Where Is the Center of Mass of a Rod That Gets Heavier? 😦

Ajouté : 2026-05-16

[00:00:00]Let's find the center of mass of a rod of length L, but this time it is non-uniform. The mass density lambda increases as we move along the rod. It equals a constant lambda naught times X over L. We slice the rod into tiny pieces. Let's look at one segment of width DX at a distance X. The mass of this tiny slice DM is its density lambda times DX. The center of mass is the integral of XDM divided by the total mass, which is the integral of DM.

[00:00:34]Substitute our expression for DM into both integrals. The constants lambda naught over L appear in both the numerator and denominator, so they cancel out perfectly. Evaluating the integrals, X squared becomes X cubed over three, and X becomes X squared over two. Applying the limits from zero to L, this simplifies beautifully to two L over three. Notice that the center of mass has shifted past the middle towards the heavier end of the rod. And as always, thanks for watching.