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Homework Help: Kinematics problem on Tetrahedron

  1. Aug 4, 2014 #1
    1. The problem statement, all variables and given/known data
    4 ants are arranged in such a way that they make up vertex of a regular tetrahedron, of side length 1m . The ants are named Calvin , Peter , David and Aron. Each ant moves at a speed 1m/s , and moves in such a way that:

    Calvin moves toward Peter,
    Peter moves toward David,
    David moves toward Arron,
    Arron moves toward Calvin

    If they continue to moves it this direction they will converge somewhere. What time in seconds will it take the ants to meet each other?



    2. Relevant equations




    3. The attempt at a solution
    The base of tetrahedron will form an equilateral triangle.
    Physics 1.png
    Length of side of equilateral triangle(a)=2cos 35.5
    It is given Peter moves toward David, David moves toward Arron.
    V[itex]_{PD}[/itex](relative velocity of Peter w.r.t.Arron) =√3V
    Physics 2.png
    Projection of V[itex]_{PD}[/itex] on line of motion of Peter=√3Vcos30[itex]^{°}[/itex]=[itex]\frac{3V}{2}[/itex]
    now
    [itex]^{0}_{a}[/itex]dl=[itex]^{t}_{0}[/itex][itex]\frac{3V}{2}[/itex]
    t=2a/3V
    =4*cos(35.5)/3
    But this is not correct.
     
    Last edited: Aug 4, 2014
  2. jcsd
  3. Aug 4, 2014 #2

    ehild

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    What do you mean with the equation above?

    The problem has a certain symmetry. What do you think where the ants meet?

    ehild
     
  4. Aug 4, 2014 #3
  5. Aug 4, 2014 #4
    pHYSICS 3.png
    I think they will meet at O.
     
  6. Aug 4, 2014 #5

    ehild

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    What do you mean? It is given that the sides of the tetrahedron are of 1 m length.

    ehild
     
  7. Aug 4, 2014 #6
    I thought 1 m length to be a bond length in tetrahedral structure as taken in chemistry.:tongue2:
    where as in this question it should be the edge length of the tetrahedron,
    According to figure in post#3 it is AC,BD,BC,AB.
    Are my further steps correct?
     
  8. Aug 4, 2014 #7

    ehild

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    Your picture with the ants is magnificent! 1e35c55ad8.df033bb1b8.CVOE2P.png

    I would like to see your real solution. I think the idea might be correct, but I am not sure. I would do it differently. What is the "official" solution?

    ehild
     
    Last edited: Aug 4, 2014
  9. Aug 4, 2014 #8
    Thank you ehild for the compliment.
    https://www.physicsforums.com/attachment.php?attachmentid=71860&d=1407155510
    From this figure
    If I put the length of side of this triangle=1m
    then t=2*1/3
    =0.67 sec.
    But the answer is 0.75 sec.
    This is my attempt at the solution.
    What is meant by "official" solution.
     
  10. Aug 4, 2014 #9

    ehild

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    I can not explain why your solution is incorrect . I think as the followed ant moves also perpendicularly to the edge, the distance to be covered is longer then a.
    I would do it differently. The centre of the tetrahedron is stationary and the ants meet there. As the ants move, they always form the vertexes of a tetrahedron which is rotated with respect to the original one and also shrinks. The velocities are along the edges of these tetrahedra which makes an angle with the radius. That is your 35.5° but rather 35.26 °as the tetrahedral angle is 109.5°. The cosine is √(2/3). The displacement is equal to the radius of the tetrahedron, and the radial velocity component is vr=vcos(35.26)=√(2/3) v. t=radius/vr
    That gives the answer.

    ehild
     
    Last edited: Aug 5, 2014
  11. Aug 5, 2014 #10
    Is this displacement equal to distance between vertex and centroid of Tetrahedral?
     
  12. Aug 5, 2014 #11
    I think that the centroid of the base triangle is just below the centroid of tetrahedral.
    Let c be the distance between the vertex and centroid of tetrahedral.
    pHYSICS 5.png
    So, by Pythagoras theorem
    1/3+b^2=c^2
    To find c I need to find another equation in terms of b and c.
    Could you please help,ehild?
     
  13. Aug 5, 2014 #12

    ehild

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    Yes.

    √(3/8) a

    ehild
     
  14. Aug 5, 2014 #13
    Doesn't the ant move along the edge?
    From the figure(in #2) Peter moves along BC and David moves along BD.
     
  15. Aug 5, 2014 #14
    Could you please explain it with figure?:redface:
     
  16. Aug 5, 2014 #15

    ehild

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    They always move along the edges of a tetrahedron, but the tetrahedron changes all time. Rotates and shrinks. The radial component of the velocity counts.

    ehild
     
    Last edited: Aug 5, 2014
  17. Aug 5, 2014 #16
    Then,what is meant by that?
    I am confused:confused:.
     
  18. Aug 5, 2014 #17

    ehild

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    Never mind. I can not explain it better. Do the other method. Here is a picture.

    I have to leave. Hoping you get the desired result.
     

    Attached Files:

  19. Aug 5, 2014 #18

    ehild

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    They start to move along those edges, but they change position and then it is not BC and BD any more. You get new positions for Peter and David and the others.

    ehild
     
  20. Aug 5, 2014 #19
    I got that.
    The triangle form by the red lines and black line is an isosceles triangle.
    So angle between vertex and centroid is (180-109.5°)/2
    V[itex]_{r}[/itex]=Vcos(35.26)
    Thank you ehild. Sorry for the inconvenience.

    But what is distance between vertex and centroid of Tetrahedron.
    I have worked on it in post#11.Please take a look.
     
  21. Aug 5, 2014 #20

    ehild

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    You need to apply Pythagoras a few times. Determine the height h of the tetrahedron from the yellow triangle.
    The blue triangle is isosceles. In the green triangle, one leg is h-r, the other is a/√3. r is the hypotenuse.

    Check the spoiler in post#12.

    ehild
     

    Attached Files:

  22. Aug 5, 2014 #21

    haruspex

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    I suspect the problem with your first method is that it supposes more symmetry than will be the case.
    Do the ants really continue to form the vertices of a regular tetrahedron?
    I modelled it in a spreadsheet and found that although the distances between successive ants are always equal, the distances between the first and third and between the second and fourth diverged from the first four distances. By the time they meet in the middle, the ratio is √2:1.

    Edit: The same modelling seems to give an answer around 0.8, so I'm not convinced 0.75 is right either.
     
    Last edited: Aug 5, 2014
  23. Aug 5, 2014 #22

    ehild

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    I have doubts, too, if the tetrahedral symmetry is maintained. The given result (t=0.75 s) suggests it was assumed. It would be nice to prove or disprove.

    ehild
     
  24. Aug 6, 2014 #23

    haruspex

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    It is clear that the four consecutive distances are always equal, and that the two non-consecutive are always equal, but there is no reason to suppose all six are always equal. My spreadsheet model indicates that the shape is asymptotically a flat square.
    I obtained the following equation:
    ##\ddot r r + 2 {\dot r}^2 + 6 \dot r v + 4 v^2 = 0##
    where r = r(t) is the distance between two consecutive ants and v is the speed of each ant.
    The evolution of this equation matches my earlier model, giving me confidence in both. Don't see how to solve it, though.
     
  25. Aug 6, 2014 #24
    Thank you ehild
    I got it √(3/8).
    t= √3/√8 *√3/√2=3/4=0.75
     
  26. Aug 6, 2014 #25

    Orodruin

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    Indeed the sides change at different rates. I solved this in cylinder coordinates with the axis through the center of the sides which are not parallel to the direction of any ant and with z=0 for the center of the tetrahedron. Doing so, I obtained
    $$
    \frac{dz}{dr} = 2\frac{z}{r}
    $$
    resulting in
    $$
    \frac{dz}{z} = 2\frac{dr}{r} \quad \ln(z/z_0) = 2\ln(r/r_0) \quad z = z_0\left(\frac{r^2}{r_0^2}\right).
    $$
    Thus, the tetrahedron is going to become flat and thus a square. The distance across the square is a factor ##\sqrt 2## larger than the distance between the ants, in accordance with haruspex's numerical findings.
     
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