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Describing the motion of a particle using polar coordinates

  1. Nov 28, 2014 #1
    1. Problem
    Consider a particle that feels an angular force only of the form:
    F_θ = 3mr'θ'. Show that r' = ± (Ar^4 + B)^(1/2), where A and B are constants of integration, determined by the initial conditions. Also, show that if the particle starts with θ' ≠ 0 and r' > 0, it reaches r = ∞ in a finite time.

    3. The attempt at a solution
    So I understand the first part of the question and can easily show r' = … using θ' = L/mr^2, where L is the angular momentum. Now my issue is with the second part. I know I probably should set up an integral to evaluate from r_0 to r = ∞. I tried starting at F_θ = 3mr'θ' = m(dθ'/dt). I think this equation is correct (although may not be what I should be using). From here I would separate variables, as 3r'dt = (1/θ')dθ'. But this doesn't seem to be right. I'm really stuck here. Any ideas ?
     
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  3. Nov 29, 2014 #2

    Orodruin

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    What do the equations of motion look like in polar coordinates?
     
  4. Nov 29, 2014 #3
    Isn't it just 3mr'θ' = m(dθ'/dt), since the radial force is 0 ?

    Edit: Or are you referring to the general form:
    F_θ = m(rθ' + 2r'θ')
     
    Last edited: Nov 29, 2014
  5. Nov 29, 2014 #4

    Orodruin

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    No, you have a problem in curvilinear coordinates and have to take into account that the basis vectors change. I suggest writing down the equations of motion in Cartesian coordinates and transform them to polar. You will have both radial and tangential acceleration in general (if the radial acceleration was zero, how would you get to infinity in finite time?).

    Also, your can see that your equation is wrong by dimensional analysis, the left hand side has units mass x length / time^2 and the right hand side mass / time^2.
     
  6. Nov 29, 2014 #5
    Okay but I assumed since the problem specified that the particle feels only an angular force that would mean the radial force is 0, an thus the radial acceleration was 0.
     
  7. Nov 29, 2014 #6

    Orodruin

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    In curvilinear coordinates, there is a difference between force and (coordinate) acceleration being equal to zero.
     
  8. Nov 29, 2014 #7
    Okay I'm a bit confused. Since the force is given in polar coordinates why would I need to go back to Cartesian ?
     
  9. Nov 29, 2014 #8

    Orodruin

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    You don't need to, but you must figure out what the equations of motion are in one way or another.
     
  10. Nov 29, 2014 #9
    okay so the radial component of the force in polar coordinates would generally be given as
    F_r = m ( r'' − r θ' ^2 ). But since F_r = 0, we have mr'' = r θ' ^2 and m(rθ' + 2r'θ') = 3mr'θ'. Am I on the right track ?
    From here i would still need to integrate from 0 to infinity I would think. This seems to be a real challenge with the above equations however.
    I guess my real issue is I'm not entirely sure how I would show that r goes to infinity in a finite time period.
     
    Last edited: Nov 29, 2014
  11. Nov 29, 2014 #10

    Orodruin

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    Your force equations look ok now. Remember, you are not tasked with computing the finite time - only with showing that it is finite. Your integral is also not from r=0 as it is difficult to have a tangential velocit in that particular point.

    And just to check: Did you use that angular momentum is conserved in your solution to the first part? Since you have a tangential force, this will not be true in general.
     
  12. Nov 29, 2014 #11
    I used L = θ'mr^2, where L is angular momentum. I arrived at the right answer, although it was only a direct substitution of θ' into the F_θ. So I didn't completely rely on the conservation property.
     
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