Question

A particle of unit mass moves under gravity on a smooth surface given in cylindrical polar coordinates \(z, r, \theta\) by \(z=f(r)\). Show that the motion is governed by the Lagrangian $$ L=\frac{1}{2} \dot{r}^{2}\left(1+f^{\prime}(r)^{2}\right)+\frac{1}{2} r^{2} \dot{\theta}^{2}-g f(r) . $$ Show that \(\theta\) is an ignorable coordinate. Write down the conserved conjugate momentum and give its physical interpretation.

Short answer

The conserved conjugate momentum in this problem is \(p_\theta = r^2\dot{\theta}\). It represents the angular momentum of the particle in the plane of motion (r, \(\theta\)).

Step-by-step solution

Kinetic energy in cylindrical coordinates

First, the velocity components in cylindrical coordinates are $$ v_r = \dot{r},\quad v_\theta = r\dot{\theta},\quad v_z = \frac{dz}{dt} = \frac{df(r)}{dr} \dot{r} = f^{\prime}(r) \dot{r}. $$ The total kinetic energy (T) of a particle of unit mass is given by $$ T = \frac{1}{2} (v_r^2 + v_\theta^2 + v_z^2). $$ Substitute the velocity components found above into the kinetic energy expression $$ T = \frac{1}{2} \left(\dot{r}^2 + r^2\dot{\theta}^2 + f^{\prime}(r)^2\dot{r}^2\right). $$

Potential energy in cylindrical coordinates

The potential energy (V) of a particle of unit mass in a gravitational field is given by $$ V = -mgz = -mg f(r), $$ where \(g\) is the acceleration due to gravity.

Write the Lagrangian

The Lagrangian (L) of a system is defined as the difference between the kinetic and potential energies $$ L = T - V $$ Substitute the kinetic and potential energies found in steps 1 and 2 into the Lagrangian definition $$ L = \frac{1}{2} \dot{r}^{2}\left(1+f^{\prime}(r)^{2}\right)+\frac{1}{2} r^{2} \dot{\theta}^{2}-g f(r) . $$ This matches the given expression for the Lagrangian, as required.

Show that \(\theta\) is an ignorable coordinate

The coordinate \(\theta\) is ignorable if the Lagrangian is independent of \(\theta\). Taking the partial derivative of the Lagrangian with respect to \(\theta\) gives $$ \frac{\partial L}{\partial\theta} = 0, $$ so \(\theta\) is an ignorable coordinate.

Calculate the conserved conjugate momentum and discuss the physical interpretation

Since \(\theta\) is an ignorable coordinate, the conjugate momentum \(p_\theta\) is conserved. The conjugate momentum is defined as $$ p_\theta = \frac{\partial L}{\partial \dot{\theta}}. $$ Differentiating the Lagrangian with respect to \(\dot{\theta}\) we get $$ p_\theta = r^2\dot{\theta}. $$ The conserved conjugate momentum \(p_\theta\) represents the angular momentum of the particle in the plane of motion. This result is expected, as the motion is happening on a smooth surface, and the effect of gravity is entirely along the \(z\)-axis, leading to conservation of angular momentum in the plane (r, \(\theta\)).

Key concepts

Cylindrical Coordinates

In physics, especially in problems dealing with particles moving along curves and surfaces, cylindrical coordinates play a vital role. They provide a convenient way to describe locations relative to a fixed axis. This system uses three parameters: radial distance \( r \), angular position \( \theta \), and height \( z \).

• \( r \) represents the radial distance from the axis of rotation.
• \( \theta \) is the angular displacement from a chosen reference direction, similar to polar coordinates in a 2D plane.
• \( z \) is the height above or below a reference plane perpendicular to \( r \).

Cylindrical coordinates simplify the analysis of rotational systems, such as the given exercise where a particle’s path is influenced by both gravity and a constraining surface curve given by \( z=f(r) \). By translating the particle's velocity components into cylindrical coordinates, its motion can be systematically analyzed in distinct radial, angular, and vertical components.

Conserved Momentum

In the realm of mechanics, conserved momentum signifies a property of a system where the sum of its momentum remains unchanged if not acted upon by external forces. This conservation law holds true due to symmetries within the system.
In the exercise, the coordinate \( \theta \) is identified as ignorable, meaning that the Lagrangian doesn’t directly depend on \( \theta \). According to Noether's theorem, if the Lagrangian does not explicitly depend on a given coordinate, the associated momentum remains conserved.

• The ignition of \( \theta \) implies that rotational movements around the axis of rotation do not influence energy directly.
• The conserved quantity due to this symmetry is the angular momentum.

Therefore, the corresponding conjugate momentum \( p_\theta \), associated with this symmetry, is conserved over time as the particle moves under the gravitational effect with no additional torque applied in the \( \theta \)-direction.

Angular Momentum

Angular momentum is a fundamental concept representing the rotational equivalent of linear momentum. It is vital when analyzing particles in motion on rotational paths or around an axis. Angular momentum \( L \) for a single particle is defined as the cross-product of the radius vector \( \mathbf{r} \) and linear momentum \( \mathbf{p} \):\[L = \mathbf{r} \times \mathbf{p}\] In the context of cylindrical coordinates and Lagrangian mechanics, the exercise illustrates this through the conjugate momentum \( p_\theta = r^2 \dot{\theta} \).

• It quantifies the 'amount of motion' the particle possesses due to its angular velocity \( \dot{\theta} \) and radial position \( r \).
• In systems free from external torque, this angular momentum remains constant, making it a conserved quantity.

For the particle governed by gravitational forces on a smooth surface, its angular momentum reflects how the constraints and forces interact. Conservation of this momentum implies that, aside from gravitational influences in the \( z \)-direction, the rotational motion remains unaffected, showcasing the symmetrical nature of its path.