# Body moving on an ellipse; find velocity, acceleration

1. Apr 14, 2008

### raul_l

1. The problem statement, all variables and given/known data

A body is moving on a trajectory $$\frac{x^2}{a^2} + \frac{y^2}{b^2} =1$$ vith a constant speed $$v_{0}$$. Find its velocity $$\vec{v}$$ and acceleration $$\vec{a}$$.

2. Relevant equations

As far as I know $$\vec{a} = \vec{a}_{\tau} + \vec{a}_{n} = \frac{dv}{dx}\vec{\tau} + \frac{d\vec{\tau}}{dx}v$$

3. The attempt at a solution

Since v=const, $$\frac{dv}{dx}\vec{\tau} = 0$$ and therefore $$\vec{a} = 0 + \frac{d\vec{\tau}}{dx}v = \frac{v^{2}_{0}}{\rho}\vec{n}$$ where $$\rho$$ is the radius. Since we are dealing with an ellipse, the radius is a function of x and y, but I don't know how to express $$\rho$$ as $$\rho(x,y)$$

I think that I can express velocity and acceleration like this:
$$\vec{v}=v_{0}(\frac{y}{b} , \frac{x}{a})$$
$$\vec{a}=\frac{v^{2}_{0}}{\rho}(\frac{x}{a} , -\frac{y}{b})$$

If so far everything has been correct (although I doubt about it) the only problem is expressing the radius $$\rho(x,y)$$.

2. Apr 14, 2008

### physixguru

HINT;

If a body moves in an elliptical path,sum of its distances from two fixed points always remains constant..

3. Apr 14, 2008

### tiny-tim

Hi raul_l!

What you need is a sensible parameter.

If you were told x and y as a function of time, you'd know what to do, wouldn't you?

Well, any parameter will do … though, of course, some make it more complicated than others.

Hint: the equation looks a lot like cos²θ + sin²θ = 1. So … ?

4. Apr 14, 2008

### raul_l

x=a*cos(t)
y=b*sin(t)
I get $$\vec{v}=(a*cos(t), b*sin(t))$$ and therefore $$\vec{a}=(a*sin(t), -b*cos(t))$$

It has to be expressed as $$\vec{v}(x,y)$$ and $$\vec{a}(x,y)$$ however, so I have to express t as t(x,y) somehow. I'm not sure how to go on from here. I can think of the following: dt=vds where ds^2=dx^2+dy^2. How do I proceed?

5. Apr 14, 2008

### tiny-tim

That's the right idea … but you should only use t for time, or you'll get very confused ("so I have to express t as t(x,y) somehow"). That's why I suggested θ.
Nooo … if (x,y) is X, then:

dX/dθ = (-a.sinθ,b.cosθ).

Now, you're told that speed is a constant, v°.

So you can get a differential equation for θ from:

v°² = (dX/dt)² = (dX/dθ)²(dθ/dt)² = … ?

(btw, an alternative useful parameter would have been the usual arc-length, s, since we know, in this case, ds/dt = vº.)

6. Apr 14, 2008

### raul_l

$$\vec{r}= (a*cos(\theta),b*sin(\theta)$$

$$\vec{v}= \frac{d \vec{r}}{dt}$$

$$v_{0}^{2} = (\frac{d \vec{r}}{dt})^2 = (\frac{d \vec{r}}{d \theta})^2 (\frac{d \theta}{dt})^2 = (-a*sin(\theta),b*cos(\theta)^2 (\frac{d \theta}{dt})^2$$

$$v_{0}^{2} dt^2 = (a^2 sin^2 (\theta) + b^2 cos^2 (\theta)) d \theta^2$$

$$v_{0}t = \int{ \sqrt{(a^2 sin^2 (\theta) + b^2 cos^2 (\theta))} d \theta }$$

Is this correct? I don't know how to take this integral.

7. Apr 14, 2008

### tiny-tim

Hi raul_l!

You don't need to take the integral. You're not asked to find r or θ in terms of t.

You're only asked for dr/dt and d²r/dt².

And you know what dθ/dt is.

So … ?

8. Apr 14, 2008

### raul_l

$$\frac{d \theta}{dt} = v_{0} \sqrt{ \frac{1}{a^2 sin^2 (\theta) + b^2 cos^2 (\theta)} }$$ I still don't know what to do with this.

$$\vec{r}= (a*cos(\theta),b*sin(\theta)$$

and therefore

$$\frac{d \vec{r}}{dt} = (a\frac{d cos(\theta)}{dt},b\frac{d sin(\theta)}{dt}) = (a\frac{d cos(\theta)}{d \theta}\frac{d \theta}{dt},b\frac{d sin(\theta)}{d \theta}\frac{d \theta}{dt}) = (-a*sin(\theta) v_{0} \sqrt{ \frac{1}{a^2 sin^2 (\theta) + b^2 cos^2 (\theta)}},b*cos(\theta) v_{0} \sqrt{ \frac{1}{a^2 sin^2 (\theta) + b^2 cos^2 (\theta)}})$$

By substituting θ with arccos(x/a) or arcsin(y/b) I get

$$\frac{d \vec{r}}{dt} = (-a\frac{y}{b} v_{0} \sqrt{ \frac{1}{a^2 \frac{y^2}{b^2} + b^2 \frac{x^2}{a^2}}},b\frac{x}{a} v_{0} \sqrt{ \frac{1}{a^2 \frac{y^2}{b^2} + b^2 \frac{x^2}{a^2}}})$$

In order to get a I need d²θ/dt². How do I do that?

9. Apr 14, 2008

### tiny-tim

ooh, that's messy.

The square-root can be simplified … $$\frac{ab}{\sqrt{ a^4 y^2\,+\,b^4 x^2}}$$ !
easy-peasy … d²θ/dt² = (d/dt)(dθ/dt) = [(d/dθ)(dθ/dt)]dθ/dt.

10. Jun 11, 2008

### raul_l

Ok, thanks a lot. That really helped.