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Centripetal Force and Roller Coasters

  1. Nov 13, 2012 #1
    1. The problem statement, all variables and given/known data

    A) A roller coaster has a vertical drop but followed by a loop. The total vertical height of the drop is 66m. It starts as a steep slope at an angle of 75 degrees relative to the horizontal and then enters into a circular path of radius R. In order to stay below 4 times the force of gravity at the bottom of the bend what is the minimum radius of this bottom section of the loop.

    B) After the bottom of the dip of radius R (from above) the track gradually transitions to the inverted (top) part of the loop. What is the force on the rider at the top of the loop if the loops height is 33 meters? Would the change fall out of the riders pockets?


    2. Relevant equations

    Fc = mV^2/r
    PEg (gravitational potential energy) = mgh
    KE = 1/2 mv^2
    Conservation of Energy : PEf-PEi = KEf-KEi

    3. The attempt at a solution

    I apologize if what I did here seems completely wrong! -- I tried!


    A)

    Fc < 4Fg
    mv^2/r = 4mg
    v^2/r = 4g
    r = v^2/4g
    v=?

    EK (bottom) = PEg (top)
    1/2mv^2 = mgh
    v = √(2x9.8m/s^2x66m)
    v = Approx 36m/s

    36m/sxcos(75°) = 9.31 m/s

    (9.31m/s)^2/9.8m/s^2 = r

    2.21m = r

    B)

    For this question I'm not sure whether or not to assume Fc = Fg.
    I found that by using the consrevation of Energy theory, I can equate the changes in Potential and kinetic energies (by subbing the v found in the previous question) to get Vf, the velocity at the top of the loop. (Right?)

    I'm stuck at this question, because there is no mass.
    Fc = mV^2/r
    F = mg
    F = ma

    ...I'm not too sure how to go about it.
    If I use the conservation of energy by equating the changes in both Kinetic/potential energy again, I was a bit stumped on how to solve for 'm'.

    I hope this is in the right section.
     
  2. jcsd
  3. Nov 13, 2012 #2

    haruspex

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    This seems inconsistent. It says vertical drop, then 75 degrees. But I can't see that it makes a difference here, so long as the 66m is from top of drop to bottom of loop.
    I would think the 4g limit includes the 1g that would apply if stationary, so the limit is an extra 3g.
    The above seems enough to give an answer.
    What is this calculating?
    That is the question to be decided - is Fc more or less than Fg.
    The masses will cancel out. Fc > Fg if and only if Fc/m > Fg/m.
     
  4. Nov 14, 2012 #3
    Yes, that is how it's supposed to be pictured.

    Yes, you are right -- My teacher has demonstrated this question, thank you very much.

    So it would have been (as demonstrated),
    Ac = V^2/r
    r = V^2/ac
    r = V^2/3g


    A mistake. The angle is unnecessary. >.>

    Can you please demonstrate how to compute this/the force?
    r = 33/2,
    v = √2xgx33m (I think)
    F = ?
    ^This portion of the question was not demonstrated; I would like to see -- I'm missing out on as to where/how the masses cancel.

    Thank you very much.
     
  5. Nov 14, 2012 #4

    haruspex

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    That's the actual speed at top of loop, √(2g(h-2R)), right? It's a bit confusing decoding from the numbers because 4R happens to equal h.
    OK, sorry, I see your problem. I was only looking at the last part, where it asks about the fate of the contents of the riders' pockets. Of course, you cannot say what the force is without knowing the mass. The question setter was probably thinking of acceleration and wrote force by mistake. I would answer it with something like "(some calculated number)m, where m is the mass of the rider".
    What is the critical speed at top of loop for the riders to keep their cash?
     
  6. Nov 20, 2012 #5
    Yes -- mgh = 1/2mv^2, where 'h' (33m) is the height difference between the top of the loop and the top of the coaster.

    Thank you!
     
  7. Nov 20, 2012 #6

    haruspex

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    So what did you get for "Would the change fall out of the riders pockets?"
     
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