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Centripetal acceleration

  1. Mar 27, 2008 #1
    the cars on a theme park ride each have a mass of 500Kg and travel round a verticle loop of diameter 20m
    i need to know the minimum speed at which the carts are traveling in order to remain in contact with the loop as they go round.
    and the maximum reaction of the track
     
  2. jcsd
  3. Mar 27, 2008 #2

    tiny-tim

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    Hi Batman! :smile:

    You need to tell us how far you've got, so we know how to help.

    What have you tried? Or considered trying? :smile:
     
  4. Mar 27, 2008 #3
    Fcf = MV(squared)
    r

    V=
    (square root) Fcf x r
    M

    Fcf = 500 (mass) x 9.81 (gravity)

    V=
    (square root) 4905 x 10
    500

    = 9.904 m/s to make the loop



    im not sure how to then work out the maximum reaction of the track.
     
  5. Mar 27, 2008 #4

    tiny-tim

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    Yes, that looks fine! :smile:

    (Though you wasted time by multiplying by M and then dividing by it again!)

    (btw, on this forum, best to write "Mv^2/r")
    (erm … you haven't actually told us whether the cars are powered, and going round the loop at a constant speed, or unpowered, and relying on their inital kinetic energy to get them to the top … which is it?)

    At what angle do you think the maximum reaction will be?

    Work out the reaction at a typical angle theta from the downward vertical (for a top-of-the-loop speed of 9.904).

    For what theta is this a maximum? :smile:
     
  6. Mar 27, 2008 #5
    the question i was given did not state weather its powered, but i belive its at the minimum to make the loop 9.904m/s. The maximum reaction is at the bottom of the loop because CFf (centrifugal force) and gravity are both acting in the same direction causing the biggest reaction from the track.
    if the reaction acts there, will i only need the CFf value, gravity and the Velocity?
     
  7. Mar 27, 2008 #6

    tiny-tim

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    Yes, that reasoning looks right! :smile:

    I would assume that it's unpowered, so you have to work out the speed at the bottom which will produce the right speed at the top, then get the cff for that. :smile:
     
  8. Mar 31, 2008 #7
    using the speed of 9.904 m/s
    the Cfc = M w^2 r
    = 490.44608X10^6 N
    the maximum reaction given by the track with a speed of 9.904 m/s is 490.4 MN, is that the correct formula to use.
     
  9. Mar 31, 2008 #8

    tiny-tim

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    mmm … I think you're out by a factor of 10.

    And where did the 10^6 come from? :confused:

    But anyway, you haven't included the effect of the gravitational force.

    Try again! :smile:
     
  10. Mar 31, 2008 #9
    M=500kg = 500X10^3
    W=9.904m/s ^2=98.0892
    r= 10m

    (500x10^3) X (98.0892) X (10) = 490446080
    In ENG form = 490.44608 X10^6 N

    490.44MN

    the gavity also acting on it would be 9.81 X mass = 4.905 X10^6


    So the total would be 490.44 +4.905 = 495.345
     
  11. Mar 31, 2008 #10

    tiny-tim

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    No … your w should be angular velocity, not ordinary velocity.

    w = v/r.

    acceleration = (w^2)r = (v^2)/r.

    And Newtons are in kg, aren't they?
     
  12. Mar 31, 2008 #11
    a mass in KG must be multiplied by gravity to get it into netwons
    M = 500 X 9.81 = 4905N
    R=10m
    W= v/r
    W=9.904 / 10 = 0.9904 m/s

    4905 X 0.9904 X 10 = 48579.12N => 48.579KN ?
     
  13. Mar 31, 2008 #12

    tiny-tim

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    In the last line, you've used W, instead of W^2 (though it hasn't made much difference, because W is nearly 1).

    And the mass only has to be multiplied by g when you want to turn it into weight (the force from gravity), not to compute a centripetal acceleration.
     
  14. Mar 31, 2008 #13
    ok so now using W^2 and not multipling the mass by g i should getthe maximum reaction from the track.
    500N X 10M X 0.9904^2 = 4904.46N
    and would i add gravity to find the max or not?
     
  15. Mar 31, 2008 #14

    tiny-tim

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    Yes! :smile:
    Oi! You tell me! :smile:
     
  16. Mar 31, 2008 #15
    50-50 chance so ill go with yes i will need to add it.
    Thanks so much for your help :>
     
  17. Mar 31, 2008 #16

    tiny-tim

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    I don't believe it! :rofl:

    You're right of course, but … don't you remember writing this:
    Just one more thing: you're assuming that the car is powered.

    As I said before, I would assume that it's unpowered, so to be safe you may like to work out the speed at the bottom which will produce the right speed at the top, then get the cff for that. :smile:
     
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