
#1
Feb107, 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? 



#2
Feb107, 07:30 PM

Mentor
P: 39,600

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.




#3
Feb107, 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.




#4
Feb107, 07:40 PM

Mentor
P: 8,287

Determining Moment of Inertia of a sphere.
How did you try to perform your integral for the sphere? Think of your choice of coordinate systemhint: spherical polar coordinates.
damn: really should have refreshed quicker! 



#5
Feb207, 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^2r^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.




#6
Feb207, 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 twothink about it. 



#7
Feb607, 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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