Center of Mass of a Circular lamina

Assume a rectangular coordinate system with origin placed at the center of the original larger disc and the $x$ -axis being the line passing through the origin and the center of the hole drilled. The $y$ -axis is thus perpendicular to this line. Exploiting some symmetry, as equal masses are taken out from upper and lower portion of the $x$ -axis, the $y$ coordinate of the center of mass remains fixed at $0$ . Thus the center of mass of the final object lies on the $x$ -axis itself.

Let the surface mass density of the disc be $\sigma$ . If we assume the hole is filled in again with the cut-out disc then the center of mass of the resultant body will be back at origin. Let $m_O$ and $m_H$ denote the masses of the final object and the hole respectively and the coordinates of their respective center of masses be $x_O$ and $x_H$ . Treating this as a two body system with center of mass at the origin and using the formula

$x_{CM} = \dfrac{m_O x_O + m_H x_H}{m_O + m_H}$

we get

$\begin{aligned} & 0 = \dfrac{\sigma \left( \pi \left( r^2 - {\left(\dfrac{r}{2}\right)}^2 \right) \right) \cdot x_O + \sigma \left( \pi {\left(\dfrac{r}{2}\right)}^2 \right) \cdot \left( - \dfrac{r}{2} \right)}{\sigma \left( \pi \left( r^2 - {\left(\dfrac{r}{2}\right)}^2 \right) \right) + \sigma \left( \pi {\left(\dfrac{r}{2}\right)}^2 \right)} \\ \implies & 0 = \dfrac{3r^2}{4} \cdot x_O - \dfrac{r^3}{8} \\ \implies & x_O = \dfrac{r}{6} = \dfrac{1}{6} \end{aligned}$

Thus, $a=\boxed{6}$ .

I placed the origin of my x,y-coordinate system at the geometric center of the large circle. Since the object is symmetric about the x-axis, I considered only the top-half of the object. To make the integration simpler, I broke the object into two pieces: the quarter circle in quadrant I, and the quarter circle with a half-circle drilled out in quadrant II.

The center of mass (on x) for the first piece can be expressed as:

$\bar{x}_{1} = \frac{4}{\pi} \int_0^1 x\sqrt{1-x^{2}}dx$

I evaluated this integral using the substitution:

$u = 1-x^{2}$

$du = -2xdx$ , $dx = \frac{du}{-2x}$

$u(0) = 1$ , $u(1) = 0$

I flipped the negative sign and the limits of integration, so the integral becomes:

$\bar{x}_{1} = \frac{2}{\pi} \int_0^1 u^{\frac{1}{2}} du = \frac{4}{3\pi}$

For the second region, I "cheated" and treated it as if it, too, were in quadrant I. Its center of mass integral is then:

$|\bar{x}_{2}| = \frac{8}{\pi} \int_0^1 x\sqrt{1-x^{2}} - \frac{8}{\pi}\int_0^1 x\sqrt{\frac{1}{4} - (x-\frac{1}{2})^{2}}dx = A - B$

The first integral "A" here is clearly twice the magnitude of $\bar{x}_{1}$ . For the second integral "B," I used a trigonometric substitution:

$x-\frac{1}{2} = \frac{1}{2} sin \theta$

$dx = \frac{1}{2} cos \theta d\theta$

$\theta(0) = -\frac{\pi}{2}$ , $\theta(1) = \frac{\pi}{2}$

So this integral becomes:

$B = \frac{1}{\pi} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} cos^{2}\theta d\theta + \frac{1}{\pi} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} sin\theta cos^{2}\theta d\theta = C + D$

By inspection, the integral " $D$ ," is $0$ , so:

$B = C = \frac{1}{\pi} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1}{2} (1+cos 2\theta)d\theta = \frac{1}{2}$

So our center of mass for the second region is:

$|\bar{x}_{2}| = \frac{8}{3\pi} - \frac{1}{2}$

Finally, the center of mass for the entire object is:

$\bar{x} = \frac{m_{1}\bar{x}_{1} - m_{2}|\bar{x}_{2}|}{M} = \frac{\frac{\pi}{2}(\frac{4}{3\pi}) - \frac{\pi}{4}(\frac{8}{3\pi} - \frac{1}{2})}{\frac{3\pi}{4}} = \frac{1}{6}$

So:

$\boxed{a = 6}$