Question

A disc of moment of inertia ; kg m2 is rotating at 600 rpm. If the
frequency of rotation changes from 600 rpm to 300 rpm, then what is
the work done?​

Answer 1

Given: Moment of inertia I = (9.8/π²) kgm²

ʋ₁ = 600 rpm = 10 rps; ʋ₂ = 300 rpm = 5 rps

.·. ω₁ = 2π ʋ₁ = 20π rad s⁻¹

.·. ω₂ = 2π ʋ₂ = 10π rad s⁻¹

Kinetic energy of rotation= (1/2) Iω²

Work done W = change in rotational kinetic energy

.·. work done W = (1/2).I [ω₂² – ω₂²]

W = (1/2) x (9.8/π²).[(10π)² – (20π)²]

= (1/2) x (9.8/π²).[-300 π²] = -1470 J