Strange center of mass equation in AP Physics C

Question

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?

Answer 1

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:

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

Answer 2

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

$\Rightarrow$

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

Answer 3

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: $\lambda=\mathrm{d}m/\mathrm{d}x$.}

Answer 4

For your uniform density rod of constant cross-section,

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 $\lambda(x) \cdot dx= \underbrace{\rho (x)\,A(x)\,dx}_{dm}$.