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Irodov kinematics question

  1. Oct 2, 2012 #1
    1. The problem statement, all variables and given/known data
    Three points are located at the vertices of an equilateral triangle whose sides equals a. They all start moving simultaneously with velocity v constant in modulus, with the first point heading continually for the second, the second for the third, and the third for the first. How soon will the points converge?


    2. Relevant equations



    3. The attempt at a solution
    I suppose this one is an easy problem, but i am still not sure if i did it the correct way. I found the original distance of a point from the centroid (where the points converge) which is a/√3 and the component of velocity along the line joining the centroid and a point is vcos(30°), therefore time taken by the points to converge is vcos(30°)/(a/√3)=2a/3v. This matches the answer given in the answer key but i don't understand why this method works here? What if it was not an equilateral triangle? Any explanation on this would help.

    Thanks!
     
  2. jcsd
  3. Oct 2, 2012 #2

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    Hi Pranav!

    Suppose all 3 points have moved a little.
    What does the resulting figure look like?
    And what is the radial speed then?
     
  4. Oct 2, 2012 #3
    Hello ILS! :smile:
    It will be still an equilateral triangle.
    Um..i have no idea on this. :uhh:
     
  5. Oct 2, 2012 #4

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    Yep!

    No? So how did you get vcos(30°)?
     
  6. Oct 2, 2012 #5
    I was confused when you said "radial speed". Sorry about that! :redface:
     
  7. Oct 2, 2012 #6

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    Ah yes, radial speed is the amount that the distance to the centroid gets shorter per unit of time.

    So... does that answer your question?
     
  8. Oct 2, 2012 #7
    Thanks, but what about my second question? What would happen if the sides were unequal?
     
  9. Oct 2, 2012 #8

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    Oh yes. Completely missed that. Sorry.

    Well, then it becomes a bit more complex doesn't it?
    Then you can't use the symmetry anymore.

    So what I'd do is draw a non-equilateral triangle - any triangle will do.
    Say, a triangle with a base of 2 cm and 10 cm high?
    Then take a step of a couple of millimeters in the proper directions and draw the new triangle.
    Then repeat a couple of times.
    That should give you a clue about what's happening...

    Alternatively, you can set up equations and try and solve those.
    But that is way more work, and you don't like a lot of work do you?

    So I'd first try to draw a couple of steps and see if you can find a pattern.
    At worst you'll have a pretty figure!
     
  10. Oct 2, 2012 #9
    I was expecting that.

    Thanks for the tips, i would be sure to use your hints if i encounter any question like this. For now, i am off to bed, getting late here. :smile:

    Yes, i hate it. :tongue2:
     
  11. Oct 2, 2012 #10

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    Ah, but you do like pretty pictures, don't you?
     
  12. Oct 2, 2012 #11
    Yes. :smile:
     
  13. Oct 2, 2012 #12

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    Oh well, sleep tight and dream of rotating triangles that becomes smaller and smaller. :zzz:
     
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