Four identical solid spheres, each of mass 2 kg and radius 10 cm , are placed with their centers at the four corners of a square of side 50 cm . The moment of inertia of the system about an axis passing through the center of the square and parallel to one of its sides is x 10⁻³ kg m ^2 . The value of x is ____.

Step-by-step solution

Let the mass of each sphere be M = 2 kg , radius R = 10 cm = 0.1 m , and the side of the square be a = 50 cm = 0.5 m . The axis of rotation passes through the center of the square and is parallel to one of its sides. The perpendicular distance from the center of each of the four spheres to this axis is d = a 2 = 0.25 m . By the parallel axis theorem, the moment of inertia of one sphere about this axis is: I₁ = I_ cm + Md^2 = 2 5 MR^2 + M ( a 2 )^2 Since there are four identical spheres at the same distance from the