question_answer

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 u, the torque, applied by the machine on the mop is - [JEE Main Online Paper (Held On 10-Jan-2019 Morning]

A) \[\mu \,FR/2\]                                        

B) \[\mu \,FR/3\]

C) \[\mu \,FR/6\]                

D)   \[\frac{2}{3}\mu \,FR\]

Correct Answer: D

Solution :

\[dN\,=\,\frac{F}{\pi {{R}^{2}}}\times 2\pi rdr\] \[dN\,=\,\frac{2Frdr}{{{R}^{2}}}\] \[{{f}_{k}}\,\,=\,\,\mu dN\] \[{{f}_{k}}\,=\frac{2Fr}{{{R}^{_{2}}}}\,\mu dr\] \[d{{\tau }_{f}}_{k}\,=\,{{f}_{k}}\,\times r\] \[{{d}_{\tau }}_{k}\,=\,\frac{2F\mu }{{{R}^{2}}}\,{{r}^{2}}dr\] \[{{_{\tau }}_{k}}\,=\,\int\limits_{0}^{R}{\frac{2F\mu }{{{R}^{2}}}\,{{r}^{2}}dr}\] \[=\,\,\frac{2F\mu }{{{R}^{2}}}\,\frac{{{R}^{3}}}{3}\,=\,\frac{2F\mu R}{3}\] Text will counter balance the torque by friction\[(\tau k)\]             \[\therefore \,\,\,{{\tau }_{ext}}\,\,=\,\,\frac{2F\mu R}{3}\]