I've been stuck on what I'm pretty sure is a simple part of a larger question. It's a cylinder (radius a) spinning in a viscous fluid. It's rotating at rate $\Omega$ .During this question we get that the motion of a uniform axisymetric flow is:

$$\displaystyle\frac{\partial u_\theta}{\partial t} = \nu \left ( \frac{\partial ^2u_\theta}{\partial r^2} + \frac{1}{r} \frac{\partial u_\theta}{ \partial r} - \frac{u_\theta}{r^2} \right )$$

We are told that the specific axial angular momentum is given by:

$$m = ru_\theta$$

Solving for steady state I then find that:

$$\displaystyle u_\theta = \frac{\Omega a^2}{r}$$

I'm then asked to "hence" obtain an expression for the viscous torque (per unit length in the axial direction) on this cylinder neglecting end effects and assuming the flow to be axismmetric (in terms of $\nu$, $\Omega$, $\rho$ and $a$.

I can see physically that there must be a toque driving this. I just can't for the life of me figure out what I'm meant to do in the next step (as by being in "steady state" there's no acceleration as such).

If I've done anything horribly wrong or you can see the next step help would be appreciated.


closed as off-topic by Kyle Kanos, ACuriousMind, Gert, CuriousOne, Daniel Griscom Mar 19 '16 at 13:57

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Kyle Kanos, ACuriousMind, Gert, CuriousOne, Daniel Griscom
If this question can be reworded to fit the rules in the help center, please edit the question.


Torque is change in angular momentum, $\frac{dm}{dt}$.

Solve this by substituting for your flow equation (consider the derivatives of the steady state solution).

  • $\begingroup$ My problem is that I found the steady state solution by setting $\frac{du_\theta}{dt} = 0$. Was that wrong? Because if I differentiate m I get either: $dm = du + dr*u$ (and don't know what to do with dr) or dr= 0 where du/dt = 0... $\endgroup$ – Pluckerpluck May 30 '14 at 12:18
  • $\begingroup$ Basically, because I have steady state I'm getting dm/dt=0. This is why I'm confused and unable to get an answer. $\endgroup$ – Pluckerpluck May 30 '14 at 12:24
  • $\begingroup$ You're quite right. I think I missed a critical factor. $u$ is a function of $r$. Therefore to get the total torque you must integrate over $r$ (all space). Then sub in for your value of $u$ and evaluate at the limits. Hopefully this won't also equal zero :). $\endgroup$ – nivag May 30 '14 at 13:42
  • 1
    $\begingroup$ Btw, I've discovered the problem. Being steady state an integral over all space will always equal 0. In other words, over the entire system there is no overall torque (because the cylinder is begin driven). As such you have to relate it to the force due to a difference in velocities. I'll post the actual answer up in a bit (to help me gain a full understanding myself). Just wanted to explain why this isn't a correct answer (why I'm not marking it as correct). $\endgroup$ – Pluckerpluck May 30 '14 at 15:01

Not the answer you're looking for? Browse other questions tagged or ask your own question.