The angular momentum of the system is given by the product of the moment of inertia and the angular velocity. Since the rod is light and rigid, we can consider the two masses as point masses located at the ends of the rod.
Let m1 and m2 be the masses at the ends of the rod, and let r be the distance from the axis of rotation to each mass. The moment of inertia of each mass about the axis of rotation is given by I = mr^2. Since the masses are symmetrically arranged about the axis of rotation, the total moment of inertia of the system is I = 2mr^2.
The angular momentum of each mass is L = Iw = 2mr^2w. The angular momentum of the system is the sum of the angular momentum of each mass, so we have:
L_total = L1 + L2 = 2mr^2w + 2mr^2w = 4mr^2w
Therefore, the angular momentum of the system is 4mr^2w