# Force exerted by rotating bar

Say we have a bar immersed in liquid, and that the bar in being rotated at a given constant angular speed. What force is exerted on the liquid by the faces of the bar moving through the liquid (so the force is tangent to the circular motion of the bar). My guess is that the expression involves the distance from the centre of the bar, $r$, and the angular velocity $\omega$. But I have no clue what exactly the relationship should be; can anybody help?

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In general, I think this is a very hard question! It has to do with how the fluid flows around the bar, and fluid flow is one of the most complicated subjects in physics. – Ted Bunn Jul 28 '11 at 20:26
If you just want the tangential force due to the bar, it should be given by $r \omega$. The rest of the forces that account for the actual motion should be due to the behavior of the fluid. No? – knucklebumpler Jul 28 '11 at 22:49
@knucklebumpler that's along the lines of what I was thinking. I'm working on a computational model, and the fluid flow is solved for numerically, but I need this additional force term. – Arpon Jul 29 '11 at 0:11

First of the sum of the forces on the bar have to be zero because the center of gravity does not move (assuming symmetry and such). The only thing you can calculate is the torque requited to maintain the constant speed. I can give an example with a rectangular bar.

Take the bar and split it into infinitesimal slices. Each slice has face area of ${\rm d}A = h\,{\rm d}r$ where $r$ is the distance from the center and varies from $-L/2$ to $L/2$, and $h$ is the height of the bar.

The liquid-solid force for each slice is ${\rm d}F =\frac{1}{2} \rho\; C_d v^2 {\rm d}A$ where $\rho$ is fluid density, $C_d$ is the coefficient of drag and $v$ is the velocity of the bar slice, which is equal to $v = \omega\;r$. Note that the force has to flip signs when $r$ flips signs and thus we have to add a ${\rm sign}(r)$ term and integrate over $r$ to get the total force

$$F = \int_{-L/2}^{L/2}\;{\rm sign}(r)\frac{1}{2}\rho\;h\;C_d\;(\omega\,r)^2\;{\rm d}r = 0$$

and to total torque

$$M = \int_{-L/2}^{L/2}\;{r\;\rm sign}(r)\frac{1}{2}\rho\;h\;C_d\;(\omega\,r)^2\;{\rm d}r = \frac{\rho C_d h \omega^2 L^4}{64}$$

The main assumption here is that the $C_d$ does not depend on the flow velocity $v=\omega\,r$ which it does. A more extensive analysis would require finding the Reynolds number and the Fanning Friction factor and deriving the pressure distribution along the bar based on the flow characteristics.

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""The main assumption here is that the Cd does not depend on the flow velocity v=ωr which it does."" And without that Your "formula" is rather irrelevant. – Georg Jul 29 '11 at 15:14
Thanks for the answer. It seems that my question was rather vague, sorry about that. So I'm looking for the force at every point (well, a discrete set of points) along the bar. Say I do know the Reynolds number, and I know about the fluid flow. Could you point me to some more details about deriving the pressure distribution? Exactly what flow characteristics are involved and such. – Arpon Jul 29 '11 at 15:54
@George, Cd usually varies slowly with velocity, so this is a reasonable first approximation. – user1631 Jul 29 '11 at 17:30
@Arpon - the answer is right there. The pressure as a function of radial position $r$ is $P(r) =\rm{sign}(r) \frac{1}{2}\rho C_d \omega^2 r^2$ – ja72 Jul 31 '11 at 4:02

The force of the bar on the liquid is the opposite of the friction force of the liquid on the bar. Except for the part close to the center (which doesn't very fast at all) or the end (hard!) you can probably treat the fluid flow as a regular orthogonal flow past every part of the bar, at speed rω. (If the fluid is non-Newtonian, you would get shear stresses in the fluid and you can't treat each part of the bar independently).

The exact forces will depend on the Reynolds number, which will be hard to predict without further information. However, the general relation is that the forces double with the square of the speed, i.e. with rω2

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Do You think the comment of Ted Bunn to the question is wrong? The most easy part is finding the Reynolds number, the things after that are much more complicated than You seem to know. – Georg Jul 29 '11 at 10:05
The question is really underspecified. For a simple case (diameter of bar significantly smaller than its length; newtonian fluid, no wall effects) you can get a reasonably accurate solution. I can easily see how this would have been a homework question when I studied fluid mechanics. Fluid mechanics only become hard when such simplifications aren't possible. – MSalters Jul 29 '11 at 13:05
Right! In such a case comment on the underspecification instead of giving an uncorrect answer. – Georg Jul 29 '11 at 13:13
Well, that's saying that a Newtonian answer is always the incorrect answer because it ignores quantum effects. Physics really is just a more or less crude approximation of reality, and in engineering one selects an approximation not on theoretical correctness but on practical applicability. – MSalters Jul 29 '11 at 13:21
No, thats saying that a oversimplified answer is misleading! The question was not about potential flow. – Georg Jul 29 '11 at 13:26