2
$\begingroup$

This is the integral for finding the centre of mass in one dimension:

$$X=\frac{1}{M}\int xdm.\tag{1}$$

But I was wondering whether we could do it by taking x as the integrating variable:

A homogenous rod of length $X$ is split into $N$ regions of width $\Delta x$.

Let $m$ be the mass of every such region.

As $N\rightarrow \infty$, $\Delta x \rightarrow dx.$

$$\text{Centre of mass}=\frac{\int mdx}{\int m}=\frac{m}{M}\int_0^xdx=\frac{mX}{2M},\tag{2}$$

where $M=\text{total mass}$ and $m=\text{some constant}.$

But it is known that

$$\text{Centre of mass}=\frac{X}{2},\tag{3}$$

$$\therefore \frac{mX}{2M}=\frac{X}{2}\tag{4},$$

$$\therefore m=M\tag{5}.$$

It is faulty ($m$ is not equal to $M$). Where have I gone wrong?

Is it possible to express the integral (in centre of mass in one dimension expression) by having $x$ as the integrating variable (like I attempted to do in the picture)? If not, then why?

$\endgroup$
2
  • 2
    $\begingroup$ $\int m$ doesn't make sense $\endgroup$ – BioPhysicist Mar 29 '20 at 4:46
  • 1
    $\begingroup$ Maybe this will be helpful for understanding the $dm$. $\endgroup$ – knzhou Mar 29 '20 at 6:54
1
$\begingroup$

If you want to go the "chopped up" route, first start with the discrete sum (ignoring the total mass term for now)

$$\sum_{i=1}^Nx_im$$ where $x_i$ is the center of section $i$. This is an important term to have here, as we will see. By the set up, $x_i=(i-1/2)\Delta x$, so we have

\begin{align} \bar x&\propto \sum_{i=1}^Nm\left(i-\frac12\right)\Delta x\\ &=m\Delta x\sum_{i=1}^N\left(i-\frac12\right)\\ &=m\Delta x\left(\frac{N(N+1)-N}{2}\right)\\ &=m\Delta x\left(\frac{N^2}{2}\right) \end{align}

Now, the total mass is given by $Nm$, and $\Delta x=X/N$, so we end up with $$\bar x=\left(m\cdot\frac XN\cdot\frac{N^2}{2}\right)/(Nm)=X/2$$ with no integral needed.

However, I did this to point out how important the position of each mass is. In your integral where you just do $m\,\text dx$ you have completely removed all information about the position of each mass, because $\text dx$ is not the position of each mass element, just like how $\Delta x$ in the above example wasn't the position $x_i$ of each section, it would be like instead starting with the sum $\sum_{i=1}^N m\Delta x$, which doesn't tell you where any mass is.

The $\Delta x$ or $\text dx$ is instead used to determine the location of each mass element. You can't just switch the integration variable without realizing this. This is why you need to specify some linear mass density function $\lambda(x)=\frac{\text dm}{\text dx}$ that can be used to indicate how much mass is located at each position as you do the integral over space. i.e. for the mass element at position $x$, $\lambda(x)\text dx=\text dm$ amount of mass is present, so then $x\,\text dm=x\lambda (x)\,\text dx\neq m\,\text dx$

In the discrete example with the uniform rod, $\lambda=M/X=Nm/X=m/\Delta x$. So to come full circle our sum of interest would have been $$\sum_{i=1}^Nx_i\lambda\Delta x$$

And so there is your full analogy $$\sum_{i=1}^Nx_im=\sum_{i=1}^Nx_i\lambda\Delta x$$ $$\int x\,\text dm=\int x\lambda\,\text dx$$

Also note that this is true for any change of variables. You cannot just exchange $m$ variables for $x$ variables and vice versa. You have to keep in mind how one variable varies with respect to the other. For example, in the usual "u-substitution" method taught in most introductory calculus classes, you have some proposed substitution $u=f(x)$, and so your differentials are related by $\text du=\frac{\text df}{\text dx}\text dx$.

$\endgroup$
4
  • $\begingroup$ Okay so if I were to do it using integration, the only way to do is by using mass density function? Converting physics problems into integrals is very confusing for me. I got my fault though so thanks, I forgot x was supposed to be a coordinate. dx would be a length and not a coordinate, correct (in terms of visualizing it)? And that we need the coordinate x for finding the centre of mass and not length dx? $\endgroup$ – Michael Faraday Mar 29 '20 at 5:39
  • 1
    $\begingroup$ @MichaelFaraday You got it :) $\endgroup$ – BioPhysicist Mar 29 '20 at 10:54
  • $\begingroup$ Okay thanks!!!! $\endgroup$ – Michael Faraday Mar 29 '20 at 14:31
  • $\begingroup$ @MichaelFaraday Notice the part I just added to the end of my answer. This can be compared to a general change of variables you might be more familiar with. $\endgroup$ – BioPhysicist Mar 29 '20 at 14:33
2
$\begingroup$

Your equation (2) is wrong. To use $x$ as the integrating variable you need to change $x\,dm$ into $x\,\dfrac{dm}{dx}\,dx$. This means we need to define $m$ as a function of $x$, and the most reasonable way to do that while keeping the original meaning of $dm$ is to let $m(x)$ be the mass of the section that goes from $0$ to $x$.

$\endgroup$
2
$\begingroup$

The integral for the center of mass, using your notation, is correct:$$ \text{CM}=\frac{1}{M} \int_{a}^{b} xdm$$

Notice that $mdx \neq xdm$ in your equation 2. Thus, you are no longer calculating the center of mass.

An easier way to think about this is defining a "linear mass density variable" $\lambda$, which is given by $$\lambda = \frac{M}{X}$$ since the rod is homogenous.

Then, in order to integrate using $dx$, you could note that $dm = \lambda dx$.

Thus, your integral for the center of mass becomes: $$\text{CM}=\frac{1}{M} \int_{a}^{b} xdm = \frac{1}{M}\int_{0}^{X} x \lambda dx = \frac{\lambda X^2}{2M} = \boxed{\frac{X}{2}}$$

$\endgroup$
1
  • $\begingroup$ Why is it that integral(m*dx)/M does not also give the centre of mass? I thought you could take either m or x as the integrating variable. How to know which is the integrating variable? $\endgroup$ – Michael Faraday Mar 28 '20 at 20:28
1
$\begingroup$

You do that because of ${\rm d}m \propto {\rm d}x$ by means of linear mass factor $\lambda$. This means you can substitute $$\boxed{{\rm d}m = \lambda \,{\rm d}x}$$

$$ M = \int {\rm d}m = \int \lambda\, {\rm d}x $$

Where $\lambda$ is in units of mass per length, and usually a function of position $x$.

So the center of mass is

$$ X = \frac{1}{M} \int x\,{\rm d}m = \frac{1}{M} \int \lambda\, x\,{\rm d}x $$

For a uniform rod with $\lambda = \text{(const)}$

$$ \left. M = \int \lambda\, {\rm d}x = \lambda\, \ell \; \right\} \; \lambda = \frac{M}{\ell}$$

and

$$ X = \frac{1}{M} \int \lambda\, x\,{\rm d}x = \frac{1}{M} \int \frac{M}{\ell}\, x\,{\rm d}x = \frac{1}{\ell} \int x \, {\rm d}x = \frac{1}{\ell} (\tfrac{1}{2} \ell^2) = \tfrac{1}{2} \ell$$

$\endgroup$
6
  • $\begingroup$ Thanks. For finding centre of mass, we need to find (sum of all mx) / M. How to make out that we need to find integral(xdm) and not integral(m*dx). I thought you can make any of the two variables as the integrating variable. $\endgroup$ – Michael Faraday Mar 28 '20 at 19:03
  • $\begingroup$ Sorry, I forgot to add the most important part. See my update now where I show how to change from ${\rm d}m$ to ${\rm d}x$ . $\endgroup$ – John Alexiou Mar 28 '20 at 23:42
  • $\begingroup$ Thanks, I have understood what you have done. But is not integral(m*dx)/M, also the expression for centre of mass? Where m is the mass of element dx? Why? $\endgroup$ – Michael Faraday Mar 29 '20 at 2:58
  • $\begingroup$ The element ${\rm d}x$ does not have a finite mass like $m$, it has an infinitesimal mass ${\rm d}m$. What you are doing is splitting the rod into small pieces and adding up each contribution. The integral $\int m {\rm d}x$ does not have a physical interpretation because $m$ is is a property of the entire rod. $\endgroup$ – John Alexiou Mar 29 '20 at 3:55
  • $\begingroup$ But I defined m as the mass of an element dx. Defined it as M/N where N is the number of dx elements. Why is it wrong? Sorry if it is a stupid question but I'm very confused. $\endgroup$ – Michael Faraday Mar 29 '20 at 4:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.