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Determining Moment of Inertia of a sphere.

by aFk-Al
Tags: determining, inertia, moment, sphere
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aFk-Al
#1
Feb1-07, 06:13 PM
P: 19
I'm having some troubles determining the moment of inertia of a sphere about it's central axis. My original question was to calculate it for a cylinder, which I've done, but I'd like to know how to find it for a sphere.
Here is the problem solved for a cylinder:

Problem:
A uniform solid cylinder has radius R, mass M, and length L. Calculate its moment of inertia about its central axis (the z axis).

Solution:
I divided the cylinder into infinitesimally small layers because I knew that [tex]dV = (2\pi*r*dr)*L[/tex]. From here I calculated the integral [tex]I = \int \rho*r^2 dV = \int_{0}^R \rho*r^2*(2\pi*r*L)dr = 2*\pi*\rho*L*R^4 [/tex]
I substituted [tex]\frac {M}{\pi*R^2*L}[/tex] (or [tex]\frac{M}{V}[/tex]) for [tex]\rho[/tex] into the equation to get [tex] I = \frac{1}{2}*\pi*(\frac {M}{\pi*R^2*L})*L*R^4 = \frac{1}{2}*M*R^2[/tex]

I understand this, but when I tried to get it as a sphere I ended up getting the wrong answer. Could anyone please show me how to start the problem with a sphere?
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berkeman
#2
Feb1-07, 07:30 PM
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Do it in spherical coordinates, being careful to take the sin(theta) for each little volume piece (since the spherical r value is from the origin, not from the axis of rotation). And be careful to use the correct value for dV in spherical coordinates.
Ja4Coltrane
#3
Feb1-07, 07:36 PM
P: 227
For the solid spere, I like to add up the moment of inertia of a bunch of disks form -R to R.

cristo
#4
Feb1-07, 07:40 PM
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Determining Moment of Inertia of a sphere.

How did you try to perform your integral for the sphere? Think of your choice of coordinate system--hint: spherical polar coordinates.

damn: really should have refreshed quicker!
aFk-Al
#5
Feb2-07, 03:14 PM
P: 19
Couldn't I divide the sphere into a hemisphere and multiply it by 2? E.G. [tex]2*\pi*\int_{0}^R (\sqrt{R^2-r^2}*r^2)*r^2 dr[/tex] Where R is the radius of the sphere. So essentially I'm adding a bunch of [tex]\pi*r^2[/tex] (circles) to get a half of a sphere. Then multiplying the whole thing times two.
Ja4Coltrane
#6
Feb2-07, 08:51 PM
P: 227
hmm, why dont you show a little more work so that it is easier to follow.
one thing though, you might want to reconsider muliplying it by two--think about it.
aFk-Al
#7
Feb6-07, 01:54 PM
P: 19
I have no idea how to integrate that, so I don't have much work to show. The idea makes sence in my head, but I dont know how to follow through with it.


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