Doesn't the applied torque in a frictionless environment continuously increase the angular velocity?

Question

Consider a sprinkle working in a no-friction universe (no air drag, no friction in the bearing). If this sprinkle ejects water with a constant flow rate, shouldn’t it be accelerating forever? A fluid-mechanics text book formulates this situation as follows:

In case of there is a retarding torque, $T_0$ at the center:

$$-T_0 = \dot{m}_\text{out}·(r·V_r).$$

$V_r$ is relative velocity between jet and nozzle where water is ejected and it is equal to $V_r = V_j - V_\text{nozzle}$ and therefore:

$$-T_0 = \dot{m}_\text{out} (r·V_{j}-r·V_\text{nozzle}).$$

Here $V_\text{nozzle} = ω·r$ then: $$-T_0 = \dot{m}_\text{out}·(r·V_{j}-ω r^2)$$

Eventually:

$$ω = \frac{V_j}{r} - \frac{T_0}{\dot{m}_\text{out}·r^2}$$

The book says if $T_0 = 0$, the angular velocity would be equal to the jet velocity, which is constant. But intuitively I expect increasing angular velocity instead of a constant value. What am I missing?

Answer 1

If $\omega = V_j/r$, then the tangential speed of the water with respect to the non-rotating "ground" frame is zero. In other words, the water coming out of the nozzle falls straight down, carrying zero angular momentum. This means the sprinkler imparts no net torque on the water, and likewise the water imparts no net torque on the sprinkler, causing the angular velocity of the sprinkler to remain constant.

Another way to think about this is as follows. Think of a small mass of water sucked into the sprinkler making its way outward in one of its arms. As this happens, the angular momentum of this mass increases, meaning the sprinkler is applying a tangential force (and hence a torque) on it in the spin direction. As the mass reaches the nozzle and is ejected, there is an instantaneous force (and torque) applied on it in the direction opposite to the spin. When $\omega = V_J/r$, these torques cancel out, so there is no net torque on either the water or the sprinkler.