Why is the mass of small elements taken as $∆m$ in center of mass of a continuous body?

A continuous body has continuous distribution of mass. Doesn't $$\Delta m$$ mean $$m_f - m_i$$? But, is the mass Changing? If yes, how is the mass varying? Why is the mass of the small elements in a body taken as $$\Delta m$$? Why isn't it taken just as $$m$$ (mass of the small element)?

• $\Delta$ means change in general, not final minus initial. Change can be with respect to anything, for example distance in case of continuous bodies, or time in case of finding speed. Nov 17 at 10:21
• I think the same symbol (in this case, the delta) can mean two different things, in this case when you divide a rigid body into small parts, $\Delta m$ is just the mass of one of those parts Nov 17 at 10:22
• For summing up or integration, we divide into number of smaller parts. Suppose mass $M$ having some volume is divided into $N$ parts all having same mass then, $\frac{M}{N}=\Delta M$. More the number of parts, less the error. Nov 17 at 10:48

It's a good question.

Normally speaking, we consider a quantity written as $$\Delta x$$ to conceptually mean a "change in a quantity". For example a "tiny change in time" or a "tiny change in position".

However, the usage in case of center of mass calculation is in a different conceptual sense. We can think of chopping up the body into some chunks with each chunk being of mass:

$$dm = \rho dV$$

Here I am considering that we have a 3-D body but it could be that we have 2d or 1d as well.

In a way, we can think of the two approaches as actually being the same. Suppose you had a rod of length L lying along the x-axis from $$x=0$$ to $$x=L$$. You can imagine a situation where tiny rods of mass $$dm$$ are placed consecutively with tail of one at head of previous as time passes to reconstruct the original rod.

• Great answer, showing how the ∆m can indeed be conceptualized as a change, and sharing the small mass picture with the better notation dm as well. Nov 17 at 15:35
• So, dm doesn't mean the mass is changing, it just refers to a small mass which has some mass m but the notation used is dm for integration purpose right? Nov 18 at 11:54
• "small mass which some mass m" ? Maybe you mean small piece which has mass "m", but yeah more or less @jsivesh Nov 18 at 15:39

You are right: it is a confusing notation. Usually it is used to "construct integrals", where it represents a finite-sized small chunk of the body, which is later assumed to go to zero size when a limit is taken. I prefer to use something like $$m_n$$ for these "chunk masses", and reserve the $$\Delta$$-notation for coordinate variables in which your "final minus initial" idea makes sense.

For example, consider a bar of length $$L$$ which has a changing linear mass density such that if you place one end of the bar at the origin of the $$x$$ axis, the density is given by $$\lambda(x) = \lambda_0 + a x \; (0 \le x \le L)$$. What is the mass of the bar?

We start by dividing the bar into a finite number of equal-length chunks, $$N$$ (which you should think of as something like 8, so you can draw a picture and identify one of the chunks, $$m_n$$). Then we give an approximate expression for the total mass, $$M$$, which will become exact once we find a Riemann sum, and take the limit as $$N \rightarrow \infty$$: $$M \approx \sum_{n=1}^N m_n$$ Notice that I have labeled the 8 chunks $$m_1, m_2, \ldots, m_8$$. These are the $$\Delta m$$ chunks that confused you. Now express the mass of an individual chunk using the density (at the position of the $$n$$th chunk, $$x_n$$) and the length of the $$n$$th chunk: $$M \approx \sum_{n=1}^N \lambda(x_n) L_n$$ The length of each chunk is $$L_n = L / N$$, but on our chosen coordinate axis, it is more useful to express it as $$L_n = \Delta x_n = x_{n+1} - x_n = \Delta x$$. This lets our approximate expression take the form of a Riemann sum, of which we can then take a limit: \begin{align} M &\approx \sum_{n=1}^N \lambda(x_n) \Delta x\\ M &= \lim_{N\rightarrow \infty} \sum_{n=1}^N \lambda(x_n) \Delta x\\ &= \int_0^L \lambda(x) dx\\ &= \int_0^L \left( \lambda_0 + a x\right) dx\\ & = \lambda_0 L + \frac{a L^2}{2} \end{align}

This idea can be generalized to higher dimensions where either $$m_n = \sigma(x,y) \Delta A_n = \sigma(x,y) \Delta x \Delta y$$ or $$m_n = \rho(x,y,z) \Delta V_n = \rho(x,y, z) \Delta x \Delta y \Delta z$$ and to other, e.g., spherical, coordinate systems where you will need to construct $$\Delta V$$ in terms of those coordinates.