# Velocity of a leak in a cylindrical water tank that spins with constant angular velocity

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

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.