Stability of equilibrium points For a spinning top, the linearised equation in the angle $\theta$ when the top is spinning about its axis of symmetry, which is vertical, is of the form $$A\ddot\theta+\left(\frac{C^2n^2}{4A}-Mgh\right)\theta=0.$$
Why should we require that the bracket coefficient be positive if we want the top to be stable, i.e. if we want small disturbances to remain small?
 A: Here's the math.  Suppose you have an equation of the form
$$
  \ddot\theta +\alpha\theta = 0
$$
If $\alpha <0$, then we can write $\alpha = -\omega^2$ for some $\omega >0$ and the general solution becomes
$$
  \theta(t) = Ae^{\omega t} + Be^{-\omega t}
$$
In particular, notice that these solutions are not oscillatory.  In fact, the solution blows up exponentially as a function of $t$.  Consider, for example, the following initial conditions:
$$
  \theta(0) = 0, \qquad \dot\theta(0) = \omega_0
$$
then we get
$$
  A+B = 0, \qquad A-B=\frac{\omega_0}{\omega}
$$
and the solution becomes
$$
  \theta(t) = \frac{\omega_0}{\omega}\frac{e^{\omega t} - e^{-\omega t}}{2}
$$
Notice, in particular, that no matter how small the initial velocity $\dot\theta(0) = \omega_0$ is, the solution has the property that $\theta(t)$ leaves the neighborhood of $\theta(0)$ for sufficiently large $t$.
On the other hand, if $\alpha >0$, then the equation we want to solve is simply the equation for simple harmonic motion with oscillatory solutions that remain in a vicinity of the initial position for all times.
A: Think about it this way, if the acceleration is in the positive theta direction when theta is positive, then the system will run away and become unstable.  
If acceleration is in the opposite direction of the sign of theta then the system will be driven back to the origin and thus be stable.
