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## Homework Statement:

- Spinning water tank with a leak and constant angular speed

## Relevant Equations:

- Bernoulli equation, constant acceleration kinematic equations, uniform circular motion equations.

We have a cylindrical water tank that spins over its axis of symmetry with constant angular velocity ω. Here's a diagram:

We wish 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 ##ω## is. But I don't know how to explain

Any help is appreciated. Thanks.

We wish 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 ##ω## 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}=Rω^2## because I'm using the center point of the water surface, which is at ##r=0##.Any help is appreciated. Thanks.