Question

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 :

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

Answer 1

The Correct Option is A

To solve this problem, we need to calculate the torque applied by the machine on the circular mop. Torque (\( \tau \)) due to frictional force is given by:

\(\tau = \int r \cdot \text{d}F \cdot \mu \cdot \sin(\theta)\),

where \( r \) is the distance from the axis of rotation, \(\text{d}F\) is the infinitesimal force, \( \mu \) is the coefficient of friction, and \( \theta \) is the angle between \( r \) and \(\text{d}F\). Since the mop is pressed uniformly and rotates horizontally, \( \theta = 90^\circ \), making \(\sin(\theta) = 1\).

The force \( F \) is distributed uniformly over the mop, which has a circular shape. The pressure distribution is uniform, so each area element \(\text{dA}\) sustains a force \(\text{dF} = \frac{F}{\pi R^2} \cdot \text{dA}\).

The element of area of the circle in polar coordinates is \(\text{dA} = r \cdot \text{d}r \cdot \text{d}\theta\). Therefore, the infinitesimal force over this area is:

\(\text{dF} = \frac{F}{\pi R^2} \cdot r \cdot \text{d}r \cdot \text{d}\theta\).

The torque due to this infinitesimal force at a distance \( r \) is given by:

\(\text{d}\tau = r \cdot \mu \cdot \text{dF}\)

\(\text{d}\tau = \mu \cdot \frac{F}{\pi R^2} \cdot r^2 \cdot \text{d}r \cdot \text{d}\theta\).

Integrating this expression over the entire area of the circular mop, with \( r \) going from 0 to \( R \) and \( \theta \) going from 0 to \( 2\pi \):

\(\tau = \int_{\theta=0}^{2\pi} \int_{r=0}^{R} \mu \cdot \frac{F}{\pi R^2} \cdot r^2 \cdot \text{d}r \cdot \text{d}\theta\).

First, solve the inner integral:

\(\int_{r=0}^{R} r^2 \cdot \text{d}r = \left[\frac{r^3}{3}\right]_{0}^{R} = \frac{R^3}{3}\).

Substitute this back into the expression for torque:

\(\tau = \mu \cdot \frac{F}{\pi R^2} \cdot \frac{R^3}{3} \cdot \int_{\theta=0}^{2\pi} 1 \cdot \text{d}\theta\)

\(\tau = \mu \cdot \frac{F}{\pi R^2} \cdot \frac{R^3}{3} \cdot [\theta]_{0}^{2\pi} = \mu \cdot \frac{F}{\pi R^2} \cdot \frac{R^3}{3} \cdot 2\pi\)

.

Simplifying the expression:

\(\tau = \frac{2}{3} \mu F R\).

Thus, the correct answer is \(\frac{2}{3} \mu F R\).