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If you know the acceleration, then Newton's Second Law does NOT tell us all the different individual physical forces acting on the particle; it only tells us the RESULTANT force.

EDIT: if you disagree, then it's really just semantics about what "force" means. There is no actual physical force pushing outwards. The fact that you "feel" a force doesn't make you correct; it just demonstrates that some physical measurements do not give complete information about the system in question. See Does centrifugal force exist?

If you know the acceleration, then Newton's Second Law does NOT tell us all the different individual physical forces acting on the particle; it only tells us the RESULTANT force.

EDIT: to be clear: any complete, quantitative answer MUST allow the rotation speed to vary, and so the usual calculation of centripetal acceleration for constant speed above does NOT apply. This is why I say that you cannot possibly understand this properly without at least the concept of differential equations with vector variables. However, if you DO know about these mathematical topics, and Newton's Second Law, then no other knowledge is required to solve this - but it's not particularly easy. It's a bit like trying to understand constant acceleration linear motion without calculus: even though it's a very special case of the general equations, there is simply NO correct way to understand it properly without calculus [or something mathematically equivalent in the special case], so if you're serious then you just have to learn it.

Sorry, no diagram, but I still hope this will be helpful.

The terminology "centrifugal force" never makes sense, to me, and only causes confusion. We should throw out the whole idea of "centrifugal force", and use only centripetal acceleration (NOT force!), combined with Newton's Second Law.

If a particle moves in a circle at constant speed, its acceleration (not force!) is directed towards the centre of the circle, by differentiating the position vector twice with respect to time.

So, if you assume that a particle is moving in a circle (at constant speed), then this tells us the acceleration. (By the way, if the speed is varying also, then the acceleration is almost always NOT towards the centre! EDIT: yes, I really do mean "almost" here, because the acceleration vector can instantaneously point towards the centre in some cases. You can see this by differentiating exp(if) for a general real-valued function f.)

Therefore, by Newton's Second Law, the RESULTANT force on the particle must also be towards the centre of the circle. Note that there are only TWO forces acting on the ball: gravity (pointing down) and the normal (reaction) force (pointing perpendicularly away from the surface pushing against the balls). THERE IS NO CENTRIFUGAL FORCE!

If you know the acceleration, then Newton's Second Law does NOT tell us all the different individual physical forces acting on the particle; it only tells us the RESULTANT force.

For example, if you spin a ball around on the end of a rope, then the physical forces are different in nature to this example, even though the acceleration is the same. You could also have magnetic fields etc. etc., but there would be no way to tell simply from the motion of the ball what forces are acting.

For a complete answer as to why the balls move upwards as you increase the rotation frequency, you have to write down the equations for the normal force, depending on the height of the balls. Assume first that the balls are instantaneously at exactly the correct rotation speed to remain at a fixed height, and consider what happens if the speed is changed by a small amount (i.e., a small perturbation from a system at equilibrium).

Since normal force is limited by the component of the ball's gravity pressing against the surface, it's not possible for the vertical component of the normal force to counteract the gravitational force fully if the rotation is too slow, so the ball must move downward.

In the other direction, if the rotation is too fast, the horizontal component of the normal force (to provide the centripetal acceleration) must be so great that the upwards vertical component of the normal force is greater than gravity, so the ball will tend to move upwards.

Of course we are neglecting friction and air resistance, and possible rotation of the balls themselves in addition to the circular motion (i.e., the ball is not a particle).

For a really good answer (even for a particle without friction/air resistance), you'd have to write down an equation for the rotational frequency, as a function of time, and solve some differential equations for the motion of the ball, but probably you don't want to do this. But strictly speaking, I don't think there's any simpler way to solve it properly - no amount of diagrams and geometry will give you the full answer without calculus.

Sorry, no diagram, but I still hope this will be helpful. As people have pointed out, this is maybe not a full answer, so you can regard this as hints for a fun exercise.

The terminology "centrifugal force" never makes sense, to me, and only causes confusion. We should throw out the whole idea of "centrifugal force", and use only centripetal acceleration (NOT force!), combined with Newton's Second Law.

If a particle moves in a circle at constant speed, its acceleration (not force!) is directed towards the centre of the circle, by differentiating the position vector twice with respect to time.

So, if you assume that a particle is moving in a circle (at constant speed), then this tells us the acceleration. (By the way, if the speed is varying also, then the acceleration is almost always NOT towards the centre! EDIT: yes, I really do mean "almost" here, because the acceleration vector can instantaneously point towards the centre in some cases. You can see this by differentiating exp(if) for a general real-valued function f.)

Therefore, by Newton's Second Law, the RESULTANT force on the particle must also be towards the centre of the circle. Note that there are only TWO forces acting on the ball: gravity (pointing down) and the normal (reaction) force (pointing perpendicularly away from the surface pushing against the balls). THERE IS NO CENTRIFUGAL FORCE!

If you know the acceleration, then Newton's Second Law does NOT tell us all the different individual physical forces acting on the particle; it only tells us the RESULTANT force.

For example, if you spin a ball around on the end of a rope, then the physical forces are different in nature to this example, even though the acceleration is the same. You could also have magnetic fields etc. etc., but there would be no way to tell simply from the motion of the ball what forces are acting.

For a complete answer as to why the balls move upwards as you increase the rotation frequency, you have to write down the equations for the normal force, depending on the height of the balls. Assume first that the balls are instantaneously at exactly the correct rotation speed to remain at a fixed height, and consider what happens if the speed is changed by a small amount (i.e., a small perturbation from a system at equilibrium).

Since normal force is limited by the component of the ball's gravity pressing against the surface, it's not possible for the vertical component of the normal force to counteract the gravitational force fully if the rotation is too slow, so the ball must move downward.

In the other direction, if the rotation is too fast, the horizontal component of the normal force (to provide the centripetal acceleration) must be so great that the upwards vertical component of the normal force is greater than gravity, so the ball will tend to move upwards.

Of course we are neglecting friction and air resistance, and possible rotation of the balls themselves in addition to the circular motion (i.e., the ball is not a particle).

For a really good answer (even for a particle without friction/air resistance), you'd have to write down an equation for the rotational frequency, as a function of time, and solve some differential equations for the motion of the ball, but probably you don't want to do this. But strictly speaking, I don't think there's any simpler way to solve it properly - no amount of diagrams and geometry will give you the full answer without calculus.