Center of Mass

To begin the problem, imagine the center of mass of the if you cut out the square and put it back in its original position. Due to symmetry, the center of mass of the disk is at its center. Since we now have two pieces of the disk (one is the square, and the other being the rest of the disk), the center of mass of the two-piece system must also be at the original center of the disk.

Since the square cut out has side length $\frac{r}{2}$ , and one of the vertices coincides with the center of the disk, the distance between the center of the square and the center of the disk is $\frac{r\sqrt{2}}{4}$ . If we let the center of the disk be $x=0$ , then the center of the square is at $x=\frac{r\sqrt{2}}{4}$ , where the x-axis is an arbitrary axis that passes through a diameter of the disk.

Due to the equality stated in the first paragraph, we now write an equation equating the center-of-mass prior to and after the square hole is made (let the uniform mass density be $\sigma$ ):

$x_{com}=\frac{m_{s} x_{s}+m_{p} x_{p}}{m_{p}+m_{p}}=0$

where the subscripts $s$ and $p$ represent the square and the remaining piece, respectively. Note that $x_{p}$ is the center of mass of the piece, which is what we are looking for (in order to find $a+b$ )! Since the mass of any part of the disk is its area multiplied the mass density $\sigma$ , this can be rewritten as:

$0=\frac{(\frac{r^{2}}{4}(\frac{r}{4}\sqrt{2})+(\pi r^{2}-\frac{r^{2}}{4})x_{p}) \sigma}{\pi r^{2} \sigma}$

Solving this down (as the algebra is not the most significant part of the problem) yields:

$\frac{r\sqrt{2}}{16}=-\frac{4\pi-1}{4}x_{p} \Rightarrow x_{p}=-\frac{\sqrt{2}}{16\pi-4}r$

The value of $x_{p}$ is negative simply because I set the position of the square as a positive x value. Thus, the value of $d$ is $|x_{p}|$ , meaning $a=16$ and $b=-4$ . Thus, our final answer is $a+b=16-4=\boxed{12}$ .

Let the center of the disk be $0, 0$ . Let the square be centered at $\left( \dfrac {r}{4}, \dfrac {r}{4} \right)$ . To find the center of mass of the new disk, we consider the disk as two objects:

$\text {(the original disk)} + \text {(the square hole)}.$

We will consider the square hole as negative mass. We find the $x$ -coordinate of the new COM. By symmetry, the $y$ -coordinate will be the same. Let $\sigma$ be the mass density of the disk. The density of the whole will be $- \sigma$ . We get: $\begin{aligned} x_f &= \dfrac {\sigma \cdot \pi r^2 \cdot 0 - \sigma \cdot \left( \dfrac {r}{2} \right)^2 \cdot \dfrac {r}{4}}{\sigma \cdot \pi r^2 - \sigma \cdot \left( \dfrac {r}{2} \right)^2} \\&= \dfrac {-\dfrac{r^3}{16}}{\pi r^2 - \dfrac {r^2}{4}} \\&= - \dfrac {r}{16\pi - 4} = y_f. \end{aligned}$ Thus: $d = \sqrt {x_f^2 + y_f^2} = \sqrt {2} x_f = \dfrac {\sqrt{2}}{16\pi - 4} \cdot r,$ so our answer is $a + b = 16 + \left( -4 \right) = \boxed {12}$ .

First define $M = \rho \pi r^{2}$ as the mass of the whole disc being $\rho$ its density. Then the mass of the square that has been taken off is $m = \rho \frac{r^{2}}{4}$ and finally the mass of the disc without the square becomes $\xi = \frac{\rho r^{2}}{4} \times ( 4 \pi - 1 )$ . The center of mass of the whole disc (say $X_{0}, Y_{0}$ ) lies in the origin while the center of mass of the square lies in the point $X_{1} = \frac{r}{4}, Y_{1} = - \frac{r}{4}$ . Then the center of mass of the disc without the square will lie in a point with coordinates $X, Y$ such that $X = \frac{M X_{0} - m X_{1}}{ \xi } = \frac{r}{16 \pi - 4}$ Because $X_{0} = 0$ . And then $Y = \frac{M Y_{0} - m Y_{1}}{ \xi } = \frac{r}{16 \pi - 4}$ Because $Y_{0} = 0$ . Then the distance from $X, Y$ to the origin is $d(O, (X, Y)) = \sqrt{X^{2} + Y^{2}} = \sqrt{2} \times \frac{r}{16 \pi - 4}$ Therefore $a = 16$ and $b = -4$ . Then $a + b = 12$ .

Let $A$ be the square piece and $B$ be the remaining part. Since the full circle is the combination of $A$ and $B$ , the center of mass (CM) of the entire circle is the weighted average of the centers of mass of $A$ and $B$

Let $\rho$ be the mass per unit area of the circle. The mass of $A$ is $\rho \frac{r^2}{4}$ , and the mass of $B$ is $\rho \left(\pi r^2 - \frac{r^2}{4}\right)$

The distance from the CM of $A$ to the disk's center is $\frac{r\sqrt{2}}{4}$ . Let the distance from the CM of $B$ to the disk's center be $x$ .

We will have:

$\rho\left(\pi r^2-\frac{r^2}{4}\right).x=\left(\rho\frac{r^2}{4}\right)\left(\frac{r\sqrt{2}}{4}\right)$

Simplifying,

$\left(\pi - \frac{1}{4}\right).x=\frac{r\sqrt{2}}{16}$

and so, solving for $x$ yields

$x=\frac{r\sqrt{2}}{16\pi -4}$

Thus $a+b=16-4=\fbox{12}$