Flywheel energy relation and rim sizing

Derivation

Let the flywheel have moment of inertia $I$, maximum speed $\omega_1$, minimum speed $\omega_2$ and mean speed $\omega = \dfrac{\omega_1+\omega_2}{2}$.

Maximum kinetic energy: $E_{1} = \tfrac12 I\omega_1^2$. Minimum kinetic energy: $E_{2} = \tfrac12 I\omega_2^2$.

$$\begin{aligned} \Delta E &= E_1 - E_2 = \tfrac12 I(\omega_1^2-\omega_2^2) \\ &= \tfrac12 I(\omega_1-\omega_2)(\omega_1+\omega_2) \\ &= I\,\omega\,(\omega_1-\omega_2) \end{aligned}$$

since $\omega_1+\omega_2 = 2\omega$. Using $C_s = \dfrac{\omega_1-\omega_2}{\omega}$, we get $\omega_1-\omega_2 = C_s\omega$, so

$$\Delta E = I\omega^2 C_s = 2E\,C_s$$

where $E = \tfrac12 I\omega^2$ is the mean kinetic energy.

Why the mass is in the rim

For a required $\Delta E$, speed and $C_s$, the value of $I$ is fixed. Since $I = mk^2$, a large radius of gyration $k$ gives the same $I$ with less mass. So the material is placed at the rim, which keeps the weight and cost low. The limit is the rim stress $\sigma = \rho v^2$, which restricts the rim speed.

Rimmed ring vs solid disc

For the same mass and outer radius, the rim stores about twice the energy of a solid disc.