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Angular Momentum of Disk Question

  1. Dec 11, 2006 #1
    1. The problem statement, all variables and given/known data
    A stationary uniform-density disk of radius 0.8m is mounted in the vertical plane. The axle is held up by supports that are not shown and is frictionless. The disk has a mass of 3.6 kg. A lump of clay with mass 0.3 kg falls and sticks to the outer edge at <-0.48, 0.64, 0>m. Before impact it has a speed 8 m/s and the disk is rotating clockwise with angular speed 0.4 radians/second. Just before impact, what is the magnitude of the angular momentum about the center. System=disk+clay, about the center (<0,0,0>)


    2. Relevant equations
    I know that it has something to do with:
    Ltot=Lrot+Ltrans
    Lrot=Iw, where I=.5*mR^2 and w is given? (the disk)
    Ltrans=mrv, where the r here equals where it sticks? (the clay)


    3. The attempt at a solution
    I tried plugging the numbers given into this equation, but the answer isn't right.. (the answer is 0.6912 kg*m^2/s) Angular momentum is a vector, so i assumed we'd use the vector position given. What am i doing wrong?
     
  2. jcsd
  3. Dec 11, 2006 #2

    OlderDan

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    Ltrans=mrv is only true when the vectors r and v are perpendicular. Your vectors are not perpendicular. You need the more general vector equation involving the cross product L = r x p
     
  4. Dec 11, 2006 #3
    how to do i find a cross product of something when the position is a vector, but the momentum is only magnitude?
     
  5. Dec 11, 2006 #4

    OlderDan

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    The momentum is also a vector. The lump of clay falls onto the disk.
     
  6. Dec 11, 2006 #5
    That clears a lot up. But now, my question is after impact, what is the angular velocity (is it w - omega)?

    Is there conservation of angular momentum? Do we still have translational angular momentum since the clay has been stuck to the wheel?
     
  7. Dec 11, 2006 #6

    OlderDan

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    There are two ways to approach the analysis. Either way rests on the principle of conservation of angular momentum. Before impact, you have two separate angular momenta, one for the disk and one for the clay that must be added to find the total system angular momentum. After that you can treat them as separate objects with a relationship between the v of the clay and the ω of the disk, or you can find the moment of inertia of the disk-clay combination with angular velocity ω. Either way, the total angular momentum after impact must be the same as before impact.
     
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