Moment of Inertia of a Sphere problem

AI Thread Summary
The discussion focuses on calculating the moment of inertia of a sphere with non-uniform density. The initial attempt used the standard formula for a uniform sphere, which was incorrect due to the varying densities. Participants suggested using the concept of treating the sphere as two separate bodies: a solid sphere and a hollow sphere, whose moments of inertia can be added together. The use of triple integrals was also discussed as a method to derive the moment of inertia from first principles. Ultimately, the problem was resolved by applying the solid sphere/shell method successfully.
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Homework Statement



Calculate the moment of inertia of a sphere of radius R and mass M about an axis through the center of the sphere. Assume that the density of the sphere is not uniform, but is given by p_1 for 0 <= r <= R_1 and by p_2 for R_1 <= r <= R.

Homework Equations



Moment of inertia of a sphere:
I = 2/5*MR^2

The Attempt at a Solution



First, I calculated the total mass in terms of densities.
M = p_2 (4/3*pi*R^3 - 4/3*pi*R_1^3) + p_1(4/3*pi*R_1^3)
Then, I plugged M into the formula and simplified a little bit to get:
I = 8/15*pi*[(p_1-p_2)R_1^3 + p_2*R^5]
The correct answer, however has R_1^5 instead of R_1^3. What did I do wrong here?
 
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That formula you are using for the moment of inertia of a sphere assumes that the density of the sphere is constant. Thus, that formula doesn't apply here.

In order to solve this you will need to go back to the definition of moment of inertia:

I=\int\int\int_V r^2\rho(r)dV
 
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I'm not sure what that triple integral notation means. How can I use that to find I?

Also, the teacher told us to use I for a sphere that he gave us in class (I = 2/5*MR^2). Is there a way that this can be adapted to fit this situation?
 
OK. Maybe you can use the formula for the moment of inertia of a hollow sphere to solve this.

I_{solid sphere}=\frac{2}{5}MR^2

and

I_{hollow sphere}=\frac{2}{3}MR^2

Now you can treat this as two bodies, a solid sphere of radius R_1 and a hollow sphere of Radius R.

Their moments of inertia should add since they are on the same axis:

I_{total}=I_{solid}+I_{hollow}

Can you find the total I now? Is this the correct answer?

If this doesn't work, then you will have to go back to the definition. BY the triple integral, I just mean integrate one time over each variable in the volume element dV. i.e.:

\int\int\int_VdV= \int_z\int_y\int_xdxdydz

or in spherical coordinates (more appropriate here):

\int\int\int_VdV= \int_{\phi}\int_{\theta}\int_r r^2 drd\theta d\phi
 
I got it to work with the solid sphere/shell method. Thanks!
 
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