# Moment of Inertia of Solid Sphere - Proof

So I have been having a bit of trouble trying to derive the moment of inertia of a solid sphere through its center of mass. Here is my working as shown in the attached file.

The problem is, I end up getting a solution of I = (3/5)MR^2, whereas, in any textbook, it says that the inertia should be equal to I = (2/5)MR^2. Is anyone able to tell me where I went wrong in my working? This is not a homework problem by the way.

Thanks.

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Orodruin
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You are using $r$ with two different meanings and mixing them up. 1) The distance from the axis of rotation (the $r$ in the definition of the moment of inertia). 2) The distance from the centre of the sphere. These are not the same.

This is not a homework problem by the way.
Regardless of whether it is actual homework or not, it should be posted in the homework forums if it is homework-like.

• Delta2
Delta2
Homework Helper
Gold Member
Just to add something to Orodruin's enlightening post,

In many moment of inertia calculations these two meanings happen to be the same thing (for example in the calculation of the Moment of Inertia of an infinitesimally thin circular disc, the distance from the axis of rotation (that passes through the center and is perpendicular to the plane of the disc) equals the distance from the center of the disc ) BUT in the general case they are not the same thing and in this case they are not the same thing.

What is the equation that relates r' (the distance from the axis of rotation) and r (the distance from the center of the sphere) in this case?

Also the dV you calculate is not the same as the dV that appears in the definition of the moment of inertia. You calculate the infinitesimal volume between a sphere with radius r and radius r+dr. But the dV in the integral in the definition of the moment of inertia is $dV=r^2\sin\theta dr d\theta d\phi$ (r is the distance from the center of the sphere). You just cant use your definition of dV because if you find the equation of r' correctly you ll see that it depends on $\theta$ and $r$ and not only $r$.

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