A disc of moment of inertia (9.8/π²) kg m² is rotating at 600 rpm. If the frequency

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A disc of moment of inertia (9.8/π²) 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?

Options

(a) 1470 J
(b) 1452 J
(c) 1567 J
(d) 1632 J

Correct Answer:

1470 J

Explanation:

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

-ve sign show that rotational kinetic energy decreases.

Related Questions:

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Topics: Motion of system of Particles and Rigid Body (73)
Subject: Physics (2479)

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