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.

Solution

Physics sir · 4 steps

Friction between the discs is internal, so angular momentum is conserved but kinetic energy is not.

1. Moment of inertia. For one disc about its axis, I = 1/2MR² = 1/2 × 5 × 2² = 10 kg m². Two identical discs together have 2I = 20 kg m².
2. Common angular velocity. The friction torques between the discs are internal and there is no external torque about the axis, so angular momentum is conserved. Iω₀ = 2Iω gives ω = ω₀/2 = 5 rad s⁻¹.

   Δ K = 1/2Iω₀² - 1/2(2I)(ω₀/2)² = 1/4Iω₀²

3. Put in numbers. Initial energy is 1/2 × 10 × 10² = 500 J. Final energy is 1/2 × 20 × 5² = 250 J. Half the energy is lost as heat in the friction while the discs slip.
4. So. The energy dissipated is 500 - 250 = 250 J.

The common slip: 500. Using I = MR² (a ring) instead of 1/2MR² for the disc doubles every energy and gives 500, not 250.

The idea to keep. When a body is placed gently on a rotating one, conserve angular momentum to find the common ω, then subtract kinetic energies. For identical bodies, exactly half the energy is lost.

Answer from the NTA final answer key. Solution by PrepWiser.