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Could someone check if this is right soon.

  1. Sep 27, 2009 #1
    Could someone plz check if this is right soon. plz

    What would the bank angle be for a circular racetrack with radius 120 m so that a car can go around the curve safely at a maximum of 25 m/s, without the help of frictional force to keep it on the road?

    (25^2)/(120)/(9.80)=.53146

    ArcSin(.53146)=32.1 Degrees

    is this right, could someone check it for me

    Thanks
     
  2. jcsd
  3. Sep 27, 2009 #2

    Delphi51

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    Re: Could someone plz check if this is right soon. plz

    I would have used arctan rather than arcsin. Maybe you should take time to draw the vector diagram to be sure.
     
  4. Sep 27, 2009 #3
    Re: Could someone plz check if this is right soon. plz

    ArcTan doesnt give you the right answer. Could you help me at all, more in depthly.
     
  5. Sep 27, 2009 #4

    Delphi51

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    Re: Could someone plz check if this is right soon. plz

    bankedCurve.jpg
     
  6. Sep 27, 2009 #5
    Re: Could someone plz check if this is right soon. plz

    So is it

    Arctan(5.208/9.81)=27.96
     
  7. Sep 27, 2009 #6

    Delphi51

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    Re: Could someone plz check if this is right soon. plz

    Yes, that's what I got.
     
  8. Sep 27, 2009 #7
    Re: Could someone plz check if this is right soon. plz

    One more thing, how did you get the 5.208
     
  9. Sep 27, 2009 #8

    Delphi51

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    Re: Could someone plz check if this is right soon. plz

    Fc = mv^2/R = m*25^2/120 = 5.208*m
    This is the centripetal or centrifugal force. From the point of view of the outside world, the force is inward needed to hold the car in circular motion. From the point of view of the car in circular motion, the force is outward, tending to push it out of circular motion into straight line motion.
     
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