Finding acceleration of a skier on parabola

[Raghav Gupta]

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?

 

[BvU]

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 ?

 

[Raghav Gupta]

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?
$\frac{1}{R}$ =$\frac{d^2y/dx^2}{[1+(dy/dx)^2]^{3/2}}$

 

[BvU]

I can help on where it comes from: a simple google search brings us here :)

 

[Raghav Gupta]

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?

 

[BvU]

Yes. Google "radius of curvature equation derivation" and end up here or here. :) I like the second one.

 

[Raghav Gupta]

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?

 

[BvU]

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

 

[ehild]

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. v_x=dx/dt, v_y=dy/dt. Apply chain rule to find v_y in terms of v_x: v_y=2x/20 dx/dt = x/10 v_x.
The magnitude of the velocity is given |v|=6, so v_x²+v_y²=36. Find vx in terms of x from here.
Once you know v_x and v_y , apply the chain rule again to derive the components of acceleration. Substitute the given coordinates and find the magnitude.

 

[Raghav Gupta]

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. v_x=dx/dt, v_y=dy/dt. Apply chain rule to find v_y in terms of v_x: v_y=2x/20 dx/dt = x/10 v_x.
The magnitude of the velocity is given |v|=6, so v_x²+v_y²=36. Find vx in terms of x from here.
Once you know v_x and v_y , 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.