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Consider a Sphere of mass $M$, and volume $V = \frac{4}{3}π(r)^3$ and uniform density $p$, If I want to get its moment of inertia around an axis running through its centroid then I shall Integrate $dm r^2$.

And If I cut the Sphere into infinitesimally small volumes of spheres I get $dV = 4πr^2 dr$, $dm = p dV$.

So I get to integrate $p.4πr^4 dr$ from zero to the whole radius if the Sphere, now the final result is $\frac{3MR^2}{5}$ and that's $1.5$ times the real moment if inertia. So what did I do wrong?

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3 Answers 3

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The moment of inertia of an object about the $z$-axis is $$\int r^2 dm = \int x^2 + y^2 dm.$$ However, for the spherical shell, you used $$\int x^2 + y^2 + z^2 dm.$$ By symmetry, all three of these terms contribute equally, and you only want two of them, so your answer should be $2/3$ as big. This gives $(2/3)(3/5)MR^2 = (2/5)MR^2$, the correct answer.

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The volume element you are choosing has symetry related to the center of the sphere, i.e., a point instead of a line. The axis around which the sphere will rotate is a line that passes by its center. This will allow you visualize the correct differential of volume you have to choose.

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You cannot do $ dI =r^2 dm$ for a hollow sphere. In fact, a hollow sphere of radius $R$ and mass $M$ has $ I = \frac{2MR^2}{3}$ Link to Wiki entry on MOI

You can use this expression, to derive a solid sphere's MOI, using integral sum over infinitesimal hollow spheres of radius $0$ to $R$. Here are the steps....

$\rho = \frac{M}{ \frac{4}{3} \pi R^3}$
and $dm = \rho \;4 \pi r^2 dr = \frac{3M}{ R^3} \; r^2 dr $

*(using the formula for hollow sphere)*

$I = \int_0^R \frac{2r^2}{3} dm $

$I = \int_0^R \frac{2r^2}{3} \frac{3M}{ R^3} \; r^2 dr$

$I = \frac{2M}{R^3} \int_0^R r^4 dr$

$I = \frac{2M}{R^3} \frac{R^5}{5} = \frac{2MR^2}{5}$

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