Center of mass of a two rod system

AI Thread Summary
The discussion centers on calculating the center of mass for a two-rod system, where one rod is thinner and lighter while the other is thicker and heavier. The center of mass for each rod is determined to be at their midpoints, with the lighter rod's center located at 8.5 inches and the heavier rod's center also at 8.5 inches, but contributing more significantly to the overall mass. The user calculates the overall center of mass to be 22.67 inches from the thinner rod, or 5.67 inches into the thicker rod, based on the weight distribution. The forum participants are encouraged to verify the calculations for accuracy. The discussion emphasizes the importance of weight distribution in determining the center of mass in composite systems.
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Homework Statement


Given that I have a system with a rod of diameter 1 1/2 inches with a length of 17 in and a weight of 10 1/3 oz and a rod with a diameter of 3 inches with a length of 17 in and a weight of 20 2/3 oz. The two rods are butted together at one end almost looking like a bottle. What is the center of mass of the system?


Homework Equations


Center of mass = 1/M Ʃ miri


The Attempt at a Solution


I calculated the center of mass of each rod for the thinner,lighter rod the center of mass is in the middle of that rod. For the thicker,heavier rod the center of mass is in the middle. I calculated the center of mass to be 2/3 of the way of the entire 34 inches from the thinner rod since the heavier rod carries 2/3 of the weight of the system. Would I be correct in saying that the center of mass of the system is 5 2/3 of the way into the bigger rod?
 
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