# Working with infinitesimal quantities and the motivation behind it

So in my freshman physics class, in classical mechanics the homework was (it's solved already, this isn't a homework thread) the following:

"A thin, spinning ring is placed on a table, that divides into 2 halves, with different coefficients of friction ($$\mu_2 > \mu_1$$). What is the initial acceleration of the ring?"

We solved it, but it required a process that i didn't have experience with, and that is working with infinitesimal quantities throughout the whole problem.
I also found some other resources where such a technique is used

Below is the solution to the spinning ring problem, but the rules we can and can't do, and the way we motivated the equations seem arbitrary. I'm looking for any resource that deals with carrying infinitesimals and getting a solution by integrating over them at the end.

Solution:
A picture that gives the idea . From this we conclude that the ring won't move in the Y direction, so we can focus only on the X. Here comes the infinitesimal part:
We take a small line-segment dl with mass dm and compute the friction force that it experiences:
$$d F_{fric} = \mu \cdot dm\cdot g$$ but since due to a uniform mass distribution $$dm = \dfrac{M}{2R \pi} \cdot dl$$ and $$dl \approx d \varphi \cdot R$$ $$d F_{fric}$$ then becomes $$d F_{fric} = \dfrac {\mu M g}{2\pi} d\varphi$$ As said above we are intrested in the X component of this, for which $$d {F_{fric}}_{X} = d F_{fric} \cdot sin(\varphi)$$
To get the net force on a semi-circle we integrated this so: $$\int_{0}^{\pi} {F_{fric}}_{X} d \varphi =\dfrac{\mu M g}{\pi}$$
and therefore the net acceleration looks like : $$a_{net} =(\mu_1 -\mu_2) \dfrac{g}{\pi}$$
Here's a picture that captures this all
But when can/should we try to approach problems with such tools? Also, as said before, it really seems that some of these equations are motivated out of thin air.

• Concerning infinitesimals, see e.g. How to treat differentials and infinitesimals? and links therein. – Qmechanic Nov 21 '19 at 13:27
• "It seems that some of these equations are motivated out of thin air"... does it? Your first equation seems to be 2nd Newton's principle which is fundamental in mechanics, am I right? Then for a force applied on an infinitesimal amount of mass, you need to look at what this mass is proportional to, which is the length of your ring. But the ring being circular it seems natural to use polar coordinates with origin centered on the ring, no? So the length is proportional to the angle and so on... Part of the exercise is also for you to grasp the reasoning behind its solution. – G.Clavier Nov 21 '19 at 13:49
• What seemed unmotivated was to use infinitesimals (because i never really saw it used in such a way before), not how we calculate the friction force or newton's 2nd law . It was surprising method of solving a problem to introduce a bunch of infinitesimal terms, establish relationships between them until we can get the quantity we were looking for with a simple integral. @G.Clavier – JohnyO42 Nov 21 '19 at 13:57
• Well that's one way of establishing the quantity which you want to integrate. As I said, part of the exercise objective is to introduce you to different way of thinking about it. This is one of them, if you have a better one, that's also great. – G.Clavier Nov 21 '19 at 14:20

Let's imagine that we want to calculate the area of a circle of radius $$R$$, and we don't know the general formula $$S = \pi R^2$$, but do know that the circumference of a circle is $$C = 2\pi R$$. Then we can say that we will divide the circle into many concentric circles with radii in the range $$(0,R]$$, and then the surface between pair of such circles is $$2\pi r dr$$ where $$r$$ is the radius of the inner circle, just because this length times 'height'. Now this is not true. The real surface between pair of such circles is $$\Delta S = \pi (r+dr)^2-\pi r^2 = 2\pi r dr + \pi dr^2$$. However, it is true to linear order in $$dr$$. So if we take infinitesimal $$dr$$, omitting the quadratic term is ok and then we can integrate $$S = \int_0^{R} 2\pi r dr = \pi R^2$$ and we got the correct result!
The idea is similar when dealing with problems in physics. If we restrict ourselves only to small changes, described by infinitesimal quantities, we tend to get linear equations, which nonetheless capture the entire physics (sometimes...). Then we can either integrate or turn this equations into a differential equation describing change over small ranges, and get the full nonlinear behavior from these solutions/integrations. The linear behavior is more easy and intuitive to derive, as it involves only small steps, just like the area was length times height when we took small $$dr$$.