Two discs of moment of inertia I₁=4 kg m² and I₂=2 kg m², about their central axes normal to their planes, rotating with angular speeds 10 rad / s 4 rad / s respectively are brought into contact face to face with their axes of rotation coincident. The loss in kinetic energy of the system in the process is _________ J.

Solution

Physics sir · 4 steps

Friction between the faces is internal, so angular momentum stays the same while kinetic energy does not.

1. Common speed. The friction torques between the discs are internal, so total angular momentum is conserved. Both discs spin in the same sense, so 4 × 10 + 2 × 4 = (4 + 2) ω, which gives ω = 48/6 = 8 rad s⁻¹.
2. Initial energy. Kᵢ = 1/2(4)(10)² + 1/2(2)(4)² = 200 + 16 = 216 J.
3. Final energy. K_f = 1/2(6)(8)² = 192 J. The loss is 216 - 192 = 24 J.

   Δ K = (I₁ I₂)/2(I₁ + I₂)(ω₁ - ω₂)² = (4 × 2)/(2 × 6)(10 - 4)² = 24 J

4. So. The shortcut formula above gives the same result. The kinetic energy lost to friction is 24 J.

The idea to keep. When two spinning bodies are coupled, conserve angular momentum to find the common speed. The energy lost is (I₁ I₂)/2(I₁+I₂)(ω₁-ω₂)², just like a perfectly inelastic collision.

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