Four identical solid spheres, each of mass 5 kg and radius 1 m , are placed such that their centers lie exactly on the four corners of a square of side length 2 m . The total moment of inertia of this system about an axis passing through the center of the square and perpendicular to its plane is kg m ^2 .

Step-by-step solution

The distance d from the center of the square to any of its corners is half the length of the diagonal. d = a 2 = 2 2 = 2 m Using the parallel axis theorem, the moment of inertia of one solid sphere about the central axis is the sum of its moment of inertia about its own centroidal axis and the product of its mass and the square of the distance to the central axis. I₁ = 2 5 MR^2 + Md^2 I₁ = 2 5 (5)(1)^2 + 5( 2 )^2 I₁ = 2 + 5(2) = 12 kg m ^2 Since there are four identical spheres symmetrically placed, the total momen