Question

A mass \(m\) moves without friction on the surface of the cone \(r=z\) under gravity acting in the negative \(z\) direction. Here \(r\) is the cylindrical coordinate \(r=\sqrt{x^{2}+y^{2}}\) Find the Lagrangian and Lagrange's equations in terms of \(r\) and \(\theta\) (that is, eliminate \(z\) ).

Short answer

The Lagrangian is \( L = \frac{1}{2} m \left( 2 \dot{r}^2 + r^2 \dot{\theta}^2 \right) - mgr \) and the equations of motion are \( \ddot{r} = r \dot{\theta}^2 - \frac{g}{2} \) and \( 2 \dot{r} \dot{\theta} + r \ddot{\theta} = 0 \)

Step-by-step solution

- Express Variables in Terms of Cylindrical Coordinates

Given that the cone's surface is defined by the relation: \[ r = z \]Thus, the height in terms of cylindrical coordinates can be expressed as \[ z = r \]

- Write Down the Kinetic Energy

The kinetic energy (T) in cylindrical coordinates \[ (r, \theta, z) \] is given by: \[ T = \frac{1}{2} m \left( \dot{r}^2 + r^2 \dot{\theta}^2 + \dot{z}^2 \right) \] Since \[ z = r \] it follows that \[ \dot{z} = \dot{r} \]Thus, the kinetic energy expression modifies to: \[ T = \frac{1}{2} m \left( \dot{r}^2 + r^2 \dot{\theta}^2 + \dot{r}^2 \right) = \frac{1}{2} m \left( 2 \dot{r}^2 + r^2 \dot{\theta}^2 \right) \]

- Write Down the Potential Energy

The potential energy (V) due to gravity acting in the negative z-direction is given by: \[ V = mgz \]Substitute \[ z = r \] into the expression: \[ V = mgr \]

- Write the Lagrangian

The Lagrangian (L) is defined as the kinetic energy (T) minus the potential energy (V): \[ L = T - V \]Thus, \[ L = \frac{1}{2} m \left( 2 \dot{r}^2 + r^2 \dot{\theta}^2 \right) - mgr \]

- Derive the Lagrange's Equations for r

The Lagrange equation is: \[ \frac{d}{dt}\left( \frac{\partial L}{\partial \dot{q}_j} \right) - \frac{\partial L}{\partial q_j} = 0 \]For \( r \): \[ \frac{d}{dt}\left( \frac{\partial L}{\partial \dot{r}} \right) - \frac{\partial L}{\partial r} = 0 \]Compute \( \frac{\partial L}{\partial \dot{r}} = 2m \dot{r} \)\( \frac{d}{dt}\left( \frac{\partial L}{\partial \dot{r}} \right) = 2m \ddot{r} \)Compute \( \frac{\partial L}{\partial r} = mr \dot{\theta}^2 - mg \)Thus, the Lagrange equation is: \[ 2m \ddot{r} - (mr \dot{\theta}^2 - mg) = 0 \]Simplify: \[ \ddot{r} = r \dot{\theta}^2 - \frac{g}{2} \]

- Derive the Lagrange's Equations for θ

For \( \theta \): \[ \frac{d}{dt}\left( \frac{\partial L}{\partial \dot{\theta}} \right) - \frac{\partial L}{\partial \theta} = 0 \]Compute \( \frac{\partial L}{\partial \dot{\theta}} = mr^2 \dot{\theta} \)\( \frac{d}{dt}\left( \frac{\partial L}{\partial \dot{\theta}} \right) = 2mr \dot{r} \dot{\theta} + mr^2 \ddot{\theta} \)Since \( \frac{\partial L}{\partial \theta} = 0 \), it simplifies to: \[ 2mr \dot{r} \dot{\theta} + mr^2 \ddot{\theta} = 0 \]Divide by \( mr \): \[ 2 \dot{r} \dot{\theta} + r \ddot{\theta} = 0 \]

Key concepts

cylindrical coordinates

Cylindrical coordinates help us describe a point in three-dimensional space by using a radius, angle, and height. Unlike Cartesian coordinates (x, y, z), cylindrical coordinates are represented by (\(r, \theta, z\)). Here, \(r\) is the radial distance from the z-axis, \(\theta\) is the angle from the x-axis in the xy-plane, and \(z\) is the height. For our problem, the mass moves on the surface of a cone defined by \(r = z\). This means the height \(z\) and the radial distance \(r\) are equal.

potential energy

Potential energy (V) is the energy stored in an object because of its position in a force field, here due to gravity. For an object at a height \(z\) in a gravitational field, the potential energy is given by: \[ V = mgz \] In our case, since the height and radial distance are equal (\(z = r\)), we update the potential energy: \[ V = mgr \] This demonstrates how the position of the mass in the gravitational field directly influences the potential energy.

kinetic energy

Kinetic energy (T) represents the energy an object has due to its motion. In cylindrical coordinates, it's given by: \[ T = \frac{1}{2} m ( \dot{r}^2 + r^2 \dot{\theta}^2 + \dot{z}^2 ) \] Here, \( \dot{r} \) is the radial velocity, \( r \dot{\theta} \) is the tangential velocity, and \( \dot{z} \) is the vertical velocity. Given \(z = r\), we get \( \dot{z} = \dot{r} \), simplifying the kinetic energy expression to: \[ T = \frac{1}{2} m ( 2 \dot{r}^2 + r^2 \dot{\theta}^2 ) \] This equation shows the components of motion in our cylindrical coordinate system.

Lagrange's equations

Lagrange's equations are used to derive the equations of motion for a system. The general form is: \[ \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}_j} \right) - \frac{\partial L}{\partial q_j} = 0 \] Where L is the Lagrangian (\( L = T - V \)). First, we write down the Lagrangian for our system: \[ L = \frac{1}{2} m ( 2 \dot{r}^2 + r^2 \dot{\theta}^2 ) - mgr \] For \(r\): \[ \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{r}} \right) - \frac{\partial L}{\partial r} = 0 \] Solving this, we get: \[ 2m \ddot{r} - (mr \dot{\theta}^2 - mg) = 0 \] Simplifying further, we find: \[ \ddot{r} = r \dot{\theta}^2 - \frac{g}{2} \] For \(\theta\): \[ \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{\theta}} \right) - \frac{\partial L}{\partial \theta} = 0 \] This results in: \[ 2mr \dot{r} \dot{\theta} + mr^2 \ddot{\theta} = 0 \] Which further simplifies to: \[ 2 \dot{r} \dot{\theta} + r \ddot{\theta} = 0 \] These equations describe how the radial distance \(r\) and the angle \(\theta\) evolve over time in the system.