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Charged particles arranged in tetrahedron

  1. May 4, 2014 #1

    utkarshakash

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    1. The problem statement, all variables and given/known data
    Four charged particles (A, B, C, D), of mass m and charge q each, are connected by light silk threads of length d forming a tetrahedron floating in outer space. The thread connecting particles A and B suddenly snaps. Find the maximum speed of particle A after that.

    3. The attempt at a solution

    I'm trying to use conservation of energy. The initial potential energy of the system will be
    [itex]\dfrac{6q^2}{4 \pi \epsilon_0 d} [/itex]

    But I can't realize when will the speed of particle A be maximum. Also, what will be the speed of other particles at that instant? Will they be equal.
     

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    Last edited: May 4, 2014
  2. jcsd
  3. May 4, 2014 #2
    I think thats not correct.
     
  4. May 4, 2014 #3

    utkarshakash

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    Sorry. My bad. Replace 4 with 6.
     
  5. May 4, 2014 #4
    The maximum speed (and maximum kinetic energy) is achieved when the particle A and B swing the furthermost from each other (minimizing potential energy).
     
  6. May 4, 2014 #5

    utkarshakash

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    Would the final arrangement of particles be like this (see attached image) ?
     

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  7. May 4, 2014 #6
    Looks right to me, so what is the final potential energy? (Notice that you look for change in potential energy, you don't have to write down the complete expression for final potential energy)
     
  8. May 4, 2014 #7
    Yes, except that the triangles ought to be equilateral.
     
  9. May 4, 2014 #8

    utkarshakash

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    This means one of the strings wouldn't be taut then, right?
     
  10. May 4, 2014 #9

    SammyS

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    Yes, the one which snapped .
     
  11. May 4, 2014 #10

    utkarshakash

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    No. I'm talking about the strings except the one which snapped. According to dauto the triangles won't be equilateral. What, then, will be the length of sides of the triangle so formed?
     
  12. May 4, 2014 #11

    TSny

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    In post #7 dauto said the triangles will be equilateral.
     
  13. May 4, 2014 #12

    ehild

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    No dauto said that the triangles should be equilateral. They did not look like that in your picture.

    ehild
     
  14. May 4, 2014 #13

    utkarshakash

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    Looks like I misunderstood his post. :redface:

    I got the correct answer.
     
  15. May 4, 2014 #14

    TSny

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    Nice. Good work!
     
  16. May 5, 2014 #15
    May be I ought not use the word ought...
     
  17. May 5, 2014 #16

    TSny

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    :smile:
    ------------

    By the way, utkarshakash. Did you find that all 4 of the particles have the same speed when particle A has maximum speed?
     
  18. May 5, 2014 #17

    utkarshakash

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    Yes. But I still have doubt why they would all have the same speed.
     
  19. May 5, 2014 #18

    TSny

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    Can you describe the motion of the center of mass of the system after the string is cut?
     
  20. May 5, 2014 #19

    utkarshakash

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    Initially the center of mass is at rest. The net external force on the system is zero as tension forces are internal to it. Thus, the COM will remain at rest even after the string is cut and so, the particles will have same speed in opposite directions. Is this explanation correct?
     
  21. May 5, 2014 #20

    TSny

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    Yes, the COM must remain at rest. The 4 particles will not have the same speed at arbitrary times, but you should be able to argue that they have the same speed at the instant of time when all 4 particles lie in the same plane.
     
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