Question

A mass m slips along the wall of a smooth semi spherical surface of radius R. The velocity at the bottom of the surface is:

• A. $\sqrt{Rg}$
• B. $\sqrt{2Rg}$
• C. $2\sqrt{\pi }Rg$
• D. $\sqrt{\pi }Rg$

Answer 1

The correct option is B $\sqrt{2Rg}$

Given, $mass=m,radius=R$

A/C to the law of conservation of energy is equal to gain in kinetic energy.

when it slides at lower point lost of potential energy $=mgh$

$PE=mgr\Rightarrow KE=\frac{1}{2}m{v}^2$

Therefore $\frac{1}{2}m{v}^2=mgr\Rightarrow v=\sqrt{2gR}$