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Lazy flea and log. Kinda tough.

  1. Oct 22, 2009 #1
    I'm given the following problem. A log sits in the woods with a radius "r". Lazy flea wants to jump over the log with the least amount of velocity necessary. If distance "d" away from the center of the log, and angle theta is the angle of velocity above the horizontal. Find theta, d, and v(min).

    I'm mostly lost as to how the hell to even begin this equation. This is the first projectile motion problem I've had where I actually need to find the derivative of something. The maximum hight is not going to be 2R, because that makes the distance farther, and thus the velocity increases. The distance can not be too close either because then the angle increases and requires a larger velocity.

    With this many unknowns, i'm a bit lost. What I think I need to do is find a relation of the radius to this odd parabola of my projectile motion. So I'm looking at it like a geometry problem.

    Any hints are appreciated!
     
  2. jcsd
  3. Oct 22, 2009 #2

    berkeman

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    Staff: Mentor

    I think the max height of the parabola is indeed 2r. When the flea jumps, the horizontal component of the velocity will determine how long it takes him to reach the peak of the log. That has to match the time in the vertical direction for the flea to reach the height of the peak of the log. Use those two equations to determine the distance away from the log where he starts his jump, based on velocity.

    You might be right that there is some geometry involved as well. Draw some of the parabolas on top of the round log cross-section, to see how close you can get to the log before you can't just clip the top going over. You should be able to use a graphing calculator or Excel to play with that a bit to get some intuition going...
     
  4. Oct 22, 2009 #3
    I know that 2R is not the peak of the parabola because I was specifically told that it was not! The reason for this is for the apex to be 2R the distance away from the log increases. As the distance increases, the more velocity in the x component necessary. The problem requires however that I find the optimal minimum velocity out of all possible distances and thetas.
     
  5. Oct 22, 2009 #4

    berkeman

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    Ah, interesting variation! So maybe you can get the minimum initial v total, even though your vertical v has to be bigger. So yeah, use geometry to fit the parabolas to the contour of the circular log you have to clear, and calculate the initial v for each (find the equation for it). Then just minimize the total initial v with respect to the takeoff distance from the log.
     
  6. Oct 22, 2009 #5
    Sounds like a start. I'll try playing around with that. Let me know if you have any other ideas that may help me out.
     
  7. Oct 23, 2009 #6
    Alright, so I tried to fit some geometries together and assumed that wherever the flea jumped over would be tangent to the circle, so the distance from the center of the log and the point I called tangent were both distance "D" (being both tangent). Then I bisected them into two congruent triangles found theta in terms of R and D and plugged into my projectile equations, then took the derivative of v with respect to D, set for min and got my values.

    I showed this to the instructor and he pointed out, assuming tangent, I drew my motion in straight lines to those points. When actually, the motion being parabolic, this is not correct.

    Any ideas?

    If this wasn't clear let me know and I can scan some of my work.
     
  8. Oct 24, 2009 #7

    berkeman

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    Nice work on a solution! That's the right way to try out an idea and see if it looks correct. Have you been able to do some plotting on a calculator or with Excel or Mathematica to see intuitively how inverted parabolas and circles fit together? That may give you some insights to help your approach.

    You could try to brute force it by writing the equation for a circle, and the equation for a parabola, and finding where they just barely intersect (so only one solution per quadrant), and see if that gives you some geometrical insight into that the angle is to the center of the circle when that happens. Be sure to have variables for both the top of the parabola and the width of the parabola in your equation for the parabola...
     
  9. Oct 24, 2009 #8
    I haven't used mathematica or a graphing calculator to do very much for me, so I don't have much experience doing so. Is it possible you can help me on how to set this up into mathematica?
     
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