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Moment of inertia of a three particle system

  1. Jun 3, 2012 #1
    1. The problem statement, all variables and given/known data
    I have three particles of mass m at (-a,-a), (a,-a) and (0,a)

    Find the moment of inertia I_z around the center of mass of the system for an axis along the z-axis.

    2. Relevant equations
    Center of mass:
    CM = Ʃmr

    Moment of inertia:
    I = Ʃmp^2

    m = mass
    r = position
    p = distance from axis

    3. The attempt at a solution
    I found that the center of mass is CM = (0,-a/3);
    CM_x = (-am + am + 0m)/3m = 0
    CM_y = (-a-a+a)m/3m = -a/3
    This is something I know is correct..

    The real problem starts with the moment of inertia:
    p_1 = 4a/3
    p_2 = p_3 = √(a^2 + (2a/3)^2)

    (p_1)^2 = (16/9)a^2
    (p_2)^2 = (a^2 + (2a/3)^2) = (13/9)a^2

    Ʃmp^2 = m(16/9)a^2 + m(13/9)a^2 + m(13/9)a^2
    = m((16/9)a^2 + (26/9)a^2) ≈ 4.67ma^2

    The problem is that the book says that the solution is 4.01ma^2. Where have I made a mistake?
     
  2. jcsd
  3. Jun 3, 2012 #2
    I got the answer that you had as well. Make sure that you've read the question correctly.
     
  4. Jun 3, 2012 #3
    Last edited: Jun 3, 2012
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