Moment of Inertia of four particles

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SUMMARY

The moment of inertia of a system comprising four particles positioned at the corners of a square with 2.0-meter edges is calculated to be 28 kg·m² about the z-axis, which runs through the origin at particle m2. The masses are defined as m1 = m3 = 3.0 kg and m2 = m4 = 4.0 kg. The calculation utilizes the parallel-axis theorem and direct computation methods, yielding a moment of inertia of 56 kg·m² before adjustment for the center of mass. The final result confirms the moment of inertia about the center of mass as 9.33 kg·m².

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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.
 
Last edited:
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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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