The flywheel of a steam engine has a radius of gyration of 1 m and mass of 2500 kg. the starting torque of the engine is 2500 N-m. The kinetic energy of such a flywheel after 10 sec from rest position will be 

Concept:

Moment of inertia is given by:

I = mk²

Torque (τ) is given by:

τ = I × α 

Kinetic energy (KE) of flywheel is given by:

\(KE = \frac{1}{2} × I × {\omega ^2}\)

Calculation:

Given:

Radius of gyration of flywheel, k = 1 m, Mass of the flywheel, m = 2500 kg, Starting Torque, T = 2500 N, Time of rotation, t = 10 s

Starting speed, ω₀ = 0 rad/sec

Moment of inertia is:

I = mk²

I = 2500 × (1)² = 2500 kg – m²

τ = I × α

2500 = 2500 × α

α = 1 rad/s²

ω = ω₀  + αt

ω = 0 + 1 × 10 = 10 rad/s²

Kinetic energy (KE) of flywheel is:

^{\(KE = \frac{1}{2} × I × {\omega ^2}\)}

\(K = \frac{1}{2} × 2500 × 100 = 125000 = 125~kN.m\)