Why center of mass constant is zero?

Question

The Center of mass frame (CM) is defined in such a way that total momentum in this frame is zero. Thus

$\vec P'=\sum (m_i\vec v_i')=\sum \big[m_i(\vec v_i-\vec u)\big] =\vec P - M\vec u=0 \tag{1} $

Where $\vec P'$ is the momentum in CM, $\vec u$ is CM's velocity WRT some inertial frame where the momentum of the system is $\vec P$. From equation (1) we get

$\vec u = \frac {\vec P}{M}\tag{2}$

Integrating this gives

$\vec R _{CM}= \frac{\sum (m_i\vec r_i)}{M}+C\tag{3}$

Now, this is the point that I am not able to grasp: Why is $C$ chosen to be $0$?

This is not the only point where momentum is zero, which was the requirement for the construction of this frame.

I believe that it has something to do with torques, but can't quite get there.

Any help would be appreciated.

Thanks

Answer 1

You write $\vec R_{CM}$, but think about what is actually meant by this?

A frame of reference has no specific position. Its positional vector can be arbitrary because different frames of reference vary in velocity or acceleration. Therefore, if you take $\vec R_{CM}$ as a position of the CM, you have the freedom of $\vec C$.

$\vec C$ can be chosen $= \vec 0$, so that $\vec R_{CM}$ is equal to the centre of mass of the set of point masses, which is nothing more than a convenient choice.

Answer 2

We are integrating over the entire system, and hence, it is a definite integral.