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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 : 

(1) $Mg\sqrt{1-\left({R-a\over R}\right)^2}$

(2) $Mg\sqrt{1-{a^2\over R^2}}$

(3) $Mg\sqrt{\left({R\over R-a}\right)-1}$

(4) $Mg{a\over R}$

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Rotational motion

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Ans. (1) $Mg\sqrt{1-\left({R-a\over R}\right)^2}$

Sol. 

$FR>mg\cos\theta R$

$F>mg\cos\theta$

$F>mg{\sqrt{R^2-(R-a)^2}\over R}\implies Mg\sqrt{1-\left({R-a\over R}\right)^2}$

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Very nice explanation
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