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# Please,please help me - To mop - clean a floor, a cleaning machine presses a circular mop of radius R vertically down with a - Rotational Motion - JEE Main

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 if coefficent of friction between the mop and the floor is $\mu$ , the torque, applied by the machine on the mop is:

• Option 1)

$\mu FR/3$

• Option 2)

$\frac{\mu FR}{6}$

• Option 3)

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

• Option 4)

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

Views

Torque -

$\underset{\tau }{\rightarrow}= \underset{r}{\rightarrow}\times \underset{F}{\rightarrow}$

- wherein

This can be calculated by using either  $\tau=r_{1}F\; or\; \tau=r\cdot F_{1}$

$r_{1}$ = perpendicular distance from origin to the line of force.

$F_{1}$ = component of force perpendicular to line joining force.

consider a strip of radius x and thickness dx

Toraue due to friction  on this  strip

= $=\int dt=\int_{O}^{R}x(\mu .F.)\left ( \frac{2\Pi xdx}{\Pi R^{2}} \right )\\\\\\\\\Rightarrow T=\frac{2\mu F}{R^{2}}\frac{R^{3}}{3}\\\\\\\\T=\frac{2\mu FR}{3}$

Option 1)

$\mu FR/3$

Option 2)

$\frac{\mu FR}{6}$

Option 3)

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

Option 4)

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

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