Question

A particle of mass m is released from rest at the top of a spherical dome of radius R.
how far below the starting point will the particle leave the surface of the dome?
how i should solve this?

Answer 1

They need to say whether the particle slides or rolls. If it slides, you have to know the friction coefficient. If it rolls, you have to know if it is a solid or hollow sphere.

If it slides without friction, we can do the problem. After descending a vertical distance H, it will have acquired a speed
V = sqrt(2gH)

Let A be the angle than it has cescended, measured from the center of the sphere

It leaves the sphere when the componemt of its weight normal to the sphere, M g cos A, is equal to the centripetal force required to make it follow the circular trajectory. When this happens, the sphere no longer needs to apply a reaction force to the particle to keep it there, and the particle leaves the surface.

So require that
M g cos A = M V^2/R = M *2g H/R

cos A = 2H/R

Geometry also tells you that
H = R (1-cos A)

Therefore
cos A = (1-cos A)
cos A = 1/2

A = 60 degrees