A The Lagrange Top: Formulas and Analysis for Non-Zero Angular Momentum Cases

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The discussion revolves around proving that the time average of the angle φ is not equal to zero in a specific case involving angular momentum and torque on a spinning top. The user struggles with this proof, noting that the integral of the torque over one period τ is non-zero, which implies a change in angular momentum. They highlight that the vectors L(t) and L(t+τ) can have the same direction but different magnitudes, indicating that the system does not return to its initial state. The conversation emphasizes the complexity of the problem and the need for a deeper understanding of the dynamics involved. Overall, the challenge lies in demonstrating the non-zero average of φ despite its seemingly straightforward nature.
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All the needed formulas are here http://hepweb.ucsd.edu/ph110b/110b_notes/node36.html
I consider the following case
$$p_\psi\ne 0,\quad p_\phi/p_\psi\in (\cos\theta_2,\cos\theta_1)$$ this case corresponds to the middle picture in the bottom of the cited page.

I can not prove that the time average of the angle ##\phi## is not equal to zero: ##\int_0^\tau\phi(t)dt\ne 0##, here ##\tau## is the period of the function ##\theta(t)##.
I know it looks like a standard simple thing but I have been thinking for three days and the result is zero, I also can not find it in books. Please help.
 
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Very interesting problem. I think I have a qualitative type proof, but it doesn't use the Lagrange equations. The top has a torque on it from gravity. This torque when integrated over time from ## 0 ## to ## \tau ## will be non-zero. (The ## \vec{r} ## and the ## \vec{F} ## don't do a cycle around the top during the time interval of one loop=they basically remain in the same general vicinity for the single loop.) This non-zero result means that the integral of ## \frac{d \vec{L}}{dt} ## (## \vec{L} ## is the angular momentum, basically from the spinning top) must be non-zero so that ## \Delta \vec{L} ## is non-zero. The top can not return to the same location after doing a small loop, because this would imply ## \Delta \vec{L}=0 ##.
 
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yes but the vector ##\boldsymbol L## is not parallel to the axis ##z_B##

UPD: the vectors ##\boldsymbol L(t)## and ##\boldsymbol L(t+\tau)## can have the same direction but different value
 
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