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I was trying to motivate the definition for the center of mass formula and I am getting a weird result. For simplicity let's assume two-point masses of mass $m$ and $2m$ connected by a rigid massless rod. Initially, they both are at rest, then if we provide an impulse perpendicular to the rod to mass $m$ let's say, the system will be in motion. The motion is restricted by a few rules: first is that according to rigid body constraint, the distance between both the mass should be constant for all times $t$. Now let's define a point called the center of mass such that the acceleration is zero of the point and we are doing it because it's a nice property. we can define the point like this $a_{cm}=\frac{ma_1+2ma_2}{m_1+m_2}$. Now this definition allows $a_{cm}$ to be zero due to newtons third laws because the force applied on the first mass is negative of the force applied on the second mass and their accelerations are inversely proportional to their masses so the numerator is zero and $a_{cm}$ is zero and if you integrate twice you get that point, our center of mass. So the motion of this system is just a rotation around the center of mass. But there is a huge problem with this. It's that the denominator can be anything it can be $m_1-m_2$ or any other expression. So what does this mean this means that according to our derivation, there is an infinite family of points such that when there is no external force to it it doesn't move. can someone help? Is there a mistake in one of my steps?

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  • $\begingroup$ But there is a huge problem with this. It's that the denominator can be anything it can be $m_1−m_2$ or any other expression. Can you elaborate on this part? Why could the denominator be anything? This would suggest that anything times $a_{cm}$ yields the numerator, but I don't think that is accurate. $\endgroup$
    – Kyle Kanos
    Commented Oct 22, 2023 at 15:21

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You just produced a very special part of constraint used to define center of mass, the general constraint is that Centre of mass is a point, for which, If all mass is assumed to be concentrated , will follow same equation of forces as if a point particle of same mass will follow, in This very specific case, you used acceleration to be zero inferring from net zero external force, which is just one part of constraint, more constraints like "even if the force is non zero, that net external force equals mass times acceleration of center of mass " are need for motivating definition of COM

Clarification to OP's Further Doubts:

Suppose There is a system of "$n$" particles, each having different accelerations as a result of different forces applied to each one, either internal, or external, Let mass of $r^{th}$ particle be $m_r$ and acceleration vector of $r^{th}$ particle be $\vec{a}_r$

what we can now algebraically analyse is as follows

$\sum \vec{F}_{external}+\sum \vec{F}_{internal}=m_1\vec{a}_1+m_2\vec{a}_2+m_3\vec{a}_3+m_4\vec{a}_4+m_5\vec{a}_5+.....m_n\vec{a}_n$

using newton's third law,

$\sum \vec{F}_{internal}=0$

now

$\sum \vec{F}_{external}=m_1\vec{a}_1+m_2\vec{a}_2+m_3\vec{a}_3+m_4\vec{a}_4+m_5\vec{a}_5+.....m_n\vec{a}_n$.......(1)

now for making it easy, we want this equation as a simple equation like

$$\sum \vec{F}_{external}=m\vec{a}.....(2)$$

to do that, we take $\sum_{1}^{n} m_r$ common on R.H.S of equation $(1)$

it becomes

$\sum \vec{F}_{external}=(\sum_{1}^{n} m_r) \cdot (\frac{m_1\vec{a}_1+m_2\vec{a}_2+m_3\vec{a}_3+m_4\vec{a}_4+m_5\vec{a}_5+.....m_n\vec{a}_n}{\sum_{1}^{n} m_r})$....$(3)$

Now compare $(2)$ and $(3)$ equation,The similarities are obvious, $(3)$ equation form has now become similar to $(2)$ equation form

where

$\sum_{1}^{n} m_r$ is now mass of complete system and and the new vector

$(\frac{m_1\vec{a}_1+m_2\vec{a}_2+m_3\vec{a}_3+m_4\vec{a}_4+m_5\vec{a}_5+.....m_n\vec{a}_n}{\sum_{1}^{n} m_r})$ acts as "some acceleration vector $\vec{a}_{new}$" similar to equation $(2)$

Now this new equation is always valid for all newton's $2^{nd}$ law applications on system, now Integrate The equation $(3)$ twice and you will get that The vector quantiity then becomes weighted average of position vectors of all different particles in system, this weighted average will still be A new Position vector, This position is what we define now "Centre of mass" after this,

"We" Define this $\vec{a}_{new}$ as Acceleration of "center of mass", why is that so, because it is double derivative of position vector that we defined as position vector of "center of mass"

this help us solve Problems involving Application of newton's second law, as if all the mass of that system is located as point particle at position vector of center of mass

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  • $\begingroup$ can you show that the center of mass equation satisfies the other constraints too? Such as if force is non zero that the net external force equals mass times acceleration. we take it for granted this is true but can you prove it? $\endgroup$
    – Hammock
    Commented Oct 23, 2023 at 6:04
  • $\begingroup$ Edited to clarify.. $\endgroup$ Commented Oct 23, 2023 at 6:46
  • $\begingroup$ Thanks. It's very helpful, with this it's also very easy to prove that an object rotates about the center of mass if you apply an impulse to it. $\endgroup$
    – Hammock
    Commented Oct 23, 2023 at 10:17
  • $\begingroup$ @Hammock,you would still require Torque analysis to do that, And For Non zero external impulse, There are two motions, the centre of mass itself will move with some velocity, so in physical situations, object will actually rotate about centre of mass as well as translate forward $\endgroup$ Commented Oct 23, 2023 at 11:32

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