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Momemnt of inertia, momentum problem

  1. Oct 23, 2011 #1
    A solid disk rotates in the horizontal plane at an angular velocity of 0.047 rad/s with respect to an axis perpendicular to the disk at its center. The moment of inertia of the disk is 0.14 kg·m2. From above, sand is dropped straight down onto this rotating disk, so that a thin uniform ring of sand is formed at a distance 0.22 m from the axis. The sand in the ring has a mass of 0.39 kg. After all the sand is in place, what is the angular velocity of the disk?

    Wo = 0.047
    Io = 0.14

    R = 0.22
    Ms = 0.39

    Lo = Lf <----conservation of momentum

    Io =If
    WoIo = WfIf

    If = MR2 = 0.39(0.222)

    0.047(0.14) = 0.39(0.222)(Wf)
    Wf = 0.35 ????

    That is what I got but the answer is wrong.
     
  2. jcsd
  3. Oct 23, 2011 #2

    Doc Al

    User Avatar

    Staff: Mentor

    The final moment of inertia is not just that of the sand--don't forget the disk itself.
     
  4. Oct 23, 2011 #3
    Would I just add the two moment of inertia together?

    IF = 0.14 +[0.39(0.222)]
    IF ~ 0.158876


    WoIo = WFIF

    WoIo / IF = WF

    WF = 0.047(0.14) / 0.158876
    WF ~ 0.414?
     
  5. Oct 23, 2011 #4

    Doc Al

    User Avatar

    Staff: Mentor

    Right!


    Everything's good except the very last line. (You misplaced a decimal point.)
     
  6. Oct 23, 2011 #5
    Thank you!!! yes I did miss a decimal point. thank you!
     
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