1. **State the problem:** We have a bar AB of mass 2 kg with moment of inertia about point A given as $I_A = 8$ kgm². 2. **Goal:** Find the moment of inertia about the center point O, where AO = OB. 3. **Use the parallel axis theorem:** Moment of inertia about point A is related to that about the center O by $$I_A = I_O + Md^2,$$ where - $M = 2$ kg (mass of the bar), - $d$ is the distance between points A and O, - $I_O$ is the moment of inertia about the center O. 4. Since O is the midpoint of AB, $d = \frac{L}{2}$, where $L$ is the length of the bar. 5. For a uniform rod about its center, moment of inertia is known: $$I_O = \frac{1}{12}ML^2.$$ 6. Substitute $I_O$ in the theorem: $$I_A = \frac{1}{12}ML^2 + M\left(\frac{L}{2}\right)^2 = \frac{1}{12}ML^2 + \frac{1}{4}ML^2 = \frac{1}{3}ML^2.$$ 7. Given $I_A = 8$ kgm²: $$8 = \frac{1}{3} \times 2 \times L^2 = \frac{2}{3} L^2.$$ 8. Solve for $L^2$: $$L^2 = \frac{8 \times 3}{2} = 12.$$ 9. Calculate $I_O$: $$I_O = \frac{1}{12}ML^2 = \frac{1}{12} \times 2 \times 12 = 2.$$ **Final answer:** The moment of inertia about the center O is $2$ kgm².