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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)?

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    $\begingroup$ $\Delta$ 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. $\endgroup$ Nov 17 at 10:21
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    $\begingroup$ 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 $\endgroup$
    – Rhino
    Nov 17 at 10:22
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    $\begingroup$ 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. $\endgroup$ Nov 17 at 10:48

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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.

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  • $\begingroup$ 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. $\endgroup$ Nov 17 at 15:35
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    $\begingroup$ 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? $\endgroup$
    – j sivesh
    Nov 18 at 11:54
  • $\begingroup$ "small mass which some mass m" ? Maybe you mean small piece which has mass "m", but yeah more or less @jsivesh $\endgroup$ Nov 18 at 15:39
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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.

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