My post is an elaboration of those here and here.

In my AP Physics C class, I was introduced to the following equation for center of mass (CM): $$x_\text{cm} = \frac 1M \int x\, dm.$$ Above, $dm$ represents a "slice" of the given object.

Problems would then be solved using the below format:

Prompt: "Find the CM of a metal rod with constant density, mass $M$, and length $L$."

Observe that $M/L = dm/dx$ so $dm = \frac{M}{L}\, dx$. Then: $$\begin{align*}x_\text{cm} &= \frac 1M \int x \, dm \\ &= \frac 1M \int \frac ML x\, dx \\ &= \int^L_0 \frac{x}{L} \tag 1 \\ &= \frac{L}{2}\end{align*}$$ The boundaries of integrations at (1) appear since the object ranges from $x = 0$ and $x = L$. They are not specified in the formula though because we don't necessarily integrate over $x$. For more examples of this form of problem-solving, see Dan Fullterton's video.

I am confused by the statement:

"Observe that $M/L = dm/dx$ so $dm = \frac{M}{L}\, dx$."

Manipulation of $dx$ and $dm$ is mathematically undefined since infinitesimals aren't variables.

Since I couldn't understand the class-provided technique, I ended up employing the equation below for an object spanning $x = a$ to $x = b$: $$x_\text{cm} = \frac 1M \int^a_b \lambda(x) x\, dx.$$ Above, $\lambda(x) = \rho (x) \cdot A(x)$ where $\rho(x)$ provides the average density at the object's intersection with the plane at $x$, and $A(x)$ is the area of that intersection.

Is this a valid alternative equation? Also, how does the earlier technique work in a technical sense?

  • 3
    $\begingroup$ Manipulation of 𝑑𝑥 and 𝑑𝑚 is mathematically undefined since infinitesimals aren't variables The fun thing about physics is that you can do so and it does work. $\endgroup$
    – Kyle Kanos
    Nov 1, 2023 at 19:26
  • $\begingroup$ See physics.stackexchange.com/q/92925/25301 and some of the linked posts on the side $\endgroup$
    – Kyle Kanos
    Nov 1, 2023 at 19:30
  • $\begingroup$ You do not have to use "undefined infinitesimals." The symbol "$\frac{dm}{dx}$" can be taken to mean the derivative of $m(x)$ with respect to $x$. If you would prefer, go ahead and write it like $m'(x)=\frac{M}{L}$. You still need to change variables in your integral. A la: $\int dm f(m) = \int dx m'(x) f(m(x))$, no "undefined infinitesimals" needed... $\endgroup$
    – hft
    Nov 1, 2023 at 22:54

4 Answers 4


Your alternative equation is valid.

A more technical way of understanding the mathematically undefined equation is by multiplying both sides by $x$ and then integrating both sides with respect to $dx$:

$$\frac{M}{L} = \frac{dm}{dx},$$

$$\int{x\frac{M}{L}} dx = \int{x\frac{dm}{dx}} dx.$$

Now, we can substitute $dx = \frac{dx}{dm} dm$ on the right side:

$$\int{x\frac{M}{L}} dx = \int{x \frac{dm}{dx} \frac{dx}{dm} dm}.$$

And then we simplify $\frac{dm}{dx} \frac{dx}{dm} = 1$, and we're left with the desired expression:

$$\int{x\frac{M}{L}} dx = \int{x dm}.$$

  • $\begingroup$ OP has expressed concerns with multiplying/dividing infinitesimals, which your answer uses repeatedly. $\endgroup$
    – Kyle Kanos
    Nov 1, 2023 at 20:25
  • $\begingroup$ Nope, this answer never multiplies or divides any infinitesimals. $\endgroup$
    – Travis
    Nov 1, 2023 at 20:44
  • $\begingroup$ Then how does $(dm/dx)(dx/dm)=1$? $\endgroup$
    – Kyle Kanos
    Nov 1, 2023 at 20:52
  • 1
    $\begingroup$ I'm not multiplying infinitesimals though, I'm multiplying total derivatives. $\endgroup$
    – Travis
    Nov 1, 2023 at 22:35
  • 1
    $\begingroup$ This answer is correct and appears to address OP's concerns, its just that Travis is writing $\frac{dm}{dx}$, which is abhorrent to OP. But Travis is taking the entire symbol "$\frac{dm}{dx}$" not as a "fraction" but as a derivative. It might be less abhorrent to OP if "$\frac{dm}{dx}$" was written as "$m'(x)$." $\endgroup$
    – hft
    Nov 1, 2023 at 23:04

the mass is equal to the density time volume

$$M=\rho\,V$$ thus if the density $~\rho~$ is constant then $$dM=\rho\,dV=\rho\,A\,dx$$

where the area $~A~$ is constant


$$\frac{dM}{M}=\frac{\rho\,A\,dx}{\rho\,V}=\frac{\rho\,A\,dx}{\rho\,A\,L}= \frac{dx}{L}$$


Observe that $M/L=\mathrm{d}m/\mathrm{d}x$ so $\mathrm{d}m=\frac{M}{L}\mathrm{d}x$

This really isn't the best way to state this. A more clear way is that for a uniformly dense rod, the linear (mass) density is $\lambda=M/L$. But since the mass density is constant, we can also say that for an infinitesimal slice of the rod, the mass of the slice is $\mathrm{d}m$ and must be equal to the product of the density and the (infinitesimal) length, $\mathrm{d}x$:1 $$\mathrm{d}m=\lambda\,\mathrm{d}x$$

You can then substitute this into your COM equation, $$\frac{1}{M}\int_{0}^{M} x\,\mathrm{d}m=\frac{1}{M}\int_{0}^{L} x\lambda\,\mathrm{d}x=\frac{1}{M}\int_{0}^{L} x\frac{M}{L}\,\mathrm{d}x=\frac{1}{L}\int_{0}^{L} x\,\mathrm{d}x=\frac{L}{2}$$ Note that in the first equality, when we made the change of variables, we had to change the limits of the integrals--it was easy in this case, but in some other changes of variables yield more difficult limit changes.

This method is essentially equivalent to the alternative method you've discovered yourself.

1. Note this would also be equivalent to saying that the mass density is the derivative of the mass with respect to the position: $\lambda=\mathrm{d}m/\mathrm{d}x$.


For your uniform density rod of constant cross-section,

enter image description here

and the relationship follows.

If you do not like the way that $dm/dx$ was used as a fraction then go back one step and use the fraction $\delta m/\delta x$, do the algebra, and then take a limit $\delta x\to 0$ for the summation which is where the integral comes from.

Your use of $\lambda(x) = \rho (x) \cdot A(x)$ is fine and note that $\lambda(x) \cdot dx= \underbrace{\rho (x)\,A(x)\,dx}_{dm}$.


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