Imagine having a circle with center of mass in orego. I need stepwise to remove a strip from that circle (i.e. removing an arc of the circle), and for each step locate the center of mass.
An analytical expression is needed. How can this be done?
EDIT: the step - or change - would preferably be in angles.
I'm assuming a circle of uniform density and with no mass in the interior of the circle. This means the center of mass is the same as the centroid of the (remainder of the) circle.
I've used the equations for finding centroids given in this document as inspiration (see Example 3.2).
Let the center of mass be $(X_c, Y_c)$, the radius of the circle be $R$ and the angle (from the positive x-axis) of the arc to be removed, be $\theta$. From the document we find that:$$X_c\int_{\theta}^{2\pi} {Rd\theta} = \int_{\theta}^{2\pi} {xRd\theta}$$
where $$x= Rcos(\theta)$$
Integrating, this gives $$X_cR(2\pi - \theta) = R^2(sin(2\pi) - sin(\theta))$$
or $$X_c = \frac{-Rsin(\theta)}{2\pi - \theta}$$
A similar argument for $Y_c$ gives $$Y_c = \frac{R(cos(\theta) - 1)}{2\pi - \theta}$$
For the unit circle $(R = 1)$, the movement of $(X_c, Y_c)$ as $\theta$ goes from $0$ to $2\pi$ gives the following nice spiral:
Notice how the center of mass starts at origo and then moves left and down as more and more of the circle from the first quadrant is removed. When all of the circle in the first and second quadrant are removed (leaving a semi-circle in the 3rd and 4th quadrant) the center of mass is at $(X_c,Y_c) = (0, -\frac{2}{\pi})$. When only the arc in the 4th quadrant is left, $(X_c,Y_c) = (\frac{2}{\pi}, -\frac{2}{\pi})$.