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Physics Tutorial: Center of Mass vs Center of Gravity
Note: this article was originally written for someone having trouble grasping the topic of center of gravity. In particular why a person's center of gravity would shift location if they extended the arms. This isn't intended to be an introduction to the subject of Center of Mass or Center of Gravity. It is hoped that it would help someone get a good grasp of the topic after it has been introduced.
The term center of gravity, cog, is very close to interchangeable with center of mass, com, here on Earth. Both terms refer to a location in a system of particles, or a uniform distribution of mass. In this discussion, I discuss a system of particles. The concept may be extended to include a uniform distribution of mass such as a person's body. If the system of particles is small enough that the gravity field is equal over the entire system, cog and com are both at the same location. (A person's body is easily small enough to consider cog and com to be in the same location. A mountain is probably not small enough.) The location of the com/cog is considered a "weighted average" of the locations of the particles. The expression "weighted average" means that some of the particles have more influence on the average than other particles. Both the location and the mass of the particle have impact on the "weighted average". The difference between com and cog is that in calculating the weighted average of the cog, the mass is multiplied by g for each of the terms. It simplifies the math and discussion if we talk about the com.
The simplest form of the formula is when the particles are found along a straight line.
Consider two point masses on the x axis of an xy coordinate system. The masses are located so that
m_{1} is at x = 1 m and m_{2} is at x = 5 m.
Various variable names are used for the com location. I'll use X_{ave}. The formula is
X_{ave} = (m_{1}*x_{1} + m_{2}*x_{2}) / (m_{1} + m_{2})
If our m_{1} and m_{2} are the same mass, then X_{ave} = 3 m. That's not at all surprising  from symmetry, you would guess that.
But what happens if m_{2} is twice as much as m_{1}? I hope to make the term "weighted average" clear with this change. If m_{2} = 2*m_{1}, then
X_{ave} = (m_{1}*1 m + 2*m_{1}*5 m) / (m_{1} + 2*m_{1}) = 11 m/3 = 3.67 m
Because m_{2} has more mass, it had more influence on the location of the com than the other mass. An exercise for you: return m_{2} to the value of m_{1} and change the value of x_{2} to 6.33 m. Compare to the result when m_{2} was double m_{1}. So moving one of the particles also moved X_{ave}.
If there were more particles along the x axis, you might write the formula
X_{ave} = (m_{1}*x_{1} + m_{2}*x_{2} + . . . + m_{n}*x_{n}) / (m_{1} + m_{2}+ . . . + m_{n})
and evaluate as n starts at 1 and increases through all the particles in the system. The math becomes more complicated when we consider a 3 dimensional object. You need the xyz coordinates of the location of each particle. And then you need to individually find X_{ave}, Y_{ave}, and Z_{ave}. But I will abandon formulas in this discussion and try to help you understand through examples and an analogy rather than through the formulas.
Consider 8 golden balls, 1 kg each. Assume a very lightweight framework holds each ball at the corners of a cube. Assume the mass of the framework can be ignored compared to the gold. Again from symmetry, you know the com will be at the center of the cube, the point where lines going from corner to opposite corner intersect. The reference system could be some point off to the side, similar to the above 2 masses on the x axis where the reference point was x=0. In this case, I'm going to believe in the suspicion that the com will be in the center and choose the center for the reference point. I'll use d_{1}, d_{2}, . . . d_{8} to mean the displacement of each ball from the reference point.
To determine the location mathematically, would require defining axes, like xyz, and finding the x, y, and z coordinates of the com. We are not going there. This discussion is about helping you understand the reaction of the location of the com when there are changes to location or mass of one or more particles.
Give the names b_{1}, b_{2}, . . . b_{8} to the balls. Let the top 4 be b_{1} through b_{4}, arranged clockwise. Arrange the balls on the bottom so that the number of each bottom ball is 4 higher than the ball above it  so a line from b_{1} to b_{7} passes through the center of the cube. The balls 5 through 8 form the base that this system sits on. Imagine that each ball has a string on it and the other end of the string tied to b_{1} (call it S_{1}) is tied to the free ends of all the other strings in a big knot. Assume the tension in string S_{1} is according to the m_{1}*d_{1} term. And the tension in the other strings is according to their contribution to the "weighted average". Since the masses are equal and the displacements are equal, so far the tension in each is equal, so the tensions reach equilibrium when the knot in the strings is at the reference point.
Now double the mass of m_{1}. The string S_{1} pulls harder than the others are pulling, so the knot is pulled toward m_{1} in setting up a new equilibrium. That's the affect of the "weighted average" principle. The new location of the knot is the new location of the com. Next return m_{1} to 1 kg and double d_{1}. This distorts the cube so m_{1} is extended along the line from m_{1} to m7 through our reference point (which was com, when it was a cube). Since the tension in S_{1} is related to the term m_{1}*d_{1}, the same thing happens to the cog. It is pulled toward m_{1}. If a vertical line down from the com falls within the square formed by b5 through b_{8}, the system is stable on its base. But I think you can see that if m_{1} were extended out farther along the line between m_{1} and m7, the system would fall over with enough extension. That's the reason a crane on a building site that tries to reach out too far and lift too heavy a load will fall over.
Now I have to admit that my analogy is flawed. That often happens to analogies when taken to extreme. The tension in string S_{1} is not truly equivalent to the term m_{1}*d_{1}. If I increase m_{1} by 100 or increase d_{1} by 100, the term m_{1}*d_{1} would have the same value. So the analogy says that the tension in S_{1} should be equal and the com would be moved by the same amount. But in the formula, the sum of the m_{n}*d_{n} terms is divided by the sum of the masses. That's why the analogy breaks down. The analogy doesn't account for the division by the sum of masses.
The analogy may still be applied in a qualitative way. Increasing d_{1} significantly could pull the knot outside the shape of the original cube. But increasing m_{1} significantly when located at its original corner can only pull the knot, which is the com, closer to the m_{1} corner of the cube. So increasing d_{1} has the potential to move the com so that a vertical line down from it will fall outside the square of its base. So it will fall over. When the system is a cube, the cube will not fall over regardless of increases in the value of m_{1}. (However as m_{1} is increased it could become marginally stable so that a small disturbance could upset it.)
In spite of the flaw in the analogy, I hope this helps you "see" the logic in the com and the cog moving when particles in a system are moved.
I have included my responses to specific questions from the person that the above was originally written for. You asked these specific questions about the shift in the cog when the boy in the chair extended his arms horizontally from the shoulder:
1) Is it because the centre of gravity depends on the distribution of its mass?
Yes, that's the "weighted average" at work.
2) Is it because you took 20 kg in each of your hand?
Yes, both moving the mass and increasing the mass increase the change in the "weighted average".
3) The tip of the centre of gravity's arrow always contacts with the ground, right?
No. The center of gravity is a location. When you make a sketch, the dimensions are distance. You can show the weight of the object as a force vector pointing down from the cog. But it doesn't have to touch the ground. The weight vector is in units of Newtons. The dimensions of the sketch are distance You might need to show the weight of more than one object. Two objects that are equal distance above the ground but with different mass would have weight vector arrows of different length, so they couldn't both touch the ground with their tip.
copyright: Steve Johnson, sojphysics@gmail.com
Steve Johnson is an electrical engineer. He earned a B. Sc. in Physics, and MS Electrical Engineering. Steve welcomes your questions.
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