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Angular momentum of a particle

  1. Apr 15, 2009 #1
    1. The problem statement, all variables and given/known data
    Determine the Angular Momentum of each particle about the origin

    Point A Located @ (-8,12), 6kg,4m/s, headed 60degrees below the horizon to the right
    Point B (2,1.5),4kg, 6m/s, headed 30 degrees above the horizon to the right
    Point C (6,-2), 2kg, 2.6m/s, headed on a 5,12,13 angled to the left and down


    2. Relevant equations

    L=r x p
    p=mv


    3. The attempt at a solution

    A. -8i+12j X 24(-i-8.66j)=
    B. 2i+1.5j X 24(i+.5j)=
    c. 6i-2j X 5.2(-12i-5j)=

    If this is right i dont know how to calculate it
     
  2. jcsd
  3. Apr 15, 2009 #2

    LowlyPion

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    Not quite right. To the right is +i isn't it?
    A. -8i+12j X 24(-i-8.66j)=
    should be
    A. (-8i + 12j) X 24*(.5i - .866j)=

    B. 2i+1.5j X 24(i+.5j)=
    should be
    B. (2i + 1.5j) X 24*(.866i + .5j)=

    c. 6i-2j X 5.2(-12i-5j)=
    should be
    c. (6i - 2j) X 5.2*(-12i - 5j -13k) =

    To take the cross product ...
    http://en.wikipedia.org/wiki/Cross_product#Computing_the_cross_product
     
    Last edited: Apr 15, 2009
  4. Apr 15, 2009 #3

    rl.bhat

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    (-8i + 12j) X 24*(.5i - 8.66j)

    It should be (-8i + 12j) X 24*(.5i - .866j)
     
  5. Apr 15, 2009 #4

    LowlyPion

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    Indeed it should. Thanks for the catch.
     
  6. Apr 16, 2009 #5
    ok i see my mistakes.

    just to clarify @ point C

    13 is not the k value, it is the third side of the triangle/hipotinuse
     
  7. Apr 16, 2009 #6
    Ok so here what i got

    A. (-8i+12j)X(12i-20.784j)=22.72k
    B. (2i+1.5j)X(2.784i+12j)=-7.176k
    C. (6i-2j)X(62.4i-26j)=280k
     
  8. Apr 16, 2009 #7

    LowlyPion

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    A. (-8 i + 12 j) X (12 i - 20.784 j)= 22.272 k

    C. (6i-2j)X(62.4i-26j)=280k
    Looking at the original direction again I think it should be -26 i -62.4 j. This yields a different result.
     
  9. Apr 16, 2009 #8
    Nope the triangle is not oriented like that


    the bottom is 12 or -12i, the side is on the right and up -5j.
     
  10. Apr 16, 2009 #9

    LowlyPion

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    OK. I can't see the picture, so going with that ... so using -62.4 i -26 j
    shouldn't it be -156 - 124.8 = -280.8 k
     
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