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Moment Of Inertia of broken disk or ring confusion

  1. Sep 5, 2012 #1
    We all know that M.I of a Uniform rigid rod about an axis perpendicular to it's length and passing through it's center is MLsquare/12.Where M is mass and L is length of the rod. If it is broken to half such that M becomes M/2 and L becomes L/2,we can't apply ML square /12 formula to it.We have to apply the equation of M.I of rod about an axis passing through one of the ends and perpendicular to the length of it.That is MLsquare/3.If we apply the condition that M/2 and L/2 ,we could get ML square /24. That is exactly half of original moment of inertia.

    I know well this much.
    But my confusion begins
    Suppose a ring of mass M and radius R.We know M.I of it is MR square.
    But if 90degree sector of it breaks,such that it's mass becomes 3/4M. My Physics teacher says the broken ring's M.I then will be 3/4MR square.
    My question is,as in the case of rigid rod,when it breaks, we just need another formula to find its MI. Similarly for this too,we have to use another formula right?
    I have the same problem with circular disk too.
  2. jcsd
  3. Sep 5, 2012 #2
    The reason you apply a different formula in the case with the split rod is because the axis of rotation changed from the center of the object's mass to the edge of the object. When you take away part of the circle, the axis of rotation is still in the same spot, and the formula is unchanged.

    These formulas cover up the real cause of the moment of inertia. A better way to understand moment of inertia is by the integral:
    [itex]\displaystyle I = \int^{m}_{0}r^{2}\delta m[/itex]
    All of the formulas for moment of inertia that you see in your textbook come directly from this integral. This integral also allows you to calculate the moment of inertia for any odd shape you like, including your 3/4-ring. The moment of inertia of the 3/4 ring is indeed still mr2 (m being the mass of the 3/4 ring, not the full ring).
  4. Sep 5, 2012 #3
    Also, I'd like to point out that when you split the rod into 2 pieces, each rod accounts for 1/2 of the total moment of inertia, so:

    [itex]\displaystyle \frac{I}{2}=\frac{(\frac{m}{2})(\frac{L}{2})^{2}}{3}[/itex]

    [itex]\displaystyle \frac{I}{2}=\frac{ML^{2}}{24}[/itex]

    [itex]\displaystyle I=\frac{ML^{2}}{12}[/itex]

    If you plug in your variables and simplify, you end up with the exact same formula as the un-split bar.
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