Irodov 1.330: Spin, Pressure and the Paraboloid – Amit & Deepali Sinha Foundation

Problem Irodov 1.330

A cylindrical vessel containing water is spun about its own vertical axis at a constant angular velocity $\omega$ . Find:

the shape taken up by the free surface of the water;
how the water pressure along the bottom of the vessel varies with distance from the axis, given that the pressure at the central point of the bottom is $p_0$ .

The cylindrical vessel of water, turning steadily about its own vertical axis at angular velocity ω. — The cylindrical vessel of water, turning steadily about its own vertical axis at angular velocity $\omega$ .

Given

$\omega$ (angular velocity)
$p_0$ (pressure on the axis at the bottom)

Find

Find (a)

z(r)

, the shape of the free surface, and (b)

p(r)

, the pressure along the bottom

Solution

Step 1Radial pressure balance in the rotating vessel

Once the water rotates rigidly with the vessel, every fluid ring at radius $r$ needs a net inward (centripetal) force $\rho\omega^2r$ per unit volume, supplied by the radial pressure gradient:

\frac{\partial p}{\partial r} = \rho\omega^2 r. \tag{1}

Vertically the fluid is in ordinary hydrostatic equilibrium, with $\partial p/\partial z = -\rho g$ as usual.

Step 2Shape of the free surface

Along the free surface the pressure is everywhere atmospheric, so $p$ is constant there and $dp=0$ ; substituting the radial gradient (1) together with $\partial p/\partial z=-\rho g$ relates $dr$ and $dz$ along the surface.

0 = \frac{\partial p}{\partial r}dr + \frac{\partial p}{\partial z}dz = \rho\omega^2 r\,dr - \rho g\,dz.

Solving for the slope $dz/dr$ and integrating from the vertex ( $r=0$ , taken as $z=0$ ) out to radius $r$ gives the surface height there.

\frac{dz}{dr} = \frac{\omega^2}{g}r \quad\Longrightarrow\quad \int_0^{z} dz' = \frac{\omega^2}{g}\int_0^{r} r'\,dr' = \frac{\omega^2}{g}\left[\frac{r'^2}{2}\right]_0^{r}.

z = \dfrac{\omega^2}{2g}r^2

(2)

The free surface in cross section: a parabola in section and a paraboloid of revolution in the vessel. Here z is measured from the lowest point of the surface, on the axis, and R is the vessel's own radius. — The free surface in cross section: a parabola in section and a paraboloid of revolution in the vessel. Here $z$ is measured from the lowest point of the surface, on the axis, and $R$ is the vessel’s own radius.

Step 3Pressure along the bottom

At any fixed height, in particular the vessel’s bottom, integrate (1) from the axis ( $r=0$ , where $p=p_0$ ) outward:

p(r) - p_0 = \int_0^r \rho\omega^2 r'\,dr' = \frac{1}{2}\rho\omega^2r^2.

z = \dfrac{\omega^2}{2g}r^2, \qquad p = p_0+\dfrac{1}{2}\rho\omega^2r^2

(3)

Check

Both results collapse to the stationary case ( $z=0$ everywhere, $p=p_0$ everywhere) as $\omega\to0$ , and both grow without bound as $r$ increases, exactly as expected for a paraboloid whose curvature is set by $\omega$ : a faster spin lifts the rim higher and raises the bottom pressure faster with radius.

Remark

The surface shape and the bottom pressure share coefficients matching up to the factor $\rho g$ , $\omega^2/(2g)$ and $\rho\omega^2/2$ respectively, both traceable to the same radial balance (1): the extra height the surface must climb at radius $r$ to keep the pressure atmospheric there is exactly the height of water whose weight would supply that radius’s own centripetal pressure. Rotation converts a spatial pressure gradient into a tilted free surface precisely because gravity is what converts pressure into height in the first place.