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Finding acceleration of a skier on parabola

  1. Feb 9, 2015 #1
    1. The problem statement, all variables and given/known data
    A skier travels with a constant speed of 6 m/s along a parabolic path y= x2/20. Find the acceleration of the skier when he is at (10,5). Neglect the size of skier.

    2. Relevant equations
    dy/dx is slope of parabola.
    In a straight line if body has distance covered=x, then velocity (v) = dx/dt and acceleration = dv/dt
    Radial acceleration is v2/R where R is radius.
    3. The attempt at a solution

    Slope is 2x/20= x/10.
    For straight line determining acceleration is easy but how about this curve case?
     
  2. jcsd
  3. Feb 9, 2015 #2

    BvU

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    For a straight line it is indeed very easy: constant speed means zero acceleration !
    For a curved trajectory I would think the tangential acceleration at constant speed is zero, so I agree your v2/R is the right answer for this exercise. v is known, so all that remains is to find R at (10,5). We need some equation for that. Any ideas ?
     
  4. Feb 9, 2015 #3
    I saw a unusual formula. From where it has come I don't know.
    Can you help on that?
    ##\frac{1}{R}## =##\frac{d^2y/dx^2}{[1+(dy/dx)^2]^{3/2}}##
     
  5. Feb 9, 2015 #4

    BvU

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    I can help on where it comes from: a simple google search brings us here :)
     
  6. Feb 9, 2015 #5
    Thanks, But in the link only formula is given with some description.
    Can you provide a derivation of it by some link or by yourself?
     
  7. Feb 9, 2015 #6

    BvU

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    Yes. Google "radius of curvature equation derivation" and end up here or here. :) I like the second one.
     
  8. Feb 9, 2015 #7
    I also liked the second link.:) Gave the derivation in a nice way. Now the question could be solved by plugging in values in formula .
    Only one question that remains is that how according to second link you have provided Curvature or kappa is inverse of radius?
     
  9. Feb 9, 2015 #8

    BvU

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    Curvature is the better word, I suppose. We associate radius with a circle. They show k = 1/R on page 5.
     
  10. Feb 9, 2015 #9

    ehild

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    You can solve the problem by yourself, by applying the chain rule.
    The equation of the path is given, y=x2/20
    You know that the components of the velocity are the time-derivatives of the coordinates. vx=dx/dt, vy=dy/dt. Apply chain rule to find vy in terms of vx: vy=2x/20 dx/dt = x/10 vx.
    The magnitude of the velocity is given |v|=6, so vx2+vy2=36. Find vx in terms of x from here.
    Once you know vx and vy , apply the chain rule again to derive the components of acceleration. Substitute the given coordinates and find the magnitude.
     
  11. Feb 9, 2015 #10
    Thanks BvU for giving wonderful link for derivation of curvature by differential geometry.

    Thanks ehild for showing me another way.
    Though it is a lengthy method comparing with curvature formula
    but is elegant because I think many do not know the curvature formula by differential geometry.
     
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