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Angular momentum of a particle

  1. Nov 7, 2009 #1
    1. The problem statement, all variables and given/known data
    The position of a particle of mass m = 2.0kg is given at time t by the equation, r = 3t i - t2 j + 2 k, where r is measured in meters from the origin, and t is in seconds. What is the angular momentum of the particle with respect to the origin at time t = 2 s?

    2. Relevant equations
    L = Iω = (mr2)(ω)

    3. The attempt at a solution
    I kind of understand the general idea of the problem, I think it's just the presence of i, j, k that is confusing me.

    From the radius equation, I took the derivative to get the angular velocity equation.
    ω = 3 i - 2t j

    From there, I calculated r(2) and ω(2).
    r(2) = 6i - 4j + 2k
    ω(2) = 3i - 4j

    Using the equation L = Iω = mr2ω
    L = (2)(6i - 4j + 2k)2(3i - 4j)

    Then, I just multiplied through to get the answer. But, the numbers look way too big and I'm not too sure I did it correctly.

    Any help is appreciated!
     
  2. jcsd
  3. Nov 7, 2009 #2

    ehild

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    Homework Helper
    Gold Member

    The angular momentum is the cross product (vector product) of the momentum of the particle and its position vector. You can determine the velocity of the particle by differentiating r(t) with respect time; multiply it by mass and then calculate the cross product with the position vector.

    ehild
     
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