Flywheel Sizing Calculator
Enter the energy to store, operating angular velocity, and radius to size your flywheel.
Result
How it is calculated
Kinetic energy in a spinning body is E = ½ · I · ω², so the moment of inertia needed is I = 2E / ω². For a solid uniform disk I = ½ · m · r², so m = 2I / r². Inputs: energy E in joules (J), angular velocity ω in radians per second (rad/s) and radius r in metres (m). Results: I in kg·m², mass in kg (two decimals) and ω echoed back. All three inputs must be above zero; zero, a negative number or a missing field gives no result. The tool assumes a solid disk of uniform density, so a rim-heavy wheel (a ring) would need less mass for the same inertia. It treats E as the energy held at speed ω, not the smaller amount released while slowing between two speeds, and it does not check stress or burst speed.
Worked example
Store E = 2000 J at ω = 200 rad/s (about 1,910 rpm) using a disk of radius 0.25 m. Step 1, inertia: I = 2 × 2000 / 200² = 4000 / 40000 = 0.1000 kg·m². Step 2, mass: m = 2 × 0.1 / 0.25² = 0.2 / 0.0625 = 3.20 kg. The tool shows 0.1000 kg·m², 3.20 kg and 200.00 rad/s. Doubling the speed to 400 rad/s cuts the required inertia to a quarter, 0.0250 kg·m², and the mass to 0.80 kg, because ω is squared.
Common mistakes
- Entering speed in rpm. The field is in rad/s; convert first with
ω = 2π × rpm / 60. Typing 3000 for 3000 rpm instead of 314.16 makes the disk look about 100 times lighter than it is. - Entering the radius in centimetres or the diameter. The radius is in metres: 25 instead of 0.25 returns 0.00 kg, and a diameter gives a disk four times too heavy.
- Treating the energy as what the flywheel can deliver. E is the total energy at that speed; the usable part is only the difference between the energy at the top and bottom operating speeds.
- Entering kWh or kJ as if they were joules. One kWh is 3.6 million J, so a unit slip is a factor of thousands.
- Applying the result to a ring or spoked wheel. Those shapes have a different inertia formula, so the mass printed here is for a solid disk only.
About this calculator
A flywheel stores rotational kinetic energy in a spinning mass to smooth out fluctuations in mechanical power delivery — the energy stored is E = ½·I·ω², where I is the moment of inertia and ω is the angular velocity. This calculator works backward from a target energy to store and an operating angular velocity to find the required moment of inertia (I = 2E/ω²), then uses the solid-disk formula I = ½·m·r² (rearranged to mass = 2I/r²) to find the mass needed at a given radius.
Mechanical and automotive engineers size flywheels for uses like smoothing the power pulses in a piston engine between combustion strokes, storing energy in kinetic energy recovery systems (KERS), and stabilizing punch presses and other machinery that draws power in short, heavy bursts — sizing the disk correctly means balancing enough stored energy against the added mass and rotational inertia the rest of the system has to support.
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Comments (1)
To size for energy you actually need to deliver, work back from the speed swing. Between top speed and a speed 20 percent lower, a flywheel gives up only 1 minus 0.8 squared, or 36 percent of its stored energy. So to deliver 1,000 J, enter about 2,777.8 J as the energy. At 300 rad/s and a 0.2 m radius that returns 0.0617 kg·m² and 3.09 kg. What speed swing does your machine allow?