I understand how spinning tops don't tip over, cf. e.g. this and this Phys.SE questions. What I'm more interested is in identifying the factors that determine the direction the spinning top moves to?
[Physics] Physics of the point of contact for a spinning top
angular momentumgyroscopesnewtonian-mechanicsrotational-dynamics
Related Solutions
Given a nonzero moment of inertia $I$, the transverse extension of the top from the rod has to be at least $\sqrt{I/M}$ because we can't have negative mass distributions. Because of this finite transverse extension, a spin-orbit cross coupling between the orbital angular momentum $L$ and the spin $S$ will have to exist. But this by itself isn't sufficient to transfer spin into orbital angular momentum.
The crucial question to ask is, is the distribution of mass of the rod rotationally symmetric about the rod axis? If yes, the kinetic and gravitational potential energy doesn't depend upon the orientation angle $\alpha$ of the top about the rod axis. However, if there are slight rotational asymmetries, we will have an $\alpha$ dependence leading to wobbles. These wobbles transfer spin into orbital angular momentum, and if sufficient angular momentum is transfered, the top will topple.
Well, the angular momentum conservation is still the essence although it may be formulated in a different language.
The top is spinning around a vertical axis and the spinning around this axis can't disappear. if the top decided to fall, the spinning would either disappear or would be replaced by a totally different spinning around a horizontal axis, and Nature doesn't allow such a change of the amount of spinning to occur quickly. One has to have a torque to change the amount of spinning, some force attempting to change the rotation, but the torque acting on the bottom tip of the top is so small that with a fast enough initial spinning, it takes a lot of time to change the spin substantially.
Moreover, energy conservation guarantees that if there's no friction, the top can't ever fall.
More practically, I would probably take a wheel from a bicycle, made the kid hold it, rotate it quickly, and then make him or her feel the forces when he tries to change the direction of the wheel. This is a pretty nice yet simple toy in various science museums, including "Techmania" we have here in Pilsen. See also this page which contains the picture above as well as some other insightful games and experiments relevant for the angular momentum.
Best Answer
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.