Why don't spinning tops fall over? One topic which was covered in university, but which I never understood, is how a spinning top "magically" resists the force of gravity. The conservation of energy explanations make sense, but I don't believe that they provide as much insight as a mechanical explanation would.
The hyperphysics link Cedric provided looks similar to a diagram that I saw in my physics textbook. This diagram illustrates precession nicely, but doesn't explain why the top doesn't fall. Since the angular acceleration is always tangential, I would expect that the top should spiral outwards until it falls to the ground. However, the diagram seems to indicate that the top should be precessing in a circle, not a spiral. Another reason I am not satisfied with this explanation is that the calculation is apparently limited to situations where: "the spin angular velocity $\omega$ is much greater than the precession angular velocity $\omega_P$". The calculation gives no explanation of why this is not the case.
 A: When it is spinning its angular momentum is quite high. By conservation of angular momentum the spinning top is then more stable against small torques like the action of gravity on the top.
The angular momentum of the top is $J = I \omega$ where $I$ is the inertia tensor and $\omega$ is the Darboux vector, whose magnitude is proportional to the rotational speed.
You can find a detailed discussion on this page of Hyperphysics.
A: The point is that conservation principles are not generally intuitive. For example, why should energy be conserved? One must have a grip of the dynamics involved in order to understand them. 
Anyway, the precession of the spinning top doesn't have to do with the conservation of angular momentum. It has to do with the strange nature of torque and its interaction with angular momentum. When a force acts on a spinning top, it excerpts a torque perpendicular to the plane defined by the axis of the top and the direction of gravity, which is a vertical plane. That direction is horizontal. 
On the other hand, the torque is the rate of change of angular momentum. That means that the direction of the torque is the direction towards which the vector of angular momentum changes. Thus, since the torque is horizontal and perpendicular to the angular momentum, it can only change the direction of angular momentum along the horizontal direction and not towards the ground. That means that the vector of angular momentum has its back on the ground, at the point that the tip of the top touches the ground, and its head is performing a circle on a plane that is parallel to the ground. That motion is the precession of the spinning top.   
Finally, I think that the reason for assuming a much faster rotation than precession for the top, is to simplify the calculations and consider the top as a gyroscope. 
A: 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,

A: when the mass is spinning it has an angular momentum pointing in a direction perpendicular to the plane it's spinning on.
The angular momentum has to be conservate: i.e. has to keep pointing in the plane-perpendicular direction. As cedric said, the gravity, works for the axis of the spinng mass to fall horizontally on the plane: if this happens also the angular momentum as to torque! and this is not convenient from an energy conservation point of view..
Then u can consider that the magnitude of the angular momentum is proportional to the spinning speed: so as the spinning velocity gets higher it gets, for lack of a better word, "easier" for the top to resist the gravity..
If u try to spin a top on an inclined plane you will need to spin it faster to obtain the same "resistance to gravity"!
A: The quick answer is that, for the top to fall over due to gravity, each fragment of the top that is moving around the spin axis has to change its individual direction of movement.
They are already changing direction around the spin axis, due the rigidity of the top keeping them moving in a circle. But gravity is operating at 90º to their direction of movement, and its effect depends on the velocity or inertia of the fragment. For a fast rotating top, this slight change of direction is what causes the top’s precession. And as it slows down, the effect of gravity has more effect, and it falls over.
It’s a similar situation to changing the orbit of a satellite with side thrusters.
A: From your linked article:

Spin a top on a flat surface, and you will see its top end slowly
  revolve about the vertical direction, a process called precession. As
  the spin of the top slows, you will see this precession get faster and
  faster. It then begins to bob up and down as it precesses, and finally
  falls over.

The drawing shows a circle instead of a spiral due to leaving out variables like friction and gravity.
A: This is a nice example which shows understanding does not come automatically after completing a calculation. But calculation still serves the (perhaps the most) important guide. Nobody in the above has mentioned the discussions given in \ittext{Landau & Lifshitz, Mechanics (BH, 3rd ed.), page 112}. I think these discussions have already elucidated the issue. Unfortunately, they proceeded using Euler angles. I have reformulated their discussions here. Hope that helps. 
