Question:
A solid cylinder of mass $M$ and radius $R$ rolls without slipping down an inclined plane of length $L$ and height $h$. What is the speed of its centre of mass when the cylinder reaches its bottom -
A solid cylinder of mass $M$ and radius $R$ rolls without slipping down an inclined plane of length $L$ and height $h$. What is the speed of its centre of mass when the cylinder reaches its bottom -
Updated On: Jun 11, 2026
- $\sqrt{2gh}$
- $\sqrt{\frac{3}{4}gh}$
- $\sqrt{\frac{4}{3}gh}$
- $\sqrt{4gh}$
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The Correct Option is C
Solution and Explanation
$mgh = \frac{1}{2} mv ^{2}\left(1+ K ^{2} / R ^{2}\right)$
$\Rightarrow v =\sqrt{\frac{2 gh }{\left(1+ K ^{2} / R ^{2}\right)}}$
$\Rightarrow v =\sqrt{\frac{2 gh }{\left(1+ K ^{2} / R ^{2}\right)}}$
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