Upward force on rotating drum top

Data: drum radius $R=0.25$ m, pipe diameter 25 mm so $r_0=0.0125$ m, water level in the pipe $h=0.6$ m above the drum top, $N=600$ rpm so $\omega=\dfrac{2\pi\times600}{60}=62.83$ rad/s, $\rho=1000\ \text{kg/m}^3$.

        |  | <- water, 0.6 m above top
        |  |
    ____|  |____   drum top
   |             |
   |   water     |   0.5 m dia
   |_____________|

The drum is full and the free surface lies in the pipe on the axis, so on the underside of the drum top (gauge pressure) at radius $r$:

$$p=\rho g h+\tfrac{1}{2}\rho\omega^{2}r^{2}$$

The upward force on the top (annulus from $r_0$ to $R$):

$$F=\int_{r_0}^{R}p\,2\pi r\,dr=\rho g h\,\pi\left(R^{2}-r_0^{2}\right)+\frac{\pi\rho\omega^{2}}{4}\left(R^{4}-r_0^{4}\right)$$

Static part:

$$F_1=1000(9.81)(0.6)\,\pi\left(0.25^{2}-0.0125^{2}\right)=1152.8\ \text{N}$$

Rotational part:

$$F_2=\frac{\pi(1000)(62.83)^{2}}{4}\left(0.25^{4}-0.0125^{4}\right)=12\,111.8\ \text{N}$$ $$F=F_1+F_2=13\,264.6\ \text{N}$$

Answer: upward force on the drum top $\approx13.3$ kN (13.27 kN if the small pipe area is ignored). Rotation contributes about 91 % of it.