center of gravity of circular sector

Consider a circular sector with central angle $2\alpha$ (in radians) and radius $R$ as shown in the diagram below. If we wish to find the distance of the center of gravity from the center of the sector, we divide the sector into elements of area $dA$ as illustrated.

In general, the mass of a lamina element is given by $dm={\delta }_{2}dA$ and the coordinates of the centers of mass are (given that the mass is evenly distributed over the area):

$\overline{x}$ $=$ ${\displaystyle \frac{\int {\int }_{A}x𝑑A}{A}},$

$\overline{y}$ $=$ ${\displaystyle \frac{\int {\int }_{A}y𝑑A}{A}}$

In this case, we will use polar coordinates as it would be much easier to carry out the integration, and the boundaries can be defined easily. In polar coordinates $dA=rdrd\theta$. The area of the sector is $A=\frac{1}{2}{R}^2(2\alpha )=\alpha R^2$. Now

$\overline{x}$ $={\displaystyle \frac{1}{\alpha R^2}}{\displaystyle {\int }_0^{2\alpha }}{\displaystyle {\int }_0^R}xr𝑑r𝑑\theta$

$={\displaystyle \frac{1}{\alpha R^2}}{\displaystyle {\int }_0^{2\alpha }}{\displaystyle {\int }_0^R}{r}^2\cos \theta drd\theta$

$={\displaystyle \frac{1}{\alpha R^2}}{\displaystyle {\int }_0^{2\alpha }}{\displaystyle \frac{R^3}{3}}\cos \theta d\theta$

$={\displaystyle \frac{R}{3\alpha }}\sin 2\alpha$

Now we follow a similar procedure for the y-coordinate:

$\overline{y}$ $={\displaystyle \frac{1}{\alpha R^2}}{\displaystyle {\int }_0^{2\alpha }}{\displaystyle {\int }_0^R}yr𝑑r𝑑\theta$

$={\displaystyle \frac{1}{\alpha R^2}}{\displaystyle {\int }_0^{2\alpha }}{\displaystyle {\int }_0^R}{r}^2\sin \theta drd\theta$

$={\displaystyle \frac{1}{\alpha R^2}}{\displaystyle {\int }_0^{2\alpha }}{\displaystyle \frac{R^3}{3}}\sin \theta d\theta$

$={\displaystyle \frac{R}{3\alpha }}(1-\cos 2\alpha )$

The center of gravity is $(\overline{x},\overline{y})$ and the distance $d$ of the center of gravity from the center of the sector is given by:

$$d=\sqrt{{\overline{x}}^2+{\overline{y}}^2}$$

We substitute for $\overline{x}$ and $\overline{y}$:

$d$ $=\sqrt{{\left({\displaystyle \frac{R\sin 2\alpha }{3\alpha }}\right)}^2+{\left({\displaystyle \frac{R(1-\cos 2\alpha )}{3\alpha }}\right)}^2}$

$={\displaystyle \frac{R}{3\alpha }}\sqrt{{\sin }^{2}2\alpha +{(1-\cos 2\alpha )}^2}$

From trigonometry, we know that:

$2{\sin }^2\alpha =1-\cos 2\alpha$

${\sin }^2\alpha +{\cos }^2\alpha =1$

$\sin 2\alpha =2\sin \alpha \cos \alpha$

Keeping these in mind, we substitute:

$d$ $={\displaystyle \frac{R}{3\alpha }}\sqrt{4{\sin }^2\alpha {\cos }^2\alpha +4{\sin }^4\alpha }$

$={\displaystyle \frac{2R}{3\alpha }}\sqrt{{\sin }^2\alpha ({\cos }^2\alpha +{\sin }^2\alpha )}$

$={\displaystyle \frac{2R\sin \alpha }{3\alpha }}$

In conclusion, the distance of the center of gravity of a circular sector with radius $R$ and angle $2\alpha$ from the center of the sector is given by:

$$d=\frac{2R\sin \alpha }{3\alpha }$$

Title	center of gravity of circular sector
Canonical name	CenterOfGravityOfCircularSector
Date of creation	2013-03-22 18:06:30
Last modified on	2013-03-22 18:06:30
Owner	curious (18562)
Last modified by	curious (18562)
Numerical id	6
Author	curious (18562)
Entry type	Topic
Classification	msc 44A99
Related topic	CentreOfMassOfHalfDisc