Flywheel Energy Storage Calculator
How much energy can a spinning rotor hold? Enter mass, radius and speed and this tool computes the stored energy E = ½·I·ω² for a solid disc or a rim rotor — or solve it backwards for the RPM or mass you need. It also checks the honest limit every flywheel faces: tip speed vs. material hoop stress.
🔒 100% client-side — your data never leaves this browser
⚙ E = ½Iω²
⤾ 3-way solver
🛡 hoop-stress limit check
Safety: this is a planning tool, not engineering advice. Real flywheel rotors store lethal energy and must be designed with real safety factors, burst containment, and vacuum enclosures — never build or spin a rotor based on this page alone.
▸ Calculator
Pick what to solve for, choose the rotor geometry and material, then enter your numbers.
⚠ Please enter positive numbers for all inputs.
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— · — specific energy
— · — specific energy
Moment of inertia—
Required speed— · —
Required mass—
Tip speed v = ωr—
Hoop stress σ ≈ ρv²—
Stress vs. tensile—
Status—
Strengths are typical reference values (steel 500 MPa, aluminum 310 MPa, titanium 950 MPa, carbon composite 1500 MPa); real grades vary. σ = ρv² is the thin-rim reference — a solid disc's peak stress differs. Design with a real safety factor.
▸ The physics in one minute
E = ½ · I · ω²
- Solid disc: I = ½·m·r². A rim/hoop puts all its mass at the edge, so I = m·r² — exactly twice the storage at the same mass, radius and speed.
- Energy grows with the square of speed but only linearly with mass. Doubling RPM quadruples stored energy — which is why practical flywheels are light, fast, and often carbon composite.
- Speed is capped by strength: the rim is pulled outward with a hoop stress of roughly σ = ρ·v² (ρ = density, v = ωr tip speed). When σ reaches the material's tensile strength, the rotor bursts.
▸ Worked examples (verified)
Each computed from first principles; the calculator reproduces them exactly.
| Case | Inputs | Stored energy |
|---|---|---|
| Solid disc | m = 100 kg, r = 0.5 m, 3000 RPM | 616,850.28 J = 0.1713473 kWh |
| Rim / hoop | same as above | 1,233,700.55 J = 0.3426946 kWh (2×) |
| Solve RPM | 3,600,000 J (1 kWh), disc, m = 100 kg, r = 0.5 m | ω = 758.95 rad/s = 7247.41 RPM |
| Solve mass | 3,600,000 J (1 kWh), disc, r = 0.5 m, 3000 RPM | m = 583.61 kg |
| Material check | steel rim at 157.08 m/s tip speed | σ = 193.69 MPa ≈ 38.7% of 500 MPa → OK |
▸ Frequently asked questions
How much energy does a flywheel store?
A flywheel stores E = ½·I·ω² joules, where I is the moment of inertia and ω is the angular velocity in rad/s. Example: a solid disc of 100 kg with radius 0.5 m spinning at 3000 RPM stores 616,850.28 J (0.1713473 kWh). Divide joules by 3,600,000 to get kWh.
Why does a rim (hoop) store twice as much as a solid disc of the same mass and radius?
Because its moment of inertia is larger: a thin rim has I = m·r² while a solid disc has I = ½·m·r², so at the same speed it stores exactly twice the energy. The same 100 kg, 0.5 m rotor at 3000 RPM stores 1,233,700.55 J (0.3426946 kWh) as a rim versus 616,850.28 J as a disc.
How do I convert RPM to rad/s?
Multiply RPM by 2π/60 ≈ 0.10472. So 3000 RPM = 314.16 rad/s, and 10,000 RPM = 1047.20 rad/s. This calculator accepts either unit.
What RPM do I need to store 1 kWh in a 100 kg disc rotor of 0.5 m radius?
Solve ω = √(2E/I). With I = ½·100·0.5² = 12.5 kg·m² and E = 3,600,000 J, you need ω = 758.95 rad/s, which is 7247.41 RPM. That is a realistic high-speed flywheel regime.
How heavy must the rotor be to store 1 kWh at 3000 RPM?
Solve m = 2E/(k·ω²·r²) with k = ½ for a disc. For 3,600,000 J, r = 0.5 m, and ω = 314.16 rad/s: m = 583.61 kg. Slowing the rotor down forces the mass up — this is why high-speed, light composite rotors are used in practice.
What limits how fast a flywheel can spin?
Material strength. The rim sees a hoop stress of roughly σ = ρ·v², where ρ is the material density and v = ω·r the tip speed. At 3000 RPM and r = 0.5 m the tip speed is 157.08 m/s, giving a steel rim (ρ = 7850 kg/m³) a stress of 193.69 MPa — about 38.7% of typical steel's 500 MPa tensile strength. Above the material limit the rotor bursts, so real designs use safety factors, vacuum enclosures, and burst containment.
How does the calculator's material-limit check work?
Pick a rotor material (steel, aluminum alloy, titanium alloy, or carbon fiber composite). The calculator computes tip speed v = ωr and hoop stress σ = ρv², then compares it to the material's typical tensile strength and flags the result: OK below 50%, warning at 50–100%, and 'over limit' at or above 100%. The strengths shown are typical reference values — real grades vary, so engineer a proper safety factor.
What is the specific energy of a flywheel, and is it better than a battery?
Specific energy is stored energy per mass. The example 100 kg disc at 3000 RPM holds 616,850 J / 100 kg = 1.71 Wh/kg — far below a lithium battery's ~250 Wh/kg. High-speed carbon-composite flywheels can reach 50–100+ Wh/kg and excel at power delivery and cycle life (millions of cycles), not at bulk energy storage.
Free tool by NRTechStudio