How this instrument works
A flywheel banks energy the way a stretched spring banks it, except the medium is rotation rather than tension. Every particle in the spinning rim moves at v = ωr, and kinetic energy is ½v² per unit mass; summing that contribution over the whole body — the integral ∫½ω²r² dm — collapses to ½Iω², where the moment of inertia I absorbs all the geometry of how far each bit of mass sits from the axle. Two rotors holding equal total mass can store very different energy if one keeps its mass out near the rim and the other keeps it close to the shaft, since I itself depends on r².
The exponent on ω is the whole design story. Doubling the mass only doubles the stored energy, but doubling the spin speed quadruples it, which is why engineers building grid-scale storage push spin rate hard and keep the rotor as light as the required stiffness allows — Beacon Power's flywheel plants are the standard reference case. The same square law shows up in a race car's kinetic energy recovery system, where a small carbon-fibre flywheel spinning past 40,000 rpm captures braking energy a battery of equal mass could not absorb quickly enough.
The formula assumes a rigid body turning about a fixed axis, and it says nothing about how fast that energy can be withdrawn or how long it survives sitting there. Real rotors hit a ceiling long before ½Iω² would suggest, because hoop stress inside a spinning ring also grows with ω²; push the rim past the material's tensile strength and it fails. That is why modern units favour carbon-fibre composites over steel, and why quoted capacity is usually the energy between a system's maximum and minimum allowed speed, not the full figure measured from a standing start.
- Enter the rotor's Moment of inertia, kg·m² — a solid disc of mass m and radius r has I = ½mr²; a thin rim carrying that same mass has I = mr².
- Set the Angular velocity, switching the unit menu to rpm if that's how your spec sheet lists spin speed rather than rad/s.
- Read the result in Stored kinetic energy; toggle between J and kJ — grid modules run in the megajoule range, benchtop rotors in the tens of joules.
- To compare rotor designs, hold Moment of inertia fixed and step Angular velocity up — the quadratic climb in stored energy shows up immediately.
Worked example — a 5 kg·m² rotor at 300 rad/s
Take a compact flywheel module with moment of inertia I = 5 kg·m², spinning at ω = 300 rad/s — about 2,865 rpm, well inside a steel rotor's safe range. The formula gives E = ½ × 5 × 300² = ½ × 5 × 90,000 = 225,000 J, or 225 kJ. That is enough to carry a 25 kW load for nine seconds, roughly the bridging window a flywheel UPS needs to cover the gap before a backup generator comes online.
Keep the same rotor and push the spin speed to 600 rad/s — double the original — and the stored energy does not double, it quadruples, to 900,000 J. That square-law relationship is why flywheel designers treat spin speed, rather than added mass, as the primary lever for more storage, right up to the point where rotor stress sets the ceiling.
Questions
Why does doubling the spin speed quadruple the stored energy?
Because angular velocity enters the formula squared. Kinetic energy per mass element is ½v², and v = ωr, so every element's contribution scales with ω². Doubling ω multiplies each contribution by four, and since moment of inertia I doesn't change, the total ½Iω² quadruples too — which is why flywheel designers chase spin speed over added mass.
What is moment of inertia, and how do I estimate it for my flywheel?
Moment of inertia measures how a rotor's mass is distributed relative to its spin axis, in kg·m². A solid disc of mass m and radius r has I = ½mr²; a thin ring or rim carrying that same mass at the same radius has I = mr², twice as much, because every gram sits farther from the axle. Manufacturers usually publish I directly on the rotor spec sheet.
Why can't a real flywheel just spin faster to store more energy?
Because the rotor has to survive the spin. Hoop stress inside a rotating ring grows with ω², the same term driving the stored energy, so past some speed the material's tensile strength is exceeded and the rim fails. That ceiling is why grid-scale and automotive flywheels increasingly use carbon-fibre composites, which tolerate far higher tip speeds than steel before failing.
Can I enter rpm instead of rad/s for the Angular velocity field?
Yes — switch the unit menu on the Angular velocity field to rpm and the instrument converts internally using ω = rpm × 2π ⁄ 60 before applying the formula. 300 rad/s and roughly 2,865 rpm describe the same spin rate and return the identical stored energy, 225 kJ.
Does ½Iω² tell me how much energy I can actually withdraw?
Not entirely. It gives total kinetic energy relative to a stationary rotor, but most flywheel systems only discharge between a maximum and a minimum allowed speed — running slower makes the power electronics inefficient — so usable energy is the difference between ½Iω² at those two speeds, smaller than the full figure measured from rest.