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A rotating container injected with a liquid

  1. Oct 14, 2012 #1
    1. The problem statement, all variables and given/known data
    A massive cylindrical container of inner radius ##R## is rotating freely with an initial angular velocity ##w_0##. A liquid of density ρ is slowly injected into the container, until the container is fully filled except the center of the container. The angular velocity of the whole system reduces to ##w## after the injection. What is the container's moment of inertia, while the gravitational acceleration is ##g##?
    rotating container.jpg

    2. Relevant equations
    The moment of inertia equation.


    3. The attempt at a solution
    I have no idea at the first place, could someone give me some advices?
     
  2. jcsd
  3. Oct 14, 2012 #2

    tiny-tim

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    hi rbwang1225! :wink:

    hint: use conservation of angular momentum (you know the density of the liquid is ρ) …

    show us what you get :smile:
     
  4. Oct 17, 2012 #3
    After I tried to solve the problem by angular momentum conservation, I got stuck on the problem of calculation the moment of inertia of the liquid.
    ##Iω_0=I'ω##,
    where ##I## is the moment of inertia of the containerm and ##I'=I+I_{liquid}##.
    I tried to calculate ##I_{liquid}=∫r^2dm##, but had a trouble in the shape of the liquid.
    ##\tan\theta=\frac{w^2r}{g}## The limit of ##r## is from the position ##r## on the curve line to ##R##, but there is a ##z## dependence of the position r. I don't know how to get the relationship.
    However, my way might be in the wrong direction.
    Could you give me some ideas?
    rotating container2.jpg

    Sincerely.
     
  5. Oct 17, 2012 #4

    tiny-tim

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    hi rbwang1225! :smile:
    you'd have to do it by integration, slicing the liquid (!) into cylindrical shells of thickness dr :wink:

    however, i wouldn't bother …

    the question doesn't tell you how tall the container is, so i reckon you're entitled to assume that the dip in the middle is too small to matter, and that the water is just a cylinder :smile:

    (or is the diagram supposed to be showing the dip actually reaching the bottom of the container? in that case, yes you need to integrate :confused:)
     
  6. Oct 17, 2012 #5
    OK. Then suppose the liquid forms a cylinder, ##I_{liquid}=\frac{ρVR^2}{2}##.
    But the problem becomes we have no height of the cylinder, how could I overcome this?

    Sincerely.
     
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