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 o…

Question

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

Accepted Answer

(3) (I₁I₂)/(2(I₁+I₂))(ω₁-ω₂)². No external torque acts about the common axis, so angular momentum is conserved while the faces rub to a common speed. I₁ω₁+I₂ω₂=(I₁+I₂)ω, so ω=(I₁ω₁+I₂ω₂)/(I₁+I₂). Initial kinetic energy: Kᵢ=1/2I₁ω₁²+1/2I₂ω₂². Final kinetic energy: K_f=1/2(I₁+I₂)ω²=((I₁ω₁+I₂ω₂)²)/(2(I₁+I₂)). Subtracting and collecting terms, the cross terms combine into a perfect square: Kᵢ-K_f=(I₁I₂)/(2(I₁+I₂))(ω₁-ω₂)². Two checks. The loss is never negative, as friction requires. And if ω₁=ω₂ there is no rubbing and no loss at all.

Step-by-step solution

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