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Moment of inertia of half disk through integration

  1. Mar 12, 2015 #1
    Hello, sorry for this stupid question.
    I struggled to find the moment of inertia of half solid thin disk (about the center of the disk) through an integration, but I couldn't get the right value.

    I'm pretty sure it has to be [tex]MR^2/4[/tex], but
    [tex]
    I=\int r^2 dm \\
    dm=(M/A)dS[/tex]

    With [tex]A=\pi R^2/2[/tex]

    I compute the integration in polar coordinates, where [tex]dS=r dr d\theta[/tex] with 0<r<R and 0<theta<pi:
    [tex] 2M/(\pi R^2) \int_0^R \int_0^\pi r^3 d\theta dr=MR^2/2[/tex]

    Where am I wrong?
     
  2. jcsd
  3. Mar 12, 2015 #2

    PeroK

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    I guess you think it should be ##MR^2/4## because the MoI of a full disk is ##MR^2/2##. But, what have you taken as the mass of your half disk? And, what would the mass of the full disk be?
     
  4. Mar 12, 2015 #3

    BiGyElLoWhAt

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    What do you mean about the center? I can think of three ways to rotate a half circle about it's center.
     
  5. Mar 12, 2015 #4
    mmh, I think I can see your point. but I'm a bit confused now.

    if I take M=(mass of full disk), I(full disk)=MR^2 /2 and I(half disk)=MR^2 /4
    if I take M=(mass of half disk) [as I did for the integral], I(full disk)=2MR^2 /2 and I(half disk)=MR^2 /2 (the result of the integral)

    So, it seems that if I have an amount M of mass and I shape it to be a full disk or a half disk, I get the same moment of inertia.

    BiGyElLoWhAt, you right, I mean about the axis perpendicular to the disk.
     
  6. Mar 12, 2015 #5

    PeroK

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    Yes, that's correct. By symmetry, each half of the disk contributes half of the MoI.
     
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