
#1
Apr2312, 06:31 PM

P: 64

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
Apr2312, 09:13 PM

P: 4,664

As you say, the moment of inertia I (see http://hyperphysics.phyastr.gsu.edu/hbase/mi.html) of the two rods where the axis of rotation passes through their center is I = M·L^{2}/12, so for two of them it is M·L^{2}/6.
For each rod parallel to the axis of rotation, the moment of inertia is M·(L/2)^{2} = M·L^{2}/4 , so for two it is M·L^{2}/2. 


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