Set the vertically standing to be initial position and horizontal to be the final position. Use energy conservation. Consider the change of potential energy: For the ball, its center of mass falls ||l+\frac d 2||. For the rod, its center of mass falls ||\frac l 2||. All these changes goes into the rotational kinetic energy.

Set y=0 at the horizontal level and use ||U_1+K_{r1}+K{t1}=U_2+K_{r2}+K_{t2}||.

$$\therefore Mg(L+\frac d 2)+mg\frac {l}{2}=\frac 1 2 I_{total}\omega^2.$$

For parts 2 and 3, you will need to calculate ||I_{total}||. Need to use parallel axis theorem to find the ball’s moment of inertia. $$I_{total}=\frac 2 5 MR^2+M(l+\frac d 2)^2+\frac {1} {3} ml^2.$$

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