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Help with angular momentum question

  1. Nov 20, 2007 #1
    A. At a particular instant, a particle of mass M = 3 kg is at the position (x,y,z) = (4,4,6) m and has velocity (2,1,-2) m/s.

    B. An identical particle is placed at (x,y,z) = (-4,-4,-6) m, with velocity (-2,-1,2) m/s.
    Find the angular momentum of the pair of particles about the origin.

    C. A thin rod of mass 9 kg is added between the two particles (Irod = M L^2/12). Since the particles are moving perpendicular to the line separating them and in opposite directions (you should convince yourself that this is true), then the particles + rod system rotates about its center of mass. Compute the angular frequency of rotation of the rod + particles system.

    D. Compute the magnitude of the total angular momentum of the system (particles plus rod).

    ok, so i got parts a,b,c pretty easily.

    A. <-42, 60, -12>
    B. <-84, 120, -24>
    C. .363

    Ok, so for part D, the main approach i used was total angular momentum = sum of all individual components in the system.

    So, i can easily find the angular momentum of the two particles, but the problem is finding the moment of inertia of the thin rod. The formula for inertia is (ML^2)/12, but i cant seem to get the length L from anywhere in the problem.

    Any help is appreciated.
    thanks alot!
  2. jcsd
  3. Nov 20, 2007 #2


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    Distance between the two particals is the length of the rod.
  4. Nov 21, 2007 #3
    how do you find the angular momentum of the particles?
  5. Nov 21, 2007 #4


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    The angular momentum of the particles = Total moment of inertia*angular velocity=(Moment of inertia of rod + moment of inertia of particals) *angular velocity.
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