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When we consider a person on a ferris wheel, the forces acting on him at the top are the normal force and gravity. The acceleration of the person is downward (towards the center), so by Newton's second law we have (for the $y$ components of the respective forces) $$n - mg = -m \frac {v^2} r \Leftrightarrow n = m\bigg(g - \frac{v^2}{r}\bigg)$$ Now when $\frac {v^2}r > g$ then $n$ becomes negatives, and supposedly you would then need a downward force (such as a seatbelt) to keep the passenger in the seat. I can't see how this follows, when $n$ is negative does the normal force even act on the person (since it's now pointing downward)?


marked as duplicate by sammy gerbil, stafusa, user191954, ZeroTheHero, Kyle Kanos Sep 27 '18 at 10:05

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  • $\begingroup$ In a typical Ferris wheel, the seat hangs from the rim of the wheel on a full rotation pivot. During the high-speed phase above, the seat would be constantly hanging "outward" from the rim, with a varying normal force supplied by the seat cushion... $\endgroup$ – DJohnM Sep 22 '18 at 18:37

You are correct in thinking that, mathematically, $n<0$ when $\frac{v^2}{r}>g$. However, this does not mean that we physically get a negative normal force (at least the same normal force as before). What this means is that we would need an inwardly acting force in order to maintain centripetal motion.

If we do not have a force like this (like as you mention, a seat belt), then the person would fly off of the Ferris Wheel.

Essentially $n$ is whatever force we need to maintain circular motion at the top of the circle. We can see this by starting with thinking of when $\frac{v^2}{r}<g$ and slowly increasing the speed. The normal force becomes smaller and smaller as the person begins to feel like they are being lifted out of the seat (you could describe this as an increasing centrifugal force in the rotating reference frame). When $\frac{v^2}{r}=g$ at the top, the person would feel weightless, as there is no normal force acting on the person. Once we go faster so that $\frac{v^2}{r}>g$, then $n$ becomes whatever force we need to maintain circular motion, which yuor math shows must now be directed towards the center of the circle. If we do not have this force, then $n=0$ ($n$ just being the force between the person and the seat) whenever $\frac{v^2}{r}\geq g$.

  • $\begingroup$ Thank you for your answer. I'm struggling to come to terms with the fact that the normal force suddenly "becomes something different". It feels unsatisfying to realize under certain circumstances it's not what you're told it is (in introductory physics, the component of the contact force orthogonal to the surface an object contacts), so to speak. Is there an explanation for this or what is the framework within which you can explain this "rigorously" (whatever this means in physics)? Hope I've made myself clear. Thanks! $\endgroup$ – firstsnow Sep 22 '18 at 14:55
  • $\begingroup$ It is the normal component of the contact force. Suppose the ferris wheel is sitting still. Gravity pushes him into the seat. He would penetrate into the seat, except the seat pushes back. The seat does not push hard enough to accelerate him upward. Just enough to stop him. Gravity plus the contact force add to the total force, 0. If the seat and rider are accelerated downward, the seat does not have to push as hard to prevent penetration. Gravity plus a smaller contact force add up to the total downward force. $\endgroup$ – mmesser314 Sep 22 '18 at 17:07
  • $\begingroup$ @firstsnow Physically the normal force does not become something else at the higher speed. It becomes $0$ when the wheel spins fast enough. Mathematically, $n$ becomes negative, but this does not mean the normal force physically changes. What it means is that your equation is assuming circular motion. Therefore, when you are spinning fast enough the math shows that you actually need an extra inward force. There is no more normal force at high enough speeds (you are no longer getting pushed into the seat). $\endgroup$ – Aaron Stevens Sep 22 '18 at 21:17
  • $\begingroup$ @firstsnow You can follow a similar line of reasoning starting with a very fast spinning wheel and a "seat belt" force the pulls you inward, and then slowing down until the seat belt force becomes $0$ and the normal force becomes non-zero. Let me know if you still need more explanation, and I will edit my answer accordingly when I can. $\endgroup$ – Aaron Stevens Sep 22 '18 at 21:19
  • $\begingroup$ @AaronStevens No need to, thanks! I really appreciate your efforts. $\endgroup$ – firstsnow Sep 23 '18 at 20:28

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