Question 3.19: Figure 3.16 shows a lawn sprinkler arm viewed from above. Th...

The entering velocity is V_{0}k, where V_{0}= Q/A_{pipe} Equation (3.59)
\sum{M_{0}}=\frac{\partial}{\partial t}\left[\int_{CV}^{}{(r\times V)\rho dv} \right] +\int_{CS}^{}{(r\times v)\rho (V\cdot n)dA}
applies to the control volume sketched in Fig. 3.16 only if V is the absolute velocity relative to an inertial frame. Thus the exit velocity at section 2 is
V_{2}=V_{0}i-R\omega i
Equation (3.59) then predicts that, for steady flow
\sum{M_{0}}=-T_{0}k=(r_{2}\times V_{2})\dot{m}_{out}-(r_{1}\times V_{1})\dot{m}_{in} (1)
where, from continuity,\dot{m}_{out}=\dot{m}_{in}=\rho Q . The cross products with reference to point O are
r_{2}\times V_{2}=Rj \times (V_{0}-R\omega)i=(R^2\omega -RV_{0})K

Equation (1) thus becomes

\omega =\frac{V_{0}}{R}-\frac{T_{0}}{\rho QR^2}
The result may surprise you: Even if the retarding torque T_{0} is negligible, the arm rotational speed is limited to the value V_{0} /R imposed by the outlet speed and the arm length