Mechanical Properties of Fluids Test - 59

As the tube rotates, the centrifugal force (as seen in the rotating frame of the tube) creates a pressure head causing the water to be jetted out of the orifice at the end of tube

Consider, an infinitesimally thin slice of water of width $$dx$$ at a distance $$x$$ from the one end of the tube. Let the area of cross-section of the tube be $$s$$. The mass of water contained in this infinitesimally thin slice of water is given by $$dm=\rho sdx$$, here $$\rho$$ is the density of water. the centrifugal force by this slice of water on the next is given by,

$$df=(dm)w^2x=\rho s \omega^2xdx$$...............(1)

the net force acting at the end of the tube just before the orifice can be obtained by integrating over all such thin slice as,

$$\int^f_0 df = \int^l _ {l-h} \rho s \omega^2xdx$$  or,

$$f=\dfrac{1}{2}\rho s \omega^2h(2l-h)$$  or,

$$P=f/s = \dfrac{1}{2}\rho \omega^2h(2l-h)$$.................(2) 

In (2), $$P$$ is the pressure head generated by the centrifugal force acting on the water. the net pressure at the end of the tube just before the orifice is given by $$P + P_o$$, when $$P_o$$ is the atmosphereic pressure. We assume that at this point the water is still i.e its velocity is zero. Right outside the tube, the water jets into an area of pressure $$P_i$$ (atmospheric pressure), say at a velocity $$v$$. Using Bernoulli's equation at point just before and just outside the orifice we have,

$$P+P_o+\dfrac{1}{2}\rho (0)^2=P_0+\dfrac{1}{2}\rho v^2$$ or,

$$P=\dfrac{1}{2}\rho v^2$$

or,  $$v=\omega \sqrt{h(2l-h)}$$