I'm asked to compute the inertia tensor for a solid sphere of radius $R$. I have done this and found it to be $$\frac{2}{5}MR^2 \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$$
I'm now asked to find the inertia tensor for a spherical shell with the same mass, where the thickness of the shell is $a \ll R$. Do I need to compute the inertia tensor all over again or can I just substitute $R-a$ into my previous answer and then find their difference?
Best Answer
Not the simple substitution. The problem changes completely. Now you have all mass in the shell, so is, almost each piece of the sphere farther than before from the axis and it mounts that the farther from the shell before, the more contributing to the moment of inertia now. But, it's true now as before that the tensor is diagonal, so it's only needed an integral about any axis passing by the center. Consider the sphere centered in the origin and calculate the moment of inertia around the $z$ axis. We can do it better in spherical coordinates.
The distance from any point of the shell to the $z$ axis is $r\sin\phi$ and the (constant) density of mass $\rho$ $$I_z=\int_{R}^{R+a}\int_0^\pi\int_0^{2\pi}r^2\sin^2\phi r^2\sin\phi\,\rho\,\mathrm d\theta\mathrm d\phi\mathrm dr=$$
$$=\rho\int_{R}^{R+a}r^4\int_0^\pi\sin^3\phi\int_0^{2\pi}\mathrm d\theta\mathrm d\phi\mathrm dr=$$
$$=2\pi\rho\int_{R}^{R+a}r^4\int_0^\pi\sin^3\phi\,\mathrm d\phi\mathrm dr=$$
$$=2\pi\rho\int_{R}^{R+a}r^4\left[(1/12)(\cos3\phi-9\cos\phi\right]_0^\pi\mathrm dr=$$
$$=\frac 83\pi\rho\int_{R}^{R+a}r^4\mathrm dr$$
Now, we can apply the condition $a\ll R$. Define $K(x)=\int_{R}^{R+x}Rr^4\mathrm dr$
$$K(x)=\int_{R}^Rr^4\mathrm dr+R^4x+O(x^2)=R^4x+O(x^2)$$
Then, for this shell:
$$I_z\approx\frac 83\pi\rho R^4a$$
Now, the shell has mass $M$ and in a symilar way as for the moment of inertia, we have the volume of the shell is in good approximation $V=4\pi R^2a$ and the mass is $M=4\pi\rho R^2a$. Finally, we can write:
$$I_z=\frac 23MR^2$$
Very different from the simple substitution.