How Do You Calculate Velocity on an Elliptical Path at a Specific Point?

[pollytree]

Homework Statement

There is an elliptical path and pegs A and B are restricted to move around it. If the link moves with a constant speed of 10m/s, determine the magnitude of velocity when x=0.6m

[PLAIN]http://users.adam.com.au/shortround/Prob.12-78.jpg

Homework Equations

\frac{x^2}{4}+y²=1

\rho=\frac{(1+(\frac{dy}{dx})^2)^\frac{3}{2}}{\left|\frac{dy}{dx}\right|}

a=\frac{dv}{dt} \vec{e}_t+\frac{v^2}{\rho} \vec{e}_n

Where \rho is the radius of curvature.

The Attempt at a Solution

I rearranged \frac{x^2}{4}+y²=1 to get x²+4y²=4

I then differentiated this to get: \frac{dy}{dx}=\frac{-x}{4y} and \frac{d^2y}{dx^2}=\frac{(x/y)-1}{4y}

Using x=0.6m, y=\sqrt{0.91}=0.9539

By substituting this into the derivative and second derivatives and then putting these into the radius of curvature equation, I found the radius of curvature. However I am not sure if this is the correct way to do it. Also once I have the radius of curvature, how do I find the velocity?

Thanks!

 

[rl.bhat]

Υou need not calculate the radius of curvature.
When x = 0.6m find y using the equation of ellipse. If O is the center of the ellipse, find the angle AOC. Then the velocity of A at that position is v*sinθ

 

[pollytree]

That doesn't work. I have an example in my book with x=1.0m and hence y=sqrt(0.75). They give v=10.4m/s, however your method gives v=6.55m/s.

 

[rl.bhat]

To find the angle, find the slope tanθ = dy/dx at x = 0.6 m, and then proceed.