Lagrangian Formalism – Lagrangian for Particle in Lower Hemisphere and Angular Velocity Consequences

Question

Having a particle to move freely inside the surface of a lower hemisphere with radius $R$ I managed to derive the Lagrangian as follows:

$$L = \dfrac{1}{2}\,m\,R^2\left(\,\dot{\theta}^2+\sin^2\theta\,\dot{\varphi}^2\right)+R\,m\,g\,\cos\theta$$

The equation of motion are obtained by the Euler-Lagrange-Equations:

for $\varphi: \quad $ $m\,R^2\,\left(2\,\sin\theta\,\cos\theta\,\dot{\theta}\,\dot{\varphi}+\sin^2\theta\,\ddot{\varphi} \right) = 0$

for $\theta: \quad m\,R^2\,\ddot{\theta}-m\,R^2\,\sin\theta\,\cos\theta\,\dot{\varphi}^2+R\,m\,g\sin\theta = 0$

$\textbf{Now the main task is to find the dependences}$ between $\omega, \quad z, \quad \text{and} \quad \rho \quad$ where $\omega$ is the angular velocity with which the particle circles around on equal hight, meaning $z = \textsf{const.}$ and $\rho = x^2+y^2$ is symbolising the straight distance to $z$-axis.

From the equation of motion and given the fact $\dot{\theta} = \ddot{\varphi} = 0$ as well as $\dot{\varphi} = \omega$ it becomes apparent that:

$-m\,R^2\,\sin\theta\,\cos\theta\,\dot{\varphi}^2+R\,m\,g\sin\theta = 0 \quad\Leftrightarrow\quad \omega^2 =\dfrac{g}{R\,\cos(\theta)}$ where $R\,\cos\theta = z \quad \text{and} \quad \rho^2 = R^2-\dfrac{g^2}{\omega^4}$ because $\rho^2+z^2 = R^2$.

The only problem I'm facing with all of it are some aberrations for some values of $\theta$. Namely it appears like $\omega$ becomes infinite for $\theta\to \dfrac{\pi}{2}$ and non zero for $\theta = 0$. Actually I'd expect the angular velocity to be zero when the particle is located at the lowest point of the sphere ($\theta = 0$) and not infinite when the particle is merely moving on the boundary of the lower hemisphere $(\theta = 90^{\circ}$). So what's wrong with my expectation?

Answer 1

Examine from the newtonian mechanics, As show in the following figure, for mass $m$ to be able to move steadily on a horizontal plane (angle $\theta$) with a constant angular frequency $\omega$.

Let examine the force diagram in the rest frame of the mass. There are 3 forces:

1. gravitational force $F_g = mg \hat z$ along the z-direction.
2. centrifugal force $F_c = m\rho \omega^2$ pointing in the $\hat\rho$ direction.
3. Normal force from the surface $N (-\hat r)$ along the radial direction inward.

These three force have to cancel in order to have a steady circular motion. We then have: \begin{align} N \sin\theta & = m \rho \omega^2 \tag{1}\\ N \cos\theta & = mg \tag{2} \end{align}

Cancel the normal force in the above two equations and remind that $\rho = R \sin\theta$ we then obtain $$ \frac{\sin\theta}{\cos\theta} = \frac{m R\sin\theta \omega^2 }{mg} $$ The angular frequency: $$ \omega^2 =\frac{g}{R \cos\theta} \tag{3} $$

Eq.(3) is the same as the OP's result. Now, we may understand the physical reason from the force analysis.

1. At the angle $\theta = \frac{\pi}{2}$, the horizontal plane, the normal force and the centrifugal force are in the $\hat \phi$ direction, and the gravitational force is alone in the $\hat z$ direction. The cancellation of forces failed, therefore, the steady circular motion cannot be achieved under this situation.
2. At the lowest point $\theta = 0$, Eq. (3) give a finite frequency $\omega = \frac{g}{R}$. But examine Eq.(1), $\sin 0 $ and $\rho$ are both vanished, Eq.(1) becomes $ 0 = 0$. Thus $\omega$ can be any value. Eq.(2) gives $N=mg$, which sheds nothing about omega. Therefore, the answer given by Eq.(3), $\omega = \frac{g}{R}$ is an artifact in the process of mathematical manipulation, since we divide $0$ in both sides of equation (1). After all, The frequncy has no meaning since the rotation radius $\rho = 0$