Since we have a compound object in both cases, we can use the parallel axis theorem to find the moment of inertia about each axis. In (a). the center of mass of the sphere is located at a distance L + R from the axis of rotation. In (b), the center of mass of the sphere is located at a distance R from the axis of rotation. In both cases, the moment of inertia of the rod Is about an axis at one end. Refer to (Figure) for the moments of Inertia for the Individual objects. 1. I_{total} = \\sum I_i = I_{Rod} + I_{Sphere}; I_{Sphere} = I_{center \\, of \\, mass} + m_{Sphere} (L + R)^2 = \\dfrac{2}{5} m_{Sphere} R^2 + m_{Sphere} (L + R)^2 ; I_{total} = I_{Rod} + I_{Sphere} = \\dfrac{1}{3} m_{Rod} L^2 + \\dfrac{2}{5} m_{Sphere} R^2 + m_{Sphere} (L + R)^2 ; I_{total} = \\dfrac{1}{3} (2.0 kg) (0.5 m)^2 + \\dfrac{2}{5} (1.0 kg)(2.0 m)^2 + (1.0 kg)(0.5 m + 0.2 m)^2; I_{total} = (0.167 + 0.016 + 0.490) kg . m^2 = 0.673 kg. m^2 .2. I_{Sphere} = \\dfrac{2}{5} m_{Sphere} R^2 + m_{Sphere} R^2 I_{total} = I_{Rod} + I_{Sphere} = \\dfrac{1}{3} m_{Rod} L^2 + \\dfrac{2}{5} m_{Sphere} R^2 + m_{Sphere} R^2; I_{total} = \\dfrac{1}{3} (2.0 kg) (0.5 m)^2 + \\dfrac{2}{5} (1.0 kg)(0.2 m)^2 + (1.0 kg) (0.2 m)^2 ; I_{total} = (0.167 + 0.016 + 0.04) kg . m^2 = 0.223 kg . m^2 .