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Flywheels

Energy and power · level 7, lesson 4 of 6

A question to start

A heavy wheel spun fast holds energy, and gives it back as it slows. Would a flywheel store as much on Mars as on Earth?

The idea

A spinning wheel has rotational kinetic energy

\[E = \tfrac12 I\omega^2,\]

where \(\omega\) is its angular speed in radians per second and \(I\) its moment of inertia. For a solid disc of mass \(m\) and radius \(r\), \(I = \tfrac12 mr^2\).

rωvacuumcase

Energy grows with \(\omega^2\), so doubling the speed stores four times as much. The limit is strength: the faster the rim moves, the harder the wheel is pulled apart, and a wheel spun too fast bursts. Real flywheels spin in a vacuum, often on magnetic bearings, to cut friction and air drag. Even so they lose energy over hours, so they suit short jobs: smoothing power for seconds or minutes, not storing a whole night.

Gravity is not in \(\tfrac12 I\omega^2\), so the same flywheel stores the same energy on either planet. What changes is the load on its bearings:

A 100 kg disc, radius 0.5 m, at 6,000 rpm Mars Earth
Energy stored 0.69 kWh 0.69 kWh
Weight the bearings carry 371 N 981 N

Worked examples

1. The disc in the table. \(\omega = 6000 \times 2\pi/60 = 628\) rad/s. \(I = \tfrac12 \times 100 \times 0.5^2 = 12.5\) kg m². \(E = \tfrac12 \times 12.5 \times 628^2 = 2.5\times10^6\) J, which is 0.69 kWh. The rim moves at \(628 \times 0.5 = 314\) m/s.

2. Per kilogram. 690 Wh ÷ 100 kg \(= 6.9\) Wh/kg, against about 200 for lithium-ion cells. Flywheels win on speed and on lasting for many charges, not on energy per kilogram.

3. Power. Giving out the 2.5 MJ in 60 s is \(2.5\times10^6/60 = 41\) kW. A flywheel can be charged and emptied fast, over and over, without wearing out the way a battery does.

Try it

  1. A ring of mass 50 kg and radius 0.4 m has \(I = mr^2\). Find its energy at 300 rad/s.
  2. If a flywheel spins three times as fast, how much more energy does it hold?
  3. Why do flywheels spin in a vacuum?
  4. Why is a flywheel a poor way to store a whole Mars night?
Answers
  1. \(I = 50 \times 0.16 = 8\) kg m², \(E = \tfrac12 \times 8 \times 300^2 = 360{,}000\) J, or 0.1 kWh.
  2. Nine times.
  3. Air drag would turn the stored energy into heat.
  4. It slowly loses energy over hours, and it stores little energy per kilogram.

Common mistake

Putting rpm into the formula. \(\omega\) must be in radians per second. Multiply rpm by \(2\pi/60\) first.

Where it leads