Question

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

• A. $\frac{\mu FR}{6}$
• B. $\frac{\mu FR}{3}$
• C. $\frac{\mu FR}{2}$
• D. $\frac{2\mu FR}{3}$

Answer 1

The correct option is D

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

Step 1: Given Data

The radius of the mop $=R$

Total force $=F$

Let angular speed $=\omega$

Coefficient of friction $=\mu$

Let the area be $A$.

Let there be an elementary circle with radius $dx$.

Let the normal be $N$.

Let the frictional force be $f$.

Step 2: Formula Used

Pressure, $P=\frac{F}{A}$

Normal force $N=$Force applied $F$ because both are acting perpendicularly.

Frictional force $f=\mu N$

Torque $\tau =fx$

Step 3: Calculate the Elementary Torque

Pressure can be given as,

$P=\frac{F}{A}=\frac{F}{\pi R^2}$

From the figure, we can see that the force applied is also equal to the normal.

Force on the elementary area,

$dN=dF=PdA=\frac{F}{\pi R^2}2\mathrm{\pi xdx}$

Therefore, elementary frictional force,

$df=\mu dN$

Therefore, the torque along the $x$ radius can be given as,

$$\begin{gathered} d\tau =dfx \\ =\mu dNx \\ =\mu .\frac{F}{\pi R^2}2\mathrm{\pi xdx}.\mathrm{x} \\ =\frac{2\mathrm{\mu F}}{R^2}{\mathrm{x}}^2\mathrm{dx} \end{gathered}$$

Step 4: Calculate the Total Torque

Therefore the total torque for the entire radius $R$ can be given as,

$$\begin{gathered} \tau =\frac{2\mathrm{\mu F}}{R^2}{\int }_0^R{\mathrm{x}}^2\mathrm{dx} \\ =\frac{2\mathrm{\mu F}}{R^2}.{\overline{)\frac{x^3}{3}}}_0^R \\ =\frac{2\mathrm{\mu F}}{R^2}.\frac{R^3}{3} \\ =\frac{2\mathrm{\mu F}R}{3} \end{gathered}$$

Hence, the correct answer is option (D).