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![[Post New]](/templates/default/images/icon_minipost_new.gif) 7 May 2007 21:13:50 IST
Accepted Answer [?]
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ok now the mass of AB=mass of DC = m/3
mass of AD=mass of BC= m/6
length of AB = length of CD = L/3
length of AD = length of BC = L/6
MOI of AB about BC = MOI of CD about BC = (m/3)*(L/3)2*(1/3)
MOI of AD about BC = (m/6)*(L/3)2 //(Use MOI=mr2 as AD is at a constant distance from BC)
Adding the 3, final MOI = 2*(m/3)*(L/3)2*(1/3) + (m/6)*(L/3)2
= 7mL2/162
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Impossible is Nothing |
this reply: 7 points
(with 1 
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