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Circular Motion involving a gravitron

  1. Sep 11, 2014 #1
    1. The problem statement, all variables and given/known data

    I have attached questions about gravitron. I'm struggling with question 2 about calculating the normal force of the gravitron.

    2. Relevant equations

    mv^2/r

    3. The attempt at a solution

    I know the answer so I can figure it out as I know it involves setting the Ncos(62) = mg (N = normal force) but I don't get how you know the angle involved is 62 as it only gives you the angle the track makes with the floor?

    Also, please could someone explain why the track is tilted as appose to being vertically upright and what this has to do with enabling the track to slide up when its rotating?
     

    Attached Files:

  2. jcsd
  3. Sep 11, 2014 #2

    Doc Al

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    Staff: Mentor

    The couch is attached to the track, so it makes the same angle. (Is that your question?) Have you drawn yourself a free body diagram of the person?

    The normal force must have a vertical component to lift the person/couch. If it were upright, the weight could not be supported or lifted.
     
  4. Sep 11, 2014 #3
    Thanks Doc Al. I have drawn a free body diagram and attached it. I have resolved the normal force into components and labeled what I think each of them do. Regarding the angle, I know from the answer that one of the angles in that vector triangle I have drawn must be 62 degrees but how do you know this? How can you show that the angle between the track and the ground is the same as the angle in that vector triangle?
     

    Attached Files:

  5. Sep 11, 2014 #4

    Doc Al

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    Staff: Mentor

    Good.

    The key is to recognize that the angle that the normal makes with the vertical is the same angle that the track makes with the horizontal. That is true because the track and the normal are always perpendicular. One way of seeing that would be to use a bit of triangle geometry or trig. If you label all the angles, I could walk you through it.

    Even easier might be this. Imagine that the track was on a hinge that could move through any angle θ, instead of being fixed at 62 degrees. Now lay the track flat so that θ = 0. What angle does the normal make with the vertical? 0, of course, since it points straight up. Now increase the angle to 1 degree. Can you see that the normal must move 1 degree to the vertical? No matter what angle the track makes with the horizontal, the normal will make that same angle with the vertical.

    Make sense?
     
  6. Sep 11, 2014 #5
    Thanks again Doc Al! Kind of making sense. I have drawn another FBD (attached) and I think I have figured out the angle using a bit of geometry, please could you check to see if it's correct? I have extended the weight force in purple. If the angle between the track and the ground is 62 then the other angle (a) must be 28. I think I am right in saying that because there are two straight lines crossing that a and b are alternate interior angles? Angle c must be 90 minus 28 which is 62 degrees. Is that how you prove it or is that a very long way of doing it?

    Also, am I right in saying that the vertical component of the normal force will be greater than the weight force when the couch slides up?
     

    Attached Files:

  7. Sep 11, 2014 #6

    Doc Al

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    Staff: Mentor

    That's pretty much it. Here's my version. Call the angle that track makes with the horizontal θ, instead of a specific number. Angles b and c must add to 90, since they make a right angle. Same for angles a and θ. But you know that a = b, so therefore c = θ.

    Symbolically:
    b + c = 90
    a + θ = 90
    thus b + c = a + θ
    but b = a, so:
    c = θ

    Yes.
     
  8. Sep 11, 2014 #7
    Thank you very much!!
     
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