Question

A ring of radius $R$ is rotating with an angular speed $\omega_{0}$ about a horizontal axis. It is placed on a rough horizontal table. The coefficient of kinetic friction is $\mu_{k}$. The time after which it starts rolling is:

• A. $\frac{\omega_{0} \mu_{k} R}{2 g}$
• B. $\frac{\omega_{0} R}{2 \mu_{k} R}$
• C. $\frac{2 \omega_{0} R}{\mu_{k} g}$
• D. $\frac{\omega_{0} R}{2 \mu_{k} g}$

Answer 1

Answer: (D)

Acceleration produced in the centre of mass due to friction
$a =\frac{ f }{ M }=\frac{\mu_{ k } Mg }{ M }=\mu_{ k } g \dots(i) $
Angular retardation produced by the torque due to friction
$\alpha=\frac{\tau}{I}=\frac{f R}{I}=\frac{\mu_{k} M g R}{I}=\frac{\mu_{k} g}{R}\, \dots(ii)$
$\therefore v =0+\mu_{ k } gt (\therefore u =0)$ $\quad$ (Using (i))
$\therefore \omega=\omega_{0}-\frac{\mu_{k} g t}{R}$ $\quad$ ( Using (ii))
For rolling without slipping
$v = R \omega$
$\frac{\mu_{k} g t}{R}=\omega_{0}-\frac{\mu_{k} g t}{R}$
$\Rightarrow t =\frac{ R \omega_{0}}{2 \mu_{ k } g }$