A sphere of mass 2 kg and radius 0.5 m is rolling with an initial speed of 1 ms-1 goes up an inclined plane which makes an angle of 30 ° with the horizontal plane, without slipping. How long will the sphere take to return to the starting point A? (a) 0.60 s (b) 0.52 s (c) 0.80 s (d) 0.57 s

Answer: (d). Solid sphere rolling on incline: a = (gsin θ)/(1 + 2/5) = 5/7gsin 30° = 5/7(9.8)(0.5) = 3.5 m/s². Same deceleration going up and acceleration coming down, so time up = time down = v/a = 1/3.5 s. Total time = 2/3.5 = 0.57 s. So the answer is (d) 0.57 s.

A sphere of mass 2 kg and radius 0.5 m is rolling with an initial… | JEE Main 2021 (17 Mar, Shift 2)

Q59JEE Main 2021 · 17 MAR · SHIFT 2Medium

Rolling Motion

A sphere of mass 2 kg and radius 0.5 m is rolling with an initial speed of 1 ms-1 goes up an inclined plane which makes an angle of 30 ${}^\circ$ with the horizontal plane, without slipping. How long will the sphere take to return to the starting point A?

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Answer: (d)

Solid sphere rolling on incline: $a = \dfrac{g\sin\theta}{1 + \frac{2}{5}} = \dfrac{5}{7}g\sin 30^\circ = \dfrac{5}{7}(9.8)(0.5) = 3.5\ \text{m/s}^2$ .
Same deceleration going up and acceleration coming down, so time up $=$ time down $= \dfrac{v}{a} = \dfrac{1}{3.5}$ s.
Total time $= \dfrac{2}{3.5} = 0.57$ s.
So the answer is (d) 0.57 s.

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