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Body moving on an ellipse; find velocity, acceleration

  1. Apr 14, 2008 #1
    1. The problem statement, all variables and given/known data

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

    2. Relevant equations

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

    3. The attempt at a solution

    Since v=const, [tex] \frac{dv}{dx}\vec{\tau} = 0 [/tex] and therefore [tex] \vec{a} = 0 + \frac{d\vec{\tau}}{dx}v = \frac{v^{2}_{0}}{\rho}\vec{n} [/tex] where [tex] \rho [/tex] 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 [tex] \rho [/tex] as [tex] \rho(x,y) [/tex]

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

    If so far everything has been correct (although I doubt about it) the only problem is expressing the radius [tex] \rho(x,y) [/tex].
     
  2. jcsd
  3. Apr 14, 2008 #2
    HINT;

    If a body moves in an elliptical path,sum of its distances from two fixed points always remains constant..
     
  4. Apr 14, 2008 #3

    tiny-tim

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    Hi raul_l! :smile:

    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 … ? :smile:
     
  5. Apr 14, 2008 #4
    x=a*cos(t)
    y=b*sin(t)
    I get [tex] \vec{v}=(a*cos(t), b*sin(t)) [/tex] and therefore [tex] \vec{a}=(a*sin(t), -b*cos(t)) [/tex]

    It has to be expressed as [tex] \vec{v}(x,y) [/tex] and [tex] \vec{a}(x,y) [/tex] 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?
     
  6. Apr 14, 2008 #5

    tiny-tim

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    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. :rolleyes:
    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 = … ? :smile:

    (btw, an alternative useful parameter would have been the usual arc-length, s, since we know, in this case, ds/dt = vº.)
     
  7. Apr 14, 2008 #6
    [tex] \vec{r}= (a*cos(\theta),b*sin(\theta) [/tex]

    [tex] \vec{v}= \frac{d \vec{r}}{dt} [/tex]

    [tex] 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 [/tex]

    [tex] v_{0}^{2} dt^2 = (a^2 sin^2 (\theta) + b^2 cos^2 (\theta)) d \theta^2 [/tex]

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

    Is this correct? I don't know how to take this integral.
     
  8. Apr 14, 2008 #7

    tiny-tim

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    Hi raul_l! :smile:

    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 … ? :smile:
     
  9. Apr 14, 2008 #8
    [tex] \frac{d \theta}{dt} = v_{0} \sqrt{ \frac{1}{a^2 sin^2 (\theta) + b^2 cos^2 (\theta)} }[/tex] I still don't know what to do with this. :confused:

    [tex] \vec{r}= (a*cos(\theta),b*sin(\theta) [/tex]

    and therefore

    [tex] \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)}}) [/tex]

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

    [tex] \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}}}) [/tex]

    In order to get a I need d²θ/dt². How do I do that?
     
  10. Apr 14, 2008 #9

    tiny-tim

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    ooh, that's messy. :frown:

    The square-root can be simplified … [tex] \frac{ab}{\sqrt{ a^4 y^2\,+\,b^4 x^2}}[/tex] ! :smile:
    easy-peasy … d²θ/dt² = (d/dt)(dθ/dt) = [(d/dθ)(dθ/dt)]dθ/dt. :smile:
     
  11. Jun 11, 2008 #10
    Ok, thanks a lot. That really helped.
     
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