How to find the centre of mass using planar polar coordinates.

Question

I have been searching everywhere online on how to approach these questions. But every example I find, it is said in the question what the density function is. That is what I'm struggling to find with this question, which is as follows:

The density $\rho$ (mass per unit area) of a semi-circular lamina $\Omega$ of radius a is proportional to the distance from the centre of the circle. Find the centre of mass $(\bar{x},\bar{y})$ of the lamina, taking $\bar{x} = 0$ by symmetry, given $$ m\bar{y} = M_y = \iint_\Omega y\rho(x,y)\mspace{4mu}dA \quad \text{and} \quad m = \iint_\Omega \rho(x,y)\mspace{4mu}dA $$ where $m$ is the mass of the lamina. Use Planar polar coordinates.

I am trying very hard to get better at these questions so a detailed explanation would be very much appreciated.

Thank you

Answer 1

The lamina can be easily described using polar coordinates with the region: $$0\le r \le a$$ $$0\le \theta \le \pi $$

Additional, the formula for center of mass (y - coordinate in this case) of a 2d object is: $$ \frac{\int\int y\rho(x,y) dA}{\int\int \rho(x,y)dA}$$ where $\rho (x,y)$ is density.
From here, all we need is a function for density, a description of y in polar coordinates, and a description of dA in polar. The density is proportion to the distance from origin. In other words $$density=k\sqrt{x^2 + y^2}$$ Where k is the proportionality constant. $x^2 + y^2 = r^2$ in polar coordinates. Hence $$density = kr$$ We can ignore the k when solving for center of mass because it cancels off in the calculation (it is in the numerator and denominator).

We also know that in polar coordinates $y=rcos(\theta)$ and $dA = rdrd\theta$.
You can thus go from there!

Answer 2

The hint tells you to use polar coordinates. So you need to write $x,y,\rho(x,y),dA,$ and $\Omega$ in terms of polar coordinates.

Assume that we use $r$ and $\theta$ as our polar coordinates, such as at $\theta=0$ we point along the positive $x$ direction, and at $\theta=\pi/2$ we point in the positive $y$ direction. Then:

$$ x=r\cos(\theta)\\ y=r\sin(\theta)\\ dA=r dr d\theta\\ \rho(r,\theta)= c r $$

Here $c$ is a constant that we will determine from he mass condition. The integration domain $\Omega$ is the $[0,R]$ interval along the $r$ coordinate and $[0,\pi]$ interval along the $\theta$ coordinate, where $R$ is the radius of the circle.

I will not continue the calculations, but I will note that the mass of the object is proportional to $cR^2$ and the $M_y$ integral must be proportional to $cR^3$, so $\bar{y}$ is going to be proportional to $R$