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 -

Rotational Motion PYQ — JEE Main 2019 · 10 Jan · Shift 1 · Q60 with Solution

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Correct answer: D

Radius of circular mop $= R$
Total downward force on mop $= F$
Coefficient of friction $= \mu$
Force is uniformly distributed over the circular area
Mop rotates with constant angular speed, so applied torque must balance frictional torque.

Since force is uniformly distributed over the circular area,

\text{Area} = \pi R^2

So pressure,

p = \frac{F}{\pi R^2}

Take an elemental ring of radius $r$ and thickness $dr$ .

Its area is

dA = 2\pi r\,dr

Normal force on this ring:

dN = p\,dA = \frac{F}{\pi R^2}(2\pi r\,dr)=\frac{2Fr}{R^2}dr

Friction on this ring is

df = \mu\, dN = \mu \frac{2Fr}{R^2}dr

This friction acts tangentially and opposes rotation.

Torque of this friction force about the axis is

d\tau = r\,df

So,

d\tau = r \left(\mu \frac{2Fr}{R^2}dr\right) = \mu \frac{2F}{R^2} r^2 dr

\tau = \int_0^R d\tau = \mu \frac{2F}{R^2} \int_0^R r^2 dr

Now,

\int_0^R r^2 dr = \frac{R^3}{3}

Therefore,

\tau = \mu \frac{2F}{R^2} \cdot \frac{R^3}{3} = \frac{2}{3}\mu FR

The torque applied by the machine is

\boxed{\frac{2}{3}\mu FR}

So the correct option is D.