Looking back at the original question, it looks like maybe Mjollnir was asking, “How long does it take for the pole to fall to the horizontal?” This is an interesting, and not altogether trivial, question. The smart-aleck answer is that it takes forever because (as has been pointed out) if the pole is vertical it is balanced and never falls. But suppose we start with the pole just a smidgen off vertical. Then how long does it take to fall? The answer depends dramatically on how small a smidgen you take!
Take p as the angle the pole makes with the vertical. Starting with an initial angle p0, how much time elapses before p=pi/2 ? If we define the zero of potential energy at p=0, then the potential energy is
U = -mg(1 - cosp)
As we’ve already seen, the kinetic energy is
T = (1/6)mh^2*w^2
Since the system is at rest at p=p0 the total energy remains constant at
T+ U = U(p0)
For small values of p0, U(p0) is also small, so approximately
T + U = 0
and
w = sqrt(3*g/h)*sqrt(1-cosp)
Since w=dp/dt, this is really a differential equation. It is variables separable, so it’s pretty easy. Integrating it from t=0 to t=T correspopnding to p=p0 to p = pi/2 , we get the time to fall is
T = sqrt((2h)/(3g)) * ln(tan(Pi/8))/tan(p0/4))
Since this equation is valid only in the regime where p0<<1, it is fair to make a further simplification to
T = sqrt((2h)/(3g)) * ln(1.66/p0)
What about the exact solution for all values of p0? The solution above involves integrating the differential dp/sqrt(1-cosp). This is easy enough [it’s sqrt(2) * ln(tan(p/4)) ]. But if you use the actual value of U(p0) for any value of p0, you have to integrate dp/sqrt(cosp0-cosp). This integral looks easy enough, but I couldn’t do it in the amount of time I had to put into this.
I wrote the above before looking at the latest few posts. Coupla comments: Ray I don’t think making the pole conical or any other weird shape makes it harder, it just changes the moment of inertia, which is a constant.
Frolix, I think you can extract a simple equation for w(t) if you keep the assumption that p0<<1, but I gotta scram right now.