Flywheel size from turning moment areas

Data: mean speed $N = 240$ rpm, $C_s = 0.03$, rim mean radius $R = 0.9$ m.

Energy at each point

Take the energy at A as zero and add the areas in order (positive area = gain of kinetic energy, negative = loss).

Maximum energy = 4.8 kJ, minimum energy = -1.1 kJ.

$$\Delta E = E_{max} - E_{min} = 4.8 - (-1.1) = 5.9\ \text{kJ} = 5900\ \text{J}$$

Moment of inertia

$$\omega = \frac{2\pi \times 240}{60} = 25.13\ \text{rad/s}$$ $$\Delta E = I\omega^2 C_s \Rightarrow I = \frac{\Delta E}{\omega^2 C_s} = \frac{5900}{(25.13)^2 \times 0.03} = 311.4\ \text{kg·m}^2$$

Mass of the rim

Treating the rim as a thin ring, $I = mR^2$:

$$m = \frac{I}{R^2} = \frac{311.4}{0.9^2} = 384\ \text{kg}$$

Answer: $\Delta E = 5.9$ kJ; mass of rim $\approx 384$ kg.