A cylinder is rolling down on an inclined plane of inclination 60°. Its acceleration during rolling down will be x/√(3) m/s², where x = ________ (use g = 10 m/s²). Numerical answer type. Work it out, then open the answer.

Answer: 10. Rolling acceleration: a = (gsin θ)/(1 + k²/R²); for a solid cylinder k²/R² = 1/2. a = 2/3gsin 60° = 2/3× 10 × √(3)/2 = (10)/(√(3)) m/s² So x = 10.

A cylinder is rolling down on an inclined plane of inclination 60°… | JEE Main 2024 (29 Jan, Shift 1)

Q27JEE Main 2024 · 29 JAN · SHIFT 1Easy

Rolling Motion

A cylinder is rolling down on an inclined plane of inclination $60^\circ$ . Its acceleration during rolling down will be $\frac{x}{\sqrt{3}}\text{ m/s}^{2}$ , where $x =$ ________ (use $g = 10\text{ m/s}^{2}$ ).

Numerical answer type. Work it out, then open the answer.

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Answer: 10

Rolling acceleration: $a = \dfrac{g\sin\theta}{1 + k^2/R^2}$ ; for a solid cylinder $k^2/R^2 = \frac{1}{2}$ .
$a = \frac{2}{3}g\sin 60^\circ = \frac{2}{3}\times 10 \times \frac{\sqrt{3}}{2} = \frac{10}{\sqrt{3}}\text{ m/s}^2$
So $x = 10$ .

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