## anonymous 5 years ago Find center of mass of a semi-circle with radius r and constant p

1. anonymous

You find the center of mass coordinates from the following definitions:$x_{cm}=\frac{1}{m}\int\limits_{A}{x{\sigma}(x,y){dA}}$and$y_{cm}=\frac{1}{m}\int\limits_{A}{y{\sigma}(x,y){dA}}$Since your density is constant, and since dA=dx.dy,$x_{cm}=\frac{\sigma}{m}\int\limits_{A}{x{dx.dy}}$and$y_{cm}=\frac{\sigma}{m}\int\limits_{A}{y{dx.dy}}$The limits of integration are determined by the order you integrate each of the variables. When we integrate over y first, we integrate from y=0 to y=sqrt(r^2-x^2). Then we integrate over x from x=-r to x=r, so$x_{cm}=\frac{{\sigma}}{m}\int\limits_{-r}^{r}xdx{\int\limits_{0}^{\sqrt{r^2-x^2}}}dy$$x_{cm}=\frac{\sigma}{m}\int\limits_{-r}^{r}x{\sqrt{r^2-x^2}}dx=-\frac{{\sigma}}{3m}(r^2-x^2)^{3/2}|_{-r}^{r}=0$

2. anonymous

So the x-coordinate is 0. Next comes y...

3. anonymous

The limits of integration are determined by the following: for some y, when integrating over the x-domain, x will roam between -sqrt(r^2-y^2) and sqrt(r^2-y^2). Our y-values will then roam from 0 to r. $y_{cm}=\frac{\sigma}{m}\int\limits_{0}^{r}ydy{\int\limits}_{-\sqrt{r^2-y^2}}^{\sqrt{r^2-y^2}}dx$$=\frac{2\sigma}{m}\int\limits_{0}^{r}y{\sqrt{r^2-y^2}}dy=-\frac{2\sigma}{3m}(r^2-y^2)^{3/2}|_{0}^{r}=\frac{2r^3{\sigma}}{3m}$Now, the surface density is constant and given by,$\sigma = \frac{m}{\frac{{\pi}r^2}{2}}=\frac{2m}{{\pi}r^2}$So your y_cm is$y_{cm}=\frac{2r^3}{3m}.\frac{2m}{{\pi}r^2}=\frac{4r}{3{\pi}}$