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I'm learning about center of mass, but I have trouble understanding the definition. How is $x_{com}=\frac{1}{M}\sum_{i=1}^{n}m_ix_i$ equal to $x_{com}=\frac{1}{M}\int xdm$?

At first I thought it should be $x_{com}=\frac{1}{M}\int mdx$ since the function of mass in terms of position sounded better to me. But then I found this question, so I understood that $\frac{1}{M}\int mdx$ equals one because mass as a function of $x$ is the density function, and the integral of the density function is always equal to $M$. But that doesn't mean that I understood why $x_{com}=\frac{1}{M}\int xdm$. Since two or more positions can have same mass, the function of position in terms of mass does not exist according to the the definition of a function, does it?

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3 Answers 3

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What is generally know is the mass density distribution $\rho(\vec{x})$, and you can write that mass differential in terms of $\rho$ and a volume differential $dm=\rho(\vec{x})dV$, so the center of mass is $$\vec{x}_{com}=\frac{1}{M}\int_V\vec{x}\rho(\vec{x})dV,$$ where $M$ is the total mass, i.e. $M=\int_V\rho(\vec{x})dV$

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    $\begingroup$ I believe that you need to divide the right side by the volume integral of the density. $\endgroup$
    – Bill N
    Mar 27, 2021 at 2:14
  • $\begingroup$ Oh I forgot that! Thanks. $\endgroup$
    – AFG
    Mar 27, 2021 at 7:26
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The concept of a small particle of mass ${\rm d}m$ existing at some location $x$ contributes to the mass torque by $x \, {\rm d}m$

$$ x_{\rm CM} \int {\rm d}m = \int x \,{\rm d}m$$

But how do we integrate over mass?

Consider mass as a result of a density field $\rho(x)$ such that $$ m = \int \rho(x) A {\rm d}x $$ and consider only a small portion to see that

$$ {\rm d}m = \rho(x) A {\rm d}x $$

The above allows us the change the variable in $\int {\rm dm}$ and make it

$$ x_{\rm CM} \int \rho A \,{\rm d}x = \int x\,\rho A \,{\rm d}x $$

or

$$ x_{\rm CM} = \frac{ \int x\,\rho A \,{\rm d}x }{\int \rho A \,{\rm d}x }$$

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The integral doesn't mean that $x = f(m)$. On the contrary, $dm = f(x)dx$ and only after knowing that function it is possible to solve the integral, with the differential as function of $x$.

A good example for a unidimensional situation is any turned part, as for example a roll for a rolling mill, as shown below. The roll diameter is a function of $x$. So, the mass $dm$ of a small slice of the roll is also a function of $x$.

enter image description here

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