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Conservation of Momentum (Linear and Angular)

  1. Apr 5, 2005 #1
    On a frictionless table, a glob of clay of mass 0.38 kg strikes a bar of mass 1.76 kg perpendicularly at a point 0.12 m from the center of the bar and sticks to it.

    If the bar is 0.66 m long and the clay is moving at 5.7 m/s before striking the bar, what is the final speed of the center of mass?

    I found this part to be 1.012 m/s, which is correct, simply using conservation of linear momentum.

    I can't figure out the next part:

    At what angular speed does the bar/clay system rotate about its center of mass after the impact?

    I tried using conservation of angular momentum, which seems to me like it should work, but the computer says it's wrong. I tried it assuming the center of mass doesn't change when the clay hits it, and then did it again figuring in the change in center of mass.
  2. jcsd
  3. Apr 5, 2005 #2
    Is the bar initially vertical?
  4. Apr 5, 2005 #3

    Doc Al

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    Staff: Mentor

    Conservation of angular momentum is the key. Using the center of mass of the composite system as your reference, calculate the angular momentum of the system before the collision. To find the angular speed after the collision, you'll need to find the rotational inertia of the "bar + clay" object about that center of mass.
  5. Apr 6, 2005 #4
    Thats what I did, the first time it was wrong, I just tried it again and now its right...figures.

    It doesnt matter if the bar is vertical or horizontal, all you need to know is that the clay is coming in perpendicular to the bar.
  6. Apr 6, 2005 #5
    Well its not clear whats going on, if the bar was standing vertically on a table and gets hit, it will rotate about its base, and theres a force of gravity adding to the angular velocity, so things would be more complicated.
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