# Moment of inertia of a solid sphere; spot the mistake [closed]

What an I doing wrong?:

$$I = \int r^2dm$$

$$M = \rho \frac{4}{3} \pi r³$$

$$dM = \rho \frac{4}{3} \pi 3r²dr = \rho 4 \pi r²dr$$

$$I = \rho 4π \int r^4dr = \rho 4π \frac{r^5}{5} = \frac{3Mr^2}{5} = \frac{3}{5} Mr^2$$

Instead of factor $$\frac{2}{5}$$.

• spot the mistake ... What an I doing wrong? Please be aware that check-my-work are usually considered off-topic on this site. I think the members who answered made an exception for your question because you made a conceptual error rather than a calculational one. – G. Smith Mar 24 at 2:46
• Yes I'm aware that this isn't really well defined for this sites but i knew my question was a conceptual error and since my question is about the method of calculation I must provide my work thus far. But I'll definitely try to avoid questions like these. – Pim Laeven Mar 26 at 23:15

You have not used the correct definition of $$I$$.

The moment of inertia is defined as $$I=\int r^2_\perp dm=\int r^2_\perp\rho dV.$$ Using spherical coordinates $$\vec r=\begin{pmatrix}x\\y\\z\end{pmatrix}=r\begin{pmatrix}\cos\varphi\sin\theta\\ \sin\varphi\sin\theta\\\cos\theta\end{pmatrix},$$ we get $$r_\perp^2=x^2+y^2=r^2\sin^2\theta\,\,\,\text{and}\,\,\,dV=r^2\sin\theta drd\varphi d\theta$$ and thus $$I=\rho\int r^2\sin^2\theta dV=\rho\int_0^R r^4dr\int_0^{2\pi} d\varphi\int_0^\pi\sin^3\theta d\theta=\rho\frac{R^5}{5}2\pi\frac{4}{3}=\frac{2}{5}MR^2$$since$$\rho=\frac{M}{4/3\pi R^3}.$$

You need to evaluate the moment of inertia about an axis through the center, and that distance is not always $$r$$ for a mass element $$dm$$ from the axis.

• Ah of course, how stupid of me, thanks a lot! – Pim Laeven Mar 23 at 23:12
• You are welcome. – John Darby Mar 24 at 12:34

While your calculation is correct, your derivation starts from the wrong premise.

A moment of inertia is calculated about a given axis of rotation, not about a point.

I assume in your case the axis of rotation runs through the Centre of Mass of the uniform sphere.

In that case, you need to slice up the sphere into infinitesimal cylinders (circular plates) along the axis, then sum by integration the moments of inertia of these plates along the axis.

You can find the full derivation here.