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Why is this method not valid-moments?

  1. Jun 1, 2008 #1
    [SOLVED] Why is this method not valid-moments?

    I'm trying to find the moment of inertia for a solid sphere, I've seen and understand the correct method, but I can't see anything wrong with this method, except the answer.
    [tex]I=\int r^2 dm[/tex]
    [tex]m=\delta V=\frac{4}{3}\delta\pi r^3[/tex]
    [tex]dm=4\delta\pi r^2 dr[/tex]
    [tex]I=4\delta\pi\int r^4dr[/tex]
    Which after some work, turns out to be, incorrectly,
    The correct answer is,
    Thanks in advance.
  2. jcsd
  3. Jun 1, 2008 #2


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    [tex]m=\delta V=\frac{4}{3}\delta\pi r^3[/tex]

    How'd you arrive at this?
  4. Jun 1, 2008 #3
    I did this by: [itex]\delta[/itex] is the density. Mass=density x volume, [itex]m=\delta V[/itex]. The volume of a sphere is: [itex]V=\frac{4}{3}\pi r^3[/itex].
    [tex]m=\frac{4}{3}\delta \pi r^3[/tex].
    Take the derivative:
    [tex]\frac{dm}{dr}=4\delta\pi r^2[/tex],
    [tex]dm=4\delta\pi r^2 dr[/tex].
  5. Jun 1, 2008 #4
    It could be smart to denote the radius of the sphere with R, and use r for the integration variable. Using the same symbol causes confusion.

    The true mistake however is that the r in the original formula is not the same r as the integration variable. The r in the original formula is the component of the [tex]\bar{r}[/tex] vector on the normal plane of the rotation axis.

    Would these look good? I cannot guarantee they are right :wink:

    I = \int (r\sin(\theta))^2 dm

    dm = 2\pi\delta r^2\sin(\theta) d\theta\; dr

    0<r<R,\quad 0<\theta<\pi
  6. Jun 1, 2008 #5

    D H

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    Here is what is wrong:
    That [itex]r[/itex] is the distance of some differential quantity of mass [itex]dm[/itex] from the axis of rotation ...
    ... while that [itex]r[/itex] is the distance of some differential quantity of mass [itex]dm[/itex] from the center of the sphere.
  7. Jun 1, 2008 #6


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    Hi gamesguru! :smile:

    You're integrating over spherical shells of constant radius.

    But Moment of Inertia is about an axis, never about a point.

    So you need to integrate over cylindrical shells of constant radius. :smile:
  8. Jun 1, 2008 #7
    Thanks D H, I see the problem now. And I see why using this method with a cylinder, by chance, works.
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