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Kinematics of a ball and track

  1. Mar 1, 2008 #1
    Hello,

    New here. I have designed a segment of a track that I am building that I need to pull some data out of. It is a high-to-low transition (down hill) that takes the form of cosine over the domain [0,pi]. Thea graph of the transition's profile is y=5.5cos(xpi/18)+5.5 It is 11 inches high and 18 inches long.

    I want to find out its exit velocity (x=18) for a ~100g ball that starts form rest at the top. The graph of the formula I used above is position-vs-position. I know that velocity is the graph of position-vs-time's first derivative.

    How do I get from the position-vs-position graph/formula to the position-vs-time graph/formula? I think I could do it if the slope was constant, but I'm not sure about the curve and I don't want to average the slope. Shouldn't it be the same graph, just compressed along the x-axis?

    Thanks,
    Chopper
     
  2. jcsd
  3. Mar 1, 2008 #2

    Dale

    Staff: Mentor

    To a first approximation just use conservation of energy. You have PE = mgh and KE = 1/2 mv² so by setting them equal you get

    mgh = 1/2 mv²
    v = sqrt(2gh)
     
  4. Mar 3, 2008 #3
    Though to get a more accurate approximation you should account for the rolling of the ball (assuming there is enough friction to prevent it from slipping):

    [tex]mgh\ =\ \frac{1}{2} mv^2\ +\ \frac{1}{2}I\omega^2[/tex]

    Where I is the moment of inertia for a sphere ([tex]\frac{2}{5}MR^2[/tex]) and [tex]\omega[/tex] is [tex]\frac{v^2}{R^2}[/tex].

    Solve that entire equation for v and you should get a more accurate approximation.
     
  5. Mar 3, 2008 #4

    rcgldr

    User Avatar
    Homework Helper

    Note that w^2 is not v^2 / R^2 if the ball running on a pair of rails as opposed to rolling along a surface. w^2 = v^w / r^2, where r is the perpendicular component of distance from center of ball to the surface of the ball where it meets the supporting track.
     
  6. Mar 3, 2008 #5
    @everyone: Thank you for the input. This will be very helpful. I had considered Inertia and that the point of contact of the ball/track would have an impact, but this is very helpful.
     
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