FLUIDS BYU 23

Solution

The air bubble has negligible mass compared to the water it displaces. Therefore, its position is determined by the pressure gradients in the fluid. The bubble will settle at a point where the pressure gradient along the tube is zero (or where the net buoyant force along the tube is zero).

There are two primary effects creating pressure gradients:

1. Gravity: Creates a pressure increase downwards. The buoyant force due to gravity pushes the bubble up (vertically).
2. Rotation (Centrifugal Force): Creates a pressure increase radially outwards ($dP/dr = \rho \omega^2 r$). The “buoyant” force due to this pressure gradient pushes the bubble radially inwards (towards the axis).

Let $x$ be the distance of the bubble from the lower end (hinge) along the tube. The radius of rotation at this point is $r = x \sin\theta$.

Resolving forces along the tube:

• Up the tube (due to gravity): Component of vertical buoyancy. $$F_{up} = (\rho_w V g) \cos\theta$$
• Down the tube (due to rotation): Component of radial buoyancy. The radial force acts horizontally towards the axis. The angle between the horizontal radius and the tube is $(90^\circ – \theta)$. Thus, the component along the tube is $\cos(90^\circ – \theta) = \sin\theta$. $$F_{down} = (\rho_w V \omega^2 r) \sin\theta$$ Substituting $r = x \sin\theta$: $$F_{down} = \rho_w V \omega^2 (x \sin\theta) \sin\theta = \rho_w V \omega^2 x \sin^2\theta$$

At equilibrium, forces balance: $F_{up} = F_{down}$

$$\rho_w V g \cos\theta = \rho_w V \omega^2 x \sin^2\theta$$ $$g \cos\theta = \omega^2 x \sin^2\theta$$ $$x = \frac{g \cos\theta}{\omega^2 \sin^2\theta}$$

The question asks for the fraction $\eta$ of the length of the tube ($l$). Therefore $\eta = x/l$.