Newton's Laws for a System of Particles - Center of Mass

Newton's 2nd Law for a system of particles is written as:

$$\vec{F}_{\rm tot} = \sum_i \vec{F}_i = \sum_i m_i \frac{d^2 \vec{x_i}}{dt^2}$$ (4.1)

If we insist that the total force act on the total mass located at a special point, that point will behave like a ``particle'':

$$\vec{F}_{\rm tot} = \sum_i \vec{F}_i$$ (4.2)

$$= \sum_i m_i \frac{d^2 \vec{x_i}}{dt^2}$$ (4.3)

$$= \left(\sum_i m_i\right) \frac{d^2 \vec{X}_{\rm cm}}{dt^2}$$ (4.4)

$$= M_{\rm tot} \frac{d^2 \vec{X}_{\rm cm}}{dt^2}$$ (4.5)

With this definition, we see that:

$$\vec{F}{\rm tot} = M_{\rm tot} \frac{d^2 \vec{X}_{\rm cm}}{dt^2} = M_{\rm tot} \vec{A}_{\rm cm}$$ (4.6)

and Newton's second law is recovered.

This now defines the position of the center of mass of an object or a collection of particles:

$$M \vec{X}_{\rm cm} = \sum_i m_i \vec{x}_i$$ (4.7)

(with $M = \sum_i m_i$) and, for continuous systems:

$$M \vec{X}_{\rm cm} = \int \vec{x} dm$$ (4.8)

(with $M = \int dm$). The latter form comes from treating every little differential chunk of a solid object like a ``particle'', and adding them all up. Integration, recall, is just a way of adding them up.

The point of this is that now a collection of particles can be treated like a single ``particle'' of mass $M = \sum_i m_i$ located at $\vec{X}_{\rm cm}$ under the action of the total force acting on all its consituents. This is a very valuable resolution, as we shall see.

Note (from the figure above) that there may be no mass at all at the position of the center of mass - it is just the geometric point that moves as if the mass were all located at that one point. For solid, homogeneous objects (like baseballs) though, it will generally be in the ``middle''. In fact, this can be interpreted as one way of precisely defining the ``middle'' of an object!

In order to use the idea of center of mass (CM) in a problem, we need to be able to evaluate it. For a system of discrete particles, the sum definition is all that there is - you brute-force your way through the sum (decomposing vectors into suitable coordinates and adding them up).

For a solid object that is symmetric, the CM is ``in the middle''. But where's that? To precisely find out, we have to be able to use the integral definition of the CM:

$$M \vec{X}_{\rm cm} = \int \vec{x} dm$$ (4.9)

(with $M = \int dm$, and $dm = \rho dV$ or $\sigma dA$ or $\lambda dl$).