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We have a system of N points $(x_i, y_i)$ with masses $m_i$ and fixed distances. I want to show that there is a center of mass and derive a formula to compute its coordinates. I have argued that we can imagine those points as crystallized in a mass-less coordinate plain and that there are a vertical and a horizontal line $x=x_s$ and $y=y_s$ where we can balance the whole plain on an infinitely long knife blade. Solving the equations for the angular momenta

$$ 0 = \sum_{i=1}^N \left(x_S-x_i\right) \cdot m_i \quad \text{and} \quad 0 = \sum_{i=1}^N \left(y_S-y_i\right) \cdot m_i$$

yields the formula

$$(x_s,y_s) = \dfrac{1}{m_{tot}} \cdot \left(\sum_{i=1}^N x_i \cdot m_i \ , \ \sum_{i=1}^N y_i \cdot m_i \right).$$

I would then like to argue that at $S = (x_S, y_S)$ we could actually balance the plane on a needle, and to show that, it would suffice to say that the plane cannot tilt in any direction, i.e. that we can balance it on ANY straight line through $S$. Any clues on how to do that, using what we have so far?

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Firstly, angular momentum is defined for moving particles, not to particles at rest. So you should state the moment of weights about $x_s$ or $y_s$ lines and not angular momenta.

Translate the coordinate plane to make $S$ as the origin. The corresponding coordinates of particles will change accordingly.

New $x_i' = x_i - x_s$ and $y_i' = y_i - y_s$.

Thus $$ 0 = \sum_{i=1}^N \left(x_S-x_i\right) \cdot m_i \quad \text{and} \quad 0 = \sum_{i=1}^N \left(y_S-y_i\right) \cdot m_i$$ can be rewritten as $$ 0 = \sum_{i=1}^N x_i' \cdot m_i \quad \text{and} \quad 0 = \sum_{i=1}^N y_i' \cdot m_i$$

Now rotate this coordinate plane by an arbitrary angle, say $\theta$ in counter clockwise direction. The equation of coordinate axes and the particles will change once again.

$x_i''= x_i'\cos\theta + y_i'\sin\theta$ and $y_i''= -x_i'\sin\theta + y_i'\cos\theta$.

But we see that the net moment about each of those axes is still zero.

For example, $$ \sum_{i=1}^N x_i'' \cdot m_i = \sum_{i=1}^N (x_i'\cos\theta + y_i'\sin\theta) \cdot m_i = (\sum_{i=1}^N x_i' \cdot m_i)\cos\theta + (\sum_{i=1}^N y_i' \cdot m_i)\sin\theta = (0)\cos\theta + (0)\sin\theta = 0$$

Similarly, $$ \sum_{i=1}^N y_i'' \cdot m_i = 0$$

Hence proved.

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In polar coordinates by the same logic it follows.

$$ \sum_{i=1}^N \left(r_S-r_i\right) \cdot m_i =0$$

yields formula for any arbitray orientation $\alpha$ passing through center of mass

$$ r_S= \sqrt{x_s^2+y_s^2}$$

EDIT1:

$$ r_s= \left(\sum_{i=1}^N r_i \cdot m_i \right )/{m_{tot}} $$

We can balance a flat weightless lamina on a point of concurrency of all straight lines through $S$. It is a physical invariant for all or any coordinate system used. Position of center of mass/gravity same for equilibrium whether the laminate is vertical or horizontal.

Horizontal or Vertical

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  • $\begingroup$ Interesting. Can you add explanation on how you concluded alpha can be arbitrary? Are you applying the equation of equilibrium in polar coordinates? $\endgroup$ – yathish Dec 7 '17 at 16:46

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