Three identical uniform rods each of length 40cm and mass 2kg form an equilateral triangle. Find the moment of inertia of this system about an axis passing through one corner ?Thanks in advance !

let ABC is the triangle formed by these rods then we have to find moment of inertia any about vertex , let vertex be A...

then , rod AB & AC will have same moment of inertia about end A ,

I_{AC = IAB} = mL²/12 + mL²/4 = mL²/3

moment of inertia of BC  = mL²/12 + md²                     (d is perpendicular distance from axis of rotation)

      from triangle ABC , d = Lcos30

         I_BC = mL²/12 + 3mL²/4 = 10mL²/12

net moment of inertia = I_AB+I_BC+I_AC

                     I_net   =18mL²/12 = 3mL² /2

                    I_net = 0.48kg-m²

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Approved

Thanks a Lot