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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
(73)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A particle of mass 100 g is thrown vertically upwards with a speed of 5 m/s. The work
- The manifestation of band structure in solids is due to
- The refracting angle of a prism is A and refractive index of the material of the prism
- A wire is stretched under a force. If the wire suddenly snaps the temperature
- When a biconvex lens of glass having refractive index 1.47 is dipped in a liquid
Topics: Motion of system of Particles and Rigid Body (73)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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