Four particles each of mass M are placed at the corners of a square of side L. The radius of gyration of the system about an axis perpendicular to the square and passing through its centre is

Step-by-step solution

The axis of rotation passes perpendicularly through the center of the square with side length L . Each particle of mass M lies at distance r = L/ 2 from the center, equal to half the diagonal length. The moment of inertia for one particle is Mr^2 = M(L/ 2 )^2 = ML^2/2 . For four identical particles, the total moment of inertia becomes I_ total = 4 (ML^2/2) = 2ML^2 . Using the relationship I_ total = (4M)K^2 for the system’s total mass 4M and radius of gyration K , we substitute 2ML^2 = 4MK^2 . Solving yields K^2 =