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Motion on a parabola

  1. Dec 3, 2009 #1
    Hi everybody



    we know that if we have an object sliding on a frictionless ramp,

    the acceleration force will be constant, and it equals to

    a=g * sin(theta)
    where theta is the ramp angle w.r.t. the ground


    so the path of motion in this problem can be written mathematically as a linear function
    y(x)=bx
    and hence the tangent tan(theta)= bx /x =b


    The Question is
    if the path of motion is parabolic and is of the form

    y(x) = a x^2

    how to solve for the acceleration with respect to time ???????

    be aware that ,in this case the acceleration is not constant , and it always equals to the tangent of the parabola at the current location of the object.
    and the tangent in this case is the first derivative of y which is
    y'=2ax

    as we see the tangent and hence the acceleration is a function of x

    so ,

    how to calculate the time as the function of position ??

    thanks
     
    Last edited: Dec 3, 2009
  2. jcsd
  3. Dec 3, 2009 #2

    rock.freak667

    User Avatar
    Homework Helper

    Using n-t coordinates you'd get that the acceleration vector is

    [tex]\vec{a}=\dot{v} \hat{e_t}+ \frac{v^2}{\rho} \hat{e_n}[/tex]

    where

    [tex]\rho =\frac{(1+(y')^2)^\frac{3}{2}}{y''}[/tex]


    Not sure if I wrote down the formula for ρ correctly though.

    then again, for a time parameter x=t.
     
  4. Dec 4, 2009 #3
    Hi

    and thanks for replying

    but can you explaine how this equation came about

    [tex]\vec{a}=\dot{v} \hat{e_t}+ \frac{v^2}{\rho} \hat{e_n}[/tex]

    or at least show me the source link
     
  5. Dec 4, 2009 #4

    rock.freak667

    User Avatar
    Homework Helper

    read http://web.mst.edu/~reflori/be150/FloriNotes/ntCoordLectureNotes1.htm" [Broken]
     
    Last edited by a moderator: May 4, 2017
  6. Dec 9, 2009 #5
    thanks :)


     
    Last edited by a moderator: May 4, 2017
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