Center of mass of a two rod system

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SUMMARY

The center of mass of a two-rod system, consisting of a thinner rod with a diameter of 1.5 inches, a length of 17 inches, and a weight of 10.33 oz, and a thicker rod with a diameter of 3 inches, a length of 17 inches, and a weight of 20.67 oz, is calculated using the formula for center of mass: Center of mass = 1/M Ʃ miri. The center of mass is determined to be located 5.67 inches from the end of the thinner rod, which corresponds to 2/3 of the total weight distribution favoring the thicker rod.

PREREQUISITES
  • Understanding of center of mass calculations
  • Familiarity with basic physics concepts related to mass and weight
  • Knowledge of the formula for center of mass: Center of mass = 1/M Ʃ miri
  • Ability to perform arithmetic operations with fractions and decimals
NEXT STEPS
  • Review the principles of static equilibrium in physics
  • Explore the effects of varying rod dimensions on center of mass
  • Learn about composite systems and their center of mass calculations
  • Investigate real-world applications of center of mass in engineering and design
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Students studying physics, particularly those focusing on mechanics, as well as educators and professionals involved in engineering and design who require an understanding of 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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