The flywheel of a steam engine has a radius of gyration of 1 m and mass 2000 kg. The starting torque of the engine is 2000 N-m. The kinetic energy of the flywheel after 10 seconds from start is

Concept:

The kinetic energy of the flywheel is given by, \(E = \frac{1}{2}I{\omega ^2}\)

Where, I = mass moment of flywheel = m × K², and ω is the angular velocity of the flywheel.

Angular velocity ω = ω₀ + αt

Where α is the angular acceleration of the flywheel

Torque, T = Iα

Calculation:

Given,

T = 2000 N-m, radius of gyration K= 1 m and m = 2000 kg, ω₀ = 0 and t = 10 sec

I = 2000 × 1² = 2000 kg.m²

So the angular acceleration \(\alpha = \frac{T}{I} = \;\frac{{2000}}{{2000}} = 1\;rad/{s^2}\)

Angular velocity ω = ω₀ + αt

ω = 0 + 1 × 10 = 10 rad/sec

The kinetic energy of the flywheel is, = \(\frac{1}{2} \times 2000 \times {10^2} = 100\;kN - m\)