Physics · Engineering Physics · A2 Level

Flywheels

Everything covered so far in this topic — moment of inertia, rotational kinetic energy, angular momentum — comes together in one of the most widely used pieces of rotating engineering hardware there is. A flywheel is, at heart, nothing more than a carefully shaped spinning mass, but the physics behind why it's shaped the way it is, and what it's actually used for, draws directly on everything you've built up across this topic.

9 min read

What a flywheel is for

A flywheel is a rotating mass, mounted on a shaft, whose job is to store rotational kinetic energy:

$$E_k = \tfrac{1}{2}I\omega^2$$

To store as much energy as possible without being unnecessarily heavy or bulky, a flywheel is deliberately shaped to maximise its moment of inertia for a given total mass. Recall from the very first lesson in this topic that, because $I = \Sigma mr^2$, mass positioned far from the rotation axis contributes far more to $I$ than the same mass positioned close to it — a thin hoop has a larger moment of inertia than a solid disc of the same mass and radius, purely because the hoop's mass all sits at maximum radius. A flywheel exploits exactly this: rather than being a uniform solid disc, most of a flywheel's mass is concentrated in a thick, heavy outer rim, connected to the central hub and shaft by comparatively light spokes. This design gives the largest possible moment of inertia — and therefore the largest possible energy storage — for the smallest possible total mass.

A flywheel viewed face-on, drawn with a noticeably thick outer rim connected to a small central hub by four thin spokes, with a curved arrow around the rim showing its spin direction and an axle line leading out from the hub. To the side, two arrows labelled energy in (spin-up) and energy out (released) show energy flowing into and out of the flywheel.
Most of a flywheel's mass sits in a heavy rim far from the axis, maximising moment of inertia for a given total mass — the flywheel spins up to store energy and spins down as it releases it.

What determines a flywheel's energy storage capacity

Two factors directly control how much rotational kinetic energy a flywheel can store, both visible directly in $E_k = \tfrac{1}{2}I\omega^2$:

  • Its moment of inertia, $I$. Maximised by concentrating as much mass as possible as far as possible from the rotation axis — the rim-heavy design described above — and by increasing the total mass of the flywheel itself.
  • How fast it can safely spin, $\omega$. Since energy storage scales with $\omega^2$, increasing the operating speed is a highly effective way to increase stored energy for a given flywheel. However, this comes with a genuine engineering trade-off: spinning any rotating rim faster increases the outward stress the rim material must withstand, since faster rotation requires a larger centripetal force per unit mass to keep the rim material moving in a circle rather than flying apart. Beyond some maximum safe speed, the rim material would fail. A flywheel's practical energy storage capacity is therefore limited not just by its size and mass, but by the strength of the material it's built from.

Two headline engineering uses

Flywheels appear throughout mechanical engineering, but their uses fall into two broad categories.

  • Smoothing torque and speed. Many machines — a piston engine is the classic example — don't deliver power smoothly and continuously, but in uneven pulses, as each piston fires in turn. Without anything to compensate, this would cause the machine's rotational speed to surge and dip in an unpleasant and inefficient way with every pulse. A flywheel attached to the rotating shaft resists these sudden changes in speed, precisely because of its large moment of inertia: during a power pulse it absorbs the excess energy, spinning up slightly, and between pulses it releases that stored energy back into the system, spinning down slightly, keeping the overall rotational speed far smoother than it would otherwise be.
  • Storing energy. A flywheel can also be used deliberately as an energy store in its own right, charged up gradually and then discharged quickly when a burst of power is needed. In vehicles, flywheel energy-storage systems can capture kinetic energy that would otherwise be lost as heat during braking, storing it as rotational kinetic energy and releasing it again to assist acceleration shortly after — a rotational counterpart to other regenerative braking approaches. In production machinery such as mechanical presses or punches, a motor spins a heavy flywheel up gradually over a longer period using a comparatively modest, steady power input, and the flywheel then delivers a short, powerful burst of energy in an instant, driving the press or punch through its brief working stroke far more powerfully than the motor alone could manage directly.
Why the rim-heavy design matters twice over

Concentrating mass in a flywheel's rim doesn't just maximise energy storage for a given mass — it also means that, for a given amount of stored energy, the flywheel needs less total mass in the first place, which is exactly why real flywheels are essentially never built as uniform solid discs.

Rotational work and power

To complete the toolkit for this topic, two further quantities describe how energy is transferred into or out of a rotating system: rotational work and rotational power. Just as linear work is $W = Fs$ (a force acting through a displacement), rotational work is a torque acting through an angular displacement:

$$W = T\theta$$

where $\theta$ is the angular displacement in radians. And just as linear power is $P = Fv$ (force multiplied by speed), rotational power is torque multiplied by angular velocity:

$$P = T\omega$$
Worked examplePower delivered by a motor
01
A motor delivers an output torque of $T = 25\text{ N m}$ while running at $\omega = 120\text{ rad s}^{-1}$.
Find the power delivered by the motor.
02
Apply $P = T\omega$ directly.
$P = 25 \times 120$
03
Evaluate.
$P = 3000\text{ W} = 3.0\text{ kW}$
P = 3.0 kW

In an idealised system, all of this power would reach whatever load the motor is driving. Real rotating machinery is never quite this clean: bearings, seals, and other moving parts inevitably generate a resisting frictional torque that must be overcome before any useful power reaches the load at all. The torque and power actually delivered usefully to the load are therefore always somewhat less than the torque and power produced internally by the motor or engine — the difference is the torque and power lost to overcoming friction within the machine itself. Any realistic analysis of a piece of rotating machinery has to account for this loss rather than assuming all of the input power arrives at the output.

Friction isn't optional in real machines

Whenever a question describes a motor, engine, or flywheel system as realistic rather than idealised, expect frictional torque to reduce the torque and power actually delivered to the load below the value the source itself produces. Quoting the source's own output torque or power as if it were what reaches the load, without accounting for friction, is a common and avoidable error.

Key takeaways

  • A flywheel stores rotational kinetic energy, $E_k = \tfrac{1}{2}I\omega^2$, and is typically built with most of its mass concentrated in a heavy outer rim to maximise $I$ for a given mass.
  • Energy storage capacity depends on moment of inertia (maximised by mass at large radius) and on how fast the flywheel can safely spin, since energy scales with $\omega^2$.
  • Spinning a flywheel faster increases stored energy but also increases stress in the rim material — a genuine engineering limit on maximum safe speed.
  • Flywheels smooth out variations in torque and speed in machinery that delivers power unevenly, such as piston engines.
  • Flywheels also store energy directly, in vehicle energy-recovery systems and in production machinery such as presses and punches that need short bursts of high power.
  • Rotational work is $W = T\theta$ and rotational power is $P = T\omega$, the direct analogues of $W = Fs$ and $P = Fv$.
  • Frictional torque in real rotating machinery, such as in bearings, means the torque and power delivered to a load are always somewhat less than that produced internally.