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Calculating moment of inertia of a system of balls

  1. Jan 9, 2006 #1
    (Resolved)Calculating moment of inertia of a system of balls

    I have a bit hard time calculating moment of inertia of a a system of balls. The system is defined as follows: 3 balls that are the same are connected with thin rods soo that the center of each ball is in the corners of a equilateral triangle with length of side to be 0.3 m. Radius of each ball is 0,08 m and mass of each ball is 1,2 kg. Now the fun part: Calculate moment of inertia of the system if the axis of rotation goes trught the center of mass of the equilateral triangle. I of ball is 2/5*m*r^2.
    I am puzzled. I know that the center of mass for a equilateral triangle is in 2/3 of its height. Soo each ball is exacly 1/3*Sqrt(3)*a -> 1/3*Sqrt(3)*0,3m away from the axis of rotation. I would simply say that the moment of inertia I = Sigma(m*r^2). I of each ball is I of b = m*(1/3*Sqrt(3)*a)^2 = 1.2 kg * 1/9 * 3 * 0.09 m^2. And then the moment of inertia of the system is simply I of system = 3 times I of ball. I am missing something. Since the radious is also given. I would apprichiate greately if somebody would explain the reasoning behind this problem and also if possible the path to solution.
     
    Last edited: Jan 9, 2006
  2. jcsd
  3. Jan 9, 2006 #2

    mukundpa

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    Go through the parallel axis theorem of M.I.
    The ball is not a point mass.
     
  4. Jan 9, 2006 #3
    Ah...
    The Moment of inertia of the ball is I = 2/5*mass*radius^2 + mass*distance from the axis of the system^2. And the moment of inertia of the system if 3 times of that of single ball?
     
  5. Jan 9, 2006 #4

    mukundpa

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    looks good !
     
  6. Jan 9, 2006 #5
    Ok, Thanks!
     
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