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Angular Speed, Glob and Rod

  1. Dec 20, 2006 #1
    1. The problem statement, all variables and given/known data

    On a frictionless table, a glob of clay of mass 0.38 kg strikes a bar of mass 0.90 kg perpendicularly at a point 0.55 m from the center of the bar and sticks to it. If the bar is 1.30 m long and the clay is moving at 8.1 m/s before striking the bar, at what angular speed does the bar/clay system rotate about its center of mass after the impact?

    2. Relevant Equations


    [tex]cm_[new]=\frac{all m*d}{total mass}[/tex]

    Parallel Axis Theorem

    3. The attempt at a solution

    M is mass of rod and m is mass of glob.



    [tex]\delta=distance from old cm to new cm[/tex]

    [tex]\delta=\frac{mb}{m+M}=.1633 m[/tex]




    I get 8.15 rad/s which is wrong, the right answer is 5.743. What did I do wrong?
    Last edited by a moderator: Apr 22, 2017
  2. jcsd
  3. Dec 21, 2006 #2

    Doc Al

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    Staff: Mentor

    This is the angular momentum of the blob about the center of the bar. What you need is the angular momentum about the center of mass of the system--that is the quantity that is conserved.
  4. Dec 21, 2006 #3
    So I should have [tex]L_i=mv(b-\delta)[/tex] ?

    Thanks, I got it.
    Last edited: Dec 21, 2006
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