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Homework Help: Minimum speed on banking road (circular motion)

  1. Dec 2, 2008 #1
    1. The problem statement, all variables and given/known data
    A car rounds a slippery curve. The radius of
    curvature of the road is R, the banking angle
    with respect to the horizontal is θ and the
    coefficient of friction is μ.

    What is the minimum speed required in order
    for the car not to slip?

    2. Relevant equations
    Fc = (mv^2) / r
    W = mg
    Ffr = μ N

    3. The attempt at a solution
    So I made the xy-plane standard, weight in the negative y-direction.


    Nx + Ffr(x) = Fc

    N = (mg) / cos (th)

    Ffr(x) = (μmg) cos (th) / cos (th)
    = μ mg

    Nx = [ (mg) sin (th) ] / cos (th)

    Nx + Ffr(x) = Fc
    μmg + (mg sin (th))/cos (th) = (mv^2) / r
    v(min) = sqrt [ gr (μ + tan (th))]

    That was my answer, but it's not an answer choice. I don't know where my thought proccess was wrong. Can someone help?
  2. jcsd
  3. Dec 2, 2008 #2
    It is good to clearly define an axis. In the next step, you have equated the forces along the x direction. What you should do next is to equate forces along the y direction, no?

    In other words, can you check again if this equation is correct?

    Finally, you should think about what direction the friction force acts. If the car is going sloooowly, which direction (upwards or downwards) will it tend to slide?
  4. Dec 2, 2008 #3
    Yeah, but I don't see why it's not correct. I don't understand what other horizontal forces are acting. I thought the net force is always the centripetal force. Weight is perpendicular to the horizontal forces, so its component would be zero...

    EDIT: oh right, so friction would act upwards? Didn't realize it would change depending on speed...
  5. Dec 2, 2008 #4
    Your equation for the horizontal forces is perfect. I'm concerned about the equation relating N to mg.

    Yes, friction opposes the direction of motion (or the direction in which the body is trying to move). If you move too fast, you can imagine that the friction acts downwards. In fact, this will mean that at some velocity, there will be no friction.
  6. Dec 2, 2008 #5
    OH! I assumed the net force in the y direction equals zero.

    N cos (th) - mg = 0
    so N = mg / cos (th)

    Yet I should've included friction's y component, right?
  7. Dec 3, 2008 #6
    Yes, good job figuring this out. You should be able to work out the answer from here.
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