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:

1. A solid sphere of mass M, radius R and having moment of inertia about an axis passing
2. A car accelerates from rest at constant rate for first 10s and covers a distance x.
3. If Q, E and W denote respectively the heat added, change in internal energy
4. Ring, hollow ring and solid sphere are rolled down from inclined plane
5. Identify the pair whose dimensions are equal

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

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