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Gravitron with greased walls - does the rider stay pinned or slide

  1. Dec 3, 2008 #1
    A gravitron is moving at great speed. When the FLOOR drops away (greased to make friction negligible), does the person stay pinned to the wall or does the person slide to the bottom?

    If the walls were not greased, we know that centrifugal force pushes an individual against the wall, and the reaction force of the wall balances this force so he is at rest with respect to the rotating frame, so that he stays pinned to the wall. The only force to balance the downward gravitional weight force is the upward frictional force between a rider and the wall. So, if the walls were greased--thereby taking away frictional force-- is there any other force/variable that keeps the rider pinned to the wall instead of sliding down? I was thinking that when a gravitron accelerates to a high enough velocity, it can escape gravity. At the gravitron speed, the riders are experiencing centripetal force equivalent to four times the force of gravity. I am wondering why this is (does the acceleration affect the weight/mass and if so, how?) and if, without the frictional force, this speed would still cause the rider to overcome gravity and stay pinned?
    Last edited: Dec 3, 2008
  2. jcsd
  3. Dec 3, 2008 #2


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    You are asking yourself great questions. Unfortunately, at this level, you have to make some boring assumptions, and those assumptions are called "freshman physics". Well, anyway, you should assume that the mass and the weight force vector are independent of the fact that the object is spinning. You nailed it in the second and third sentences of your suggestion, at least for freshman physics' sake.

    However, just as a side note, the Gravitron that I remember had slanted sides so that the patrons would actually slide up the tapered walls away from the center. What would you think about this case?
  4. Dec 3, 2008 #3
    I've seen gravitrons with vertical walls so tapered walls shed a whole new arena of possibilities. I am really not sure-- depends on the extent to which the velocity is escaping the gravity???

    What I feel is really stumping me in my endeavors for an answer here is the need for some fundamental, CAUSAL explanations on how gravity and horizontal acceleration in general is working in relations to each other. For example, how does the magnitude of the speed allow for the break from gravity, and what determines the point at which it would break free? How exactly does this work when the gravitron speed overcomes the gravity? What is happening when it breaks free? Yes, this is quite elementary physics, and I've tried going over some fundamental physic laws and research but without the causal understandings, I am stumped in making any applications to the gravitron scenario.
  5. Dec 3, 2008 #4


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    For the sake of freshman physics, horizontal and vertical vectors have no effect on each other. (However, they can combine in the cross-product to produce a new third vector). There is no notion of "breaking free of gravity" in the gravitron. Perhaps you are confusing this situation with the launch trajectory situation and the escape velocity. Escape velocity has nothing to do with the gravitron. Regardless of what happens in the gravitron, for your freshman physics purposes, the weight that you put on the free-body diagram is unaffected.
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