A hemispherical bowl or radius $R$ is kept on a horizontal table. A small sphere of radius $r(r \ll R)$ is placed at the highest point at the inside of the bowl and let go. The sphere rolls without slipping. Its velocity at the lowest point is

Question

Dr. Abubacker · Accepted Answer

(d) As is clear from figure,
[image:7_image146.emf]
On reaching the bottom of the bowl, loss in PE=$mgR$, and
Gain in KE$= \frac{1}{2}mv^{2} + \frac{1}{2}I\omega^{2}$
$= \frac{1}{2}mv^{2} + \frac{1}{2} 	imes \left( \frac{2}{5}mr^{2} ight)\omega^{2}$
$= \frac{1}{2}mv^{2} + \frac{1}{5}mv^{2} = \frac{7}{10}mv^{2}$
As again in KE =loss in PE
$	herefore\ \frac{7}{10}mv^{2} = mgR$
$v = \sqrt{\frac{10gR}{7}}$

Q. A hemispherical bowl or radius $R$ is kept on a horizontal table. A small sphere of radius $r(r \ll R)$ is placed at the highest point at the inside of the bowl and let go. The sphere rolls without slipping. Its velocity at the lowest point is

Explanation

Reviewed by Dr. Abubacker

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