Assume we are provided with a perfectly spherical and frictionless ball of radius $R$. A point is placed on the top of the ball. A velocity $u$ is imparted on the ball to one side of the ball. How can we determine the angle subtended from the centre at which the ball will leave the surface of the ball as it rolls down?
A point is placed on the top of the ball.
For a frictionless ball/particle hypotheis, there is only one point on the sphere in the gravitational field of the earth where the particle is stable, and it is a metastable point, defined as a point on the sphere being in a higher potential energy than all the neighboring points. A particle sitting on this point just needs a random impulse in any direction to fall to a lower state. In the random impulse case with no friction the angle in the horizontal plane is indeterminate because very small impulses (brownian motionn of air for example) can start the fall. The angle of leaving the sphere should be very small as it will be sliding off an incline. It will depend on the random impulse.
A velocity u is imparted on the ball to one side of the ball. How can we determine the angle subtended from the centre at which the ball(particle) will leave the surface of the ball as it rolls down?
It will not roll down because the ball will leave underneath the particle. If it is friction less still, the particle will stay put in the air before falling . With respect to the ball it is moving with momentum -p (the momentum of the ball).
If there is friction between particle and ball and u is very slow with respect to the acceleration the particle would get from gravity, it will roll down in the direction of -p , the angle of losing contact will depend on all the parameters, including friction which is the only thing keeping it connected to the ball. For a very sticky particle the angle would be 90degrees from the vertical, where only gravity would be acting.