Rotational Motion Question 5
Question 5 - 2024 (29 Jan Shift 1)
A cylinder is rolling down on an inclined plane of inclination $60^{\circ}$. It’s acceleration during rolling down will be $\frac{\mathrm{x}}{\sqrt{3}} \mathrm{~m} / \mathrm{s}^{2}$, where $\mathrm{x}=$ (use $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}$ ).
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Answer: (10)
Solution:
For rolling motion, $a=\frac{g \sin \theta}{1+\frac{I_{c m}}{\mathrm{MR}^{2}}}$
$a=\frac{g \sin \theta}{1+\frac{1}{2}}$
$=\frac{2 \times 10 \times \frac{\sqrt{3}}{2}}{3}$
$=\frac{10}{\sqrt{3}}$
Therefore $\mathrm{x}=10$
