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 ?

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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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