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In the Cartesian system, the position vector of a particle at $\left( x,y,z \right) $ is given by $\vec { p } =x\hat { x } +y\hat { y } +z\hat { z } $ and the centre of mass vector by $$\vec { C } =\frac { \Sigma \vec { { p }}_{ i } { m }_{ i } }{ \Sigma { m }_{ i } } $$

Similarly, will the centre of mass vector in polar coordinates be given by the same $$\vec { C } =\frac { \Sigma \vec { { p }}_{ i } { m }_{ i } }{ \Sigma { m }_{ i } }.$$ In that case the position vector of the particle at $\left( r,\theta \right) $ will be $\vec { p } =r\hat { r } $ and has no component in $\hat { \theta } $. How do we determine the angle with respect to the axis.

I'm trying to derive the centre of mass of a semicircular disk of constant surface density in polar coordinates using double integrals.

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closed as unclear what you're asking by Kyle Kanos, Jon Custer, John Rennie, Yashas, sammy gerbil Sep 13 '17 at 17:55

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  • $\begingroup$ $\overrightarrow { p } =r\hat { r }( \theta ) $ $\endgroup$ – Sanya Sep 4 '17 at 14:05
  • $\begingroup$ I don't quite understand. Or maybe I'm not familiar with the notation. As I understand, should the magnitude of the position vector also depend on the angle? Thanks. $\endgroup$ – Alpha7200 Sep 4 '17 at 14:25
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    $\begingroup$ Perhaps I'm missing something, but isn't this a straight-forward usage of coordinate transformations? $\endgroup$ – Kyle Kanos Sep 4 '17 at 15:14
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    $\begingroup$ You just need to use the fact that $\hat{r} = \cos\theta\ \hat{x} + \sin\theta\ \hat{y}$, then you can perform the integrations over $r$ and $\theta$. $\endgroup$ – Paul G Sep 4 '17 at 19:34
  • $\begingroup$ So does it mean that the centre of mass is defined only in Cartesian coordinates and to convert it into polar form, I'll have to use coordinate transformation? So there exists no direct formula to locate the COM directly using Polar Coordinates? $\endgroup$ – Alpha7200 Sep 5 '17 at 0:25