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Conservation of angular momentum

  1. Jul 27, 2012 #1
    1. The problem statement, all variables and given/known data

    A 0.5-kg particle is located at the point r = 3m i + 4m k and is moving with a velocity v = 5m/s i -2m/s k . What is the angular momentum of this particle about the origin?
    A) 13 kgm^2/s k
    B) 13 kgm^2/s j
    C) 26 kgm^2/s k
    D) 26m^2/s j
    E) 13 kgm^2/s i



    2. Relevant equations
    L = RxP
    where L is angular momentum, R is the distance from the origin and P is the momentum and the x stands for the cross product
    P = mv



    3. The attempt at a solution

    P = 0.5(5m/s i - 2m/s k)
    = 5/2 kgm^2/s i - 1kgm^2/s

    L = (3i + 4i) x (5/2 kgm^2/s i - 1kgm^2/s)
    = -3j - 10j
    = -13j

    I'm fairly certain that the answer is B because I believe that the cross product should give me a value that is perpendicular to the two vectors, which in this case are i and k, thus the cross product should, I think, include j. Also, I believe that I can eliminate D because the units are incorrect. What I don't understand is why my answer is negative.
     
  2. jcsd
  3. Jul 27, 2012 #2
    [itex]L=m\cdot r\times v=\frac{1}{2}\cdot \det\begin{bmatrix}i&j&k\\3&0&4\\5&0&-2\end{bmatrix}=13j[/itex]
     
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