A uniform cylinder of mass M and radius R is | vedclass

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(D) To pull the cylinder over the step,the torque produced by the applied force $F$ about the edge of the step must be greater than or equal to the to

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$M g \sqrt{1-\left(\frac{R-a}{R}\right)^{2}}$

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(D) To pull the cylinder over the step,the torque produced by the applied force $F$ about the edge of the step must be greater than or equal to the torque produced by the gravitational force $Mg$ about the same edge.Let the edge of the step be point $P$. The perpendicular distance from $P$ to the line of action of force $F$ is $R$.The perpendicular distance from $P$ to the line of action of the weight $Mg$ is $x$.From the geometry of the cylinder,we have a right-angled triangle with hypotenuse $R$ and one side $(R-a)$. Thus,$x = \sqrt{R^2 - (R-a)^2}$.For the cylinder to just start moving over the step,the torques must balance:$F 	imes R = Mg 	imes x$$F 	imes R = Mg 	imes \sqrt{R^2 - (R-a)^2}$$F = \frac{Mg}{R} \sqrt{R^2 - (R-a)^2}$$F = Mg \sqrt{\frac{R^2 - (R-a)^2}{R^2}}$$F = Mg \sqrt{1 - \left(\frac{R-a}{R}ight)^2}$

$A$ uniform cylinder of mass $M$ and radius $R$ is to be pulled over a step of height $a$ $(a < R)$ by applying a force $F$ at its centre $'O'$ perpendicular to the plane through the axes of the cylinder on the edge of the step (see figure). The minimum value of $F$ required is

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