Find centre of mass of a circle when one half is heavier than the other half?

Question

I have a problem which simply states:

Consider a circle (lamina) of radius 1 with centre (0,0) where the left half is twice as heavy as the right. Find its centre of mass. Extend your solution to consider the left half being $n$ times as heavy as the right.

So I know that the formulae for the coordinate of centre of mass (for uniform density are):

$$x = \frac{\int_{A} x\rho dA}{\int_{A} \rho dA}$$ $$y = \frac{\int_{A} y\rho dA}{\int_{A} \rho dA}$$

I'm quite stuck but the only thing I can think of is to convert to polar coordinates to compute the actual integration but as for setting up the problem, I'm not really sure. Should I compute each half separately and just compute the weighted average? Or just compute for one set of coordinates, and then multiply the coordinate by 2 (or $n$) and find an average from there?

EDIT: so I've gone ahead and used polar coordinates to find the centre of mass for each half, here they are:

Using $\frac{\pi}{2} < \theta < \frac{3\pi}{2}$ and $0 < r < 1$

right half:

$$ x = \frac{1}{\pi2\rho}$$$$y = 0$$

and left half I could easily derive by looking at the above and dividing by the factor of two and reversing the sign:

$$x = \frac{-1}{\pi\rho}$$$$y = 0$$

What can I do now?

I'm still stuck

Answer 1

Using calculus, you can derive the following formula for the distance, along the line of symmetry, from the centre of the semicircle to the centroid: $$\bar{x}=\frac{4r}{3\pi}=\frac{4}{3\pi}$$

Assuming this result for the moment, let the area of the semicircles be $A$, the mass of the lighter one be $A\rho$, and the heavier one (on the left) be $nA\rho$.

Applying Varignon's Principle, we have $$A\rho\bar{x}-An\rho\bar{x}=A(n+1)\rho\bar{X},$$ where $\bar{X}$ is the combined centroid distance from the centre.

so this gives the general result $$\bar{X}=\frac{4(1-n)}{3\pi(n+1)}$$