Two discs have moments of inertia I₁ and I₂ about their respective axes perpendicular to the plane and passing through the centre. They are rotating with angular speeds, ω₁ and ω₂ respectively and are brought into contact face to face with their axes of rotation coaxial. The loss in kinetic energy of the system in the process is given by :

Solution

Physics sir · 4 steps

Friction between the faces is internal, so angular momentum is conserved while kinetic energy is not.

1. Common angular speed. The friction torques between the discs are internal, so total angular momentum stays the same. Both discs end up turning together at ω = (I₁ω₁ + I₂ω₂)/(I₁ + I₂).
2. Energy before and after. Before contact, Kᵢ = 1/2I₁ω₁² + 1/2I₂ω₂². After contact, K_f = 1/2(I₁ + I₂)ω² = (I₁ω₁ + I₂ω₂)²/2(I₁ + I₂).
3. Subtract. Over the common denominator 2(I₁ + I₂), the numerator is (I₁ + I₂)(I₁ω₁² + I₂ω₂²) - (I₁ω₁ + I₂ω₂)². The I₁²ω₁² and I₂²ω₂² terms cancel, leaving I₁I₂(ω₁² + ω₂² - 2ω₁ω₂).

   Δ K = Kᵢ - K_f = (I₁ I₂)/2(I₁ + I₂)(ω₁ - ω₂)²

4. So. The loss in kinetic energy matches option (c). It is zero only when ω₁ = ω₂, because then the faces never slip against each other.

The idea to keep. When two bodies lock together, conserve momentum first and then compare energies. The loss always has the form 1/2μ(Δ v)², and for rotation μ = I₁I₂/(I₁+I₂).

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