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More Angular Momentum

  1. Nov 21, 2005 #1
    A thin, uniform rod of length L = 0.850 m and mass M rotates horizontally about an axis through its center, at an angular velocity 18.5 rad/s. A particle with mass M/3.00 is attached to one end. The particle is then ejected from the rod by a small explosion. The explosion sends the particle along a path that is perpendicular to the rod at the instant of the explosion, with a speed that is 6.00 m/s greater than the speed of the end of the rod after the explosion. What is the speed of the ejected particle?

    I attempted to find the speed by solving for v in the following equation:
    [(1/12)ml^2]*(w) = (M/3)(v)(l/2)

    then i added 6. this gives me 8.62, which is wrong.

    Please tell me what I'm doing wrong! Thanks!
  2. jcsd
  3. Nov 21, 2005 #2


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    On what principles did you base your equation on? The conservation of momentum principle states that the momenum of a system cannot change with no external forces acting on it. Hint: an internal explosion is not considered an external force.
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