Finding Center of Mass of Two Rods

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Discussion Overview

The discussion revolves around finding the center of mass of a system consisting of two rods with different weights and diameters. Participants explore the methodology for calculating the center of mass and moment of inertia for this configuration, considering both theoretical and practical aspects.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant proposes averaging the centers of mass of the two rods, suggesting that the larger rod should be weighted more due to its greater mass.
  • Another participant questions whether averaging is appropriate for calculating the moment of inertia, indicating that the correct approach is to add the moments of inertia of each rod rather than averaging.
  • A participant calculates the center of mass to be 2/3 of the way up the larger rod, based on the assumption that uniform rods have their center of mass at their midpoint.
  • Another participant confirms that if the rods are uniform and laid end to end, the center of mass will indeed be within the heavier rod, but emphasizes the need for an exact calculation.

Areas of Agreement / Disagreement

Participants express differing views on the methodology for calculating the center of mass and moment of inertia, indicating that there is no consensus on the averaging approach for these calculations.

Contextual Notes

Assumptions about the uniformity of the rods and their arrangement are critical to the discussion, and the exact calculations for the center of mass and moment of inertia have not been fully resolved.

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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