Question

If a vertical cylinder of water (or any other liquid) rotates about its axis, as shown in FIGURE CP8.71, the surface forms a smooth curve. Assuming that the water rotates as a unit (i.e., all the water rotates with the same angular velocity), show that the shape of the surface is a parabola described by the equation$z=\left(\frac{{\omega }^2}{2g}\right)r^2$

Hint: Each particle of water on the surface is subject to only two forces: gravity and the normal force due to the water underneath it. The normal force, as always, acts perpendicular to the surface.

Short answer

Shape of the parabola is described by the equation $z=\left(\frac{{\omega }^2}{2g}\right)r^2$

Step-by-step solution

Given information

The figure shows the detail

Explanation

First draw Free Body diagram as below

From the diagram equate the vertical and radial force

Gravitational force, F = mg ………………..(1)

Radial Force F_radial=mrω²

Lets consider a very small element at radius dr and the height of the water above the surface is dh

The angle given as

$$\begin{gathered} \tan \varphi =\frac{mr{\omega }^2}{mg}=\frac{r{\omega }^2}{g}\dots \dots \dots \dots \dots .\dots \left(2\right) \\ \tan \varphi =\frac{dz}{dr}\dots \dots \dots \dots .....\dots \dots \dots \left(3\right) \end{gathered}$$

From equation (2) and (3) we get

$$\begin{gathered} \frac{dz}{dr}=\frac{r{\omega }^2}{g} \\ dz=\frac{r{\omega }^2}{g}dr \end{gathered}$$

Upon integration we get

$z=\left(\frac{{\omega }^2}{2g}\right)r^2$