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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

Answers (1)

best_answer

 

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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