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,
It depends on the friction of the contact. With a frictionless plane the top would precess around its center of gravity and the contact point will prescribe a circle.
Add friction, and the friction force translates the center of gravity the same way tire traction translates a car. Here you have the cases of a) pure rolling, or b) rolling with slipping.
With pure rolling the motion is similar to a spinning coin rolling on its edge, or a spinning glass which precesses in a circle smaller than the contact ball radius.
With slipping there isn't enough force for such a tight circle, so the precession yields wide circles that become progressively smaller and smaller.
Best Answer
The torque that rotates a top upright, as happens in that youtube video, is due to sliding friction between the top and its supporting surface.
Crucial to this effect is the fact that the top in that youtube video has a rounded bottom, instead of coming to a sharp point at the bottom like some tops do. The effect is more pronounced and dramatic in tops that have a larger radius of curvature on their bottom, such as in the extreme case of a tippe top, which has a radius of curvature so large that it's possible for the top's center of mass to be at a height that's smaller than the radius of curvature. Indeed, the papers I've seen that show how sliding friction causes a top's center of mass to rise are specifically doing an analysis of a tippe top.
The analysis of a top in full generality, including the effects of friction, is quite complicated. To simplify the analysis enormously, I'll just look at the top at an instant in time at which the top has no linear momentum, and has a very large angular momentum that lies precisely along the top's axis of symmetry.
I'll also consider gravity to be negligible in this simple explanation. Gravity causes a purely horizontal torque on the top, but we're only interested in torque that has a vertical component, which will cause the top to become increasingly upright. In reality, if it weren’t for gravity holding the top and the table together, there wouldn’t be any sliding friction at the point of contact between the two, but we will simply assume that the sliding friction exists, without considering how the sliding friction is related to gravity.
The diagram above shows a vertical cross section through the top, that contains the top's axis of symmetry. The point $P$ lies on the axis of symmetry, as does the top's center of mass $O$. The top's angular momentum $\vec{L}$ points in the direction of the axis of symmetry.
Because the top has a rounded bottom instead of a pointed bottom, the top's point of contact isn't at $P$, but rather at some point $C$. From the assumptions stated above, at the instant of interest $P$ is stationary. In contrast, from the direction of $\vec{L}$, at $C$ the surface of the top is moving towards the viewer, straight up out of the plane of the diagram. The sliding friction is a force $\vec{F}_k$ (not shown) on the top at $C$, in the direction opposite to the top's motion at that point, i.e., straight down away from the viewer.
The position vector of $C$ from $O$ is $\vec{X}_C$. The force $\vec{F}_k$ on the top produces a torque on the top around the top’s center of mass,
$$\vec{\tau} = \vec{X}_C \times \vec{F}_k \,\, .$$
The torque $\vec{\tau}$ can be written as
$$\vec{\tau}=\vec{\tau}_{\parallel}+\vec{\tau}_{\perp} \,\, ,$$
where $\vec{\tau}_{\parallel}$ is parallel to $\vec{L}$, and $\vec{\tau}_{\perp}$ is perpendicular to $\vec{L}$.
The torque $\vec{\tau}$ is how the top's angular momentum $\vec{L}$ changes with time,
$$\frac{d\vec{L}}{dt} = \vec{\tau}=\vec{\tau}_{\parallel}+\vec{\tau}_{\perp} \,\, .$$
$\vec{\tau}_{\parallel}$ points in the opposite direction as $\vec{L}$, so the effect of $\vec{\tau}_{\parallel}$ is to reduce the magnitude of $\vec{L}$, i.e., to slow the top down.
If the top was in empty space, the effect of $\vec{\tau}_{\perp}$ would be to rotate the top around $O$ clockwise in the diagram. However, due to the constraint that the top remains in contact with the table, the effect of $\vec{\tau}_{\perp}$ is instead to raise $O$ away from the table, and to make $O$ closer to being above $C$.
For a much more detailed analysis of how sliding friction on a top's bottom causes the top's center of mass to rise, see pretty much any paper on the tippe top, such as this one.