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Next: General Properties of a Up: Fluids and Nonmagnetic Plasmas Previous: Hydrostatic Liquids and Atmospheres

Rotating Liquids

What happens when you swing an object that is attached to a string? The object does not move in a straight line but instead moves in a circular orbit. So what stops the object moving in a straight line as predicted by Newton? The answer is the force due to the tension in the string which accelerates the object. We can see this in Figure 2.4.

Figure 2.4: If the radius is fixed as $R$, then the velocity changes direction as the object rotates. Hence, there is a change in velocity and so an acceleration.
\includegraphics [scale=0.7]{fundfig24.ps}

The speed of the object is $v = \omega R$, where

\begin{displaymath}
\omega = {d\phi \over dt}, \hbox{ where } \phi = \phi (t).
\end{displaymath}

$\phi$ is the angle. The velocity is

\begin{displaymath}
\omega R {\bf e}_{\phi},
\end{displaymath}

and there is a change in the direction of the velocity with time (e.g. when $\phi = 0$, ${\bf e}_{\phi} = {\bf j}$ but when $\phi =
\pi/2$, ${\bf e}_{\phi} = - {\bf i}$). Since there is a change in the velocity, there must be an acceleration and it is in the negative radial direction. The acceleration is due to the tension in the string where

\begin{displaymath}
T_{string} = m\omega^{2}R = m{v^{2}\over R}.
\end{displaymath}

For a uniform, imcompressible liquid in a cylindrical can rotating steadily about the $z$-axis, what is the equation of its surface? Set up cylindrical polars. Figure 2.5 illustrates the situation.

Figure 2.5: Forces acting on a fluid in a rotating cylindrical can.
\includegraphics [scale=0.7]{fundfig25.ps}

The basic equation is
\begin{displaymath}
\rho \left ({\bf v}\cdot \nabla \right ) {\bf v} = - \nabla p - \rho
g{\bf e}_{z}.
\end{displaymath} (2.14)

If the velocity is only in the $\phi$ direction, then (2.14) reduces to

\begin{displaymath}
-\rho{v_{\phi}^{2}\over R}{\bf e}_{R} = - {\partial p\over ...
...\partial p \over \partial z}{\bf e}_{z} - \rho
g{\bf e}_{z}.
\end{displaymath}

Thus, considering the radial and vertical direction separately we have

\begin{displaymath}
{\partial p \over \partial R} = \rho {v_{\phi}^{2}\over R} = \rho
\omega^{2}R,
\end{displaymath}

and

\begin{displaymath}
{\partial p \over \partial z} = - \rho g.
\end{displaymath}

$\omega$ is constant (the can is rotating at a constant rate). Integrating the radial component gives

\begin{displaymath}
p = \rho \omega^{2}{R^{2}\over 2} + f(z),
\end{displaymath}

where $f(z)$ is an arbitrary function obtained by integrating with respect to $R$. Substituting into the vertical component gives

\begin{displaymath}
{\partial p \over \partial z} = f^{\prime}(z) = - \rho g, \hbox{
}\Rightarrow \hbox{ } f = -\rho g z + \hbox{ const.}
\end{displaymath}

and so

\begin{displaymath}
p {1\over 2} \rho \omega^{2}R^{2} - \rho g z + C.
\end{displaymath}

At the surface of the liquid the pressure, $p$, is constant and so the equation of the surface is ($z=0$, $R=0$)
\begin{displaymath}
z = {\omega^{2}\over 2g}R^{2}.
\end{displaymath} (2.15)

The equation is a parabola. Note that as $\omega \rightarrow 0$, $z \rightarrow 0$ and that as $\omega$ increases the slope increases.
next up previous
Next: General Properties of a Up: Fluids and Nonmagnetic Plasmas Previous: Hydrostatic Liquids and Atmospheres
Prof. Alan Hood
2000-11-06