anonymous
  • anonymous
Find center of mass of a semi-circle with radius r and constant p
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • 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\[x_{cm}=\frac{{\sigma}}{m}\int\limits_{-r}^{r}xdx{\int\limits_{0}^{\sqrt{r^2-x^2}}}dy\]\[x_{cm}=\frac{\sigma}{m}\int\limits_{-r}^{r}x{\sqrt{r^2-x^2}}dx=-\frac{{\sigma}}{3m}(r^2-x^2)^{3/2}|_{-r}^{r}=0\]
anonymous
  • anonymous
So the x-coordinate is 0. Next comes y...
anonymous
  • 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\[y_{cm}=\frac{2r^3}{3m}.\frac{2m}{{\pi}r^2}=\frac{4r}{3{\pi}}\]

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