Moment of Inertia of a Sphere problem

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Homework Help Overview

The problem involves calculating the moment of inertia of a sphere with a non-uniform density distribution. The sphere has a radius R and mass M, with different densities defined for two regions within the sphere.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of the standard moment of inertia formula for a uniform sphere and question its validity given the non-uniform density. There is an exploration of using integrals to define the moment of inertia and suggestions to adapt existing formulas for solid and hollow spheres.

Discussion Status

The discussion is active, with participants exploring different methods to approach the problem. Some guidance has been offered regarding the use of integrals and the combination of moments of inertia for different geometries. There is no explicit consensus on a single method yet.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the methods they can employ. The original poster has noted a discrepancy in their calculations compared to expected results, prompting further inquiry into the assumptions made.

Frillth
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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:

[tex]I=\int\int\int_V r^2\rho(r)dV[/tex]
 
Last edited:
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.

[tex]I_{solid sphere}=\frac{2}{5}MR^2[/tex]

and

[tex]I_{hollow sphere}=\frac{2}{3}MR^2[/tex]

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:

[tex]I_{total}=I_{solid}+I_{hollow}[/tex]

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.:

[tex]\int\int\int_VdV= \int_z\int_y\int_xdxdydz[/tex]

or in spherical coordinates (more appropriate here):

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

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