Moment of Inertia of four particles

AI Thread Summary
The discussion focuses on calculating the moment of inertia for a system of four particles positioned at the corners of a square with 2.0-meter edges. The particles have masses of 3.0 kg and 4.0 kg, and the moment of inertia is calculated about the z-axis, which runs through one of the particles. Using the parallel-axis theorem, the initial moment of inertia is found to be 56 kgm², and the moment of inertia about the center of mass is computed as 28 kgm². However, a direct computation method yields a different result of 9.33 kgm² for the moment of inertia about the center of mass. The calculations illustrate the complexities involved in determining the moment of inertia for multi-particle systems.
Rubidium
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Homework Statement


Four particles, one at each of the four corners of a square with 2.0-m-long edges, are connected by massless rods. The masses are m1=m3=3.0 kg and m2=m4=4.0 kg. Find the moment of inertia of the system about the z axis. (the z axis runs through m2, which is at the origin, m1 is on the y axis, and m3 is on the x axis.
Use the parallel-axis theorem and the result for Problem 41 to find them moment of inertia of the four-particle system about an axis that passes through the center of mass and is parallel with the z axis. Check your result by direct computation.

Homework Equations


I don't know what they mean by direct computation but I get two different answers.

The Attempt at a Solution


I=m1r1^2+m2r2^2+m3r3^2+m4r4^2=(3kg)(2m)^2+(4kg)(0)+(3kg)(2m)^2+(4kg)(2sqrt(2))^2=56kgm^2
Icm=I-Mh^2=56-(14kg)(4)=28kgm^2
Then integrating for Icm I get 9.33 as a result.
 
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Since the four mass are at equal distances from the axis passing through cm, you can find Icm by using the formula sigma(m)*r^2
 

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