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NEET 1992 (AIPMT) · Rolling Motion

Rolling Motion question from NEET 1992 (AIPMT)

Q93NEET 1992 · AIPMTEasy

Rolling Motion

The speed of a homogenous solid sphere after rolling down an inclined plane of vertical height $h$ form rest without sliding is

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Answer: (a)

$mgh = \frac{1}{2}mv^2\left(1 + \frac{k^2}{R^2}\right)$ with $\frac{k^2}{R^2} = \frac{2}{5}$ for a solid sphere.
$mgh = \frac{7}{10}mv^2 \Rightarrow v = \sqrt{\frac{10}{7}gh}$
So the answer is (a) $\sqrt{\frac{10}{7}gh}$ .

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