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Although it might be a simple question, I'm struggling with something here:

Suppose you have 3 mass-points along the X axis, each one having masses $M_1, M_2, M_3$, at positions $x_1, x_2, x_3$, respectively ($x_1 < x_2 < x_3$).

Now I wish to find the position of the center of the mass, and to do that I must find the place ($x$ value) in which the Torque is null. Now comes the problem: I have four options here. The first is to consider that the center of mass is to the left of $x_1$. The second is to consider it is between $x_1$ and $x_2$. The third, between $x_2$ and $x_3$. The fourth, to the right of $x_3$. How am I supposed to know which one to pick? Also, even if I work out all possible options, at the end I will find that only one of them is not absurd, and that is surely the center of mass of the system, but how this is equivalent to the well-known mathematical definition for the center of mass (ponderating the masses and $x$ coordinates) ?

leleo222
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  • Take the weighted average of the positions, i.e. $x_{cm}=\frac{x_1M_1+x_2M_2+x_3M_3}{M_1+M_2+M_3}$. – John Douma Apr 26 '18 at 04:40
  • This is the well-known mathematical definition I was referring to. I want to know how this is obtained by considering the physical (and intuitive) approach by the Torque – leleo222 Apr 26 '18 at 04:48
  • Have you tried to search for "derivation of center of mass"? – John Douma Apr 26 '18 at 05:00
  • I did, John, and I found a similar question here on math.exchange, apart from the Torque formulation. Still no answer for that question. Thanks – leleo222 Apr 26 '18 at 05:45

1 Answers1

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If I understand you correctly you would like to see how the coordinate of the center of mass arises from considering the torques.

For any point $x$ each mass $M_1,M_2,M_3$ creates a torque with respect to that point $x$ which sum up to a resulting torque $T_r$: $$(x-x_1) M_1 g + (x-x_2) M_2 g + (x-x_3) M_3 g = T_r$$

The center of mass is at $x_M$ such that $T_r = 0$: $$(x_M-x_1) M_1 g + (x_M-x_2) M_2 g + (x_M-x_3) M_3 g = 0 \Rightarrow x_M =\frac{x_1M_1+x_2M_2+x_3M_3}{M_1+M_2+M_3}$$ I hope that is what you wanted to see.

trancelocation
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  • Yes. The $(x_M - x_i)$ already grasps the sign of the torques. That's what I was missing. Thank you ! – leleo222 Apr 26 '18 at 11:53