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Centripetal Forces and Speed Question

  1. Oct 9, 2015 #1
    1. At a velodrome, bicycle riders race around the inside of a course shaped like a cup whose angle increases as the rider moves up the side. At one point the radius of the path is 50.0m and the track makes an angle of 45.0 degrees with the horizontal. If a rider is perpendicular to the track at this point, what is the speed of the rider?

    mv2 / r = Fnx
    ma = Fny - Fg

    3. So my attempt was to break it into components and solve for Fn in the y and sub that into the x.

    ∑Fy = Fny - Fg
    ma = cosθ(Fn) - Fg
    0 = cosθ(Fn) - mg
    mg = cosθ(Fn)
    mg/cosθ = Fn

    Now with that I attempted plugging that into the x direction.

    ∑Fx = Fnx
    mv2 / r = sinθFn
    mv2 / r = (sinθmg) / cosθ
    mv2 / r = tanθmg
    v2 / 50.0m = tan45°x9.81m/s2
    v = 22.1m/s

    I was just wondering if someone can proof read and make sure that this is correct or incorrect and why.
  2. jcsd
  3. Oct 9, 2015 #2
    Looks good!
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