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Angular momentum cross product - Please help!

  1. Dec 1, 2012 #1
    1. The problem statement, all variables and given/known data

    A 1.47kg particle moves in the xy plane with a velocity of v = (4.59i - 3.28j)m/s. Determine the magnitude of the particle's angular momentum when its position vector is r = (1.35i + 2.57j)m.


    2. Relevant equations

    p = mv

    L = r x p (the x is supposed to be a cross product and not a variable)

    L = r x mv

    3. The attempt at a solution

    First I scaled the velocity vector: v = (4.59i - 3.28j)m/s by the mass, 1.47 kg, to get a new momentum vector (6.75i - 4.82j)kg*m/s.

    Then I took the cross product of the r vector with the new momentum vector:
    (1.35i + 2.57j)m x (6.75i - 4.82j)kg*m/s (I let a=1.35, b=2.57, c=0, d=6.75, e=-4.82, and f=0, the got the k vector cross product by doing k=ae-bd)

    The answer I got was -23.9 kg*m^2/s, which wasn't right.

    What did I do wrong? Am I even anywhere near the correct solution/answer?
     
    Last edited: Dec 1, 2012
  2. jcsd
  3. Dec 1, 2012 #2

    gneill

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    Staff: Mentor

    Did you specify the direction of the angular momentum? (It's a pseudo-vector, so it has a magnitude and direction).
     
  4. Dec 1, 2012 #3
    I didn't, but I just went back and tried it and it picked up the k as being part of the unit. So, it's not that, I don't think. The question also just asks for magnitude.
     
  5. Dec 1, 2012 #4

    gneill

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    Staff: Mentor

    Okay, just the magnitude. What answer did you type in for that?
     
  6. Dec 1, 2012 #5
    -23.9 kg*m^2/s
     
  7. Dec 1, 2012 #6

    gneill

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    Staff: Mentor

    Are magnitudes ever negative?
     
  8. Dec 1, 2012 #7
    Nope. Wow, I feel a little stupid now. Thanks!
     
  9. Dec 1, 2012 #8

    gneill

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    Staff: Mentor

    Heh. We've all been there, done that, got the T-shirt. Cheers.
     
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