I Find CM of a Rod w/ Length & Mass Attached | 3L/4

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To find the center of mass (CM) of a rod of length L and mass m with an additional mass m attached, the CM of the rod is located at L/2. When another mass of the same size is attached at one end, the combined CM shifts to 3L/4. This is calculated by averaging the positions of the rod's CM and the attached mass. The discussion clarifies that both the rod and the attached mass are of equal mass m. The final position of the CM is determined by their respective distances and masses.
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How do you find the CM of a rod of length L and mass m that has been attached by another mass m?

It is supposed to be 3L/4
 
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Cm of rod is at L/2. Attach (assume at the end) another mass of the same size. Cm of rod plus mass is at average position = 3L/4..
 
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mathman said:
Cm of rod is at L/2. Attach (assume at the end) another mass of the same size. Cm of rod plus mass is at average position = 3L/4..
Could you be more explicit? I was imagining a mass of negligible size being attached to it. Is there a mathematical formula?
 
The original question said that the rod and the attached mass were both mass m.
 
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mathman said:
The original question said that the rod and the attached mass were both mass m.
I got it!
 

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