# Appendix A: Centre of Mass

Consider an object consisting of two masses m1 and m2 connected by light rod. Where would be centre of mass (CM) of this object be?

This is same as asking where the balance point of this object is. So the equation below allows us to calculate the position of the CM along the rod.

If we use an arbitrary reference point, then the CM can be calculated using through

$\displaystyle ({{m}_{1}}+{{m}_{2}}){{x}_{{CM}}}={{m}_{1}}{{x}_{1}}+{{m}_{2}}{{x}_{2}}\text{ }\Rightarrow \text{ }{{x}_{{CM}}}=\frac{{{{m}_{1}}{{x}_{1}}+{{m}_{2}}{{x}_{2}}}}{M}$

In general, for a system consisting of a collection of masses, the CM’s position can be calculated from

$\displaystyle {{x}_{{CM}}}=\frac{{\sum\limits_{{i=1}}^{N}{{{{m}_{i}}{{x}_{i}}}}}}{M},\text{ }{{y}_{{CM}}}=\frac{{\sum\limits_{{i=1}}^{N}{{{{m}_{i}}{{y}_{i}}}}}}{M},\text{ }{{z}_{{CM}}}=\frac{{\sum\limits_{{i=1}}^{N}{{{{m}_{i}}{{z}_{i}}}}}}{M}$

As long as the gravitational field is uniform throughout the object, where the mass appears to be concentrated is also where the weight appears to act. That’s why for all practical purposes, the CM an object is also its CG.

Demonstration

Projectile Motion of a Two-Ball Object

The Balancing Bird