# Derivation of the equation for flow of a viscous liquid on a rotating disk

In this paper,

Emslie, Alfred G., Francis T. Bonner, and Leslie G. Peck. “Flow of a Viscous Liquid on a Rotating Disk.” Journal of Applied Physics 29, no. 5 (1958): 858. doi:10.1063/1.1723300.

the authors derive an equation for the $z$ dependence of the radial velocity of a viscous liquid. The question is how did the authors arrive at equation (1) and how they arrived at the second derivative?

Experimental setup: there is a rotating disk with a certain angular velocity $\omega$ with some viscous liquid on top of the disk. Upon spinning, the liquid spreads over the disk, lowering in thickness. There is a centrifugal force acting in the outward direction, and a viscous force acting in the opposite direction.

Given this setup, equation (1) of the paper says,

$$-\eta \frac{\partial^2v}{\partial z^2} = \rho \omega^2r$$

where the variables are defined in cylindrical polar coordinates $(r,\theta,z)$, as follows: $\eta$ is the viscosity; $v$ is the radial velocity of the liquid, which is dependent on $z$; $\rho$ is the density of the fluid and $\omega$ is the angular velocity of the rotating disk. The authors equate the viscous force and the centrifugal force per unit volume.

I found here that the viscous force is $F_V=\mu A \frac{\partial u}{\partial y}$, where $\mu$ is the dynamic viscosity of the fluid ($\eta$ can also be used), $A$ is the plate area, $u$ is the speed of the fluid moving between two parallel plates and $y$ is the separation between plates. Converting those variables to the rotating disk case, $u$ becomes $v$ and $y$ becomes $z$. Therefore, the same formula becomes: $F_V=\eta A \frac{\partial v}{\partial z}$.

I found here that the centrifugal force is $F_C=m \omega ^2/r$.

By equating the two formulas, $$-\frac{F_V}{V}=\frac{F_C}{V}$$ where $V$ is the volume, $$-\eta \frac{A}{V}\frac{\partial v}{\partial z}=\frac{m}{V}\omega^2r$$ $\frac{m}{V}$ is the density $\rho$ and $V=Ah$, so $\frac{A}{V}$ becomes $\frac{1}{h}$, where $h$ is the height.

Therefore, $$-\eta \frac{1}{h}\frac{\partial v}{\partial z}=\rho\omega^2r$$

How did the authors arrive at the second derivative? Also, $h$ could be $z$?

## 1 Answer

Textbook derivation is for a fluid element, and you need to find net force on it. If you consider the fluid element to be cubical in shape, then considering only z-direction, viscous forces act on both top and bottom face of the cubical element. These are given respectively by $\eta\frac{\partial u}{\partial z}\bigr|_z$ and $\eta\frac{\partial u}{\partial z}\bigr|_{z+\delta z}$, where $\delta z$ is the width of the cube in z-direction. Net force is the difference between the two forces above, which by using Taylor expansion and in the limit $\delta z\rightarrow0$, gives the second derivative in the textbook formula.

• In addition to the important analysis that Zero provided, there are some additional approximations involved here. The authors are assuming locally steady and parallel flow, with a zero shear stress at z = h(r). This is very much similar to viscous flow of a fluid film of thickness h down a vertical flat plate under the action of gravity. In the disk problem, the artificial gravity is radially dependent, and equal to $\omega^2 r$ – Chet Miller Jul 20 '16 at 12:11