Finding Center of Mass of Two Rods

AI Thread Summary
To find the center of mass of two rods, one must consider their weights and lengths, as the heavier rod will influence the overall center of mass more significantly. The first rod weighs 20 2/3 oz and the second weighs 10 1/3 oz, both measuring 17 inches long. The center of mass will be located closer to the heavier rod, and it is essential to calculate the exact position rather than simply averaging the centers of mass. For the moment of inertia, it is necessary to add the individual moments of inertia of each rod rather than averaging them. Accurate calculations are crucial for both center of mass and moment of inertia in this system.
schaafde
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I am currently trying to find the center of mass of two rods. One rod is 20 2/3 oz and is 17 inches long with a diameter of 3. The other rod is 10 1/3 oz and 17 inches long with a diameter of 1 1/2 The bigger rod is touching the smaller one if they were both standing up. It looks almost like a bottle. Would I just take the center of mass of each rod and average them together with the bigger rod accounting for more or is that too simple?
 
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schaafde said:
Would I just take the center of mass of each rod and average them together with the bigger rod accounting for more or is that too simple?
That's all there is to it.
 
Would that be the same for finding the moment of inertia for that system? just find the moment of inertia for each individual rod and average them together?
 
schaafde said:
Would that be the same for finding the moment of inertia for that system? just find the moment of inertia for each individual rod and average them together?
No, you wouldn't average them. You'd add them. (Make sure you have the moment of inertia of each about the same axis.)
 
So i calculated the center of mass to be 2/3 of the way up the "object" which would make it in the bigger rod? Would I be correct in assuming that? I am going off the fact that a uniform rod would have its center of mass directly in the middle.
 
schaafde said:
So i calculated the center of mass to be 2/3 of the way up the "object" which would make it in the bigger rod? Would I be correct in assuming that?
Did you assume it or calculate it? Yes, assuming the rods are uniform and laid end to end, the center of mass of the system will be within the heavier rod. Calculate the exact position.
I am going off the fact that a uniform rod would have its center of mass directly in the middle.
Good.
 
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