JEE Main · 2024 · 30 Jan 2024 · Shift 1 · Rotational Mechanics

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Consider a Disc of mass 5 kg, radius 2 m, rotating with angular velocity of 10 rad/s about an axis perpendicular to the plane of rotation. An identical disc is kept gently over the rotating disc along the same axis. The energy dissipated so that both the discs continue to rotate together without slipping is ______ J.

Answer: 250

Solution

Initial:

I = \frac{M R ^{2}}{2} = \frac{5 \times 4}{2} = 10 kg m^{2}

, so

L_{i} = I ω = 10 \times 10 = 100 kg m^{2} /s

and

E_{i} = \frac{1}{2} I ω^{2} = \frac{1}{2} (10) (100) = 500 J

.

When the identical disc is dropped on, the moment of inertia doubles; by conservation of angular momentum,

100 = 2 I ω_{f} \Rightarrow ω_{f} = 5 rad/s

E_{f} = \frac{1}{2} (2 I) ω_{f}^{2} = \frac{1}{2} (20) (25) = 250 J

.

Energy dissipated

= E_{i} - E_{f} = 500 - 250 = 250 J

JEE Main · 2024 · 30 Jan 2024 · Shift 1 · Integer answer

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JEE Main · 2024 · 30 Jan 2024 · Shift 1 · Rotational Mechanics

Solution

More JEE Main questions on Rotational Mechanics