# Centre of mass of an annular sector

If I have an annular sector with inner radius $r_1$ and outer radius $r_2$ and angle $\theta$, how can I calculate its centre of mass?

If you know the formula for the center of mass of a circular sector and you know the way centers of mass add, you can work it out without any calculus.

For two centers of mass we have: $$M_i*x_i + M_j*x_j = (M_i+M_j)*x_{combined}$$

where $x_{combined}$ is the center of mass in the $x$ direction for the combined masses.

For the annular sector we know that adding the sector with radius $r_1$ to the annular sector will give the sector with radius $r_2$. See figure below:

Assuming uniform density, we therefore have: $$M_1*x_1 + M_a*x_a = (M_1+M_a)*x_2$$

We know that $M_a=M_2-M_1$ which means the above equation becomes $$x_a=\frac{M_2 \, x_2 - M_1 \, x_1}{M_2-M_1}$$

With uniform density we know that mass is proportional to surface area and therefore $M_1=\frac{\theta}{2}r_1^2$ and $M_2=\frac{\theta}{2}r_2^2$. Lastly, we know the formula for the center of mass of a circular sector which is $$x_i = \frac{4r_i \, sin(\frac{\theta}{2})}{3\theta}$$

Inserting these formulas into the equation for $x_a$ gives $$x_a=\frac{4 \, sin(\frac{\theta}{2})(r_2^3-r_1^3)}{3\theta (r_2^2-r_1^2)}$$