This is a rather old topic, but I felt I might have what you're looking for.
In response to some of the answers, you write:
Since the angular acceleration is always tangential, I would expect that the top should spiral outwards until it falls to the ground.
Absolutely, that is what you should expect to happen. And it does ... momentarily. The final solution is a little more involved than just being uniform rotation around the vertical axis.
In order to understand this, imagine that you take a spinning top which you have just set down at time $t = t_0 $ on the ground. Now, what happens in the next instant is exactly what you intuitively expect - the top begins to fall under gravity's influence and $ \phi $ (see figure for notation) starts to increase going from $ \phi \rightarrow \phi + \delta \phi $ at time $t_1$. Consequently the angular momentum $ \mathbf{L} $ of the top changes.
This is similar to what happens in the 2nd figure on the hyperphysics page, where $\delta \mathbf{L}$ is in the direction of $ \delta \theta $, only now $ \delta \mathbf{L} $ in the direction $ \delta \phi $ and lies in the plane containing the longitudinal axis $L_A$ of the top and the central vertical axis $V_A$.
Increasing $\phi$ lowers the center of mass of the top and thus its potential energy by an amount $ -\delta U $. Assuming energy conservation, this translates to an increase in the kinetic energy $\delta K$ of the top. Since the top is constrained to have zero linear momentum, this $ \delta K$ contributes entirely to the top's rotational energy.
Keep in mind, however, that the top is now rotating around two different axes. One component is the original spinning motion around its own longitudinal axes and the other is the rotation induced by gravity around the direction $N_A$ normal to the plane containing $L_A$ and $V_A$. Therefore, the $\delta K$ must be appropriately portioned between these two motions. Let's see how this happens.
The moment of inertia of the top ($I_A$) around the axis $L_A$ is clearly less than that ($I_V$) around the axis $N_A$. This is true for all but the most oddly shaped tops. Convince yourself that this is the case. In circuits more current flows through paths with lower resistance. Likewise in mechanics more energy is transferred to the component with lesser inertia. Thus the greater portion of $\delta K$ will go to increasing the angular momentum of the top around its longitudinal axis $L_A$ by some amount $\delta L'$
Now, conservation of angular momentum requires that there be a torque corresponding to this increase. The effect of this induced torque is to cause the falling top to start swinging back upwards. In this way, instead of a spiral, the tip of the top traces out something like a cycloid as it precesses around the central axis.
However, the diagram seems to indicate that the top should be precessing in a circle, not a spiral.
The circular trajectory is an idealization only achieved in the limit that $\omega_s / \omega_p \rightarrow \infty$, where $\omega_s$ is the spin angular velocity and $\omega_p$ is the precession angular velocity. Any top with realistic values of $\omega_s$ and $\omega_p$ will have finite "wobble".
I would not have known of this rather elaborate dynamics if not for one of Feynman's lecture volumes (Part I, I think) where this question is considered in great detail!
The above write-up is a little on the hand-wavy side and there probably are errors in my reasoning. For the full kahuna look up the Feynman lectures !
Cheers,
