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I'm studying Classical Mechanics by Goldstein. I solved a problem but I have a question.

Pro 2.18
A point mass is constrained to move on a massless hoop of radius a fixed in a vertical plane that rotates about its vertical symmetry axis with constant angular speed ω. Obtain the Lagrange equations of motion assuming the only external forces arise from gravity. What are the constants of motion? Show that if ω is greater than a critical value ω0, there can be a solution in which the particle remains stationary on the hoop at a point other than at the bottom, but that if ω < ω0, the only stationary point for the particle is at the bottom of the hoop. What is the value of ω0?

So I proceeded like this solution here.

So here, when we choose only one generalized coordinate $\theta$(polar angle), energy function $h$ is not same as the energy. But in the text (chapter about Lagrangian) it says that if potential $V=V(q)$, $h=E$. For this problem $V=mga \cos \theta$, (or negative, according to how define $\theta$ or axis) so it satisfies the condition that potential only depends on generalized coordinate, not on generalized velocity. So $h$ should be $E$, but apparently not. What is wrong here?

I know that if I set the azimuthal angle as an independent variable, such a contradiction doesn't appear. But I cannot see why I should do that ( the problem says that azimuthal angle is not independent variable, and derivation of $h=E$ says nothing about that.)

Surely something must be wrong with my reasoning, because if the Lagrangian of a system is $L=\frac{1}{2}my'^2+mgy$, we can insert the constant horizontal kinetic energy $\frac{1}{2}mx'^2$ (x is not generalized coordinate here), but that would destroy $h=E$. Can someone explain this to me?

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The fact that the potential $V$ does not depend on velocities is not enough for the energy function to be equal to the energy.

In general, the energy function, $$h\equiv\sum_i\frac{\partial L}{\partial\dot q_i}\dot q_i-L,$$ equals the mechanical energy $E=T+V$ if

  1. The potential does not depend on velocities;
  2. The kinetic energy is homogeneous function of degree two on the velocities.

On the other hand, systems which are both holonomic and scleronomous, i.e., the positions can be written as $\vec r_a=\vec r_a(q_1,\ldots,q_n)$, satisfy condition 2. This is so because in this case $$T=\frac 12\sum_{i,j}a(q)_{ij}\dot q_i\dot q_j,$$ which satisfies $T(q,\lambda\dot q)=\lambda^2T(q,\dot q)$.

Clearly, the system you are considering is not scleronomous. The hoop is actually a time dependent constraint so that the position of the point mass is described by $\vec r=\vec r(\theta,t)$. Hence, $h\neq E$.

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  • $\begingroup$ Thanks. Actually I reread the derivation of the textbook myself and found that there is the additional assumption which you pointed out. I'm very glad to see that my reasoning was correct. $\endgroup$ – Septacle Oct 11 '17 at 23:17
  • $\begingroup$ Uh... but there arose an another problem. So finding equilibrium angle was to set θ''=0. We get θ=cos^-1(-sqrt(g/aw^2)). But when I use the effective potential (Veff=T+V-(ma^2θ^2)/2), and finding the point where dVeff/dθ=0, it gives another value, θ=cos^-1(sqrt(g/aw^2)). (I don't like this effective potential theme, but this is how our professor did.) I'm not sure which is real equilibrium angle from intuition. Because our professor used E=h, (which is wrong), his answer magically change sign to make the answer coherent, but my answer seems to give contradiction. $\endgroup$ – Septacle Oct 12 '17 at 1:21
  • $\begingroup$ There is no sqrt. It was mistake. $\endgroup$ – Septacle Oct 12 '17 at 1:34

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