Flywheel (Solid Rotating Disk)

energy-storagestressmass-cost

A flywheel is a battery made of motion: spin mass up to store work, let it back down to release it. Punch presses bank energy between strokes, engines smooth the gaps between cylinder firings, satellites trade spin with reaction wheels to point themselves, and grid-scale flywheel farms soak up seconds-long power surges. The catch is that the flywheel loads itself — every gram needs a centripetal pull of $\rho\omega^2 r$ per unit volume to stay on its circle, and that pull has to come through the disk as stress:

$E_k = \tfrac{1}{2} I_z\,\omega^2 \qquad \sigma_{max} = \frac{3+\nu}{8}\,\rho\,\omega^2 R^2 \qquad e_{max} = \frac{2\,\sigma_y}{(3+\nu)\,\rho}$

Three things to notice, one per formula:

• Energy and stress are the same knob. Both grow with the rim speed squared, $(\omega R)^2$. Storing more energy in a given disk means stressing it harder — there is no way to have one without the other, and the derivation below shows the exchange rate is fixed.
• The stress moves with the material here. On the torsion shaft the stress famously ignores the material; a flywheel is the opposite case, because density is the load. Swap steel for Ti-6Al-4V in the widget and watch $\sigma_{max}$ drop by nearly half at the same speed — the lighter disk pulls on itself less (and stores less too; nothing is free).
• Size cancels. Eliminate $\omega$ between the first two formulas and $R$ and $t$ vanish: energy per kilogram at the strength limit is $2\sigma_y/((3+\nu)\rho)$ — a pure material index, essentially $\sigma_y/\rho$. You cannot build your way past it with a bigger disk; a big slow flywheel and a small fast one of the same material store the same energy per kilogram when pushed to the same margin. That is why modern storage flywheels are wound from carbon fibre rather than machined from steel, and it is the flywheel’s ticket into the Ashby chart way of thinking.

The energy-in configuration runs the same relations the other way: name the energy you need, and the widget finds the speed that stores it — then tells you what that speed costs in stress and margin. Watch $\omega_y$, the yield-onset speed: it is the model’s own ceiling, and it corresponds to a fixed rim speed $\omega_y R = \sqrt{8\sigma_y/((3+\nu)\rho)}$, another pure material number.

How long does charging take? Spinning the disk up from rest with a constant drive torque $T_d$ takes time, and angular impulse–momentum sets the clock — the torque’s angular impulse $T_d\,t$ must supply the angular momentum $I_z\,\omega$ the flywheel carries at speed:

$t_{spin} = \frac{I_z\,\omega}{T_d}$

More torque spins it up faster; more inertia — a bigger or denser disk — or a higher target speed draws it out (swap in a lighter alloy and $t_{spin}$ falls with the inertia, invariant 3 again). This is the number the chain builder’s headline example holds in tension with shaft stress: the gear ratio ahead of the flywheel sets $T_d$, so cranking the ratio up shortens $t_{spin}$ while winding the driveshaft that carries $T_d$ correspondingly tighter — storing energy quickly is not free. A negative $T_d$ brakes rather than accelerates, so the widget withholds $t_{spin}$ while the energy, stress, and margin readouts still stand — the spin-up time is the only thing that depends on how the disk is driven. (At exactly $T_d = 0$ the formula divides by zero, and the whole readout set refuses.)

Try it

Governing relations

Derivation

Steps marked modeling step are where physics enters by citation — every other line is machine-proven to follow from them, and the cited models are independently re-derived in the test pipeline where possible. See what is and isn't machine-verified.

How it fails

The widget’s margin is first yield at the disk centre, under a steady spin, for a solid uniform disk. Real flywheels die harder and uglier:

• Burst. The defining flywheel failure: past its limit the disk tears into two or three large sectors that leave tangentially at rim speed, each carrying its share of the stored energy as shrapnel. The energy that made the flywheel useful is exactly what makes it dangerous — industrial flywheels and turbine disks live inside containment housings sized for a full burst, and spin-pit testing (deliberately bursting disks in a bunker) is a routine acceptance step for turbine rotors.
• First yield is not burst — the margin hides on both sides. A ductile disk that yields at the centre just redistributes stress outward and keeps spinning; actual burst arrives at the higher speed where the average hoop stress across a diameter approaches the material’s ultimate strength. So the widget’s SF is conservative for ductile metals. For brittle materials it is the opposite kind of wrong: no yield, no redistribution, no warning — the first crack is the burst. That is why grey cast iron, the classic Victorian flywheel material, killed so many operators when steam engines overspun: it publishes no yield strength at all (which is exactly why it is absent from this page’s material dropdown), and old engineering handbooks are full of flywheel-explosion case studies.
• A centre hole doubles the stress. Put even a small bore through the disk for a shaft — as nearly every real flywheel must — and the peak hoop stress at the bore edge jumps to twice the solid-disk centre value (exactly twice, in the vanishing-bore limit). The solid-disk numbers here are the optimistic textbook case; a bored disk reaches yield at about 70% of the solid disk’s speed. That story is now its own page — the rotating disk with a central bore, where the factor of 2 is derived, machine-checked, and yours to drag a knob through. Real designs use shrink-fitted hubs, integral shafts, or profiled (constant-stress) disks to claw the penalty back.
• Overspeed is squared. Because stress grows with $\omega^2$, a 20% overspeed is a 44% stress increase; 40% over is nearly double. Flywheel accidents are overwhelmingly governor/controller failures, not material defects — the protection that matters is the one that limits speed.
• Fatigue from charge–discharge cycling. A storage flywheel is stress-cycled every time it banks and releases energy. Tens of thousands of deep cycles of centre stress make crack initiation at the bore or hub fillet the life-limiting mechanism long before any static margin is consumed.
• Everything off-axis. Imbalance, gyroscopic moments when the vehicle turns, and shaft critical speeds all load the bearings, not the disk — and bearing failure at speed releases the rotor just as effectively as a burst. High-speed flywheels run on magnetic bearings in vacuum partly for efficiency, partly because nothing mechanical survives the interface for long.

Sources

• Timoshenko, S. P., & Goodier, J. N., Theory of Elasticity, 3rd ed., McGraw-Hill, 1970 — ch. 4 (two-dimensional problems in polar coordinates), §32 "Rotating Disks": plane-stress field of a solid uniform disk, σ_r = σ_θ = (3+ν)ρω²R²/8 at the centre, σ_r(R) = 0 at the free rim.
• Hibbeler, R. C., Engineering Mechanics: Dynamics, 14th ed., Pearson, 2016 — §17.1 (mass moments of inertia; the uniform solid disk I = mR²/2), ch. 18 (planar kinetics, work and energy: E = ½Iω² for rotation about a fixed axis), and ch. 19 (planar kinetics, impulse and momentum: the principle of angular impulse and momentum, I_O·ω₁ + Σ∫M_O dt = I_O·ω₂, §19.1–19.2) — the basis for the constant-torque spin-up time T_d·t = I_z·ω.
• Budynas, R. G., & Nisbett, J. K., Shigley's Mechanical Engineering Design, 10th ed., McGraw-Hill, 2015 — ch. 16 (flywheels: energy exchange and sizing) and ch. 1 (design factor against first yield).
• Ashby, M. F., Materials Selection in Mechanical Design, 4th ed., Butterworth-Heinemann, 2011 — the "materials for flywheels" case study: maximum kinetic energy per unit mass is governed by the material index σ_y/ρ, independent of the disk's size.