We have a cylindrical water tank that spins over its axis of symmetry. Here's a diagram:

Water tank

The objective is to find:

$1$ - The tangential and radial components of the velocity of the water as it leaves the tank.

$2$ - The radius $r$ reached by the water.

I'm not sure at all about how to handle the fact that the tank spins, as well as the deformation in the water surface.

I chose to apply Bernoulli's equation and compare the point at the center of the surface, and the point where water leaves the tank:

$$P_{atm} + \delta g_{ef} H = P_{atm} + \frac{1}{2} \delta v_{r}^2$$

So: $$ v_{r} = \sqrt{2Hg_{ef}}$$

Where $g_{ef} = g + a_{c} = g + R\omega ^2$ is the effective acceleration of the water.

Then we have:

$$ v_{r} = \sqrt{2H(g + R \omega ^2)}$$

The tangential component of the water's velocity is:

$$ v_{t} = R \omega $$

The time it takes the water to reach the floor is $t = \sqrt{\frac{2d}{g}}$, and $r = R + v_{r}t$. Thus:

$$ r = 2\sqrt{Hd(1 + \frac{R \omega ^2}{g})} + R $$

My issue is that I have no idea if my $g_{ef}$ is correct. It seems to make sense because I expect $r$ and $v_{r}$ to be higher the larger $\omega$ is. But I don't know how to explain why it makes sense to use it. That is, if it's even correct to begin with.

I'm also not sure about using $a_{c} = R\omega ^2$ because I'm using the center point of the water surface, which is at $r = 0$.

Any help is appreciated. Thanks.


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