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 $\mu$, the torque, applied by the machine on the mop is:
Answer: (A) $\dfrac{2\mu FR}{3}$
At constant angular speed the machine's torque equals the friction torque. The normal force on a ring of radius $r$ and width $dr$ is $\dfrac{F}{\pi R^2}\,2\pi r\,dr$, so the friction torque on it is
$$d\tau = \mu\frac{F}{\pi R^2}\,2\pi r\,dr\cdot r = \frac{2\mu F}{R^2}r^2\,dr$$
$$\tau = \frac{2\mu F}{R^2}\cdot\frac{R^3}{3} = \frac{2\mu FR}{3}$$
Solution by Sreeraj P, M.Sc Physics