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I´m a little confused with a fluid-dynamics exercise.

We have a cylinder of radius $R$ rotating with respect to the vertical axis with angular velocity $\omega $.

The cylinder has a small hole in the wall almost in the bottom. The hole is small, thus we can neglect the velocity of the surface of the fluid.

We have as data that the lower part of the surface is at $z=H$.

Now we have to calculate the radial velocity of the fluid when it leaves the cylinder.

My attempt to solve the exercise was: as a first step, I calculated the shape of the surface using a non-intertial frame of reference with a centripetal force.

Doing this I obtained that the surface is paraboloid shaped. More precisely it is a revolution paraboloid defined by this equation:

$$z=\frac{\omega^2}{2g}r^2\,.$$

Now, I'm confused about the next step. How can I calculate the exit velocity using Bernoulli and the maximum altitude of the fluid in the parabola? I know that Bernoulli is applicable only to streamlines, but I can't even imagine the course of these.

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  • $\begingroup$ Your equation tells you what the hydrostatic pressure at the hole should be. Use Toricelli law (for inviscid fluids) to calculate the exit velocity. $\endgroup$ – Deep Apr 9 at 5:28
  • $\begingroup$ @Deep My question is: is correct use Torricelli law in presence of the centrifugal force? And, if is this possible, what is the correct justification of that. $\endgroup$ – Ezequiel Saidman Apr 9 at 14:21
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The radial velocity energy will be equal to the total liquid height above the hole, thus the radial velocity $v=\sqrt {2g(H+z)}$. Detailed energy balance: The problem statement indicated the hole size is very small. This is meant to indicate that the rate of mass exit from the cylinder is very small as to allow neglecting change of liquid level and also change of momentum and energy content when mass moved to a different radius. Also the cylinder would be perceived to have zero thickness such that a particle can be considered at the same radius before and after it exits the cylinder. With no energy transfer, the total energy (per unit mass)of the particle would be constant. Before exit: pressure energy = H+z and kinetic energy = tang vel squ / 2g (radial velocity = zero) After exit: pressure energy = zero and kinetic energy = total vel squ /2g = (tang vel squ + radial vel squ) / 2g

Resolving the balance shows the pressure energy before exit is equal to the radial velocity energy after exit.

Please excuse my long hand statement of math symbols. My keyboard is limited.

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  • $\begingroup$ How you can justify this? $\endgroup$ – Ezequiel Saidman Apr 9 at 1:41

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