Finding acceleration of a skier on parabola

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Homework Statement


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.

Homework 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.

The Attempt at a Solution



Slope is 2x/20= x/10.
For straight line determining acceleration is easy but how about this curve case?
 

Answers and Replies

  • #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 ?
 
  • #3
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so all that remains is to find R at (10,5). We need some equation for that. Any ideas ?
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}}##
 
  • #4
BvU
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I can help on where it comes from: a simple google search brings us here :)
 
  • #5
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I can help on where it comes from: a simple google search brings us here :)
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?
 
  • #6
BvU
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Yes. Google "radius of curvature equation derivation" and end up here or here. :) I like the second one.
 
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  • #7
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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?
 
  • #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.
 
  • #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.
 
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  • #10
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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.
Thanks BvU for giving wonderful link for derivation of curvature by differential geometry.

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.
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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