What is the height at which the small object loses contact with the spheroid?

[vabamyyr]

hi ppl! I have a problem in physics, if u could help me then id be glad.

There is a spheroidal object with radius R. on top of it there is a small object. And the object starts to move (fall) on the spheroid. There is no friction between the spheroid and small object. The question is on what height does the small object leave the spheroid (loses physical contact).

 

[airbuzz]

you must write the dynamic equation of the ball on the spheroid dividing normal and parallel forces to the surface. When the centrifugal acceleration equals the normal force to the surface made by the normal component of the weight of the ball then the small object leaves the spheroid

 

[siddharth]

Let the velocity at time it falls of be v. Let the angle the radius vector makes from the center of the spheroid to the vertical be alpha.
Use energy conservation to find the velocity of the object as it leaves
it will be
$$v= \sqrt{2gR(1- \cos \alpha)}$$
Now, the Force along the radius vector to the object is $$\frac{mv^2}{R}$$
This must be equal to force towards center of sphere due to its weight and the force away from the center due to the Normal Reaction.
ie,
$$mg \cos \alpha - N= \frac{mv^2}{R}$$
When the object does not touch anymore, N=0. Using this find the angle alpha and thus the height.

 

[ramollari]

I had solved this problem a couple of times a few years ago, but I just didn't recall the condition that the normal force has to be zero when the ball leaves the surface.

since (mv^2)/R = 2KE/R and KE = mgR(1-cosa)
then,
mgcosa - N = 2mg(1-cosa)

when N = 0:
3mgcosa = 2mg
cosa = 2/3
from which results that h = R/3.