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Angular momentum of three particles of the same mass

  1. Dec 9, 2007 #1
    [SOLVED] angular momentum

    Figure 11-26 shows three particles of the same mass and the same constant speed moving as indicated by the velocity vectors. Points a, b, c, and d form a square, with point e at the center. Rank the points according to the magnitude of the net angular momentum of the three-particle system when measured about the points, greatest first (use only the symbols > or =, for example a>b>c=d=e).


    Okay, so angular momentum is equal to mass (radius cross product velocity). Since velocity is constant and the mass is the same, this leads only the radius or distance to compare. However, I am not quite sure how I should judge the distance of the particles. I mean what is the origin or the reference point that should use to judge these particles from?
     

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  3. Dec 9, 2007 #2

    Doc Al

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    Find the net angular momentum about each point (a, b, c, d, & e) then rank order them.
     
  4. Dec 9, 2007 #3
    But how am i able to find the net angular momentum.
     
  5. Dec 9, 2007 #4

    Doc Al

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    All you need is the relative net angular momentum. Just label the distance between a and b by some symbol (how about D, for distance?) and crank it out.
     
  6. Dec 9, 2007 #5
    I am still not understanding this problem. I came up with the answer e>a=d>b=c. I am sure that c is the lowest value but i do not know what the rank of the other values are.
    thanks for your help
     
  7. Dec 9, 2007 #6

    Doc Al

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    Just calculate the net angular momentum about each point and compare. (Your answers will be in terms of the distance D that I defined earlier, as well as m and v.)

    For example, what do you get when you compute the net angular momentum about point a?
     
  8. Dec 9, 2007 #7
    i understand momentum to be m(RxV). i do not understand how to calculate this around three points. Do i use all three or do i use the closest velocity to athe distance beteww point a and the velocity vector would be .5d, right, but do i have to compare a to the other velocity vectors
     
  9. Dec 9, 2007 #8

    Doc Al

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    Using point a as the reference, what's the angular momentum of each particle? (Too bad they don't have names. Let's call them 1, 2, and 3, from left to right.) The angular momentum of particle 1 is zero (since r X v is zero); the angular momentum of particle 2 is +mvD (since r X v = Dv); the angular momentum of particle 3 is also +mvD (since r X v = Dv).

    Does this make sense? So what's the net angular momentum about point a? (Add them up.)

    Do the same for each point.
     
  10. Dec 9, 2007 #9
    when i did that i came up with the answers a=+2mvD, b=-mvD, c=+mvD, d=+2mvD and e=+.5mvD. Is this what you get this gives me the answer a=d>c>e>b
     
  11. Dec 9, 2007 #10

    Doc Al

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    Recheck your value for d. (Also, even though only magnitude matters here, recheck the signs for b & c.)

    But that's the right idea. Just take the magnitudes and rank order them.
     
  12. Dec 9, 2007 #11
    so c is a -mvD and b is +mvD. d would be +mvD. Is this needed to be changed.
     
  13. Dec 9, 2007 #12

    Doc Al

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    Right.
    No. What's the angular momentum (about point d) for each particle? (Careful with signs.)
     
  14. Dec 9, 2007 #13
    so from particle 1 (closest to a) d=-mvD from particle 2 (between a and b) d=+mvD and from particle 3 d =0

    this will make the answer a>b>e>d>c
     
  15. Dec 9, 2007 #14

    Doc Al

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    So what's the total angular momentum about d?

    When you rank order, ignore the signs.
     
  16. Dec 9, 2007 #15
    Thank you so much for your quick replies i was able to get the answer. Thank you for your help
     
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