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Moment of inertia of a squARE lamina

  1. Apr 23, 2012 #1
    a square lamina is made of 4 uniform thin rods each of which was a moment of inertia Ml^2 /12 about an axis perpendicular to their length and passing through their centres.

    My way of understanding it is the following : the two rods to whom the axis of rotation is perpendicular and passes through their centres have a total moment of inertia:

    2*(1/12)*M*(L/2)^2 so ML^2 / 24 .

    The rest two thin rods who are parallel to the axis of rotation are again rotating "forming a cylinder" and are L/2 away from the axis of rotation ( perpendicular distance ).So

    the total is : 2 * ML^2/24 = ML^2 / 12 which is the moment of inertia of each rod .

    Is my thinking correct ? If not could you please help me understand it ?

    Thank you very much :)
  2. jcsd
  3. Apr 23, 2012 #2
    As you say, the moment of inertia I (see http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html) of the two rods where the axis of rotation passes through their center is I = M·L2/12, so for two of them it is M·L2/6.
    For each rod parallel to the axis of rotation, the moment of inertia is M·(L/2)2 = M·L2/4 , so for two it is M·L2/2.
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