= Flywheel
A \b[flywheel] stores <kinetic energy> in a rotating body. For fixed-axis rotation with <moment of inertia> $J$ and <angular velocity> $\omega$, its energy and axial <angular momentum> are
$$
E=\frac12J\omega^2,\qquad L=J\omega.
$$
An applied <torque> changes the angular momentum; a drive can put energy into the flywheel, and a load can receive energy as it slows. Energy varies quadratically with speed, while angular momentum varies linearly.
For an original idealized calculation, take $J=0.20\,\mathrm{kg\,m^2}$. Slowing from $300$ to $200\,\mathrm{rad\,s^{-1}}$ releases
$$
\Delta E=\frac12(0.20)(300^2-200^2)=5000\,\mathrm J.
$$
The wheel still retains4000joules at the lower speed. Delivering the5000joule difference uniformly over10seconds would supply an average <power> of500watts before losses. This example distinguishes usable energy over a speed interval from the energy at the initial speed.
A NASA laboratory demonstration combined two flywheels to regulate electrical power while supplying a commanded <torque>. Its G2 and D1 units used magnetically levitated rotating wheels, were mounted on an air-bearing table, and operated between20000and60000revolutions per minute. A load cell measured table force so that output torque could be inferred. The apparatus connects energy exchange with the reaction torque exerted on the surrounding structure.<Flywheel Integrated Power and Attitude Control System Demonstrated by NASA Glenn Research Center>[opening paragraphs and load-cell/table-torque measurement]{ref}
A flywheel can also produce <mechanical inertance> through a <rack and pinion>. If relative terminal speed $v$ gives $\omega=v/r$, the ideal stored energy is $Jv^2/(2r^2)$ and the equivalent inertance is $b=J/r^2$. Bearing losses, <friction> and material stresses constrain real devices; the ideal energy equation alone is not an operating-speed specification.
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