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Angle for a banked curve (friction-less surface)

  1. Oct 26, 2011 #1
    1. The problem statement, all variables and given/known data

    At what angle should the roadway on a curve with a 50 m radius be banked to allow cars to negotiate the curve at 12 m/s even if the roadway is frictionless?

    a. 0°
    b. 12.2°
    c. 17.1°
    d. 35.0°
    e. 73.2°


    2. Relevant equations

    β=arctan(v^2/r*g)


    3. The attempt at a solution

    im using β=arctan(v^2/r*g) but the key says the answer is b?
    can anyone help me figure out why its not 16.4??
     
  2. jcsd
  3. Oct 26, 2011 #2
    Why not try to derive the formula?
     
  4. Oct 27, 2011 #3
    i did...i though of the triangle as if the gravity was pointing downward the centripetal force outward and then the angle should be the same as the angle for the curve right?
     
  5. Oct 27, 2011 #4
    If your acceleration is towards the centre the force will be too. You need to find the angle of the road that gives a normal force with the horizontal component equal to the force needed to keep the circular motion described by v = 12 m/s and r = 50m

    I'm aware that my response is a mouthful :smile:, feel free to ask me for clarification.
     
  6. Oct 27, 2011 #5
    do you mean like a triangle with the vertical 9.81*m and the horizontal (12^2/50)*m ??
     
  7. Oct 27, 2011 #6
    Yes. But the horizontal component will be a fraction of the normal force, which depends on the angle.
     
  8. Oct 27, 2011 #7

    I like Serena

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    Welcome to PF, physgrl! :smile:

    I derived the same formula as you have.
    When I fill in the values I also get 16.4 degrees.
    So I believe your answer is correct (assuming you did not make a typo).
     
  9. Oct 27, 2011 #8
    I believe your answer is incorrect, as I got 17.1 as my answer. You should investigate why you are using arctan.
     
  10. Oct 27, 2011 #9

    I like Serena

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    Here's my interpretation of the forces.

    attachment.php?attachmentid=40391&stc=1&d=1319734802.gif

    We have:
    [tex]F_{resultant} = {m v^2 \over r}[/tex]
    [tex]\tan \beta = {F_{resultant} \over m g }[/tex]
    The OP's formula follows from this.
     

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  11. Oct 27, 2011 #10

    ehild

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    16.4° is correct, the answer key is wrong. It happens quite often.

    ehild
     
  12. Oct 27, 2011 #11
    Oooh I see where I went wrong. My apologies to all. I just always have trouble with the normal force being more than gravity. Its very unintuitive to me.
     
  13. Oct 27, 2011 #12
    Thanks to everyone!!
     
  14. Oct 27, 2011 #13

    I like Serena

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    You're welcome! :smile:
     
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