Moment of Inertia for a Solid Sphere

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SUMMARY

The discussion centers on the calculation of the moment of inertia for a solid sphere, specifically addressing the discrepancy between the derived formula I=3/5 m b^2 and the correct formula I=2/5 m b^2. The user, Norman, initially applies the formula I=r^2dm with a constant mass density assumption, leading to confusion regarding the variable 'r' in different contexts. The clarification provided highlights the importance of distinguishing between the distance from the axis of rotation and the distance from the center of the sphere, which is crucial for accurate integration.

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Norman
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So I am having an issue getting the correct moment here. Although this isn't homework, I thought this would be the best forum for the issue.
So... I=r^2dm Assuming a const mass density: I=p r^2 dV, (p=mass/volume). For a sphere the natural choice of
dV = r^2 dr d(omega). So this leads to:
I=4(pi) p r^4 dr . Let's say that the sphere has a radius b.
Therefore: I=4(pi) p b^5/5, with p=m/(4/3 pi b^3)
This gives me a final answer of I=3/5 m b^2, which is according to ever text I have looked at (Serway, Halliday/Reznik) incorrect.
The supposed correct answer would be: I=2/5 m b^2
I should be able to do this but I seem to be unable to find the assumption that is leading to my error. Something is tickling the back of my mind from my undergrad years telling me there is a subtle point I am missing.
Any help would be greatly appreciated.
 
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Originally posted by Norman

So... I=r^2dm Assuming a const mass density: I=p r^2 dV, (p=mass/volume). For a sphere the natural choice of
dV = r^2 dr d(omega).
The r means different things in each of your formula; don't mix them up. In I=r^2dm, r means distance from the axis of rotation; in dV = r^2 dr d(omega), r means distance from the center.
 
Ahhhhhh...

Yes of course. So now this is one ugly looking integral. Thanks a lot for the help.
 

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