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?
 A: 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.
A: 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
