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Inward Acceleration for Ununiform Slope

  1. Jul 19, 2006 #1
    I started doing some physics today for the first time since school got out a few months ago. The period of inactivity was painfully apparent throughout the attempt.

    I'm trying to find the apparent instantenous "r" of an object moving along an ununiform slope where "r" is the radius of a circle that would have the same inward acceleration. Example of possible path: y=x^2-x.
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  3. Jul 19, 2006 #2


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    I assume you are referring to the radius of a circle that would have the same acceleration as described by the equations of rotational motion? Is the example function representative of the displacement the velocity or the acceleration?
  4. Jul 19, 2006 #3
    The function describes the vertical displacement, y, compared to the horizontal displacement, x.
  5. Jul 20, 2006 #4


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    Would'nt it be

    [tex]r = \frac{V_{tan}^2}{a_{in}}[/tex]

    where one uses the tangential speed in the calculation?
    Last edited: Jul 20, 2006
  6. Jul 20, 2006 #5
    I'm trying to find the equivalent "r" in terms of x and y for the acceleration. Yes, if it was a circle the acceleration would be v^2/r...but this surface we're talking about is not as nice as a circle.
  7. Jul 20, 2006 #6


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    What is this "inward" acceleration you are talking about? I thought you were talking about the normal acceleration component. The other component that goes with it is the tangential acceleration component. That is the components normal and tangential to the object's path.

    The normal acceleration component is directed towards the inner curvature of the path and is given by

    [tex]a_n = \frac{v^2}{\rho}[/tex]

    where [tex]\rho[/tex] is the curvature of the path. Do you need to determine it for a given path?
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