Composite body moment of inertia

Given: rod $m_r = 3$ kg, $L = 1.2$ m; sphere $m_s = 2$ kg, $R = 0.1$ m, centre on the rod axis at $1.2 + 0.1 = 1.3$ m from the free end O.

 O                               ___
 |============================( o )
 |<------- 1.2 m ------------>|<R>
 |<------ 1.3 m to sphere centre

Moment of inertia about O (transverse axis)

Rod about its end: $I_r = \dfrac{m_rL^2}{3} = \dfrac{3(1.2)^2}{3} = 1.440$ kg·m²

Sphere about its own centre: $\tfrac25 m_sR^2 = 0.4 \times 2 \times 0.01 = 0.008$ kg·m²

Parallel axis shift: $m_s d^2 = 2(1.3)^2 = 3.380$ kg·m²

Centre of mass from O

$$\bar x = \frac{3(0.6) + 2(1.3)}{5} = \frac{1.8 + 2.6}{5} = 0.88\ \text{m}$$

Radius of gyration about O

$$k_O = \sqrt{\frac{I_O}{m}} = \sqrt{\frac{4.828}{5}} = 0.983\ \text{m}$$

(As a check, the moment of inertia about the centre of mass is $I_G = 4.828 - 5(0.88)^2 = 0.956$ kg·m².)

Answer: I_O = 4.828 kg·m²; centre of mass 0.88 m from the free end; k_O = 0.983 m.