The water exits the nozzle with relative velocity $u$. The reaction force (thrust) exerted by the exiting water on the nozzle is given by the rate of momentum flux.

Thrust Force on one nozzle: $F = \dot{m} v_{rel} = (\rho S u) u = \rho S u^2$.

There are two such nozzles, creating a force couple. The torque $\tau$ is:

$$\tau = 2 \times F \times l = 2 \rho S u^2 l$$

Using Newton’s Second Law for rotation $\tau = I \alpha$:

$$I \alpha = 2 \rho S u^2 l$$ $$\alpha = \frac{2 \rho S u^2 l}{I}$$

As the tube rotates with angular velocity $\omega$, we must account for the angular momentum imparted to the water flowing inside the tube. The tube accelerates the water from zero angular momentum (at the center) to an angular momentum corresponding to tangential speed $\omega l$ at the tip.

Torque Equation:

$$ \tau_{net} = \tau_{thrust} – \tau_{resist} = I \frac{d\omega}{dt} $$

1. Thrust Torque: Remains constant $\tau_{thrust} = 2 \rho S u^2 l$.
2. Resistive Torque (Coriolis term): This is the rate at which the tube transfers angular momentum to the flowing water.
   Mass flow rate (total) $\dot{M} = 2 \rho S u$.
   Angular momentum carried away per unit time $= \dot{M} (\omega l) l = 2 \rho S u l^2 \omega$.

Equation of Motion:

$$ 2 \rho S u^2 l – 2 \rho S u l^2 \omega = I \frac{d\omega}{dt} $$ $$ \frac{2 \rho S u l}{I} (u – l\omega) = \frac{d\omega}{dt} $$

Rearranging and integrating:

$$ \int_0^\omega \frac{d\omega}{u – l\omega} = \int_0^t \frac{2 \rho S u l}{I} dt $$ $$ -\frac{1}{l} \left[ \ln(u – l\omega) \right]_0^\omega = \frac{2 \rho S u l}{I} t $$ $$ \ln\left( \frac{u – l\omega}{u} \right) = – \frac{2 \rho S u l^2}{I} t $$ $$ 1 – \frac{l\omega}{u} = \exp\left( – \frac{2 \rho S u l^2}{I} t \right) $$ $$ \omega(t) = \frac{u}{l} \left[ 1 – \exp\left( – \frac{2 u S \rho l^2}{I} t \right) \right] $$