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Centripetal force

  1. Jun 23, 2008 #1
    1. What is the maximum speed with which a bike can move at a curve, and on which angle from the vertical should he turn, so that he doesn't fall off the road, if the coefficient of friction is 0,4 and the radius of the curve is 100m?

    Now, I easily calculated the maximum speed by equating the centripetal force and the force of friction, but I don't understand what exactly is meant by an angle. Could someone help?
     
  2. jcsd
  3. Jun 23, 2008 #2
    Do you know the relation between banking angle & velocity?
     
  4. Jun 23, 2008 #3
    To be honest, since English is not my native language, I don't even know what a banking angle is :blushing: Is it important in this case? Is that the angle that the problem asks for? Further information is appreciated.
     
  5. Jun 23, 2008 #4
    You got the first part of your problem right. :smile:
    i.e. [tex]\mu mg = \frac{mv^2}{r}[/tex]

    The banking angle refers to the angle of incline and it is the angle the problem is asking for. It can be calculated by taking the ratio of the horizontal component to the vertical component. You are very close to the answer.
     
  6. Jun 23, 2008 #5
    Well, I think the formula for vmax should be (according to what I think you mean):
    v^2=Rg(sinA+kcosA)/(cosA-ksinA), where A is the desired angle, and k is the coefficient of friction. But, according to these calculations, the wanted angle in my calculations is about 80,5 degrees, and my textbook says it should be 22 degrees. Could you compare these to your results? Is there any other possible meaning for the wanted angle?
     
  7. Jun 23, 2008 #6

    tiny-tim

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    Hi guardians! :smile:

    The principle you need is that the motorcycle will only balance at a fixed angle if the torque about the wheel is zero (if it isn't zero, the motorcycle will rotate left or right until it hits the road :cry:).

    So the torque of gravity (about the point of contact of the wheel with the road) must be equal to the torque of the centripetal acceleration. :smile:

    Friction is not involved (because the friction goes through the point of contact, so its torque is zero).
     
  8. Jun 24, 2008 #7
    So is this line of thought right:

    If the inclination is [tex]\varphi[/tex], then the Ffriction=m*g*sin[tex]\varphi[/tex], i.e. k*m*g*cos[tex]\varphi[/tex]=m*g*sin[tex]\varphi[/tex], so [tex]\varphi[/tex]=atan(k)
    ?
     
  9. Jun 26, 2008 #8
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