This question was previously asked in

GATE PI 2020 Official Paper

**Concept:**

\(Δ E =\frac{1}{2}I\left ( ω^2 _{max}-ω^2_{min} \right )\)

\(ω =\frac{2\pi N}{60}\)

Where Δ E = maximum fluctuations of energy (N-m), I = Mass moment of Inertia (kg-m^{2}), ω = angular speed of the flywheel (rad/s), N = RPM

**Calculation:**

**Given:**

N = 500 RPM, ΔE = 10000 Nm, Nmax - Nmin = 30 RPM

\(ω =\frac{2\pi N}{60}\)

\(ω =\frac{2\pi \times 500}{60}=52.36\;rad/s\)

\(ω_{max} =\frac{2\pi N_{max}}{60}\)

\(ω_{min} =\frac{2\pi N_{min}}{60}\)

\(ω_{max} -ω_{min} =\frac{2\pi }{60}(N_{max}-N_{min})\)

\(ω_{max} -ω_{min} =\frac{2\pi }{60}\times 30\)

ωmax - ωmin = 3.1415 rad/s

\(Δ E =\frac{1}{2}I\left ( ω^2 _{max}-ω^2_{min} \right )\)

\(\Delta E = I\left ( \frac{\omega_{max}+\omega_{min}}{2} \right )\left ( \omega_{max}-\omega_{min} \right )\)

\(\Delta E = I\omega\left ( \omega_{max}-\omega_{min} \right )\)

10000 = I × 52.3598 × 3.1415

I = 60.7945 kg-m2