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Moment of Inertia for a Solid Sphere

  • Thread starter Norman
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  • #1
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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 . Lets 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.
 

Answers and Replies

  • #2
Doc Al
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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.
 
  • #3
887
2
Ahhhhhh.....

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

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