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To mop – clean a floor, a cleaning machine presses 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 the 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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$\frac{μ F R}{2}$

$\frac{μ F R}{3}$

$\frac{μ F R}{6}$

$\frac{2 μ F R}{3}$

answer is D.

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Detailed Solution

The force per unit area P is $P = \frac{F}{π R^{2}}$

Torque due to frictional force is

$τ = μ \int (d N) x = \frac{μ F}{π R^{2}} \int_{0}^{R} 2 π x^{2} d x \Rightarrow τ = \frac{μ F \times 2 π R^{3}}{3 π R^{2}} = \frac{2}{3} μ F R$

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