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Dynamics Question: What is the velocity of a car traveling on a known curvature

  1. Mar 1, 2012 #1
    Hi all,

    I'm trying to write a Matlab simulation that determines the velocity of a car of known mass and moment of inertia which travels on a track whose curvature is also known.

    To say the least, I'm at a loss as to what approach I should take to create my simulation. I'm finding it somewhat difficult to grasp the physics of the situation. The free body diagram is easy enough, but I can't recall how to couple this information with the constraint that the cart is confined to roll on the path.

    If someone could provide some hints or a resource that I can read, I'd really appreciate it.


  2. jcsd
  3. Mar 2, 2012 #2
    I've given this some thought, was wondering if some higher up could check my reasoning/math:

    Coordinate system: x,y conventional.

    Say we have a ramp given by the cubic function: [itex]y(x)[/itex] = a[itex]x^{3}[/itex] + b[itex]x^{2}[/itex] + [itex]cx[/itex] + d

    If we start the cart at x = 0 and y = [itex]y_{0}[/itex], where [itex]y_{0}[/itex], is a maximum, the kinetic energy at any point is

    Kinetic Energy T = [itex]\frac{1}{2}[/itex]M[itex]v_{x}(t)^{2}[/itex] + [itex]\frac{1}{2}[/itex]I[itex]ω^{2}[/itex] = mg([itex]y_{0}[/itex] - [itex]y(t)[/itex])

    Assuming roll without slip condition: ω = [itex]\frac{v_{x}}{R}[/itex] and some simplification we get:
    T = C[itex]v_{x}(t)^{2}[/itex] = mg([itex]y_{0}[/itex] - [itex]y(t)[/itex]) where C is some constant.

    In order to get the equation solely in terms of [itex]x(t)[/itex] and [itex]v_{x}(t)[/itex], we can substitute [itex]y(t)[/itex] for the cubic function [itex]y(x)[/itex], which is implicitly a function of time through x.

    By integrating both sides of the equation now, we can get a function for [itex]x(t)[/itex], which we can plug back into [itex]y(x)[/itex] to get [itex]y(t)[/itex].

    Alright, how far off am I?

    Thanks in advance,

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