Rotation Ques 11
11 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
(2019 Main, 10 Jan I)
(a) $\frac{2}{3} \mu F R$
(b) $\frac{\mu F R}{6}$
(c) $\frac{\mu F R}{3}$
(d) $\frac{\mu F R}{2}$
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Answer:
Correct Answer: 11.( a )
Solution:
Torque applied to move this strip is
$d \tau=$ Force on strip
$\times$ Perpendicular distance from the axis
$\Rightarrow d \tau=$ Force per unit area $\times$ Area of strip $\times$ Perpendicular distance from the axis.
$ =\frac{\mu F}{\pi R^2} \cdot 2 \pi x d x \cdot x \Rightarrow d \tau=\frac{2 \mu F x^2}{R^2} \cdot d x $
So, total torque to be applied on the mop is
$ \begin{aligned} \tau & =\int_{x=0}^{x=R} d \tau=\int_0^R \frac{2 \mu F x^2}{R^2} \cdot d x \\ & =\frac{2 \mu F}{R^2} \times \frac{R^3}{3}=\frac{2}{3} \mu F R(\mathrm{~N}-\mathrm{m}) \end{aligned} $
BETA
