Micropolar fluids are fluids with microstructures. They
belong to a class of fluids with a nonsymmetric stress tensor.
Micropolar fluids consist of rigid, randomly oriented
or spherical particles with their own spins and microrotations,
suspended in a viscous medium.
Physical examples of micropolar fluids can be seen in
ferrofluids, blood flows, bubbly liquids, liquid
crystals, and so on, all of them containing intrinsic polarities.
The following (including notations) is based on textbook
G. Lukaszewicz, Micropolar Fluids: Theory and Applications, Birkhauser, Boston, 1999
http://books.google.ru/books?id=T3l9cGfR9o8C
We start with Cauchy momentum equation
$$
\rho \frac{D\vec{v}}{Dt} = \rho \vec{f} + \nabla \cdot \hat{T},
$$
where $\vec{f}$ is body force and $\hat{T}$ is stress tensor.
If we assume that fluid is polar, that is it has its own internal angular momentum (independent of the motion of fluid as a whole) then we need an additional
equation expressing conservation of angular momentum (for nonpolar fluid conservation of angular momentum is a consequence of Cauchy equation):
$$
\rho \frac{D}{Dt}(\vec{l} + \vec{x}\times\vec{v}) = \rho \vec{x} \times \vec{f} + \rho\vec{g} + \nabla \cdot (\vec{x} \times \hat{T} + \hat{C}).
$$
Here $\vec{l}$ is an intrinsic (internal) angular momentum per unit mass, $\vec{g}$ is body torque and $\hat{C}$ is a new object called couple stress tensor
(this equation could be seen as its definition).
Now in order to close this system of equation we need to express the stress tensor and couple stress tensor through the characteristics of fluid dynamics.
For this we need to make certain assumptions: the absence of preferred direction and position, reduction to hydrostatic pressure in the absence of deformations and
linear dependence on the velocity spatial derivatives $v_{i,j}$ (deformation). For polar fluid we also define the vector field $\vec{\omega}$ -- microrotation
which represents the angular velocity of rotation of particles of the fluid. We further assume that the fluid is isotropic and $\vec{l} = I \,\vec{\omega}$ with $I$ a scalar called the microinertia coefficient. Couple stress tensor should be a linear function of the spatial derivatives of the microrotation field: $\omega_{i,j}$. All this assumptions allow us to specify the general form for stress tensor:
$$
T_{ij}=(- p +\lambda v_{k,k})\,\delta_{ij}+\mu\,(v_{i,j}+v_{j,i})+\mu_r\,(v_{j,i}-v_{i,j}) - 2 \mu_r\, \epsilon_{mij}\omega_m,
$$
and couple stress tensor:
$$
C_{ij} = c_0\, \omega_{k,k} \delta_{ij} + c_d\, (\omega_{i,j}+\omega_{j,i}) + c_a \,(\omega_{j,i}-\omega_{i,j}).
$$
Note: we have three parameters $c_0$, $c_d$ and $c_a$ (called coefficients of angular
viscosities) because there are three irreducible representations for the action of $SO(3)$ group on rank 2 tensor: scalar (times $\delta_{ij}$), traceless symmetric tensor, antisymmetric tensor. The same for the triplet $(\lambda,\mu, \mu_r)$ (which are called second viscosity coefficient, dynamic
Newtonian viscosity and dynamic microrotation viscosity).
Substituting this expressions for $T_{ij}$ and $C_{ij}$ into Cauchy equation and angular momentum equation we obtain the equations from the question.
The notations correspondence is: $$ \vec{\omega} \to \vec{N}^{*}, \qquad \mu_r \to k_1^{*}, \qquad I\to j^{*} \qquad c_a+c_d \to \gamma^{*}.$$
Additionally we see that the equations in the question assume $\mathrm{div} \vec{v} =0 $ (incompressible flow) and $\mathrm{div} \vec{\omega} =0 $ -- this
one is generally not true, but symmetries of the system could make it so. Also we see that there are no body torque and the body force term corresponds to Lorentz force.
I agree with user3823992 that it was incorrect to neglect the pressure differential. With the steady sinusoidal body force that's given, it's basically a hydrostatics problem with the pressure differential balancing the body force. Consider the Navier-Stokes momentum equation:
$$
\frac{\partial \mathbf{v}}{\partial t}+(\mathbf{v} \cdot \nabla)\mathbf{v}=-\frac{\nabla p}{\rho}+\mathbf{f} +\nu \nabla^2 \mathbf{v}
$$
If we assume the velocity $\mathbf{v}$ is zero then it reduces to the hydrostatic case, where:
$$
\frac{\nabla p}{\rho}=\mathbf{f}
$$
$\mathbf{f}$ only has a component in the s-direction, therefore so will $\nabla p$:
$$
\begin{eqnarray*}
\frac{dp}{ds} \cdot \hat s&=&\rho q \sin(s) \cdot \hat s\\
\int_{p_0}^pdp&=&\rho q \int_0^s \sin(s) ds\\
p-p_0&=&\rho q (1- \cos (s))
\end{eqnarray*}
$$
($p_0$ is just an arbitrary reference pressure that may have been present before the force was applied).
$\mathbf{v}=0$ obviously satisfies the continuity equation, although I think any other solution with a constant $\mathbf{v}$ would also satisfy the equation. This would just be bulk fluid motion that was present before the force was applied, and would tend to zero in steady-state if viscous drag on the walls is included.
In the case where the tube is closed like a torus, the flow is still governed by the Navier-Stokes equations. The momentum equation in polar coordinates (r, $\theta$, z) can be reduced to:
$$
\begin{eqnarray*}
\theta: f_\theta&=&-\nu (\frac{1}{r} \frac{\partial}{\partial r}(r \frac{\partial V_\theta}{\partial r})+\frac{\partial^2 V_\theta}{\partial z^2}-\frac{V_\theta}{r^2})\\
r: \frac{\partial p}{\partial r}&=&\rho \frac{V^2_\theta}{r}
\end{eqnarray*}
$$
The only component of velocity is in the $\theta$ (circumferential) direction. The first line is the body force balanced by the wall friction and the $\frac{\partial p}{\partial r}$ in the second line is necessary to provide the centripetal force for the curved streamlines. However, this is now a 2-dimensional PDE and I think it's pretty unlikely that you'd be able to integrate or find a simple function to satisfy it - to find the velocity profile you would probably have to resort to CFD at this point.
The Poiseuille equation has a nice, straightforward solution because it is axisymmetric and effectively 1-dimensional.
Best Answer
The equation, $$ \nabla\cdot (\rho \textbf v \otimes \textbf v), $$ can be written in index notation as, $$ \partial_i (\rho v_i v_j), $$ where the dot product becomes an inner product, summing over two indices, $$ \textbf a \cdot \textbf b = a_i b_i, $$ and the tensor product yields an object with two indices, making it a matrix, $$ \textbf c \otimes \textbf d = c_i d_j =: M_{ij}. $$ Now we differentiate using the product rule, $$ \partial_i (\rho v_i v_j)=(\partial_i \rho) v_i v_j + \rho (\partial_i v_i) v_j + \rho v_i (\partial_i v_j). $$ Let’s look at the terms separately:
$\bullet (\partial_i \rho) v_i v_j $: assuming $\rho=\rho(\textbf x)$, the expression within the brackets is the vector $(\partial_x\rho, \partial_y\rho, \partial_z\rho)$, which then gets dot multiplied with the vector $\textbf v$. This yields a number, say $c_1$, which gets multiplied to every component of the vector $v_j$. So the result here is a vector. If $\rho$ is constant, this term vanishes.
$\bullet\rho (\partial_i v_i) v_j$: Here we calculate the divergence of $\textbf v$, $$ \partial_i a_i = \nabla \cdot \textbf a = \text{div }\textbf a, $$ and multiply this number with $\rho$, yielding another number, say $c_2$. This gets multiplied onto every component of $v_j$. The resulting thing here is again a vector.
$\bullet\rho v_i (\partial_i v_j)$: Here we construct a matrix with the composition rule, $$ M_{ij} := \partial_i v_j, $$ that is for example $M_{13}=\partial_x v_z$. We then multiply a (row)vector $v_i$ to this matrix, yielding a different vector. Finally, every component of this new vector gets multiplied by $\rho$, so we have a vector again.