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Angular Momentum

  1. Oct 14, 2008 #1
    1. The problem statement, all variables and given/known data

    Consider a particle of mass m and a particle of mass 2m. They are connected by a horizontal, massless, rigid rod of length 2a. The rod is fixed to a vertical stick that connects to the rod's midpoint. The stick is spinning with constant angular speed.

    Consider a point P, located on the stick a distance d below the rod. Show that angular momentum of the two particle system, when taken about point P, is not conserved.

    2. Relevant equations

    L = r X p

    theta = 90 degrees

    3. The attempt at a solution

    So what I've done is find angular momentum for either mass, separately.
    Doing the following:

    L_1 = R x P
    = |r||p| sin(theta)
    = (a^2+d^2)^(1/2)* ma(omega)

    L_2 = R x P
    = |r||p| sin(theta)
    = (a^2+d^2)^(1/2)*2ma(omega)

    What I want to do... by intuition... is just add these two together. but they're magnitudes not vectors... so i am a bit confused if I can.

    If not, what the heck do I do?
     
    Last edited: Oct 14, 2008
  2. jcsd
  3. Oct 15, 2008 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    angular momentum is a vector

    Hi Kites! :smile:

    Angular momentum is a vector (strictly, a pseudovector), not a scalar …

    it's the cross-product of two vectors!! :smile:

    The moment form of good ol' Newton's second law is Net Torque = rate of change of angular momentum … and both sides of the equation are (pseudo-)vectors. :biggrin:

    Most exam questions on angular momentum are two-dimensional, so the angular momentums are all perpendicular to the plane of the exam paper, and you can just add them!

    But this question is three-dimensional … so you must treat them like the vectors they really are. :smile:

    (btw, applying Newton's second law, can you see why it's not conserved? :wink:)
     
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