: 1.3846153846153844; 4.4375: 1.352112676056338; 4.541666666666666: 1.3211009174311927; 4.645833333333334: 1.2914798206278024; 4.75: 1.263157894736842; 4.854166666666666: 1.2360515021459229; 4.958333333333334: 1.2100840336134453; 5.0625: 1.1851851851851851; 5.166666666666667: 1.161290322580645; 5.270833333333333: 1.1383399209486167; 5.375: 1.1162790697674418; 5.479166666666667: 1.0950570342205324; 5.583333333333333: 1.0746268656716418; 5.6875: 1.054945054945055; 5.791666666666667: 1.0359712230215827; 5.895833333333333: 1.0176678445229683; 6: 1

PAUSE & TRY IT

Which axis can simplify a sticking collision on a fixed disk?

Reveal answer

The axle, if external axle forces exert negligible torque about it.

06

Choose the conservation law from the interactions

Before choosing an equation, identify the system, axis, external forces, and external torques. Linear momentum requires negligible external impulse; angular momentum about an axis requires negligible external angular impulse; mechanical energy requires suitable energy-transfer conditions. One can hold while another does not.

A good explanation states why the selected quantity is conserved and why another may change. Conservation is not a property attached permanently to a type of problem; it follows from the system boundary and interval being analyzed.

07

Split rolling energy into two contributions

For rolling without slipping, center-of-mass speed satisfies v=Rω. Total kinetic energy is translational plus rotational . Both contributions come from the same energy budget. Omitting rotation predicts too large a speed for an object rolling down a height under an energy-conserving model.

Substituting I=βmR2 gives kinetic energy mv2. For otherwise comparable objects released from the same height, a smaller β gives a larger center-of-mass speed. The result depends on mass distribution, not simply on which object is heavier. Static friction can enforce rolling without dissipating energy in the ideal stationary-surface model.

PAUSE & TRY IT

Can angular momentum be conserved while rotational kinetic energy increases?

Reveal answer

Yes. Internal work can increase kinetic energy while net external torque remains negligible.

08

Distinguish zero torque from zero work

If a rotating person pulls masses inward with negligible external torque, angular momentum Iω remains constant and angular speed increases as I decreases. Rotational kinetic energy can increase because internal muscular work is done. Conserved angular momentum does not imply conserved rotational kinetic energy.

For a point mass moving relative to an axis, angular momentum magnitude can be written mvr_perpendicular. A particle moving in a straight line can have nonzero angular momentum about an off-line point. State the axis before calculating; changing the reference point can change the result.

PAUSE & TRY IT

Why can an axle force fail to change angular momentum about the axle?

Reveal answer

Its line of action passes through that axis, so its torque about the axis is zero.

09

Angular collisions and choosing a system

When a small object sticks to a rotating platform, the collision can conserve angular momentum about the axle if external torque is negligible. The final inertia includes the added object, and a shared final angular velocity follows. Kinetic energy generally decreases during sticking, so it should not be imposed as a second conservation equation.

An axle can exert an external force while producing negligible torque about its own axis. This is why angular momentum may be conserved about the axle even when linear momentum of the platform–object system is not. The conservation law follows the relevant external interaction, not a blanket assumption that every quantity is conserved in the same event.

10

Use rotational energy and angular momentum for their own purposes

Rotational kinetic energy is ½Iω2, while angular momentum for the appropriate fixed-axis model is Iω. They depend differently on angular speed. Net external torque changes angular momentum; zero external torque can conserve it even while rotational kinetic energy changes because internal work is performed.

When a rotating person pulls masses inward, inertia decreases. If external torque is negligible, angular speed increases to preserve angular momentum. Rotational kinetic energy increases because the person does work; conservation of angular momentum does not require kinetic-energy conservation.

Specify the axis and system. An off-center force can exert torque and change angular momentum even if the same force is central about a different origin. A conservation claim without an axis and an external-torque argument is incomplete.

PAUSE & TRY IT

If a rotating system halves I with negligible external torque, what happens to ω and rotational kinetic energy?

Reveal answer

ω doubles to conserve Iω. K= doubles; internal work supplies the energy change.

11

Account for both parts of rolling motion

A rigid body rolling without slipping has translational motion of its center and rotation about its center. Its kinetic energy is ½Mv2+½Icmω2, with v=Rω under the no-slip constraint. Leaving out the rotational term overpredicts speed down a ramp.

Objects with different mass distributions can reach different speeds after the same drop. A larger directs a larger share of mechanical energy into rotation, leaving less translational kinetic energy. This comparison does not require their masses to be different.

Static friction can provide the torque needed for rolling and does not necessarily dissipate mechanical energy in the ideal rigid no-slip model on a stationary surface. If sliding occurs, kinetic friction can convert mechanical energy into thermal energy and the no-slip relation cannot be imposed throughout.

12

Separate rotational collisions from later energy changes

A mass sticking to a rotating platform is an angular-momentum problem during the short impact when external torque is negligible. The final inertia includes the added mass at its final radius. Kinetic energy generally decreases during the sticking event, so imposing both kinetic-energy conservation and sticking can overconstrain the problem incorrectly.

After the collision, a motor, brake or friction torque can change angular momentum. Rotational work and power relate torque to angular displacement and angular speed. Keep the interval of approximate isolation distinct from later driven motion.

Use units to check which quantity you computed. Angular momentum has units kg , whereas energy has kg . Although torque and energy share dimensions, torque describes a turning influence and energy a scalar accounting quantity; they are not interchangeable measurements.

13

Rolling combines translation and rotation

For rolling without slipping on a stationary surface, center-of-mass speed satisfies v=ωR. The total kinetic energy is ½mv2+½Iω2. Using only the translational term misses part of the energy, while using only rotation about the center misses the center’s motion.

The relation v=ωR is a constraint, not a universal equation for every rotating object. A slipping wheel can rotate too quickly or too slowly for its translational speed. Identify whether the no-slip assumption is stated or justified before substituting it.

Objects with larger devote a larger fraction of energy to rotation at a given rolling speed. Released from equal heights under ideal no-slip conditions, they generally reach different translational speeds. Equal mass alone does not determine which reaches the bottom faster.

14

Angular momentum needs a specified axis and external torque

Angular momentum conservation applies when net external torque about the chosen axis is negligible. Linear momentum and angular momentum are separate conservation questions; a system can satisfy one without the other. A pivot force can provide external linear impulse while having zero torque about that pivot.

When rotational inertia decreases at fixed angular momentum, angular speed increases. Rotational kinetic energy can also increase because internal work is done, as when a person pulls masses inward. Conservation of angular momentum does not require conservation of kinetic energy.

For a collision involving rotation, define the before and after angular momenta about the same axis. After an object sticks to a rotating body, include it in the final rotational inertia. A mass landing farther from the axis can have a larger effect on final angular speed.

PAUSE & TRY IT

Why can a pivot force allow angular momentum conservation about the pivot but not linear momentum conservation?

Reveal answer

Its lever arm about the pivot is zero, so its torque there is zero, but it can still exert a nonzero external force or impulse.

FROM IDEA TO APPLICATION

Worked examples

EXAMPLE 1

A rolling solid cylinder

A solid cylinder with Icm = ½MR2 rolls without slipping from rest through vertical drop h. Find its center-of-mass speed, neglecting losses.

Reveal worked solution
  1. Mgh = ½Mv2 + ½(½MR2)ω2.
  2. Use ω = to get Mgh = ¾Mv2.
  3. Solve for v.
Result & interpretation

v = √(), smaller than the speed of a nonrotating frictionless sliding particle through the same drop.

EXAMPLE 2

Pulling inward

A rotating system’s inertia decreases from 4.0 to 2.0 kg·m2 with negligible external torque. Initial angular speed is 3.0 rad/s. Find final speed and kinetic energies.

Reveal worked solution
  1. I1ω1 = I2ω2 gives ω2 = 6.0 rad/s.
  2. Initial energy = ½(4)(32) = 18 J.
  3. Final energy = ½(2)(62) = 36 J.
Result & interpretation

Final speed is 6.0 rad/s. The 18 J energy increase comes from internal work, not an external angular impulse.

EXAMPLE 3

Changing rotational inertia

A rotating system has I=4.0 kg·m2 and ω=3.0 rad/s. Its inertia decreases to 2.0 kg·m2 with negligible external torque. Find final angular speed and compare kinetic energies.

Reveal worked solution
  1. Initial angular momentum is 4.0(3.0)=12 kg·.
  2. Final ω==6.0 rad/s.
  3. Initial K=()(4)(32)=18 J; final K=()(2)(62)=36 J.
Result & interpretation

Angular speed becomes 6.0 rad/s and kinetic energy increases by 18 J, supplied by work during the redistribution.

EXAMPLE 4

Compare rolling energy shares

A solid cylinder with I=½MR2 rolls without slipping at speed v. What fraction of its total kinetic energy is rotational?

Reveal worked solution
  1. Rotational energy is ½(½MR2)()2=¼Mv2.
  2. Translational energy is ½Mv2, giving total ¾Mv2.
  3. The rotational fraction is =.
Result & interpretation

One-third is rotational and two-thirds is translational in this model.

EXAMPLE 5

Rolling energy partition

A solid cylinder with I=½mR2 rolls without slipping at speed v.

Reveal worked solution
  1. Rotational energy is ½(½mR2)()=¼mv2.
  2. Total kinetic energy is ½mv2+¼mv2=¾mv2.
Result & interpretation

One-third of total kinetic energy is rotational.

EXAMPLE 6

Inertia change at conserved angular momentum

A rotor’s inertia decreases from 4 to 2 kg·m2 while initial angular speed is 3 rad/s and external torque is negligible.

Reveal worked solution
  1. Initial L=4(3)=12 kg·.
  2. Final ω==.
Result & interpretation

Final angular speed is 6 rad/s; any energy change needs separate work accounting.

MAKE THE DISTINCTION

Common mistakes, clearer reasoning

The trapConserved angular momentum means conserved rotational kinetic energy.

The better explanationInternal work or inelastic interactions can change energy while angular momentum remains constant.

The trapRolling always dissipates energy through friction.

The better explanationIdeal static contact on a stationary surface need not dissipate energy; slipping or deformation can.

RETRIEVE BEFORE YOU REVEAL

Practice checkpoints

Revisit the quick checks from this guide without looking back. Explain why, then reveal the answer.

1. Why must rolling energy include two terms?

Reveal answer

The center of mass translates while the body rotates about it.

2. Which axis can simplify a sticking collision on a fixed disk?

Reveal answer

The axle, if external axle forces exert negligible torque about it.

3. Can a straight-moving particle have angular momentum?

Reveal answer

Yes, about a point not on its line of motion.

4. Can angular momentum be conserved while rotational kinetic energy increases?

Reveal answer

Yes. Internal work can increase kinetic energy while net external torque remains negligible.

5. Why can an axle force fail to change angular momentum about the axle?

Reveal answer

Its line of action passes through that axis, so its torque about the axis is zero.

6. If a rotating system halves I with negligible external torque, what happens to ω and rotational kinetic energy?

Reveal answer

ω doubles to conserve Iω. K= doubles; internal work supplies the energy change.

7. Why can a pivot force allow angular momentum conservation about the pivot but not linear momentum conservation?

Reveal answer

Its lever arm about the pivot is zero, so its torque there is zero, but it can still exert a nonzero external force or impulse.

Key language

Angular momentum
A rotational motion quantity defined relative to a point or axis.
Angular impulse
Net torque accumulated over time.
Rolling without slipping
Motion satisfying the no-relative-slip condition at contact.
Rotational kinetic energy
Energy associated with angular motion.
Connect it to the course

Rotational conservation unifies force, torque, energy, and impulse reasoning.

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Written for ScienceHub · Original instructional material. Course framework reference ↗. These notes are independently authored and are not College Board materials. External photographs retain their credited licenses.

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