To mop-clean a floor, a cleaning machine a circular mop of radius R vertically down with a total force F and rotates it with a constant angular speed about its axis If the force F is distributed uniformly over a mop and if coefficient of friction between the mop and the floor is $μ$ . The torque, applied by the machine on the mop is

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$\frac{μ F R}{3}$

$\frac{μ F R}{6}$

$\frac{μ F R}{2}$

$\frac{2}{3} μ F R$

answer is D.

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Detailed Solution

Force on the small area element $\frac{F}{π R^{2}} (d A)$

Frictional torque on then elements area $d T = μ (\frac{F}{π R^{2}}) d A r = r μ (\frac{F}{π R^{2}}) 2 πr d r$

On integrating $T = \frac{μ F}{π R^{2}} 2 π \frac{R^{3}}{3} = \frac{2 μ F R}{3}$

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