How Do You Calculate the Moment of Inertia for a Sphere Using Shell Integration?

AI Thread Summary
To calculate the moment of inertia of a sphere using shell integration, the sphere is divided into infinitesimal shells, with the mass of each shell expressed as dm = ρ(4πr²)dr. The integration setup involves I = ∫(2/3)dm r² dr from 0 to R. The user initially miscalculated by integrating dm incorrectly, leading to an incorrect result of I = mr². The correct approach requires proper substitution for dm and careful integration to arrive at the correct moment of inertia for a sphere.
Zamba
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Homework Statement


calculate the moment of inertia of a sphere of mass M and radius R by integrating over thin shells

Homework Equations


Ishell=(2/3)mR2

The Attempt at a Solution


this is what i have so far
the sphere is decomposed into infinitesimal shells with surface area 4\pir2
the mass of each shell is dm=\rho(4\pir2)dr
after expanding rho and canceling terms I get
dm=(3m/r)dr

I=\int(2/3)(dm)r2dr from 0 to R.
I=2mrdr from 0 to R
this gives me I=mr2

does anyone see what I did wrong?
PS. those "pi" are not supposed to be powers. sorry, I'm not sure how to change them
 
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You did not replace dm with ρ(4πr2) under the integral sign. Instead, you integrated dm into m and then integrated over dr once more.
 
i'm sorry, i didn't write it all out

I=\int(2/3)dmr2 dr
from here I replaced dm with (3m/r)dr so i got

I=\int(2/3)(3m/r)r2 dr
then i got
I=\int2mr dr from 0 to R
which gave me mr2
 
Zamba said:
... from here I replaced dm with (3m/r)dr...
And why is that correct? What expression did you use for the density?
 

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