Where Did I Go Wrong in Calculating the Moment of Inertia for a Disk?

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SUMMARY

The calculation of the moment of inertia for a disk with uniform density was initially miscalculated by using the area differential dA instead of the volume differential dV. The correct approach involves recognizing that the density ρ is defined as mass divided by volume, leading to the integral I = ∫ ρ r² dA. The error was identified as using r twice in the integral, rather than differentiating between the inner radius r and the outer radius R. The correct moment of inertia for a disk is established as I = (1/2)mr².

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amcavoy
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I am having a bit of trouble deriving the moment of inertia for a disk with uniform density:

[tex]I=\int r^{2}\,dm=\int \rho r^{2}\,dV[/tex]

For a disk, I just used dA instead of dV. Now, to calculate the density:

[tex]\rho = \frac{\text{mass}}{\text{volume}}=\frac{m}{\pi r^{2}}[/tex]

So now we have:

[tex]I=\int \rho r^{2}\,dA=\frac{m}{\pi}\int \,dA=\boxed{mr^{2}}[/tex]

However, I know that the moment of inertia for a disk is [itex]\frac{1}{2}mr^{2}[/itex]. Where did I go wrong?

Thank you.
 
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Nevermind I see where I went wrong (I used r twice, whereas I should have used R and r, where r ranges from 0 to R).
 
In the integral, one would have [itex]r^2\,\frac{m}{\pi\,r^2} r dr d\theta[/itex] the r2 terms cancel, and that leaves dA = r dr d[itex]d\theta[/itex].
 

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