How Does a Particle Reach Infinity in Finite Time with Angular Force?

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Ertosthnes
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Homework Statement


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.


Homework 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])

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?
 
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Ertosthnes said:
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?

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)
 
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?
 
Ertosthnes said:
Thanks Tim! Okay, obviously the integral as is would be pretty tough to solve. Could I say that dt = dr/√(Ar^4 + B) ≤ dr/(Ar^2), and then integrate to show that t<infinity?

Hi Ertosthnes! :smile:

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

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