Torque on a rotational cylinder in viscous fluid [closed]

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 GriscomMar 19 '16 at 13:57

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

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Torque is change in angular momentum, $\frac{dm}{dt}$.
• 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... – Pluckerpluck May 30 '14 at 12:18
• 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 :). – nivag May 30 '14 at 13:42