# Fluid rotating in theta direction in a tank

A Newtonian fluid of constant density $$\rho$$ is in a vertical cylinder of radius R with the cylinder rotating about its axis at angular velocity $$\omega$$. Find the shape of the free surface at steady state.Consider the cylindrical coordinate system for analysis. Consider the pressure (P) to be a function of two coordinate system r and z. Refer to the figure below for more details.

I used Navier Stoke's equation in the angular direction since the principal motion is in $${\theta}$$ direction.

\begin{aligned} &\theta \text { -component: }\\ &\rho\left(\frac{\partial u_{\theta}}{\partial t}+u_{r} \frac{\partial u_{\theta}}{\partial r}+\frac{u_{\theta}}{r} \frac{\partial u_{\theta}}{\partial \theta}+\frac{u_{r} u_{\theta}}{r}+u_{z} \frac{\partial u_{\theta}}{\partial z}\right)\\ &=-\frac{1}{r} \frac{\partial P}{\partial \theta}+\rho g_{\theta}+\mu\left[\frac{1}{r} \frac{\partial}{\partial r}\left(r \frac{\partial u_{\theta}}{\partial r}\right)-\frac{u_{\theta}}{r^{2}}+\frac{1}{r^{2}} \frac{\partial^{2} u_{\theta}}{\partial \theta^{2}}+\frac{2}{r^{2}} \frac{\partial u_{r}}{\partial \theta}+\frac{\partial^{2} u_{\theta}}{\partial z^{2}}\right] \end{aligned}

Reducing it with assumptions I get

\begin{aligned} 0=\frac{1}{r} \frac{\partial}{\partial r}\left(r \frac{\partial u_{\theta}}{\partial r}\right)-\frac{u_{\theta}}{r^{2}} \end{aligned}

\begin{aligned} c=\left(r \frac{\partial u_{\theta}}{\partial r}\right)-\int\frac{u_{\theta}}{r} dr \end{aligned}

Now it's clear that in order to solve it I have to integrate it twice, i.e., consider $$u_{\theta}$$ not a function of $$r$$. But physics bites me from inside. Can someone provide me a logic to this? Another thing is why $$u_{\theta} = r \omega$$ i.e., ( $$sin (\theta) = 1$$) not valid here?

• Is $u_{\theta}$ the theta component of velocity? – Buraian Oct 26 '20 at 18:03
• – Vishesh Mangla Oct 26 '20 at 18:04
• I have a feeling that your diagram is inaccurate because usually it looks like a parabola cavity once it starts rotating – Buraian Oct 26 '20 at 18:06
• I am not dismissing your problem but am suggesting edits for clarity's sake – Buraian Oct 26 '20 at 18:08
• Talking about the diagram, well how will you know that the profile is parabolic without actually performing the expt?(I know how to prove it parabolic). – Vishesh Mangla Oct 26 '20 at 18:11

Your fluid is roting as a solid body so $$v_\theta =r\omega$$. You can find the shape of the surface by observing that in the fluid is stationary in the rotating frame, and in that frame the potential energy is $$V(r,z)=\rho g z- \frac 12 \rho \omega^2 r^2,$$ a sum of the gravitational and centrifugal potentials. The surface must be an equipotential, so $$z(r)= \frac 1{2g} \omega^2 r^2,$$ which is a parabola of revolution.
Your original problem askes you to use the pressure. Euler (divided by $$\rho$$) tells us that $$\frac{\partial {\bf v}}{\partial t}+ \left({\bf v}\cdot \nabla \right) {\bf v}= - \frac 1 \rho \nabla P- {\bf g}$$ Now $${\bf v}=(-\omega y, \omega x, 0)$$ so $$\left({\bf v}\cdot \nabla \right) {\bf v}= -\frac 12 \omega^2 \nabla (x^2+y^2)$$ so $$\nabla\left(\frac P\rho - \frac 12\omega^2 r^2 +gz\right)=0.$$ Thus again $$P$$ is constant on $$z= \frac 1{2g}\omega^2 r^2.$$
• thanks for the answer but if you see it's a homework problem. I have to follow the instructions. On the other hand in the description, I have written about why $v_{\theta} = r \omega$ is wrong. I know how to do it by Newton's ways. Also an answer which I do not require is here too physics.stackexchange.com/questions/88340/… – Vishesh Mangla Oct 26 '20 at 19:09
• If $v$ were not $\rho \omega$ you are not rotating a s rigid body. Viscosity will come into play until it is rotating rigidly. – mike stone Oct 26 '20 at 19:11
• I would not have answered in so detailed a manner if I known it was an hw problem. But anyway: what is the "$\sin \theta$ tha ask about? – mike stone Oct 26 '20 at 19:13