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Roller Coaster Loop

  1. Aug 17, 2014 #1
    1. The problem statement, all variables and given/known data

    Hi, I am stuck on this thought experiment:

    A roller coaster of mass m starts on a inclined plane at a certain height, and then enters a circular loop, with radius r. At what height, h on the plane, must the trolley start in order to stop at the exact top of the loop, and then what happens?

    2. Relevant equations

    gpe=g*h*m
    total energy = KE + PE

    3. The attempt at a solution

    In the beginning the trolley has only PE which is ghm (for h unknown)
    At the top of the loop, again KE is zero, and PE is g(2r)m.

    So h=2r (ie same height as the loop), and the trolley falls vertically down.

    Is this correct? Thanks,
     
  2. jcsd
  3. Aug 17, 2014 #2
    Welcome to PF!

    Looks alright .
     
    Last edited: Aug 17, 2014
  4. Aug 17, 2014 #3
    What prevents the trolley from falling down before reaching the top point?
     
  5. Aug 17, 2014 #4
    Before it reaches the top (say at 1 oclock), it is moving in the direction of the track, so it keeps going round. As it moves upwards it losses KE to GPE, I think? Though I am not really sure how centripetal acceleration comes into this?
     
  6. Aug 17, 2014 #5
    Imagine your are inside a spherical cavity. Can you just walk up to its top?
     
  7. Aug 17, 2014 #6
    Assuming you have enough kinetic energy to begin with, then yes?
     
  8. Aug 17, 2014 #7
    No, don't assume that. You are a human, a mere mortal. Can you walk up to the top of say a 2-meter radius sphere, on its internal surface? And can you, a mere mortal, walk up the stairs to a platform that is 4 m higher? Assuming you answer honestly, can explain the difference?
     
  9. Aug 17, 2014 #8
    I suppose the difference is the direction of the contact force which opposes gravity only in the stairs case?
     
  10. Aug 17, 2014 #9
    Very well. Now let's say you have a powerful motorcycle. You can make it to the top with that contraption. Why is suddenly the contact force complication no longer a complication?
     
  11. Aug 17, 2014 #10
    You now have some sort of inertia in the upwards direction? sorry I don't know.
     
  12. Aug 17, 2014 #11
    Even if you start at the bottom?
     
  13. Aug 17, 2014 #12
    The track pushes you towards the middle??
     
  14. Aug 17, 2014 #13

    Orodruin

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    Voko, there are roller coasters that do experience negative g. Those typically are constructed such that the track can give accelerations in both directions. It is here reasonable to assume this is the case, since the problem does not have a solution otherwise - you cannot stop on top without the track providing a force toward the outside of the loop.
     
  15. Aug 17, 2014 #14
    #1 says the trolley falls vertically down in the end. That clearly assumes the track can only push, but not pull.

    Either way, the complete answer in #1 is incorrect.
     
  16. Aug 17, 2014 #15

    Orodruin

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    Agreed, the problem and proposed solution make incompatible assumptions.

    It could be reformulated to ask for the minimum height for which the full loop is performed before falling. This was already discussed here (for example)
    https://www.physicsforums.com/showthread.php?t=750306
     
  17. Aug 17, 2014 #16
    Ah, so is it not possible that the trolley can get to the exact top and stop? It either keeps going round, or never gets there? For the original question lets assume that the track cannot pull.
     
  18. Aug 17, 2014 #17
    Yes, if the trolley gets to the top, on a push-only track, it will then complete the loop.
     
  19. Aug 17, 2014 #18
    But surely there is a starting height and equivalent speed of entry to the loop at which it just makes it to the top, it's speed drops to zero, and it falls vertically down? Can you explain why this is not possilbe?
     
  20. Aug 17, 2014 #19

    Orodruin

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    No, this is not possible. I suggest you read the thread I linked above.

    If you are at the top and want to continue along the loop, the centripetal acceleration must be v^2/r. Anything bigger than that and you fall. The least acceleration occurs when the force from the track is zero and the acceleration is then due to gravity only, so g. This gives you the minimum velocity the coaster must have at the top to reach the top. Anything less and it falls earlier.
     
  21. Aug 17, 2014 #20
    Ah, I see now, thanks!
     
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