First slide

Rolling motion

Question

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

Difficult

A

$\frac{ω_{o} μ_{k} R}{2 g}$
B

$\frac{ω_{o} g}{2 μ_{k} R}$
C

$\frac{2 ω_{0} R}{μ_{k} g}$
D

$\frac{ω_{o} R}{2 μ_{k} g}$

Solution

Acceleration produced in the centre of mass due to friction
a = $\frac{f}{M} = \frac{μ_{k} Mg}{M} = μ_{k} g$
where M is the mass of the ring ------(i)
Angular retardation produced by the torque due to friction
$α = \frac{τ}{I} = \frac{fR}{I} = \frac{μ_{k} MgR}{I} - - - - - - (ii)$
As v = u+at
$v = 0 + μ_{k} gt (u = 0) (Using (i))$
As $ω = ω_{0} + αt$
$ω = ω_{o} - \frac{μ_{K} {MgR}_{}}{I} t (Using (ii))$
For rolling without slipping
v = R $ω$
$\frac{v}{R} = ω_{o} - \frac{μ_{k} MgR}{I} t$
$\frac{μ_{k} gt}{R} = ω_{o} - \frac{μ_{k} M_{g} R}{I} t \Rightarrow \frac{μ_{k} gt}{R} [1 + \frac{{MR}^{2}}{I}] = ω_{0}$
$\frac{μ_{k} gt}{R} = \frac{ω_{o}}{1 + \frac{{MR}^{2}}{I}} \Rightarrow t = \frac{{Rω}_{0}}{μ_{k} g (1 + \frac{{MR}^{2}}{I})}$
$t = \frac{{Rω}_{o}}{μ_{k} g (1 + \frac{{MR}^{2}}{{MR}^{2}})} = \frac{{Rω}_{0}}{2 μ_{k} g}$

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