Finding the centre of mass in polar coords with double integrals

Question

The centre of mass of a body can be found using the general formula:

$$ \bar{\boldsymbol{r}} = \frac{1}{M} \int \boldsymbol{r} \ \mathrm{d}M $$

(RHB, p. 195)^*. When I try to use this method with polar coordinates however, it fails.

Consider a circular disc with radius $a$ and uniform mass per unit area, $\mu$. We know immediately the centre of mass should be at $(0, 0)$ in Cartesian coords or $r = 0$ in polar. Let's try to prove this:

Firstly, we know we have:

$$ M = \pi a^2 \mu $$

And a little bit of mass is given by:

$$ \mathrm{d}M = \mu r \ \mathrm{d}r \mathrm{d}\phi $$

Then

$$ \bar{\boldsymbol{r}} = \frac{1}{\pi a^2 \mu} \int_{0}^{2\pi} \int_{0}^{a} \pmatrix{ r \\ \phi } \mu r \ \mathrm{d}r \mathrm{d}\phi $$

and

$$ \bar{\boldsymbol{r}} = \frac{1}{\pi a^2} \int_{0}^{2\pi} \pmatrix{ \frac{a^3}{3} \\ \frac{\phi a^2}{2} } \ \mathrm{d}\phi $$

and thus

$$\begin{align} \bar{\boldsymbol{r}} &= \frac{1}{\pi a^2} \pmatrix{ \frac{a^3 \pi}{3} \\ \frac{\pi^2 a^2}{4} } \\ &= \pmatrix{ \frac{a}{3} \\ \frac{\pi}{4} } \end{align}$$.

Ignoring the meaningless $\phi$ part of the result, we see that $r \neq 0$, unlike what might have been expected?

So, what is wrong with my method? I'm really struggling to see it myself. It can't be that it's due to integrating radially surely, since if that were the case I might expect to see $r = \frac{1}{\sqrt{2}}$ (giving equal areas and thus masses inside and outside the radius).

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^*This is derived from the idea that an equivalent point mass could be placed at the centre of mass, i.e.:

$$ M\bar{\boldsymbol{r}} = \int \boldsymbol{r} \ \mathrm{d}M $$

Answer 1

The 2-d vector $\vec{r}$

$$ \vec{r} = \hat{x} x + \hat{y} y \ne \hat{r} r + \hat{\phi} \phi. $$

Instead, the vector in polar coordinate: $$ \vec{r} = \hat{r} r. $$

Therefore, the center of mass $\vec{r}_c$:

$$ \vec{r}_c = \frac{1}{M} \int \vec{r} dM = \frac{\mu}{M} \int_0^R r dr \int_0^{2\pi}d\phi \hat{r}. $$

The unit vect $\hat{r}$ depend only on $\phi$ not $r$, $\hat{r} = \hat{x} \cos\phi + \hat{y} \sin\phi $. ( Suggested by Dan Pollard)

$$ \vec{r}_c = \frac{\mu R^2}{2M} \int_0^{2\pi} \hat{r} d\phi. $$

The integral on $\phi$ vanished.

Answer 2

The integral you worte down is wrong. The right one would be $$ \vec r _{CM} = \frac \mu M\int_0^{2\pi}\int_0^a\begin{pmatrix} r\cos(\phi)\\r\sin(\phi) \end{pmatrix} r \text d r \text d \phi = \begin{pmatrix}0\\0\end{pmatrix}, $$ since $$ \int_0 ^{2\pi}\sin\phi\,\text d\phi = \int_0 ^{2\pi}\cos\phi\,\text d\phi = 0 $$