A ring of radius R is first rotated with an angular velocity $ω_{0}$ and then carefully placed on a rough horizontal surface. The coefficient of friction between the surface and the ring is $μ$ . Time after which its angular speed is reduced to half is

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$\frac{ω_{0} μ R}{2 g}$

$\frac{ω_{0} g}{2 μ R}$

$\frac{2 ω_{0} R}{μ g}$

$\frac{ω_{0} R}{2 μ g}$

answer is D.

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Detailed Solution

$Taking τ about CM$

$\begin{array}{l} μ m g R = M R^{2} α \\ μ g = R α \\ α = \frac{μ g}{R} \end{array}$

$\begin{array}{l} ω = ω_{0} - \frac{μ g}{R} t = \frac{ω_{0}}{2} \\ \Rightarrow \frac{ω_{0}}{2} = \frac{μ g}{R} t \\ \Rightarrow t = \frac{ω_{0} R}{2 μ g} \end{array}$

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