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Moment of Inertia of a Disk

  1. Dec 1, 2009 #1
    1. The problem statement, all variables and given/known data
    Figure (a) shows a disk that can rotate about an axis at a radial distance h from the center of the disk. Figure (b) gives the rotational inertia I of the disk about the axis as a function of that distance h, from the center out to the edge of the disk. The scale on the I axis is set by IA = 0.010 kg·m2 and IB = 0.210 kg·m2. What is the mass of the disk?
    Here's a link to the image --> http://www.webassign.net/hrw/10-35.gif


    2. Relevant equations

    Inertia=Icom + Mh^2
    =(mL^2)/2 + Mh^2


    3. The attempt at a solution

    I set .010= (mL^2)/2 and solved for L^2 which turns out to be (.02/m). I then plugged this term into L^2 variable of the equation I=(mL^2)/2 + Mh^2. Then I plugged in .210 for I and .2 for h and solved for m which came out to be about 5.25. I thought I had it right but I don't. Help would be great I appreciate it.
     
  2. jcsd
  3. Dec 2, 2009 #2

    kuruman

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    Homework Helper
    Gold Member

    You know that at the edge of the disk the moment of inertia is 0.210 kg m2 and that h = L = 0.2 m. Put everything in the parallel axes theorem and solve for the mass. You don't have to guess what the moment of inertia at h = 0 is.
     
  4. Dec 2, 2009 #3
    Yes but in this case h does not equal L because the axis of rotation is not at the edge of the disk it is somewhere between the center of mass and the edge of the disk.
     
  5. Dec 2, 2009 #4
    Oh okay I just read the problem over again and that makes sense now. Thanks a lot for your help.
     
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