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Question: A mass \(m\) slips along the wall of a semispherical surface of radius \(R\). The velocity at the bo...

A mass mm slips along the wall of a semispherical surface of radius RR. The velocity at the bottom of the surface is

A

Rg\sqrt{Rg}

B

2Rg\sqrt{2Rg}

C

2πRg2\sqrt{\pi Rg}

D

πRg\sqrt{\pi Rg}

Answer

2Rg\sqrt{2Rg}

Explanation

Solution

By applying law of conservation of energy

mgR=12mv2⇒v=2RgmgR = \frac{1}{2}mv^{2} \Rightarrow v = \sqrt{2Rg}