Frictionless sphere, falling particle

[professor]

Homework Statement

A particle initially sits on top of a large smooth sphere of radius R. The
particle begins to slide down the sphere without friction. At what angle .
does the particle lose contact with the sphere?

Homework Equations

g=9.8m/s₂ perhaps? (Is this dependent on the acceleration?)

The Attempt at a Solution

It seems like the angle that the particle loses contact with the sphere is possibly dependent on the acceleration of the particle. If so, I don't know how to mathematically explain this relationship. If the angle is equal to x, sinx will give a tangent vector of a magnitude proportional to the size of the angle, but I don't know if that gets me anywhere.

My second guess is that the particle either loses contact with the sphere at 90 degrees, or 45 degrees. I know that the particle cannot possibly hold on to the sphere without an adhesive property after 90 degrees. I do not know if a particle would be able to stay on the sphere past 45 degrees though.

 

[LowlyPion]

Homework Statement

A particle initially sits on top of a large smooth sphere of radius R. The
particle begins to slide down the sphere without friction. At what angle .
does the particle lose contact with the sphere?

Homework Equations

g=9.8m/s₂ perhaps? (Is this dependent on the acceleration?)

The Attempt at a Solution

It seems like the angle that the particle loses contact with the sphere is possibly dependent on the acceleration of the particle. If so, I don't know how to mathematically explain this relationship. If the angle is equal to x, sinx will give a tangent vector of a magnitude proportional to the size of the angle, but I don't know if that gets me anywhere.

My second guess is that the particle either loses contact with the sphere at 90 degrees, or 45 degrees. I know that the particle cannot possibly hold on to the sphere without an adhesive property after 90 degrees. I do not know if a particle would be able to stay on the sphere past 45 degrees though.

As the particle slides down it will gain speed according to mgh being converted to kinetic energy. You may recognize that h is a function of θ.
You may also want to recognize that when mv²/R exceeds a certain point that it will lose contact with the sphere.

 

[professor]

Thanks, I got it now, the angle is the arccos of 2/3. I just used conservation of energy for the particle, and get mgh+mv^2/r =mgh+mv^2/r, then solved fo V^2 later by equating centripetal force with mgcos(x) so the net force would be 0, where the particle fell off of the sphere.