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Superelevation circular motion

  1. Sep 10, 2008 #1


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    A question about when a car take a superelevated curve. I don't know how to be more precise in English. Take a look at this photo, maybe you'll understand better : http://www.fmciclismo.com/noticias/bmx/imagenes/peralte%20bmx%20copia.jpg [Broken]
    My question is : when you draw the free body diagram, you see that the normal force (N) has a component in direction of the center of the radius of the curve that describe the movement of the car. Does this component is precisely the centripetal force? That would mean that the centripetal acceleration is worth [tex]\frac{F_c}{m}[/tex].
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Sep 10, 2008 #2


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    Yes, that component of the normal force creates the centripetal acceleration. Nice picture, by the way.
  4. Sep 11, 2008 #3


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    Thanks for your answer. And for the picture, thanks to google :rolleyes:.
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