Moment of inertia of 3-mass system

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SUMMARY

The moment of inertia for a rigid 3-mass system was calculated with respect to a rotation axis located 2.3 m from the leftmost 3 kg mass. The formula used was I = Σ(m_i * r_i²), where the individual masses are 3 kg, 2 kg, and 6 kg, and their respective distances from the axis are 2.3 m, 0.7 m, and 3.7 m. The final computed moment of inertia is 98.99 kg·m², confirming the accuracy of the calculations presented in the discussion.

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  • Understanding of moment of inertia calculations
  • Familiarity with rigid body dynamics
  • Knowledge of mass distribution and distance measurements
  • Proficiency in applying the formula I = Σ(m_i * r_i²)
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  • Study the effects of changing the axis of rotation on moment of inertia
  • Explore parallel axis theorem applications in rigid body dynamics
  • Learn about the implications of moment of inertia in rotational motion
  • Investigate the relationship between mass distribution and stability in physical systems
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DrMcDreamy
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Homework Statement


This is a two part problem, I answered the first but was wondering if I did the second part right.

a) Consider a rigid 3-mass system (with origin at the leftmost mass 3 kg) which can rotate about an axis perpendicular to the system. The masses are separated by rods of length 3 m, so that the entire length is 6 m. 3.8

b) Now consider a rotation axis perpendicular to the system and passing through the point x0 at a distance 2.3 m from the leftmost mass 3 kg.
Find the moment of inertia of the 3-mass system about the new axis. Answer in units of kg · m2.

Homework Equations



I= \summiri2

The Attempt at a Solution



I= (3 kg)(2.3)2 + (2 kg)(.7)2 + (6 kg)(3.7)2 = 98.99

Is this correct?
 
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Forget it, I realized my work is right :smile:
 

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