How long does it take for a stirred cup of water to come to a complete stop? Given that the kinematic viscosity of water $\eta \approx 10^{-2}$ cm$^2$/s, and that the cup has a radius $R\approx 3$ cm, simple diffusion arguments give $t \sim R^2/\eta \sim 10^3 $s $\sim 15$ min, which is obviously not the case.
How would you improve on this estimate?
 A: This is a difficult problem indeed. First question that arises is does the body of liquid attain solid body rotation at all. As @Gert's answer shows, attainment of solid body rotation requires that the cup itself be rotated (for long enough). At first sight stirring by spoon may not seem to achieve the same effect. Below is a 2-dimensional figure in which a one dimensional spoon is driving a flow in concentric circles. If stirred long enough the fluid within radial distance $R_2$ will be set into solid body rotation. If the spoon moves near the edge of the cup (so that $R_2=R_3$) then the whole body of liquid will be set into solid body rotation.

Now the estimate of $R^2/\eta$ is good enough if diffusion were the only mechanism that dissipates momentum of the fluid. However this is not the case in a stopped solid body rotation. Recall that surface of the liquid attains a parabolic shape during solid body rotation. Therefore hydrostatic pressure is higher at greater radial distance, and is lowest on the axis of rotation. This is necessary because a fluid element at radial distance $r$ experiences centrifugal force per unit volume of $\rho r \omega^2$, which increases with $r$ necessitating larger pressure gradients with increasing $r$ to maintain equilibrium.
When the solid body rotation is stopped, motion begins to cease from the outer wall towards inner axis of the cup. As velocity of liquid in the outermost layers decreases, which results in a decrease of centrifugal force experienced by a fluid element located there, the pressure gradient it experiences (due to parabolic shape of the free surface) is in excess of what is required for equilibrium. This drives an inward flow which mixes momentum more efficiently than what diffusion alone could achieve. This is analogous to convection achieving greater heat transfer than diffusion alone.
If $u$ is the typical velocity scale for the aforementioned inward flow, then time scale for mixing of outer fluid with inner fluid is $R/u$. This is like having an effective diffusion coefficient of $\eta_{flow}\sim Ru$ due to the flow. Of course whether diffusion or flow decides the time in which the liquid comes to rest is determined by comparing $\eta_{flow}$ and $\eta$. If Reynolds number of the inward flow $\eta_{flow}/\eta\sim Ru/\eta\gg 1$, which is usually the case, then you may neglect molecular diffusion altogether. So the question is how to estimate velocity scale $u$ for inward flow given solid body rotation speed $\omega$.
Now $\Delta h=R^2\omega^2/(2g)$ is the height of the parabolic free surface from its center to the point where it meets the cup's wall, then a pressure difference of $\Delta p=\rho g\Delta h$ exists in the liquid between outer wall and rotation axis. If the inward flow were inertial to a good approximation (which is saying that pressure gradient is much stronger than viscous forces that act upon the inward flow) then $u\sim\sqrt{\Delta p/\rho}=\sqrt{g\Delta h}\sim R\omega$. Of course we could have arrived at this result more quickly by noting that if viscosity does not determine the inward flow then $u$ is determined by $R$ and $\omega$ alone and then matching dimensions. Thus for convection to dominate momentum transfer upon stopping of solid body rotation we need the Reynolds number $R^2\omega/\eta\gg 1$. If this condition is satisfied then the time scale for rotating liquid to come to rest is $R/u\sim 1/\omega$.
Now some numbers. $\eta\sim 10^{-2}$ cm$^2/$s is small enough that Reynolds number $R^2\omega/\eta\gg 1$ even for slow stirring speeds $\omega$, say 1 rotation in 10 seconds which gives $\omega=0.2\pi$ rad/s. For $R=3$ cm, Reynolds number is $\sim 100$. The time scale for complete stopping of the rotating fluid is $1/\omega\sim 2$ s. This number is only an estimate, the message being that the liquid stops rotating in a matter of few seconds rather than minutes.
A: Assume a vertically mounted tube, with a high length to radius ratio (this way we can neglect end effects), filled with water:

Now we start spinning the object about its central axis at an angular velocity $\omega$. If the liquid was completely inviscid no forces would be acting on it and it would remain stationary. 
But water isn't inviscid and immediately the difference in velocity between the inner surface of the tube and the outermost concentric layer of fluid causes a shear stress to arise and the outermost concentric layer starts to accelerate as a result of the torque (caused by the shear stress).
For a tube of radius $R$ and length $L$ ($L\gg R$), the net force acting on the thin outermost concentric layer of fluid (thickness $\delta r$) can be approximated as, from the shear stress:
$$F\approx2\pi RL\eta \frac{\omega R}{\delta r}$$
Its torque causes the layer to start rotating.
Similarly, the velocity difference between the outermost concentric layer and the "next" concentric layer causes shear stress and torque acting on the latter, causing also acceleration. At some time $t$ the tangential velocities of the concentric layers will be a function of both time and distance from the centre line:
$$v(r,t)$$
Given enough time, all of the fluid will rotate at the same angular velocity, i.e. $\omega$ and:
$$v(r,\infty)=\omega r$$

Now we reset the clock and at $t=0$ we abruptly stop spinning the tube. Inertia means the rotation of the liquid will continue but as in the above case, shear stresses will cause torques on the various layers, 'diffusing' from the outside to the inside. Here the shear stresses point in the opposite direction as above and cause deceleration. Here too a velocity distribution establishes itself:
$$v(r,t)$$
The force acting on a concentric layer at distance $r$ is given by:
$$F(r,t)=2\pi rL\eta \frac{\partial v(r,t)}{\partial r}$$

Although I hope the above provides some useful insights, I've not been able to derive an expression for $v(r,t)$.
One way forward maybe to only consider the force acting on the outermost concentric layer and not the shear stresses between the inner layers but I think that would simply be too approximate.
A: As far as I understand the problem and as far I understand the use of order of magnitude estimates, your calculated 15 [min] should apply only for a some kind of "Taylor-Couette" (without inner cylinder in this case) flow to stop only by means of viscosity and not by stirring it. In other words: if you don´t stir the fluid, but rotate the cup, it will take  ~15 [min] to generate a solid body rotation of the fluid. Once you stop rotating the cup, it should take other ~15 [min] for the fluid to stop by it own means (viscosity). 
Stirring is a convective phenomena, not a diffusive one. 
