Rotating Top: Euler Angles?

Hi all,

I've just started studying the motion of a rigid heavy (i.e. with gravity) spinning top using Euler's angles. It mostly makes sense, but there's one point in my notes where we've derived the equations of motion using Euler-Lagrange and one of them is (theta, phi are the usual Euler angles)

$\displaystyle \dot{\phi} = \frac{P_{\phi} - I_3 \omega_3 \cos{\theta}}{I_1 {\sin}^2 \theta}$

But P, I_1, I_3 and w_3 are independent constants so when theta gets really small (or close to pi) this is going to blow up. It obviously shouldn't since it's modelling a real object which can't spin infinitely fast, so why doesn't it?

Thanks!

Re: Rotating Top: Euler Angles?

$\displaystyle \theta$ equal to 0 or $\displaystyle \pi$ would mean the top is lying on its side. If $\displaystyle \theta$ is close to 0 or $\displaystyle \pi$, the top would have to spin very, very fast to stay spinning!