Three identical particles each o mass m are placed at the three corner of a planar equilateral triangle ABC of side L calculate M oment of inertia about the line XX' as axis

Since, it is an equilateral triangle, AB = BC = CA = L

So, AD = AB sin60 = (√3/2)L

Thus, masses at B and C are at distance AD from the axis XX/.

So, MI of the system about XX/ is,

I = 0 + m[(√3/2)L]2 + m[(√3/2)L]2

=> I = (3/2)mL2

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