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I'm looking at sample calculations of moment of inertia of a sphere here.

In the first example (disc method), it has the integral as $dI = \frac{1}{2}r^2 \,dm$, while in the second example (shell method), $dI = r^2 \,dm$. Why is this so? Where does the 1/2 come from?

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  • $\begingroup$ It comes from the fact that all of the mass of a shell is concentrated on the outside, while some of the mass of a disc is also located closer to the axis of rotation, giving it less inertia. $\endgroup$
    – Danu
    Commented Apr 13, 2014 at 9:24

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The factor comes from the moment of inertia of the infinitesimal piece. In the disc method, each piece is a filled flat circle (a disc) of radius $r$, and the moment of inertia of a flat circle is $\frac{1}{2}mr^2$. The $\frac{1}{2}$ accounts for the fact that the mass of the circle is distributed between the center and the edge. But in the shell method, each piece is a hollow cylinder in which all the mass is on the edge, at radius $r$. The moment of inertia when all the mass is at the same radius is $mr^2$.

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  • $\begingroup$ That makes sense, thank you. However, I was taught that the general integral is $I = \int r^2 \,dm$ is this incorrect? $\endgroup$
    – Dylan
    Commented Apr 13, 2014 at 9:23
  • $\begingroup$ @DylanDang that is correct when each individual piece (with mass $\mathrm{d}m$) has all its mass at the same radius. In the disc method, the mass is distributed from radius 0 to r, so you get a factor that accounts for that distribution. But it is possible to derive the factor of $\frac{1}{2}$ from the general formula. (Calculate the moment of inertia of a flat disc of mass $\mathrm{d}m$.) $\endgroup$
    – David Z
    Commented Apr 13, 2014 at 9:36
  • $\begingroup$ If I apply this for a thin rod and slice it along the length, those pieces will not have the same radius, distributed from 0 to L. Does that formula work because they are treated as point masses? $\endgroup$
    – Dylan
    Commented Apr 13, 2014 at 9:48
  • $\begingroup$ @DylanDang Yes, what David Z is saying is that $dm$ in the notes is not infitesimal mass in dimension 3, the notation is confusing. If you write this as $dI = 2\pi \rho dz \int_0^r r^{\prime 3} d r'$, it should be clear. $\endgroup$
    – auxsvr
    Commented Apr 13, 2014 at 11:15

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