Solid-body rotation of fluid in polar coordinates: How to compute the stress tensor In a course on continuum mechanics, we are given an exercise concerning solid-body rotation of a fluid in polar coordinates. In the first parts (feel free to correct any errors here) we are tasked with finding the velocity field (which we have simply stated to be $\vec v=\begin{bmatrix}v_r \\ v_\theta \end{bmatrix}=\begin{bmatrix}0 \\\omega_0 r \end{bmatrix}$ where $\omega_0$ is the angular velocity), and find out if the solid-body rotation features shear strain (no, since the strain rate tensor is zero), expansion (no, same reason) or rotation (yes, this is given by the antisymmetric part of $\nabla\vec v$ which in this case is identically equal to $\nabla\vec v$).
The next part of the exercise would have us compute the stress tensor for this flow, and this is where we're stuck. How could we go about doing this? We are not given any specific details about the fluid in question, so we don't know if we could just use the constitutive relations for e.g. a Newtonian fluid, $\sigma_{ij}=-p\delta_{ij}+2\mu d_{ij}$.
(The last part of the exercise asks if this flow satisfies the Navier-Stokes or Euler equations, and we're also stuck here, feel free to help out.)
 A: You need to solve the equilibrium equations as well. 
$$
\rho\frac{D\mathbf{v}}{Dt} = div(\mathbf{\sigma}) + \mathbf{b}
$$
For an incompressible Newtonian fluid, these are the Navier-Stokes equations. Fortunately, we can make MAJOR simplifications. N-S equations are derived from this balance of linear momentum with the appropriate constitutive law for stress, so any solution that satisfies one, satisfies the other. 
As you noted, the strain rate tensor $\mathbf{D}$ is zero, so there are no contributions from the viscous terms. Second, although there are nonzero components of the velocity gradient, the convective part of the material time derivative is still zero since the velocity is zero in the $r$ direction.
At this point, the reduced equations are (neglecting gravity since it's not specified in problem):
$$
\rho \frac{d \mathbf{v}}{d t} = -\nabla P
$$
Next, notice that the accelerations in the $r$ direction are not zero due to centripetal acceleration. Rather, they are
$$
\left(\frac{d \mathbf{v}}{d t} \right)_r = 
-\frac{v_\theta^2}{r}
$$
Substituting this into the problem and using your velocity field gives
$$
\rho \frac{v_\theta^2}{r} = \rho \omega^2r = \frac{dP}{dr}
$$
Integrating this will give you the pressure field that is quadratic in $r$ plus an integration constant.
$$
P(r) = \frac{\rho\omega^2r^2}{2} + C
$$
Note: Extending this analysis to 3D with gravity would give the well-known parabolic free surface solution.
