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Finding the minimal speed of

  1. Oct 18, 2007 #1
    Basically, we were told that from (mv^2/r)(cos [tex]\theta[/tex]) = [tex]\mu[/tex][(mv^2)/r) sin [tex]\theta[/tex] + (mg cos [tex]\theta[/tex])] + mg sin[tex]\theta[/tex]

    you could rearrange for the max speed of a car going in a circular path on a banked road at a [tex]\theta[/tex] angle. From that equation above, I derived v max, but now I need to get v minimal. I really don't even understand the visual concept of the equation above - I drew a free-body diagram but I still don't understand it.

    Could someone help me figure out the minimal speed the car has to go without falling off?

    Also in the equation, he substituted a variable N for normal force with the equation in the bracket. So it's basically

    (mv^2/r)(cos [tex]\theta[/tex]) = [[tex]\mu[/tex]N + (mg cos [tex]\theta[/tex])] + mg sin[tex]\theta[/tex]
    Last edited: Oct 18, 2007
  2. jcsd
  3. Oct 18, 2007 #2
    Can anyone help me with this?
  4. Oct 22, 2007 #3
    Well, it's been a few days, so... *BUMP* :)

    I can't get credit for the answer, but I am still curious as to what it is.
  5. Oct 22, 2007 #4
    I think it's the speed of the car, which will provide a centripetal force, which is equal in magnitude to the x component of the normal/support force.

  6. Oct 23, 2007 #5
    No, I don't think that works. I'm suppose to get the result 8. something
    But instead I get 44 with that solution.
    The max speed was 16.1 m/s, which means the minimal speed needs to be under that.
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