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Homework Help: Minimum possible period

  1. Dec 8, 2018 at 1:47 PM #1
    1. The problem statement, all variables and given/known data
    A particle moves periodically around an ellipse of equation ##\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1##. You can assume ##a>b##. The ##x## and ##y## components of the particle's velocity can never exceed ##v## at any point. What is the minimum possible period of the periodic motion of the particle?

    2. Relevant equations

    Motion in an ellipse. Kinematics.
    3. The attempt at a solution

    It seems I could divide the ellipse into sectors and come up with constraints for the time the particle takes to travel in those sectors. I am not sure though how?
     
    Last edited: Dec 8, 2018 at 1:58 PM
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  3. Dec 8, 2018 at 1:52 PM #2

    haruspex

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    On a given portion of the path, what attribute determines which constraint limits the overall speed?
     
  4. Dec 8, 2018 at 1:59 PM #3
    In the limit both components of the velocity would be ##v## for a speed of ##\sqrt{2}v##.
     
  5. Dec 8, 2018 at 2:01 PM #4

    haruspex

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    Right, but where on the path does that happen?
     
  6. Dec 8, 2018 at 2:08 PM #5
    The two components would be equal to each other when a line connecting the particle to the origin is ##45^{\circ}## with the x-axis?
     
  7. Dec 8, 2018 at 2:12 PM #6

    haruspex

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    Yes.
    Does this allow you to make progress?

    Edit: blunder by me... fortunately picked up by gneill.
     
    Last edited: Dec 8, 2018 at 4:45 PM
  8. Dec 8, 2018 at 2:18 PM #7
    Is it always true that the components of the velocity are equal at 45 degrees? Or only for an ellipse/circle? (I am a bit confused about the basics here, I intuitively said that at 45 degrees the components are equal but not sure why)
     
  9. Dec 8, 2018 at 2:43 PM #8

    haruspex

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    If the speed is v and the direction is at 45 degrees to the axes, what are the x and y components?
     
  10. Dec 8, 2018 at 2:49 PM #9
    Ok got it. So I think you are suggesting that I divide the ellipse into 2 regions using lines at 45 degrees. If the angle is smaller than 45 then the y component of the velocity is always bigger than the x component. So I can assume that at its maximum the y component is ##v## in all the range were the angle is smaller than 45 degrees. I need to relate this to time and arc length somehow.
     
  11. Dec 8, 2018 at 2:54 PM #10

    gneill

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    You need to find where the slope of the ellipse is 45°. For an ellipse this will not be at the same location as where a 45° line passing through the origin intercepts the ellipse.
     
  12. Dec 8, 2018 at 3:01 PM #11
    Good point, I was missing this.
     
  13. Dec 8, 2018 at 6:00 PM #12
    For the region where the angle is less than 45, do I assume the vertical velocity is a constant equal to the maximum of ##v##, find the vertical distance the particle moved and find the time by dividing that distance by v?
     
  14. Dec 8, 2018 at 6:06 PM #13

    haruspex

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    It depends what angle you mean.
    Where the magnitude of the slope exceeds 1 the limiting constraint is the y component.
     
  15. Dec 8, 2018 at 6:06 PM #14

    gneill

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    Yes, sounds like a good plan. At the critical point where the velocities are equal, you "hand off" the maximum velocity to the x-component.

    Edit: I just realized (thanks for the heads-up @haruspex) that by the angle you were referring to the slope of the ellipse. It will be greater than 45° w.r.t. the horizontal until the hand off point. Vertical (so infinite) at the major axis ends, horizontal (so zero) at the minor axis ends.
     
    Last edited: Dec 8, 2018 at 6:15 PM
  16. Dec 8, 2018 at 6:54 PM #15

    Ray Vickson

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    You do not say how the velocity varies as you go around the ellipse. Your subsequent workings seem to be assuming that at any point, either ##|v_x| = v## or ##|v_y|= v##, or both. Of course, it is intuitive that such a velocity profile will be needed to minimize the cycle time, but you should at least state that explicitly. It never hurts to point out such things so that the person marking the question can see that you have thought about it.
     
  17. Dec 8, 2018 at 8:33 PM #16

    gneill

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    I see that you've marked the question as solved. I hope that you appreciate the mathematical elegance of the final solution :smile:
     
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