Body moving on an ellipse; find velocity, acceleration

[raul_l]

Homework Statement

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}.

Homework 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

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).

 

[physixguru]

HINT;

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

 

[tiny-tim]

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}.

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 … ?

 

[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?

 

[tiny-tim]

x=a*cos(t)
y=b*sin(t)

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.

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

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.)

 

[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.

 

[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 … ?

 

[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?

 

[tiny-tim]

\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}}})

ooh, that's messy.

The square-root can be simplified … \frac{ab}{\sqrt{ a^4 y^2\,+\,b^4 x^2}} !

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

easy-peasy … d²θ/dt² = (d/dt)(dθ/dt) = [(d/dθ)(dθ/dt)]dθ/dt.

 

[raul_l]

Ok, thanks a lot. That really helped.