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?


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.


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

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


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

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

  • $\begingroup$ 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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