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Conservation of angular momentum in an inelastic collision?

  1. Nov 13, 2013 #1

    mot

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    1. The problem statement, all variables and given/known data
    On a frictionless table, a glob of clay of mass 0.380 kg strikes a bar of mass 0.9 kg perpendicularly at a point 0.550 m from the center of the bar and sticks to it.
    a) a) If the bar is 1.300 m long and the clay is moving at 8.100 m/s before striking the bar, what is the final speed of the center of mass? (Got this, 2.405m/s)
    b)At what angular speed does the bar/clay system rotate about its center of mass after the impact (in rad/s)?

    2. Relevant equations

    Li=Lf
    L=r x p =mrv (when theta = 90, as in this case)
    L=Iw
    I of a rod = (ML2)/12
    xcenter of mass=Ʃmx/Ʃm

    3. The attempt at a solution
    m = mass of clay
    M=mass of bar
    u=clay's initial speed
    r=the distance from the clay to the new CM

    So I set one end of the bar as x=0 and the other as x=1.3 to calculate CM
    CM=((0.38)(0.65+0.55)+(0.65)(0.9))/(0.9+0.38)=0.81328m from the end of the bar
    r=1.2-0.81328=0.3867m

    Initial angular velocity=mru
    Final " "=((1/12)ML2+mr2
    so ω=(mru)/((1/12)(ML2+mr2)
    When I plug in my values, I get the wrong answer, it is supposed to be 5.73. Where am I going wrong? I have spent hours at this :( I'm so frustrated!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
    Last edited: Nov 13, 2013
  2. jcsd
  3. Nov 13, 2013 #2

    ehild

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    Homework Helper
    Gold Member

    Where are the data in red come from?

    You need the moment of inertia with respect to the new CM. (1/12)ML2 is the moment of inertia with respect to the centre of the rod.


    ehild
     
  4. Nov 14, 2013 #3

    mot

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    Sorry I made a typo in the question, 0.43 should have been 0.55. I've fixed it now. (1.2 is 0.65+0.55, the "position" of the clay on the rod with respect to the end).

    That's what I was thinking with the moment of inertia...but doesn't adding the mr^2 take that into account?
    I could also do I=Ʃmr^2, where r is the distance from the new CM
    =(0.9)(0.81328-0.65)^2+(0.38)(0.3867)^2=0.0808

    and my old I=((1/12)(0.9)(1.3^2)+(0.38)(0.3867^2))=0.1836

    The new I gives me 14.7rad/s when I plug it into the equation:(
     
  5. Nov 14, 2013 #4

    ehild

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    Last edited: Nov 14, 2013
  6. Nov 14, 2013 #5

    D H

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    Staff Emeritus
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    No. That's the blob of clay's contribution to the moment of inertia of the bar+clay system. The bar's contribution is the moment of inertia of the bar about the combined center of mass.

    Hint: Parallel axis theorem.
     
  7. Nov 14, 2013 #6

    mot

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    Thanks so much guys, I got it! It seemed so obvious this morning, I guess that's why you shouldn't do physics at 2am. I was comparing it to a question I had done with a mass on a rotating disc, forgetting that the disc was still rotating about the central axis
     
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