Moment of inertia of a hollow sphere wrt the centre? [closed]

I've been trying to compute the moment of inertia of a uniform hollow sphere (thin walled) wrt the centre, but I'm not quite sure what was wrong with my initial attempt (I've come to the correct answer now with a different method). Ok, here was my first method:

Consider a uniform hollow sphere of radius $R$ and mass $M$. On the hollow sphere, consider a concentric ring of radius $r$ and thickness $\text{d}x$. The mass of the ring is therefore $\text{d}m = \frac{M}{4\pi R^2}\cdot 2\pi r\cdot\text{d}x$. Now, use $r^2 = R^2 - x^2:$ $$\text{d}m = \frac{M}{4\pi R^2}\cdot 2\pi \left(R^2 - x^2 \right)^{1/2}\text{d}x$$ and the moment of inertia of a ring wrt the centre is $I = MR^2$, therefore: $$\text{d}I = \text{d}m\cdot r^2 = \frac{M}{4\pi R^2}\cdot 2\pi\left(R^2 - x^2\right)^{3/2}\text{d}x$$ Integrating to get the total moment of inertia: $$I = \int_{-R}^{R} \frac{M}{4\pi R^2} \cdot 2\pi\cdot \left(R^2 - x^2\right)^{3/2}\ \text{d}x = \frac{3MR^2 \pi}{16}$$

which obviously isn't correct as the real moment of inertia wrt the centre is $\frac{2MR^2}{3}$.

What was wrong with this method? Was it how I constructed the element? Any help would be appreciated, thanks very much.

closed as off-topic by ACuriousMind♦, Kyle Kanos, John Rennie, Gert, Sebastian RieseDec 13 '15 at 22:02

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The mass of the ring is wrong. The ring ends up at an angle, so its total width is not $dx$ but $\frac{dx}{sin\theta}$

You made what I believe was a typo when you wrote

$$\text{d}m = \frac{M}{4\pi R^2}\cdot 2\pi \left(R^2 - x^2 \right)\text{d}x$$

because based on what you wrote further down, you intended to write

$$\text{d}m = \frac{M}{4\pi R^2}\cdot 2\pi \sqrt{\left(R^2 - x^2 \right)}\text{d}x$$

This problem is much better done in polar coordinates - instead of $x$, use $\theta$. But the above is the basic reason why you went wrong.

In essence, $sin\theta=\frac{r}{R}$ so you could write

$$\text{d}m = \frac{M}{4\pi R^2}\cdot 2\pi \frac{r}{sin\theta} \ \text{d}x \\ = \frac{M}{4\pi R^2}\cdot 2\pi \frac{r}{\frac{r}{R}} \ \text{d}x\\ = \frac{M}{4\pi R^2}\cdot 2\pi R \ \text{d}x\\ = \frac{M}{2 R} \ \text{d}x$$

Now we can substitute this into the integral:

$$I = \int_{-R}^{R} \frac{M}{2 R} \cdot \left(R^2 - x^2\right)\ \text{d}x \\ = \frac{M}{2R}\left[{2R^3-\frac23 R^3}\right]\\ = \frac23 M R^2$$

• Thanks for answering, and I've arrived at the correct answer using the polar coordinates method, I was wondering what was wrong with this method. Your answer makes sense, but I'm not quite sure now why a similar technique (as in the OP) works for a solid sphere and a plate of thickness $\text{d}x$ but a similar doesn't work for a hollow sphere with a ring. Could I ask you to clarify this, please? Thank you. – Felix Felicis Apr 24 '14 at 2:03
• The difference is that when you have a solid object, the angle at the edge is insignificant; but when you have a thin shell, that angle becomes hugely significant. I have expanded my answer to show how to finish the derivation using the method you started (without resorting to polar coordinates). It turns out that things cancel quite nicely... – Floris Apr 24 '14 at 2:10