Center of mass for one-dimensional objects is given by $\displaystyle\frac{\int x \, dm}{M}$ or $\displaystyle\frac{\int x \rho \, dx}{M}$, where $\rho$ is density. Now, the center of mass of a rod with uniform density, $\rho$, and a length of $L$ is simply $\displaystyle\frac{\int_0^L x \rho \, dx}{M}$ $\implies$$\displaystyle \frac{\rho \frac{x^2}{2}\big|_0^L}{M}$ $\implies$ $\frac{L}{2}$, or half its length. How do I perform these operations for, say, three rods of various lengths connected at the origin, and placed non-axially? I guess my principle failure of understanding is how to "encode" the geometry of "combinations" of 1-dimensional objects into these integrals.
[Math] Center of Mass of “Combinations” of 1-Dimensional Objects.
calculusphysics
Best Answer
In terms of integrals, what you want is $$X=\frac{\int_0^{L_1} x_1\rho_1\,\mathrm dx_1 + \int_0^{L_2} x_2\rho_2\,\mathrm dx_2 + \cdots}{M_1+M_2+\cdots}.$$ As @anorton's comment suggests, this simplifies to a linear combination of the centers of mass because $X_i=\dfrac1{M_i}\int_0^{L_i} x_i\rho_i\,\mathrm dx_i$, so $$X=\frac{M_1X_1 + M_2X_2 + \cdots}{M_1+M_2+\cdots}.$$