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Motion in a Plane Problem

  1. Oct 30, 2008 #1
    1. The problem statement, all variables and given/known data
    Consider a particle that feels an angular force only, of the form F[tex]_{\theta}[/tex] = 3m[tex]\dot{r}[/tex][tex]\dot{\theta}[/tex]. Show [tex]\dot{r}[/tex]=[tex]\pm[/tex][tex]\sqrt{Ar^{4}+B}[/tex], where A and B are constants of integration, determined by the initial conditions. Also, show that if the particle starts with [tex]\dot{\theta}[/tex][tex]\neq[/tex]0 and [tex]\dot{r}[/tex]>0, it reaches r=[tex]\infty[/tex] in a finite time.


    2. Relevant equations
    F[tex]_{r}[/tex]=m([tex]\ddot{r}[/tex]-r[tex]\dot^{\theta}[/tex]^2)=0
    F[tex]_{\theta}[/tex]=m(r[tex]\ddot{\theta}[/tex]+2[tex]\dot{r}[/tex][tex]\dot{\theta}[/tex])

    3. The attempt at a solution
    I've already shown that [tex]\dot{r}[/tex]=[tex]\pm[/tex][tex]\sqrt{Ar^{4}+B}[/tex]. What I need to do now is show that it reaches r=[tex]\infty[/tex] in a finite time. I'm not sure what I need to do here... any thoughts?
     
  2. jcsd
  3. Oct 31, 2008 #2

    tiny-tim

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    Hi Ertosthnes! :smile:

    (have a theta: θ and a square-root: √ and an infinity: ∞ :smile:)

    (ooh … and use dashes rather than dots on this forum … they're easier to read!)


    You need to solve dr/√(Ar4 + B) = dt. :wink:

    (or you could "sandwich" it between two integrals that are easier)
     
  4. Oct 31, 2008 #3
    Thanks Tim! Okay, obviously the integral as is would be pretty tough to solve. Could I say that dt = dr/√(Ar^4 + B) [tex]\leq[/tex] dr/(Ar^2), and then integrate to show that t<infinity?
     
  5. Oct 31, 2008 #4

    tiny-tim

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    Hi Ertosthnes! :smile:

    (you could have used the ≤ a also :wink:)

    … and it's always positive, so … yes, that's fine! :smile:
     
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