Three bodies , a ring, a solid cylinder and a | vedclass

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Question

According to conservation of mechanical

energy

$\mathrm{mgh}=\frac{1}{2} \mathrm{mv}^{2}\left(1+\frac{\mathrm{k}^{2}}{\mathrm{R}^{2}}\right)$

Vedclass · Accepted Answer

Solid sphere

Vedclass · Answer

According to conservation of mechanical

energy

$\mathrm{mgh}=\frac{1}{2} \mathrm{mv}^{2}\left(1+\frac{\mathrm{k}^{2}}{\mathrm{R}^{2}}ight)$

or $v^{2}=\left(\frac{2 g h}{1+\frac{k^{2}}{R^{2}}}ight)$

$Note $$:$ It is independent of the mass of the rolling body.

For a ring. $\mathrm{K}^{-2}=\mathrm{R}^{2}$

$v_{ring}=\sqrt{\frac{2 g h}{1+1}}=\sqrt{g h}$

For a solid cylinder, $\mathrm{k}^{-2}=\frac{\mathrm{R}^{2}}{2}$

$V_{	ext {cylinder }}=\sqrt{\frac{2 \mathrm{gh}}{1+\frac{1}{2}}}=\sqrt{\frac{4 \mathrm{gh}}{3}}$

For a solid sphere, $\mathrm{k}^{-2}=\frac{2}{5} \mathrm{R}^{2}$

$V_{	ext {sphere }}=\sqrt{\frac{2 g h}{1+\frac{2}{5}}}=\sqrt{\frac{10 g h}{7}}$

Among the given three bodies the solid sphere has the greatest and the ring has the least velocity at the bottom of the inclined plane.

Three bodies , a ring, a solid cylinder and a solid sphere roll down the same inclined plane without slipping. They start from rest. The radii of the bodies are identical. Which of the bodies reaches the ground with maximum velocity?

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