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JEE Main 2015 (10 Apr, Shift 1) · Rolling Motion

Rolling Motion question from JEE Main 2015 (10 Apr, Shift 1)

Q75JEE Main 2015 · 10 APR · SHIFT 1Medium

Rolling Motion

A uniform solid cylindrical roller of mass 'm' is being pulled on a horizontal surface with force F parallel to the surface and applied at its centre. If the acceleration of the cylinder is 'a' and it is rolling without slipping then the value of 'F' is:

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Answer: (c)

$F - f = ma$ and $fR = \frac{1}{2}mR^2\cdot\dfrac{a}{R} \Rightarrow f = \frac{1}{2}ma$ .
$F = ma + \frac{1}{2}ma = \frac{3}{2}ma$ .
So the answer is (c) $\frac{3}{2}ma$ .

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