Question

A particle moves without friction under gravity on the surface of the paraboloid \(z=x^{2}+y^{2} .\) Find the Lagrangian and the Lagrange equations of motion. Show that motion in a horizontal circle is possible and find the angular velocity of this motion. Use cylindrical coordinates.

Short answer

The Lagrangian is \( L = \frac{1}{2}m(5\dot{r}^2 + r^2 \dot{\theta}^2) - mgr^2 \). Motion in a horizontal circle is possible with an angular velocity of \( \sqrt{2g} \).

Step-by-step solution

- Define the coordinates and convert to cylindrical coordinates

Convert the Cartesian coordinates to cylindrical coordinates. Use the conversion equations: \[ x = r \, \cos(\theta) \]\[ y = r \, \sin(\theta) \]\[ z = r^2 \] where \( r \) is the radial distance and \( \theta \) is the azimuthal angle.

- Express the kinetic energy

The kinetic energy in Cartesian coordinates is given by \[ T = \frac{1}{2}m(\dot{x}^2 + \dot{y}^2 + \dot{z}^2) \]. Convert this to cylindrical coordinates using the relations from Step 1: \[ T = \frac{1}{2}m(\dot{r}^2 + r^2 \dot{\theta}^2 + \dot{z}^2) \]. Since \( z = r^2 \), we get \( \dot{z} = 2r\dot{r} \). Substitute \( \dot{z} \): \[ T = \frac{1}{2}m(\dot{r}^2 + r^2 \dot{\theta}^2 + (2r\dot{r})^2) \] \[ = \frac{1}{2}m(5\dot{r}^2 + r^2 \dot{\theta}^2) \].

- Express the potential energy

The potential energy due to gravity is \[ V = mgz \]. Using the cylindrical coordinate equation for \( z \), \[ V = mgr^2 \].

- Write down the Lagrangian

The Lagrangian \( L \) is given by \[ L = T - V \]. Substitute the expressions from Steps 2 and 3: \[ L = \frac{1}{2}m(5\dot{r}^2 + r^2 \dot{\theta}^2) - mgr^2 \].

- Derive the Lagrange equations

Use the Euler-Lagrange equations: \[ \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}} \right) - \frac{\partial L}{\partial q} = 0 \], where \( q \) represents the generalized coordinates \( r \) and \( \theta \). For \( r \): \[ \frac{d}{dt}(m(5\dot{r})) - (mr\dot{\theta}^2 + 2mgr) = 0 \]. Simplify: \[ 10m\ddot{r} - mr\dot{\theta}^2 - mg2r = 0 \] \[ \Rightarrow 10\ddot{r} - 2gr - r\dot{\theta}^2 = 0 \].For \( \theta \): \[ \frac{d}{dt}(m r^2 \dot{\theta}) - 0 = 0 \] \[ \Rightarrow m r^2 \dot{\theta} = \text{constant} \].

- Prove horizontal circular motion is possible

Consider the case where \( \dot{r} = 0 \) and \( r = R \) (constant). This means \( \ddot{r} = 0 \). Substitute into the radial equation: \[ -2gR - R\dot{\theta}^2 = 0 \] Solve for \( \dot{\theta}^2 \): \[ \dot{\theta}^2 = 2g \].

- Find the angular velocity

Since \( \dot{\theta} \) represents the angular velocity, we have: \[ \dot{\theta} = \sqrt{2g} \].

Key concepts

Euler-Lagrange equations

The Euler-Lagrange equations are fundamental in Lagrangian mechanics. They are derived from the principle of least action and are used to find the equations of motion for a system. Given a Lagrangian function, which is defined as the difference between kinetic energy (T) and potential energy (V), the Euler-Lagrange equations take the form: \[ \frac{d}{dt} \frac{\text{∂}L}{\text{∂}\text{dot{q}}} - \frac{\text{∂}L}{\text{∂}q} = 0 \] Here, L represents the Lagrangian, and q represents the generalized coordinates. For example, to find the motion of a particle in cylindrical coordinates, we write down the Lagrangian and apply these equations to obtain the equations of motion for each coordinate.

cylindrical coordinates

Cylindrical coordinates ( r, θ, z) are a three-dimensional coordinate system that are useful for problems with cylindrical symmetry. They are defined as follows: \[ x = r \text{cos}(\theta) \] \[ y = r \text{sin}(\theta) \] \[ z = r^2\] where r is the radial distance, θ is the azimuthal angle, and z is the height. In the context of this exercise, converting from Cartesian to cylindrical coordinates simplifies the description of a particle’s motion on a paraboloid. This change of variables allows for an easier formulation of kinetic and potential energies, which are central to writing the Lagrangian.

potential energy

Potential energy (V) represents the stored energy in a system due to its position, configuration, or state. In gravitational systems, potential energy is given by V = mgh , where m is mass, g is the acceleration due to gravity, and h is the height above a reference level. In our exercise, the particle moves under gravity on a surface defined by \( z = r^2 \), so the potential energy becomes \[ V = mgz = mgr^2 \]. This term is then used in the Lagrangian to account for the effect of gravity on the motion of the particle.

kinetic energy

Kinetic energy (T) is the energy of an object due to its motion. In Cartesian coordinates, it's given by: \[ T = \frac{1}{2} m(𝑥˙^2 + 𝑦˙^2 + 𝑧˙^2) \] After converting to cylindrical coordinates, and noting \( z = r^2 \) and thus \( \dot{z} = 2r\dot{r}\), the kinetic energy becomes: \[ T = \frac{1}{2} m(𝑟˙^2 + r^2θ˙^2 + 4r^2 r˙^2) = \frac{1}{2} m (5r˙^2 + r^2θ˙^2)\] Kinetic energy reflects how different types of motion (radial and angular in this case) contribute to the total energy of the particle's system. Combining this with potential energy allows for a complete description of the system's dynamics through the Lagrangian.