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JEE Main 2019 (11 Jan, Shift 2) · Rolling Motion

Rolling Motion question from JEE Main 2019 (11 Jan, Shift 2)

Q71JEE Main 2019 · 11 JAN · SHIFT 2Medium

Rolling Motion

A string is wound around a hollow cylinder of mass 5 kg and radius 0.5m. If the string is now pulled with a horizontal force of 40 N, and the cylinder is rolling without slipping on a horizontal surface (see figure), then the angular acceleration of the cylinder will be (Neglect the mass and thickness of the string):

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Answer: (a)

String pulled from the top: $a = \dfrac{2F}{(1+k)m}$ ; hollow cylinder $k = 1$ .
$a = \dfrac{2 \times 40}{2 \times 5} = 8\ \text{m/s}^2$ .
$\alpha = \dfrac{a}{R} = \dfrac{8}{0.5} = 16\ \text{rad/s}^2$ .
So the answer is (a) 16 rad/s $^2$ .

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