Double pendulum angular momentum, m1 = 1, m2 = 2 kg

Take the origin at the fixed pivot $O$, $x$ vertically downward, $y$ to the right and $\hat k$ out of the plane ($+\hat k$ = counter-clockwise). Assume both arms lie on the same (right-hand) side of the vertical, and that $\omega_2$ is the absolute angular velocity of the lower arm (as is usual in this problem). Angular momentum of the system about $O$:

$$\vec H_O = m_1(\vec r_1 \times \vec v_1) + m_2(\vec r_2 \times \vec v_2)$$

Position vectors

Upper arm ($a=0.5$ m at $30^\circ$ to the $x$-axis), lower arm ($b=0.6$ m at $60^\circ$ to the vertical):

$$\begin{aligned} \vec r_1 &= a(\cos 30^\circ\,\hat i + \sin 30^\circ\,\hat j) = (0.433\,\hat i + 0.250\,\hat j)\ \text{m}\\ \vec r_{2/1} &= b(\cos 60^\circ\,\hat i + \sin 60^\circ\,\hat j) = (0.300\,\hat i + 0.520\,\hat j)\ \text{m}\\ \vec r_2 &= \vec r_1 + \vec r_{2/1} = (0.733\,\hat i + 0.770\,\hat j)\ \text{m} \end{aligned}$$

Velocities

$$\begin{aligned} \vec v_1 &= \omega_1\hat k \times \vec r_1 = (-0.500\,\hat i + 0.866\,\hat j)\ \text{m/s}\\ \vec v_2 &= \vec v_1 + \omega_2\hat k \times \vec r_{2/1} = (-2.578\,\hat i + 2.066\,\hat j)\ \text{m/s} \end{aligned}$$

Angular momentum

$$\begin{aligned} H_1 &= m_1(x_1 v_{1y} - y_1 v_{1x}) = 1(0.433\times 0.866 - 0.250\times -0.500) = 0.50\ \text{kg m}^2/\text{s}\\ H_2 &= m_2(x_2 v_{2y} - y_2 v_{2x}) = 2(0.733\times 2.066 - 0.770\times -2.578) = 7.00\ \text{kg m}^2/\text{s} \end{aligned}$$ $$\vec H_O = H_1 + H_2 = 7.50\,\hat k\ \text{kg m}^2/\text{s}$$

Answer: $\vec H_O \approx 7.50\,\hat k$ kg m²/s (counter-clockwise, about the axis perpendicular to the plane).