JEE MAIN - Physics (2019 - 10th January Morning Slot - No. 17)
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 -
$$\mu $$FR/2
$$\mu $$FR/3
$$\mu $$FR/6
$${2 \over 3}$$$$\mu $$FR
Explanation
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Consider a strip of radius x & thickness dx,
Torque due to friction on this strip.
$$\int {d\tau = \int\limits_0^R {{{x\mu F.2\pi xdx} \over {\pi {R^2}}}} } $$
$$\tau = {{2\mu F} \over {{R^2}}}.{{{R^3}} \over 3}$$
$$\tau = {{2\mu FR} \over 3}$$
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