How do I intuit viscosity in a rotating fluid?

I think the right thing here is to note that we are dealing here not with some gradient in the sense of vector (covetor, 1-form or whatever) but with a deformation tensor, some higher order object. The flaw in the reasoning that the proper cause of viscosity is not some gradient, but shear strain rate tensor.

Roughly speaking in your both cases you take a single coordinate of a tensor and call it "gradient". However just a single coordinate has no physical meaning. Thus it is illegal even to reason about a single coordinate. With parallel plates we we lucky to take the only non-zero tensor coordinate which represents the whole tensor, but with cylinders we forgot the others.

The proper thermodynamic force which causes dissipation due to shear deformations in Cartesian coordinates is:

$$V_{ij} = frac{1}{2} left[ frac{partial u_i}{partial x_j} + frac{partial u_j}{partial x_i}right] - frac{1}{3} frac{partial u_k}{partial x_k}, delta_{ij}$$

the last term is $operatorname{div} boldsymbol v$, here double $k$ means summation. You can easily see that there are other non-zero components apart from the .

The shear stress tensor is $$P_{ij} = mu V_{ij},$$ the dissipation is proportional to $$P_{ij} V_{ij}.$$

For those who wonders why was $operatorname{div} boldsymbol v$ subtracted --- that's because it is another thermodynamic force and it is corresponded by another coefficient rather than $mu$.

Let's look at the stationary solutions in 2D for incompressible fluids. I'm using the Wiki Couette flow page as a basis. Let $boldsymbol{u} = (u,v)$ be the velocity field. The stationary Navier-Stokes equations are given by $$ rho (boldsymbol{u}cdotnabla)boldsymbol{u} = -nabla p +mu Delta boldsymbol{u},\ nabla cdot boldsymbol{u}=0.$$

In the case between two flat plates, I plant my reference frame with the $x$-axis parallel to the the channel, right in the middle. The $y$-axis is then perpendicular to the channel (of width $L=1$), the plates are at $y=0$ and $y=1$. The symmetries of the problem lead me to suppose $v=0~$ and $partial_x u = 0$. This way, the incompressibility condition is automatically satisfied. Then we are left with (after simplifications), $$ 0 = rho(upartial_xu + vpartial_yu)= -partial_x p + mu partial_y^2 u.$$ Suppose then that we prescribe the pressure gradient as a constant, say $partial_x p = delta P$. The solution respecting your boundary conditions is $$ u(x,y) = frac{delta P}{2mu} y^2 + frac{1}{L}(v_1-v_0- frac{delta P L^2}{2mu}) y + v_0.$$ If $delta P = 0$, the solution is $$ u(x,y) =(v_1-v_0) y + v_0.$$ The viscous damping at the top plate per unit area is given by $$ sigma_mu cdot boldsymbol{n} = mubegin{pmatrix}2partial_x u & partial_y u + partial_x v\ partial_y u + partial_x v & 2partial_y vend{pmatrix}begin{pmatrix}-1\0end{pmatrix} \ =mubegin{pmatrix}0 & partial_y u \ partial_y u & 0end{pmatrix}begin{pmatrix}-1\0end{pmatrix}= begin{pmatrix}- frac{delta P}{2} - frac{1}{L}(v_1-v_0- frac{delta P L^2}{2})end{pmatrix}$$ Your intuition is thus correct: the plates experience a viscous damping force.

In the concentric shells scenario, we'll use cylindrical coordinates $(r,theta)$. Symmetries of the problem leads to $u_r = 0, partial_theta u_theta = 0$ and $partial_theta p =0$. Writing the Navier-Stokes equations in cylindrical coordinates and simplifying with the symmetry constraints we get $$ rho u_theta^2 = r partial_r p,\ 0 = mu(rpartial_r (rpartial_r u_theta)-u_theta)$$ The solution to the second equation is of the form $u_theta = C_1cdot r + frac{C_2}{r}$, and the boundary conditions you chose (i.e., the two shells have the same angular velocity) are compatible with that if $C_2=0$ and $C_1$ is the angular velocity of the shells. The viscous damping at the outer shell per unit area is given by (from Batchelor's Introduction to Fluid Dynamics) $$ sigma_mu cdot boldsymbol{n} = mu begin{pmatrix}2partial_r u_r & rpartial_r(frac{u_theta}{r}) + frac{1}{r}partial_theta u_r\ rpartial_r(frac{u_theta}{r}) + frac{1}{r}partial_theta u_r & frac{2}{r}partial_theta u_theta + frac{u_r}{r}end{pmatrix}begin{pmatrix}-1\0end{pmatrix} \ = mu begin{pmatrix}0 & rpartial_r(frac{u_theta}{r})\ rpartial_r(frac{u_theta}{r}) & 0end{pmatrix}begin{pmatrix}-1\0end{pmatrix} = begin{pmatrix}0\-mu rpartial_r(frac{u_theta}{r})end{pmatrix}$$ Indeed, if $C_2=0$, then there is no viscous damping. However, this is tied to the fact that both shells have the same angular velocity, otherwise, you get a non-zero force damping force.

As such, zhermes's intuition seems sound to me: in your particular case, the fluid indeed rotates along with the shells in a "rigid body motion" as you say. The velocity gradient in the $y$-axis is non-zero between the two plates, but your choice of boundary conditions in the case of the two shells is precisely the one that makes the gradient along the radius of the shells vanish.

I think the answer has to do with the velocity gradient in the radial direction being zero in the rotating frame (i.e. consider a Lagrangian fluid element). Conceptually, it should be clear that two radially separated parcels of fluid have no relative motion in the cylindrical case, while they do in the planar case. In other words, there is no gradient in the angular velocity between cylinders.