A ring of radius R is rotating with an angular speed ${\mathrm{\omega }}_0$ about a horizontal axis. It is placed on a rough horizontal table. The coefficient of kinetic friction is ${\mathrm{\mu }}_{\mathrm{k}}$. The time after which it starts rolling is

detailed solution

Correct option is D

Acceleration produced in the centre of mass due to frictiona = fM = μkMgM = μkgwhere M is the mass of the ring ------(i)Angular retardation produced by the torque due to frictionα = τI= fRI= μkMgRI------(ii)As v = u+atv = 0 + μk gt (u = 0)     (Using (i))As ω = ω0+αtω = ωo-μKMgR It      (Using (ii))For rolling without slippingv = RωvR = ωo-μkMgRItμkgtR = ωo-μkMgRIt   ⇒  μkgtR[1+MR2I] = ω0μkgtR = ωo1+MR2I⇒t = Rω0μkg(1+MR2I) t = Rωoμkg(1+MR2MR2) = Rω02μkg