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Maximum speed on banked corner

  1. Jul 31, 2008 #1
    Hi, not sure if this is the right section for this question, but I had better try. Anyway, the question is:

    Show that the maximum speed on a banked corner where the coefficient of friction between the road and tyres is 0.8 for dry roads and 0.3 when wet:

    v=sqrt[(rg(sin(theta) + mu cos (theta)))/(cos(theta)-mu sin(theta))]

    where r= the corner radius, mu = the coefficient of friction, theta = the angle of banking

    I am unsure of how to approach this. Any advice would be appreciated.

    Thanks
     
  2. jcsd
  3. Aug 1, 2008 #2

    Hootenanny

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    Re: Cornerning

    I would start with some basic assumptions. Firstly, model the car as a particle; secondly, assume that whilst cornering the particle follows a circular path. And finally, assume that the car's vertical position remains constant.

    Now, identify all the forces acting on the car and use the assumptions stated above to determine what the constrains are on those forces.
     
  4. Aug 1, 2008 #3
    Re: Cornerning

    Alright so you know that the friction provides your centripetal force. So what condition will your system be in, such that the centripetal force is overcome and the car skids?

    I think you should get it ;)
     
  5. Aug 1, 2008 #4
    Re: Cornerning

    I have got expressions for the magnitude of both the velocity and acceleration. But I am confused when it comes to finding Nsin(theta) [normally equals ma] and Ncos(theta) [normally equals [mg]. I am just unsure of how to work friction into there. Any ideas?
     
  6. Aug 1, 2008 #5

    Hootenanny

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    Re: Cornerning

    White Space
     
  7. Aug 1, 2008 #6
    Re: Cornerning

    Nsin (theta) > Ncos (theta)?
     
  8. Aug 2, 2008 #7
    Re: Cornerning

    I'm guessing I am wrong? Any help would be appreciated thanks
     
  9. Aug 2, 2008 #8

    Hootenanny

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    Re: Cornerning

    You're on the right lines, but not quite there yet. As Physicsnoob93 has said, the frictional force provides the force required for the centripetal acceleration. In other words, the centripetal force must be equal to the frictional force for the car to continue travelling around the corner without skidding.

    Can you write the above in the form of an equation?
     
  10. Aug 2, 2008 #9
    Re: Cornerning

    If a=v^2/r and the frictional force provides the required force for the centripetal acceleration, then:
    mu.N=v^2/r

    Is that what you meant?
     
  11. Aug 2, 2008 #10
    Re: Cornerning

    mu.N is the force provided by friction whereas (v^2)/r is an acceleration. Remember the centripetal force is a force.
     
  12. Aug 2, 2008 #11
    Re: Cornerning

    i solved the equation, many thanks to physicsnoob93 and hootenanny
     
  13. Aug 3, 2008 #12
    Re: Cornerning

    Not a problem.
     
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