The moment of inertia of a square loop made of four uniform solid cylinders, each having radius R and length L( R < L ) about an axis passing through the mid points of opposite sides, is (Take the mass of the entire loop as M ) :

Step-by-step solution

Each cylinder has mass M/4 , radius R , length L . Axis through midpoints of top and bottom sides (along y-direction). Top/bottom cylinders (perpendicular to axis, axis through midpoint): I = M_c R^2 4 + M_c L^2 12 each. Left/right cylinders (parallel to axis, distance L/2 from it): I = M_c R^2 2 + M_c(L/2)^2 each. Total = 2 ( M_c R^2 4 + M_c L^2 12 ) + 2 ( M_c R^2 2 + M_c L^2 4 ) = M 2 ( 3R^2 4 + L^2 3 ) = 3MR^2 8 + ML^2 6 .