# Motion Around An Ellipse

## 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

## 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!!

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rl.bhat
Homework Helper
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

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