A sphere of radius 'a' and mass 'm' rolls along a horizontal plane with constant speed v0. It encounters an inclined plane at angle θ and climbs upward. Assuming that it rolls without slipping, how far up the sphere will travel? (a) v₀²2gsin θ (b) 7v₀²10gsin θ (c) 2/5v₀²gsin θ (d) v₀²5gsin θ

Answer: (b). KE of rolling solid sphere = 1/2mv₀²(1+2/5) = 7/10mv₀². mgh = 7/10mv₀² ⇒ h = (7v₀²)/(10g). Distance along incline s = (h)/(sin θ) = (7v₀²)/(10gsin θ). So the answer is (b).

A sphere of radius 'a' and mass 'm' rolls along a horizontal plane… | JEE Main 2021 (25 Feb, Shift 2)

Q60JEE Main 2021 · 25 FEB · SHIFT 2Medium

Rolling Motion

A sphere of radius 'a' and mass 'm' rolls along a horizontal plane with constant speed v0. It encounters an inclined plane at angle $\theta$ and climbs upward. Assuming that it rolls without slipping, how far up the sphere will travel?

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Answer: (b)

KE of rolling solid sphere $= \frac{1}{2}mv_0^2\left(1+\frac{2}{5}\right) = \frac{7}{10}mv_0^2$ .
$mgh = \frac{7}{10}mv_0^2 \Rightarrow h = \dfrac{7v_0^2}{10g}$ .
Distance along incline $s = \dfrac{h}{\sin\theta} = \dfrac{7v_0^2}{10g\sin\theta}$ .
So the answer is (b).

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