Flywheel sizing: mass versus radius at a fixed rim-speed limit
A cycle-share company is prototyping a regenerative braking unit: a solid steel disc of mass 20 kg and radius 0.30 m that spins up to 3000 rpm. The steel is rated so that rim speed must never exceed 100 m/s, and the design team has taken 3000 rpm as the working ceiling. The disc stores energy E = ½Iω². The workshop can either (Option A) machine a second identical disc and stack it on the same shaft, doubling the mass at the same radius, or (Option B) fit a disc of the same 20 kg mass but 0.60 m radius, running at whatever speed the rim-speed limit allows. Bearings and housing are unchanged in either case. A junior engineer argues that Option B must be better because a larger radius squares the moment of inertia. The project lead wants the numbers, the reason the two options differ, and a safety view before committing to tooling.
1. Compute the moment of inertia, angular speed, stored energy and rim speed of the existing disc at 3000 rpm.
Read the worked answer
I = ½mr² = ½(20)(0.30)² = 0.90 kg m².
ω = 2π(3000)/60 = 314 rad/s.
E = ½Iω² = ½(0.90)(314)² = 0.45 × 98 700 = 4.44 × 10⁴ J ≈ 44 kJ.
Rim speed v = ωr = 314 × 0.30 = 94.2 m/s, inside the 100 m/s rating.
Headroom note. If the disc were taken to exactly 100 m/s, the energy would rise by the factor (100/94.2)² = 1.13, to about 50 kJ. So the present operating point already uses about 89% of the stress-limited energy, and the remaining margin is small. The rating is a stress limit, not a round number: hoop stress scales as v², so any overspeed is a burst-safety issue rather than a performance tweak.
2. Evaluate Option A and Option B quantitatively and adjudicate the junior engineer's claim.
Read the worked answer
Key identity: for a uniform disc, E = ½Iω² = ½(½mr²)(v/r)² = ¼mv², with v = ωr the rim speed. Energy depends on mass and rim speed only - the radius cancels.
Option A (m = 40 kg, r = 0.30 m, 3000 rpm): I = ½(40)(0.09) = 1.80 kg m², E = ½(1.80)(314)² = 88.8 kJ, v = 94.2 m/s - still legal. Energy doubles.
Option B (m = 20 kg, r = 0.60 m): the stress limit caps v at 100 m/s, so ω = 100/0.60 = 166.7 rad/s and I = ½(20)(0.36) = 3.60 kg m². E = ½(3.60)(166.7)² = 50.0 kJ. Only about 13% more than the original, not double.
Adjudication. The junior engineer is wrong. I does rise four-fold, but the achievable ω falls by the same factor because the rim is now twice as far from the axis, and the two effects cancel exactly at fixed rim speed. Option A is the only one of the two that doubles storage.
Caveats on A. Twice the mass means higher bearing loads, a longer shaft and more unsprung mass; the stacked discs must be balanced as one assembly; and the second disc adds cost without improving power capability, only capacity.
3. The specification is later raised to double the stored energy with no increase in mass. What would that require, and what must be re-examined?
Read the worked answer
From E = ¼mv², doubling E at fixed m needs v → v√2 = 94.2 × 1.414 = 133 m/s. Hoop stress goes as v², so the stress on the material roughly doubles - far outside the stated 100 m/s rating for steel.
Options. (i) Change material: a carbon-fibre composite rim rated above 130 m/s keeps mass low but changes cost, containment design and failure mode. (ii) Accept mass, i.e. fall back to Option A. (iii) Raise the speed ceiling only after re-rating the containment ring, bearings and shaft critical speed, and after a burst test - at 133 m/s a disc failure is a fragment-containment problem, not a spreadsheet adjustment.
Losses to re-check. Windage on the rim grows roughly as v³ and bearing drag also rises with speed, so the extra stored energy is not all recoverable during a stop; at higher speed the round-trip efficiency drops. Also confirm the shaft's first critical speed is well above the new running speed to avoid resonance during spin-up.