Question

a mass m slips along the wall of semispherical surface of radius r. the velocity at the bottom of the surface is

Answer 1

ANSWER

Let r be the radius of the bowl.

According to the law of conservation of energy,

loss is potential energy= gain in kinetic energy.

When bowl moves from rim at lowest point,

loss in potential energy= mgH

where H is the height which is equal to the radius= r

Potential energy=mgr

Kinetic Energy=

$\frac{1}{2} mv {}^{2}$

therefore,

$= > \frac{1}{2} mv {}^{2} = mgr \\ = > v = \sqrt{2gr}$

HOPE it helps ✔️