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Conservation of Angular Momentum of a Disk

  1. Nov 11, 2006 #1
    A solid disk rotates at an angular velocity of 0.094 rad/s with respect to an axis perpendicular to the disk at its center. The moment of inertia of the disk is 0.12 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 0.48 m from the axis. The sand in the ring has a mass of 0.36 kg. After all the sand is in place, what is the angular velocity of the disk?

    I just have no clue why I'm not getting the correct answer on this problem.

    Here are my steps

    I know that since Momentum is convserved L0 = Lf or I0W0 IfWf

    W = angular velocity

    so I set (0.12 kg m ^2)(.094 rad/s) = (If)(Wf)

    Since we were to treat the Inertia of it to be a hoop the equation of Inertia is I = MR ^2

    So this is my final equation (0.12 kg m ^2)(.094 rad/s) = (If)(Wf)

    I think where I'm going wrong is when I solved for Inertia (I) for If I used (.36 kg)(.48 ^2)

    Please point out what I need to do to correct this problem

  2. jcsd
  3. Nov 11, 2006 #2


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    The sand increases the moment of intertia of the disk/sand system, but it does not replace the disk. It looks like you have not included the disk in your I_f.
  4. Nov 11, 2006 #3
    ok am I suppose to include the inertia of the disc by adding the I0 to (.36 x .48 ^2).

    I'm still not getting the answer

    nvm I got it it was just an error in calculation.

    you are awsome DOC :)

    thx a lot
    Last edited: Nov 11, 2006
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