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Moment of Inertia of four particles

  1. Oct 30, 2007 #1
    1. The problem statement, all variables and given/known data
    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.

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


    3. 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: Oct 30, 2007
  2. jcsd
  3. Oct 31, 2007 #2

    rl.bhat

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    Homework Helper

    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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