Centre of mass of sector.

Question

How do I find the centre of mass (COM) of a sector of angle $\alpha$?

Attempt:

Go to an angle $\theta$ from $0$ rad and then choose a triangular strip of angle $d\theta$ and base $Rd\theta$. The centre of mass of this triangular strip lies at a height $\dfrac R3$ from the base. What do I do after this? How do I express the $R/3$ in terms of $\theta$? After that I just need to integrate from $0$ to $\alpha$

Answer 1

If you set up the sector $-\alpha/2 \le \theta \le \alpha/2$ and $0 \leq r \leq R$ then it is clear that the COM lies on the $x$-axis. You need to find the $x$-coordinate. $$ x_{\text{COM}} = \frac{\int_{\text{sector}} x\,dM}{\int_{\text{sector}} \,dM} = \frac{\int_{\text{sector}} x\,dA}{\int_{\text{sector}} \,dA} = \frac{\int_{r=0}^{R} \int_{\theta = -\alpha/2}^{\alpha/2} r^2\cos\theta\,d\theta dr}{R^2\alpha/2} =\frac{4R\sin(\alpha/2)}{3\alpha} $$

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ Hereafter, $\ds{d}$ is the sector radius: \begin{align} \vec{\mc{R}}_{cm} & \equiv {\ds{\int_{sector}\vec{r}\,\dd^{2}\vec{r}} \over \ds{\int_{sector}\dd^{2}\vec{r}}} = {\ds{\int_{0}^{\alpha}\int_{0}^{d}\bracks{r\cos\pars{\theta}\,\hat{x} + r\sin\pars{\theta}\,\hat{y}}r\,\dd r\,\dd\theta} \over \ds{\int_{0}^{\alpha}\int_{0}^{d}r\,\dd r\,\dd\theta}} \\[5mm] & = {\ds{\pars{d^{3}/3}\braces{\bracks{% \int_{0}^{\alpha}\cos\pars{\theta}\,\dd\theta}\hat{x} + \bracks{\int_{0}^{\alpha}\sin\pars{\theta}\,\dd\theta}\hat{y}}} \over \ds{\pars{d^{2}/2}\int_{0}^{\alpha}\dd\theta}} \\[5mm] & = {2 \over 3}\,d\,{\sin\pars{\alpha}\,\hat{x} + 2\sin^{2}\pars{\alpha/2}\,\hat{y} \over \ds{\alpha}} = \bbx{{2 \over 3}\,{\sin\pars{\alpha} \over \alpha}\,d\,\hat{x} + {2 \over 3}\,{\sin\pars{\alpha/2} \over \alpha/2}\,d\,\hat{y}} \end{align}