Question

Why is center of mass defined in the way that it is?\[x_{\text{cm}}=\frac{\sum_i m_i x_i}{\sum_i m_i}\]How would we have known, for instance, that the relationship is not through \(x_i^2\)? I know obviously that this relation is supported through experimentation... I am rather looking for an argument that derives from simpler notions about mechanics.

Answer 1

One such explanation can be obtained through momentum.

Centre of mass is the point that goes straight if no external forces act on a system of mass m.
So we have to try to find a point so that momentum of the system is total mass times velocity of that point.

\(\vec p=\sum_i m_i \:\vec v_i =\sum_i m_i \:\Large \frac {d\vec r_i}{dt}\normalsize = \Large \frac{d}{dt} \normalsize \sum_i m_i \:\vec r_i = (\sum_i m_i )\; \Large \frac{d}{dt} ( \frac{\sum_i m_i \:\vec r_i }{\sum_i m_i } ) \)

So the point that meets this condition is C such as

\( \vec r_C= \Large \frac{\sum_i m_i \:\vec r_i }{\sum_i m_i }\)

Answer 2

Well, if we define the center of mass that way, then the question subsequently becomes why is momentum conserved? And of course the question then subsequently becomes why is the universe a closed system? Alternatively, what if we were to define the center of mass of a one-dimensional rod as its balance point. How do we know the center of mass is linear with respect to \(x\) (as we know), or through another relationship?

Answer 3

Centre of gravity, if gravitation field is uniform, happens to coincide with centre of mass.

To prove this you have to sum all weights of elements making up the system, then find a point through which net torque is zero.
This defines the point of action of the (unique) force that will fully mimic the actions of gravity spread across the whole system. This point is the centre of gravity.

When you do that, you end up with the same expression for this point as that of the centre of mass.