The Balance Point
For a collection of masses, the Center of Mass is the average position weighted by mass. For a continuous object with density \(\rho(x)\), we use integrals:
\[\bar{x} = \frac{1}{M} \int x \rho(x) dx\]
where \(M = \int \rho(x) dx\) is the total mass.
Worked Examples
Example 1: Uniform Rod
Find the center of mass of a uniform rod of length \(L\).
- \(\rho(x) = \text{constant}\).
- \(M = \int_0^L \rho dx = \rho L\).
- \(\bar{x} = \frac{1}{\rho L} \int_0^L x \rho dx = \frac{\rho}{\rho L} [\frac{x^2}{2}]_0^L = \frac{L^2}{2L} = L/2\).
- Result: \(L/2\). (Exactly in the middle).
The Bridge to Quantum Mechanics
In Quantum Mechanics, we don't have "mass density," but we do have Probability Density \(P(x) = |\psi(x)|^2\). The average position of a particle, \(\langle x \rangle\), is calculated using the exact same formula as the center of mass: \(\int x P(x) dx\). This is why we can treat a spread-out quantum wavepacket as if it were a single particle located at its "center of mass" when we look at its classical behavior. The "First Moment" of probability is the particle's position.