EngineerBenchEnergy available from a slowing flywheel
A flywheel releases energy as its rotational speed falls. Usable energy is the difference between its energy at the high and low speeds, not its total energy at maximum RPM. The model treats the wheel as a uniform solid or annular disk.
Describe the mass distribution
Enter an inner radius of zero for a solid disk. An annular disk needs both inner and outer radii. Wheels with spokes, attached weights, or a heavy rim can have a different mass moment of inertia even when their mass and outside diameter match.
Energy is not a speed rating
The required-mass result answers an energy question only. It does not check material stress, balance, bearings, containment, or an allowable operating speed. Those checks are essential before building or operating rotating energy-storage equipment.
Comparing energy between two speeds
Rotational energy is one half of mass moment of inertia multiplied by angular speed squared. A wheel slowed to half its original speed retains one quarter of its original kinetic energy, so three quarters has been released. A machine that must maintain nearly constant speed can use only a smaller fraction of the total stored energy. Entering a zero final speed assumes a complete stop is acceptable.
Mass moment of inertia depends on where the material lies relative to the axis. Moving the same mass farther outward increases inertia, which is why a solid disk and a rim-heavy wheel cannot be treated as equivalent by weight alone. For the uniform annular model, both radii contribute to the inertia calculation. Convert diameters to radii before entering radius fields. Compare energy demand over the actual load event with the usable energy result, while allowing for losses and the motor's contribution separately. Do not infer a safe construction or operating speed from the energy balance.
Formula
I = m(ro² + ri²)/2; ΔE = I(ωhigh² − ωlow²)/2.