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Moment of Inertia for a Disk with Hole

  1. Jun 14, 2010 #1
    1. The problem statement, all variables and given/known data

    A uniform circular disk has radius 36 cm and mass 350 g and its center is at the origin. Then a circular hole of radius 7.2 cm is cut out of it. The center of the hole is a distance 10.8 cm from the center of the disk. Find the moment of inertia of the modified disk about the origin.

    2. Relevant equations

    I=.25mr^2 (I think...!)

    3. The attempt at a solution

    I tried finding the moment of inertia for both the hole and the disk and subtract the hole, but that didn't work.
    I'm not sure if that equation I posted is correct, I just googled it.
    How am I supposed to find the mass for the moment of inertia equation?
    At this point I just wish there was a straight up equation for it! Please help!
  2. jcsd
  3. Jun 14, 2010 #2


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    Homework Helper

    I = 1/2 MR2

    You need to subtract the inertia of the hole about the axis through the center of the disk.

    (Parallel Axis theorem: I = Ic +md2)
  4. Jun 14, 2010 #3
    How would I find the mass of the hole in that case?

    If I use I=.5MR^2 for the entire disk (not counting the hole), I get I = 226800 g cm^2.

    EDIT: Okay I figured out the mass of the hole. For the parallel axis theorem, would Ic be the moment of inertia for the whole disk? I'm just not sure how to apply it. Thanks again!
    Last edited: Jun 14, 2010
  5. Jun 14, 2010 #4
    Wait, never mind I got it!

    Once I got the mass by using the proper ratio: (350/[Area of disk - Area of hole]) = Mhole/[Area of hole], I just did the following:

    I (of disk without hole) - .5(Mhole)(Rhole)^2
    and got my answer!!

  6. Jun 15, 2010 #5
    Are you sure it is the correct answer? Remember you need to consider the moment of inertia of the hole relative to the origin of the disc.
  7. Jun 15, 2010 #6
    It's correct, it's an online homework program that tells you whether it is or not =)
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