Pequaide, obviously you (and others) do not understand the concept of radius of gyration. When you have weights of different mass located at different radii, it becomes complex and difficult to figure out what will happen. Your writing about this shows that it is a complex problem. The more outward weight will end up moving faster due to its greater radius. You cannot simply move this 2000 gram weight to the more inward radius and expect it to act the same.
Many years ago, people who worked hands on with machinery learned how to accurately calculate the expected results. They knew that if all of the weight in a rotating system were moved to a specific radius, then the system would function exactly like when the weights were spread out. They devised a formula for calculating this fictitious radius. So if you move the 500 grams outward to this radius and you move the 2000 grams inward to the same radius, then the whole assembly will act exactly the same as before. And with all of the weight located as a single radius, the amount of force and the time involve can be calculated using simple formulas.
The formula for calculating this radius of gyration is rather simple....
Two weights at the same radius can be considered as one weight.
Thus the two 1 Kg weights can be figured as a single 2 Kg weight.
So now you have a 2 Kg Weight at 1 Meter and a 1/2 Kg weight at 0.10 meter.
Assign R and W to be the Radius and Weight of one weight.
Assign r and w to be the radius and weight of the second weight.
The formula for Radius of Gyration is:
RoG = (R²×W+r²×w)÷(R×W+r×w)
Filling in the values:
RoG = (100²×2000+10²×500)÷(100×2000+10×500)
Radius of Gyration result:
RoG = 97.804878 cm
Simply stated, the assembly of the two weights at different radii (the machine) will function exactly as if all 2.5 Kg is located at 97.804878 cm
This assumed both weights are point mass weights. In reality a large spread-out weight will act different from a small compact weight even when both weigh the same. Radius of Gyration formulas can be used to compensate for such discrepancies. These are formula that real hands on engineers know and use out in the work place. The ivory tower engineers seldom bother with such mundane details of accuracy.
Pequaide, in your experiments, I've often seen you using wheels and then attempt to use their mass as if it were all located at the outer rim. This screws up the results. There are available specific formulas for specific shapes of rotating objects. A solid wheel acts as if all of its weight mass is located at 0.7071 times its radius. Thus if you have a solid 2 meter diameter cast iron wheel, you can assume that all of its weight mass is located at a radius of 0.7071 meters. If you don't understand this effect and assume all of the weight is at the 1 meter rim, or assume all the weight is half way out at 0.5 meter, then all the rest of your calculation will be wrong. This was the point I was attempting to make in that post from a year and a half ago.
