Question

A uniform cylinder isrotating aboutis own axis with a constant angular speed ω. Very gently it is placed on a rough horizontal surface at t₁ . At t = t₂ it starts rolling without slipping. The rotational kinetic energy of cylinder at t₁ , t₂ and t₃ (>t₂ ) is K₁ , K₂ and K₃ respectively. Then

(1) K₁ = K₂ = K₃

(2) K₁ = K₂ ; K₂ = K₃

(3) K₁ > K₂ > K₃

(4) K₂ > K1 > K₃

Answer 1

(2) K₁ = K₂ ; K₂ = K₃

When rotating cylinder is gently placed on rough horizontal surface at t = t₁ ; it initally slips. But at t = t₂ , slipping stops. There is loss of energy due to friction between horizontal surface and cylinder i.e K₂ < t₂ ; the cylinder moves with speed it has accquired and its energy remains same as at t = t₂ i.e K₂ = K₃ .