Posted by **Anonymous** on Saturday, April 17, 2010 at 9:53am.

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?

- physics -
**drwls**, Saturday, April 17, 2010 at 10:49am
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

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