How this calculation works
A flywheel usually cannot be discharged to zero speed while still serving its load. Usable energy is the difference between two rotational-energy states. This page combines that speed band with an entered efficiency and constant output demand.
ΔE = I(ωhigh² − ωlow²)/2; delivered duration = efficiency × ΔE / output powerA worked example
An inertia of 2 kg m² dropping from 3000 to 1500 rpm releases 74022.03 J mechanically. At ninety-percent delivery efficiency, 66619.83 J reaches the load. A constant 1000 W output would consume that budget in about 66.62 seconds.
Good to know
This is a lumped energy budget with a constant entered efficiency. It excludes rotor stress, bearing loss, windage, containment and overspeed protection. No physical flywheel construction, operating speed or containment specification is recommended.
Reference: Illinois mechanics reference — dynamics (opens in a new tab) ↗
What this model answers
A flywheel usually cannot be discharged to zero speed while still serving its load.
Usable energy is the difference between two rotational-energy states. This page combines that speed band with an entered efficiency and constant output demand.
The Kinetic energy calculator estimates annular-contact torque. An energy budget also needs the speed range, since torque alone is not stored energy.
Keep the input conventions consistent
Enter mass moment of inertia in kg m², not an area moment in mm⁴.
Speeds use revolutions per minute and are converted to radians per second. Output power means delivered useful power, so efficiency reduces available energy rather than reducing that demand.
Read the result carefully
Half the upper speed still retains one quarter of the upper-speed kinetic energy.
The usable fraction is therefore seventy-five percent, not fifty percent. Energy alone also does not establish the torque a motor or transmission can deliver throughout the band.
Energy removed between two speeds depends on the difference of their squared angular speeds. A fixed RPM drop therefore releases more energy when it starts at a higher speed. Compare the full speed band rather than multiplying stored energy by the fractional RPM change as if energy were linear in speed.
Work through the numbers
An inertia of 2 kg m² dropping from 3000 to 1500 rpm releases 74022.03 J mechanically.
At ninety-percent delivery efficiency, 66619.83 J reaches the load. A constant 1000 W output would consume that budget in about 66.62 seconds.
Where the calculation stops
This is a lumped energy budget with a constant entered efficiency.
It excludes rotor stress, bearing loss, windage, containment and overspeed protection. No physical flywheel construction, operating speed or containment specification is recommended.
A few questions, answered
Does half speed mean half the stored energy?
No. Rotational energy is proportional to speed squared when inertia is constant.
What should I change in the comparison?
Compare the full mathematical speed range with a restricted usable band.