Two identical solid spheres, each of mass 5 kg and radius 1 m , are fixed with their centers at the ends of a uniform rod of mass 3 kg and length 4 m . The entire system rotates freely about an axis passing through the midpoint of the rod and perpendicular to its length with an angular speed of 2 rad/s . The rotational kinetic energy of the system is J .

Step-by-step solution

The total moment of inertia of the system is the sum of the moment of inertia of the rod and the moments of inertia of the two spheres. The moment of inertia of the uniform rod about its perpendicular bisector is: I_ rod = 1 12 M_ rod L^2 = 1 12 (3)(4)^2 = 4 kg m ^2 The distance from the axis of rotation to the center of each sphere is half the length of the rod: d = L 2 = 4 2 = 2 m Using the parallel axis theorem, the moment of inertia of one solid sphere about the axis of rotation is: I_ sphere = 2 5 mR^2 + md^2