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Consider a rod in three dimensional space where $y$ is the height axis. $h$ is the height of the rod and $a$ is the radius of the rod. The function $\rho(r, \theta, y)$ is the density function. The mass of the rod can be calculated with

$$ m =\int_{y=0}^{h}\int_{\theta=0}^{2\pi}\int_{r=0}^{a}\rho(r, \theta, y)rdrd\theta dy $$

similarly the center of mass in the y direction is

$$ C_y = \frac{\int_{y=0}^{h}\int_{\theta=0}^{2\pi}\int_{r=0}^{a}y\rho(r, \theta, y)rdrd\theta dy}{m} \label{1} $$

How can the center of mass in the $x$ and $z$ direction and or in terms of $r$ and $\theta$ be expressed?

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  • $\begingroup$ Definitely an issue. For example if you have a rod thats radially uniform, then the center of mass would be 0 but the integral with respect to r would be positive $\endgroup$ – Saketh Malyala Jun 12 at 0:31
  • $\begingroup$ A "rod" is generally a linear instrument. You say that $a$ is the (uniform?) radius of the rod, so presumably the rod has a circular cross-section but perhaps density varies as a function of $r\in [0,a]$ and $\theta\in [0,2\pi)$. So while the cross-sections are uniformly circular, their density can vary arbitrarily? I'm guessing at your meaning, but a (right circular) cylinder might be a better term for the figure. $\endgroup$ – hardmath Jun 12 at 0:32
  • $\begingroup$ @hardmath The density can vary with all three variables $r, \theta, y$ within their domain however it wants, continuously, continuously whatever. Otherwise yes, I think you are correct. $\endgroup$ – fullnitrous Jun 12 at 0:36
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You can convert these coordinates back to $r$ and $\theta$ using $r=\sqrt{x^2+z^2}$ and $\theta=\displaystyle \arctan\left(\frac{z}{x}\right)$.

$\displaystyle C_x = \frac{\int_{y=0}^{h}\int_{\theta=0}^{2\pi}\int_{r=0}^{a}r\cos(\theta)\rho(r, \theta, y)rdrd\theta dy}{m}$

$\displaystyle C_z = \frac{\int_{y=0}^{h}\int_{\theta=0}^{2\pi}\int_{r=0}^{a}r\sin(\theta)\rho(r, \theta, y)rdrd\theta dy}{m}$

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    $\begingroup$ damn that's a large angle expression, good thing it works $\endgroup$ – user251865 Jun 12 at 2:10
  • $\begingroup$ Wait I think this solution is a bit problematic. I am numerically solving these triple integrals so the limits need to be constant. The limit for $x$ expressed as $\sqrt{a^2 - z^2} = \sqrt{a^2 - (rsin(\theta))^2}$ varies with $\theta$ right? That does not work for my case. Also are you saying that it would be more accurate if I were to use the larger expression for $\theta$ instead of $arctan(\frac{z}{x})$ as argument for $\rho$? $\endgroup$ – fullnitrous Jun 12 at 9:40
  • $\begingroup$ yes, about the other expression, because arctan only works for positive x $\endgroup$ – Saketh Malyala Jun 12 at 10:36
  • $\begingroup$ oh no, i'm literally dumb. the other method (for x, for example) would be to use polar integration, but instead of multiplying by x, multiply by rcostheta, and divide by mass as usual $\endgroup$ – Saketh Malyala Jun 12 at 10:38
  • $\begingroup$ for z, use polar coordinates again, but multiply by rsintheta, and divide by mass $\endgroup$ – Saketh Malyala Jun 12 at 10:39
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Calculate center of mass along radial direction. Then put $\theta=0$ for x direction and $\theta=\pi/2$ for z direction

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  • $\begingroup$ By radial do you mean in the $y$ direction? Also do i use the same expression for $C_y$ when getting the center of mass for $x$ and $z$ only having the limits changed? $\endgroup$ – fullnitrous Jun 12 at 0:40
  • $\begingroup$ Actually I am not sure about the answer. I was trying to say that, you calculate $C_r$, then use different values of $\theta$ to find the center of mass the corresponding direction. $\endgroup$ – Vishal Tripathy Jun 12 at 0:53

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