How to devise moment of inertia formula of solid sphere?

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SUMMARY

The moment of inertia formula for a solid sphere about its central axis is derived using the equation I = ∫r²dm, resulting in I = 2/5 MR². The discussion highlights a common error in the integration process where an extra factor of r² was introduced, leading to confusion in the calculations. The correct approach involves treating the solid sphere as a collection of infinitesimally thin disks and accurately applying the mass element dm = ρdv = πρr²dx. The final result confirms that the moment of inertia is I = 4/5 MR².

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  • Ability to visualize three-dimensional shapes as collections of two-dimensional cross-sections
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imadrea
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Homework Statement


how to divide moment of inertia of solid sphere about its central axis?. Solid sphere has radius R, mass M.

Homework Equations



I=∫r2dm
2/5 MR^2

The Attempt at a Solution


https://photos.google.com/search/_tra_/photo/AF1QipPoXyad0q1Y3yisc0LeeJHGApkIrGbitK6kAk5p
i try to imagine that solid sphere is a group of infinite disk. a disk have volume dv=πr2dx.
dm=ρdv=πρr2dx.
I=R-Rr2πρr2dx
I=πρR-Rr4dx
I=πρR-R(R2-x2)2dx
I=πρR-RR4-2R2x2+x4dx
I=πρ[2R5-4/3R5+2/5R5]
!=πρR5[30-20+6]/15
I=16/15 πρR5
I=4/5 (4/3πρR3)R2
I=4/5MR2

where my eror? i have spent 2 days to solve it but i am failed until now.
 

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In your first integral, you introduce an extra factor r2. I assume this is related to the moment of inertia of a disk about its axis. Have you forgotten something there?
 
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I assume this is related to the moment of inertia of a disk about its axis. Have you forgotten something there?
i think this is just subtitution for dm=ρπr2dx. I'm confused
 
imadrea said:
i think this is just subtitution for dm=ρπr2dx. I'm confused
No, in the next line you have another r2 factor.
 

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