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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.
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Topics: Motion of system of Particles and Rigid Body
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Subject: Physics
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