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Energy and Momentum of Rotating Systems
Calculus-based mechanics: connect the physical model, its equations, and the evidence.
What you’ll learn
- Account for rotational kinetic energy.
- Use angular momentum conservation with stated external torques.
- Analyze rolling and rotational collisions.
- Set up and interpret derivatives or integrals with physical initial conditions.
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.
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.

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.PAUSE & TRY IT
Why must rolling energy include two terms?
Reveal answer
The center of mass translates while the body rotates about it.
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.
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Which axis can simplify a sticking collision on a fixed disk?
Reveal answer
The axle, if external axle forces exert negligible torque about it.
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.
PAUSE & TRY IT
Can a straight-moving particle have angular momentum?
Reveal answer
Yes, about a point not on its line of motion.
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.
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
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.
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.
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.
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.
Rotational work and kinetic energy
The work done by torque about a fixed axis is ∫τ dθ. Rotational kinetic energy is when I is constant about that axis. Power is τω for aligned torque and angular velocity; signs indicate energy delivery or removal. A brake can have negative power while the wheel still rotates in the positive direction.
For rolling without slipping, include both translational and rotational energy: K=Mv_+I_. The contact point is instantaneously stationary on a stationary surface, so ideal static friction does no work there. Nevertheless, friction may provide the torque required to change ω.
PAUSE & TRY IT
A hoop and solid disk roll from the same height without slipping. Which arrives faster?
Reveal answer
The disk: less of its available energy must go into rotation for a given center-of-mass speed.
Choose the reference point for angular momentum
For a particle, L=r×p about a specified origin; for planar motion its signed magnitude is the perpendicular lever arm times momentum. For fixed-axis rotation of a rigid body, L=Iω. External torque changes angular momentum through ΔL=∫τ_external dt.
Conservation requires negligible external torque about the chosen point, not zero external force. During a brief impact against a hinged object, the hinge force can give a large linear impulse but no torque about the hinge. Angular momentum about the hinge may be conserved even when linear momentum and kinetic energy are not. If I changes without external torque, ω changes to preserve Iω; energy can change through internal work.
Read figure values as text
L=3t: 0: 0; 5: 15
PAUSE & TRY IT
If I halves while angular momentum is conserved, what happens to ω and rotational K?
Reveal answer
ω doubles and K= doubles; internal work supplies the energy increase.
Orbits connect energy and angular momentum
In a central gravitational field, r×F=0 about the center, so angular momentum is conserved. In a noncircular orbit, radius and speed change while the perpendicular component of r×mv stays constant. At a closest or farthest point, velocity is tangential, so mrv is especially convenient.
For a circular orbit, combine the radial force equation with energy to get K=, U=−GMm/r, and E=. Raising the circular orbit makes total energy less negative while speed decreases. Escape from radius r with zero final speed at infinity requires v_escape=√(), neglecting atmosphere and other bodies.

A satellite’s motion connects central force, energy, and angular momentum. The ideal two-body model neglects propulsion, atmospheric drag, and other bodies.
Photo: NASA · Source · Public domain · Unmodified.PAUSE & TRY IT
Why does an orbiting satellite not need a tangential force to keep moving?
Reveal answer
In ideal motion no tangential force is required for continuing velocity; gravity changes its direction.
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.
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.
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.
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.
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
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
- Mgh = ½Mv2 + ½(½MR2)ω2.
- Use ω = to get Mgh = ¾Mv2.
- Solve for v.
v = √(), smaller than the speed of a nonrotating frictionless sliding particle through the same drop.
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
- I1ω1 = I2ω2 gives ω2 = 6.0 rad/s.
- Initial energy = ½(4)(32) = 18 J.
- Final energy = ½(2)(62) = 36 J.
Final speed is 6.0 rad/s. The 18 J energy increase comes from internal work, not an external angular impulse.
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
- Initial angular momentum is 4.0(3.0)=12 kg·.
- Final ω==6.0 rad/s.
- Initial K=()(4)(32)=18 J; final K=()(2)(62)=36 J.
Angular speed becomes 6.0 rad/s and kinetic energy increases by 18 J, supplied by work during the redistribution.
A sticking rotational collision
A particle of mass m moving tangentially at speed v sticks to the rim of a stationary disk of inertia I and radius R. Find final angular speed.
Reveal worked solution
- Choose the fixed disk axle as the origin; assume negligible external angular impulse.
- Initial angular momentum is mvR.
- Final inertia is I+mR2, so (I+mR2)ω=mvR.
ω=. Kinetic energy is not conserved in the sticking collision.
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
- Rotational energy is ½(½MR2)()2=¼Mv2.
- Translational energy is ½Mv2, giving total ¾Mv2.
- The rotational fraction is =.
One-third is rotational and two-thirds is translational in this model.
Rolling energy partition
A solid cylinder with I=½mR2 rolls without slipping at speed v.
Reveal worked solution
- Rotational energy is ½(½mR2)()=¼mv2.
- Total kinetic energy is ½mv2+¼mv2=¾mv2.
One-third of total kinetic energy is rotational.
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
- Initial L=4(3)=12 kg·.
- Final ω==.
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. A hoop and solid disk roll from the same height without slipping. Which arrives faster?
Reveal answer
The disk: less of its available energy must go into rotation for a given center-of-mass speed.
7. If I halves while angular momentum is conserved, what happens to ω and rotational K?
Reveal answer
ω doubles and K= doubles; internal work supplies the energy increase.
8. Why does an orbiting satellite not need a tangential force to keep moving?
Reveal answer
In ideal motion no tangential force is required for continuing velocity; gravity changes its direction.
9. 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.
10. 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
- Rolling without slipping
- Motion satisfying the no-relative-slip condition at contact.
- Rotational kinetic energy
- Energy associated with angular motion.
- Angular impulse
- The time integral of external torque.
- Angular momentum
- A vector measured about an origin, r×p for a particle.
- Rolling kinetic energy
- The sum of center-of-mass translation and rotation about the center of mass.
- Escape speed
- The minimum launch speed needed to reach infinity with zero remaining speed in an ideal gravitational model.
Rotational conservation unifies force, torque, energy, and impulse reasoning.