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I am grateful for any help offered:

Ignoring friction (and hence rolling), what are the forces in play on the ball in the middle frame??

3 force diagram frames of a ball meeting the floor after travelling along a slope

I am attempting to mirror this diagram to make a half hexagonal shape. With this I should be able to visualise what would happen when the slope becomes more and more curved until it becomes a circle. This is an attempt to explain to myself how the normal force at point C exists (see below) in the first place.

How does the normal force exist in centripetal motion?

"A normal force will always be only as much as is needed to prevent the two object from occupying the same space." Since there is no force directed at the track at the exact point C, there should be nothing making them occupy the same space, and thus no force. I know that circular motion means there must be some centripetal force, but there is no force for the track to react to, so it simply cannot push the ball!

Some answers on the internet mention the centrifugal force, however all my teachers have strongly instructed I ignore any mention of the centrifugal force. Instead, my teacher's response to this question is that the ball is in fact exerting a kind of outward force since its momentum is tangential and the track is attempting to change its momentum.

Investigating his response lead me to draw the above diagram to explain it to myself and to this question.

[EDIT] Additionally, this dialogue below may help anyone with a similar query regarding how the track could "know" that in the next instant the ball will hit the next molecule of the track, and that it needs to act pre-emptively and exert a normal force to prevent this.Explanation of "Pre-emptive nature" of the normal force in Circular Motion

ThePhysicsOverthinker
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"A normal force will always be only as much as is needed to prevent the two object from occupying the same space."

That is true. But that does not mean that $N=mg$ always. Let us consider a hypothetical situation when the normal force is not acting (as you have assumed). Then, since no force is acting, the velocity will remain constant and tangential to the circular track.

This means that the very next instant, the object collapses into the track, which is precisely what should not happen. So, the only plausible conclusion is that the normal force is pushing the ball inward towards the centre to prevent the ball from going into the track.

So, there is no other force, but just the normal force $N$, where, $$N= \frac{mv^2}{R}$$

As you can see from the hexagon you have drawn (which is a highly efficient way to clear a problem involving circles, used by Archimedes once), there are two points of contact. The sum of the normal forces, will give a resultant that acts perpendicular to the tangent drawn at the point of contact, as in the illustration below

enter image description here

Elendil
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  • Thanks for your help. I see what you mean, however, how does the track at C know the future? How does it know that in the next instant it is going to hit the track? Additionally, at that very instant, the ball isn't pushing into the track and thus the track cannot push back even if it knew the future? – ThePhysicsOverthinker May 19 '20 at 15:20
  • @ThePhysicsOverthinker, the object is touching the part it is about to collapse into rather than the part it is at now. In the wonderful diagram of the hexagon, you can see that the contact points are on the side, not on the edges. If you try to figure out the net force due to the points on the sides, you will find them directed away from the edge, though the edge is not providing the normal force at all – Elendil May 19 '20 at 15:35
  • I see! That's why I asked this question to see if anyone could show me how the forces end up when there are 2 contact points. Any chance you could edit a diagram into your answer so I can see how the net force would be inward towards the centre? I would be very grateful. – ThePhysicsOverthinker May 19 '20 at 16:04
  • Sorry just saw your edit. Ok, so weight acts downwards at both contact points. Thus the net force on the track is radial. Additionally on the left contact point, the floor experiences more force as the ball's momentum is changed. This therefore is responsible for the Normal Force > mg. Am I right?! – ThePhysicsOverthinker May 19 '20 at 17:58
  • @ThePhysicsOverthinker, Yes. – Elendil May 19 '20 at 18:18