# Center of Mass of “Combinations” of 1-Dimensional Objects.

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$$\displaystyle \frac{\rho \frac{x^2}{2}\big|_0^L}{M} \implies \frac{L}{2}, or half its length. How do I perform these operations for, say, three rods of various lengths connected at the origin, and placed non-axially? I guess my principle failure of understanding is how to "encode" the geometry of "combinations" of 1-dimensional objects into these integrals. - One technique is to "lump" the mass of each rod at its center of mass, and then compute as point masses. This works for sets of any shape, not just rods. – apnorton Jan 17 '13 at 3:57 ## 1 Answer In terms of integrals, what you want is$$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}.$\$

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I've also been told that this can just be done vectorially, and am a bit confused as how to "jump" from the notation to the calculation. – jyoungs Jan 17 '13 at 20:58
@Jyoung, I can't say I understand your comment. Can you elaborate? – Rahul Jan 19 '13 at 18:23