A recording disc rotates steadily at 45 rev/min on a table. When a small mass of 0.02 kg is dropped gently on the disc at a distance of 0.04 m from its axis and stucks to the disc, the rate of revolution falls to 36 rev/min. The moment of inertia of the disc about its centre is
Answers
The moment of inertia of the disc about its centre is 1.3 * 10⁻⁴ kg m²
Explanation:
=> It is given that,
Initial rate of revolution of a disc = 45 rev/min
The rate of revolution fall due to small mass, = 36 rev/min
∴ v₁ = 45/60 rev s⁻¹
= 3/4 rad s⁻¹
v₂ = 36/60 rev s⁻¹ = 3/5 rev s⁻¹
=> Using conservation of angular momentum, we get:
I *3/4 = [I + 0.02 * 0.04 * 0.04] * 3 / 5
5I = 4I + 4 * 0.02 * 0.04 * 0.04
∴ I = 1.3 * 10⁻⁴ kg m²
Thus, the moment of inertia of the disc about its centre is 1.3 * 10⁻⁴ kg m².
Learn more:
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Q:2 A gramophone disc rotates at 60 rpm. A coin of mass 18 g is placed at a distance of 8 cm from the centre . Calculate the centrifugal force on the coin. Take pi-square= 9.87.