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Moment of Inertia for Half Disk

  1. Jul 30, 2014 #1
    1. Question:
    What is the moment of inertia of a thin, flat plate in the shape of a semicircle rotating about the straight side (Fig. 12.28)? The mass of the plate is M and the radius is R

    The diagram is attached for reference.

    2. Relevant equations
    Moment of Inertia: I = ∫R2dm
    Perpendicular Axis Theorem: Iz = Ix + Iy

    3. The attempt at a solution
    I made attempts to solve this is a couple of ways...

    First attempt: Using I = ∫R2dm

    I chose the y-axis to be parallel to the distance from the axis of rotation to mass elements on the disk.

    I defined the distance from the rotation axis y = rsinΘ, where r is the distance of the mass element from the center.

    In my integral for I, I substituted for R2, r2sin2Θ

    dm = (M/A)*dA where A=(πR2)/2 and dA=drdΘ

    Plugging into the integral, I get I = 2M/(πR2) ∫r2dr from o to R * ∫sin2ΘdΘ from 0 to π

    Solving, I get MR/3

    Now, I know there is something fundamentally wrong with process because I realize I should have some multiple of R2... Was hoping someone could help shed some light

    Attempt 2
    I believe the problem becomes much simpler if I use the perpendicular axis theorem but I still failed to get the correct answer this way.

    Iz = Ix + Iy

    Ix would mean the half disk coincides with the radial center of the disk. Ix (for a full circle) = (MR2)/2 but we have of a disk, so I believe Ix= (MR2)/4

    Iy would mean the half disk rotation about the y axis intersecting the radial center. Iy (for a full circle) = (MR2)/4 but this is a half disk to Iy = (MR2)/8

    Ix + Iy = Iz, for which I get (3MR2)/8

    Book says the solution is (MR2)/4 so I'm a little frustrated at this point.

    Any help would be great, thanks!
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution

    Attached Files:

  2. jcsd
  3. Jul 30, 2014 #2

    Doc Al

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    Staff: Mentor

    That's the way to go, but let Iz be that of a full disk about its axis. Then you'll solve for Ix and Iy, for a full disk. Then getting half a disk should be easy.
  4. Jul 30, 2014 #3


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    Gold Member

    Not sure I understand what you wrote there, but bear in mind that if you want M to be the mass of the half disk then the full disk has mass 2M.
  5. Jul 31, 2014 #4
    Thanks for the tips. I think I'm missing something in my understanding...

    If I let Iz be that of a full disk about its axis, you mean through its center and perpendicular to its faces, correct?

    The math works out in that case and I get Iz = (1/4)MR2 but it seems to me that that is answering a different question. Shouldn't Iz be through the center and parallel to its faces, as the question asks? This confuses me

  6. Jul 31, 2014 #5

    Doc Al

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    Staff: Mentor


    How did you get that?

    No. Iz should be known. You'll solve for Ix and Iy. Ix (or Iy) will be through the center and parallel to the face.
  7. Jul 31, 2014 #6
    If Iz is through the center and perpendicular to the faces, Ix is though the center and parallel to the faces and Iy is through the center and parallel to the faces (but at 90 degrees to Iy)

    So Ix is what I want to find

    Iz = (1/2)MR2

    Iy = (1/4)MR2

    Ix = Iz - Iy = (1/2)MR2 - (1/4)MR2 = (1/4)MR2

    But this is only a half disk, so Ix = (1/2)(1/4)MR2 = (1/8)MR2
  8. Jul 31, 2014 #7

    Doc Al

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    Staff: Mentor

    Excellent. (You didn't have to solve for Ix by subtracting. By symmetry, you know that Ix and Iy must be equal, so Ix = Iy = (1/2)Iz.)

    Now recall that "M" is the mass of a full disk. Rewrite in terms of the mass of just the half disk and you are done.
  9. Jul 31, 2014 #8
    k. Got it

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