# Pressure at depth $h$ of a spinning bucket at top of swing

This was a question posed to me some time ago and it has haunted me ever since. Hints and suggestions would be awesome.

Say someone is spinning a bucket of water in a vertical circle of radius $$r$$ with constant velocity $$v$$. Let the atmospheric pressure be $$P_0$$. When the bucket is at the top of the circle, what is the pressure $$P(h)$$ in the water at depth $$h$$.

Let $$h=0$$ correspond to the water-bucket surface, and let $$h$$ increase as we move toward the center of the circle.

For any parcel of water, when the bucket is at the apex of the arc, both gravity and the bucket exert of a force on it. Additionally, we have atmospheric pressure $$P_0$$, which we will ignore for the moment. While both forces act in the direction of the center of the circle, intuitively it seems that the force from the bucket is more like the (standard) normal force.

Let me clarify. When you have water in an upright stationary bucket, the normal force opposes that of gravity and thereby allows for pressure to increase with depth. In my mind, it seems that, in a similar way, the force exerted by the bucket would be this "normal force", and thus the pressure should increase as you move to the bottom of the bucket (top of the circle).

However, if we were to release the bucket at precisely this moment, it would tangentially to this force, not in the opposite direction. So that poses a significant issue.

If we consider only the forces, devoid of the situation, and kind of sweep the (standard) normal force under the rug, it seems that:

$$P(h)=P_0 + \rho gh + \frac{\rho v^2}{r-h}$$.

Surely, this is wrong. Any guidance would be incredible.

This is how I approached it, the force needed for centrepetal acceleration of the water is $$F = ma = P A$$. But $$m = \rho V$$ and $$a = v^2/r$$ so $$P = \rho *\frac{V}{A}*\frac{v^2}{r}$$ assuming the bucket has a constant cross section $$V/A = h$$ so the final equation is $$\frac{\rho hv^2}r$$. Note here that the particles tend to push outwards and so in the mentioned configuration the particles in the bottom will be weighing on the particles above them and pressure builds up from 0 to maximum at the top. so $$h$$ here should be adjusted to be $$r-h$$ and the final pressure is adjusted to contain the static pressure and should be $$P_o + \rho g h + \frac{\rho (r-h)v^2}r$$ This assumes that the bucket is rotating around its top so that length of the bucket = r. This perhaps provides a more reasonable answer as when h = r, the added pressure is zero, which makes sense because the centre of rotation is not affected by the rotation.