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Please read this passage, from section I-18-1 of The Feynman Lectures on Physics:

We may consider any object as being made of lots of little particles, the atoms, with various forces among them. Let $i$ represent an index which defines one of the particles. (There are millions of them, so $i$ goes to $10^{23}$, or something.) Then the force on the $i$th particle is, of course, the mass times the acceleration of that particle: $$ \mathbf{F}_i=mi(d^2\mathbf{r}_i/dt^2). \tag{18.1} $$

In the next few chapters our moving objects will be ones in which all the parts are moving at speeds very much slower than the speed of light, and we shall use the nonrelativistic approximation for all quantities. In these circumstances the mass is constant, so that $$ \mathbf{F}_i=d^2(m_i\mathbf{r}_i)/dt^2. \tag{18.2} $$ If we now add the force on all the particles, that is, if we take the sum of all the $\mathbf{F}_i$’s for all the different indexes, we get the total force, $\mathbf{F}$. On the other side of the equation, we get the same thing as though we added before the differentiation: $$ \sum_i \mathbf{F}_i = \mathbf F = \frac{d^2(\sum_i m_i \mathbf{r}_i)}{dt^2} \tag{18.3} $$ Therefore the total force is the second derivative of the masses times their positions, added together.

Now the total force on all the particles is the same as the external force. Why? Although there are all kinds of forces on the particles because of the strings, the wigglings, the pullings and pushings, and the atomic forces, and who knows what, and we have to add all these together, we are rescued by Newton’s Third Law. Between any two particles the action and reaction are equal, so that when we add all the equations together, if any two particles have forces between them it cancels out in the sum; therefore the net result is only those forces which arise from other particles which are not included in whatever object we decide to sum over. So if Eq. $(18.3)$ is the sum over a certain number of the particles, which together are called “the object,” then the external force on the total object is equal to the sum of all the forces on all its constituent particles.

Now it would be nice if we could write Eq. $(18.3)$ as the total mass times some acceleration. We can. Let us say $M$ is the sum of all the masses, i.e., the total mass. Then if we define a certain vector $\mathbf R$ to be $$ \mathbf R= \sum_i m_i \mathbf{r}_iM, \tag{18.4} $$ Eq. $(18.3)$ will be simply $$ \mathbf F=d^2(M\mathbf R)/dt^2=M(d^2\mathbf{R}/dt^2), \tag{18.5} $$ since M is a constant. Thus we find that the external force is the total mass times the acceleration of an imaginary point whose location is $\mathbf{R}$. This point is called the center of mass of the body. It is a point somewhere in the “middle” of the object, a kind of average r in which the different $\mathbf{r}_i$’s have weights or importances proportional to the masses.

My question is: We can say vector $R$ to be $R=\sum_{i} \frac{m_ir_i}{[\mathrm{any\:mass}]}$

If we put this $R$ in the final formula of force given in image 2, the $[\mathrm{any\:mass}]$ will cancel and the formula of force will remain the same. So, why do we only divide by $M(=m_i)$

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    $\begingroup$ Please do not post images of texts you want to quote, but type it out instead so it is readable for all users and so that it can be indexed by search engines. For formulae, use MathJax instead. This is particularly the case for texts like the Feynman lectures, which are openly available online and can be seamlessly copy-pasted. I have formatted this post for you, but you should do it yourself in future posts. $\endgroup$ – Emilio Pisanty Aug 20 '20 at 9:50
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My question is: We can say vector $R$ to be $R=\sum_{i} \frac{m_ir_i}{[\mathrm{any\:mass}]}$

If we put this $R$ in the final formula of force given in image 2, the $[\mathrm{any\:mass}]$ will cancel and the formula of force will remain the same. So, why do we only divide by $M(=m_i)$

You could, but then the point in question is no longer the center of mass. In particular, the property mentioned by Feynman,

It is a point somewhere in the “middle” of the object, a kind of average r in which the different $\mathbf{r}_i$’s have weights or importances proportional to the masses

is no longer guaranteed. The center of mass is a convex combination, which gives it several special geometrical properties. If you mess with the definition, then things will start going wrong.

In a sense, your modification is not too bad if this total-force property is where you want to stop. However, the center of mass is much more useful than that: in the following lectures in the book, you can see that the COM of a rigid body has the special property that the motion of the entire body can be reduced to

  • the total force $\mathbf F = \sum_i \mathbf F_i = \sum_i \mathbf F_{i,\mathrm{ext}}$ on the body, seen as acting on the position of the center of mass, coupled with
  • a rigid rotation of the body about its center of mass, caused by the total torque of the external forces on the body when taken about the center of mass.

The second property breaks completely if you take any other definition for the COM.

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  • $\begingroup$ +1, Good answer. However, I would suggest you to also add the following reason (for using the total mass in the denominator) in your answer: The center of mass, when calculated the standard way (with the total mass in the denominator), is a fixed point in space (here, I am specifically talking about a system with a static center of mass). Now, iff you apply any coordinate transformations, the location of that center of mass does not change. However, if you used any other mass in the denominator, the center of mass' location would not be fixed, and would be different for [contd.] $\endgroup$ – user258881 Aug 21 '20 at 17:34
  • $\begingroup$ [Contd.] different set of coordinates. This can easily be seen by shifting the origin in the case of the non-standard COM. The location of the non-standard COM changes, but the standard COM does not. A short mathematical proof explaining this would also be a good addition to the answer (however, that's completely up to you). But do consider stating the above facts explicitly innyour answer. Thank you. $\endgroup$ – user258881 Aug 21 '20 at 17:37

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