Chapter 9: Problem 21

Write Lagrange's equations in cylindrical coordinates for a particle moving in the gravitational field \(V=m g z\).

Short Answer

Expert verified

The equations are: \(\ddot{r} = r\dot{\theta}^2\), \(r^2\dot{\theta} = \text{constant}\), and \(\ddot{z} = -g\).

Step by step solution

- Define the Problem

Identify the system and the coordinates. In this case, we have a particle with mass m moving in a gravitational field in cylindrical coordinates \(r, \theta, z\). The potential energy is given by \(V=mgz\).

- Write the Kinetic Energy

The kinetic energy in cylindrical coordinates (r, \theta, z) is given by \(T = \frac{1}{2}m(\dot{r}^2 + r^2\dot{\theta}^2 + \dot{z}^2)\).

- Write the Lagrangian

The Lagrangian \(L\) is the difference between the kinetic and potential energies: \(L = T - V\). Substituting the expressions from the previous steps, we get \[L = \frac{1}{2}m(\dot{r}^2 + r^2\dot{\theta}^2 + \dot{z}^2) - mgz\].

- Compute the Euler-Lagrange Equations

The Euler-Lagrange equations are given by \(\frac{d}{dt}(\frac{\partial L}{\partial \dot{q}_i}) - \frac{\partial L}{\partial q_i} = 0\), where \(q_i\) are the generalized coordinates. Apply these to each coordinate \(r, \theta, z\).

- Apply to Coordinate r

For \r\: \(\frac{d}{dt}(\frac{\partial L}{\partial \dot{r}}) - \frac{\partial L}{\partial r} = 0\). We have \(\frac{\partial L}{\partial \dot{r}} = m\dot{r}\) and \(\frac{\partial L}{\partial r} = mr\dot{\theta}^2\). Therefore, \(m\ddot{r} - mr\dot{\theta}^2 = 0\), or \(\ddot{r} = r\dot{\theta}^2\).

- Apply to Coordinate theta

For \theta\: \(\frac{d}{dt}(\frac{\partial L}{\partial \dot{\theta}}) - \frac{\partial L}{\partial \theta} = 0\). We have \(\frac{\partial L}{\partial \dot{\theta}} = mr^2\dot{\theta}\) and \(\frac{\partial L}{\partial \theta} = 0\), so \(\frac{d}{dt}(mr^2\dot{\theta}) = 0\), or \(\frac{d}{dt}(r^2\dot{\theta}) = 0\). Thus, \(r^2\dot{\theta} = \text{constant}\).

- Apply to Coordinate z

For \z\: \(\frac{d}{dt}(\frac{\partial L}{\partial \dot{z}}) - \frac{\partial L}{\partial z} = 0\). We have \(\frac{\partial L}{\partial \dot{z}} = m\dot{z}\) and \(\frac{\partial L}{\partial z} = -mg\), so \(m\ddot{z} + mg = 0\), or \(\ddot{z} = -g\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kinetic Energy in Cylindrical Coordinates

Kinetic energy measures the energy that a particle possesses due to its motion. In cylindrical coordinates \(r, \theta, z\), the kinetic energy \(T\) can be calculated using the following formula: \[T = \frac{1}{2}m(\dot{r}^2 + r^2\dot{\theta}^2 + \dot{z}^2)\]\
This is a bit different from Cartesian coordinates because cylindrical coordinates include radial and angular components. Here's a quick breakdown of each part:

\( \dot{r} \): The velocity in the radial direction.
\( r^2 \dot{\theta}^2 \): Represents the tangential or angular component.
\( \dot{z} \): The velocity in the vertical direction.

Together, these parts account for the complete motion of a particle along the radial, angular, and vertical axes.

Potential Energy in a Gravitational Field

Potential energy is the energy stored due to an object's position in a force field, in this case, gravity. For a particle of mass \(m\) in a gravitational field, the potential energy \(V\) in cylindrical coordinates is given by: \[V = mgz\]\
Here, 'g' is the acceleration due to gravity, and 'z' represents the height or the vertical coordinate. As the particle moves along the z-axis, it gains or loses potential energy.

Euler-Lagrange Equations

The Euler-Lagrange equations help us derive the equations of motion for a system. Given a Lagrangian \(L = T - V\), the Euler-Lagrange equation for each generalized coordinate \(q_i\) is: \[\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}_i}\right) - \frac{\partial L}{\partial q_i} = 0\]\
In our problem, we apply this to each coordinate \(r, \theta, z\). For \(r\), \[m\ddot{r} - mr\dot{\theta}^2 = 0\]\For \(\theta\), \[\frac{d}{dt}(mr^2\dot{\theta}) = 0\]\and for \(z\), \[m\ddot{z} + mg = 0\] These lead to equations of motions that describe how the particle moves under the influence of forces in the system.

Understanding Cylindrical Coordinates

Cylindrical coordinates \(r, \theta, z\) are a three-dimensional extension of polar coordinates. They are particularly useful when dealing with problems involving symmetry around an axis. The coordinates represent:

\(r\): The radial distance from a chosen axis.
\(\theta\): The angular position around the axis.
\(z\): The height or position along the axis.

These coordinates simplify the calculations of kinetic and potential energies in systems with cylindrical symmetry over using Cartesian coordinates.

Particle Motion in a Gravitational Field

In this problem, we analyze a particle's motion in a gravitational field. The gravitational acceleration \(g\) acts downward along the z-axis, and the potential energy due to gravity is given by \(V = mgz\).
The Lagrange equations help us determine how the particle moves.
Therefore, \( \ddot{r} = r\dot{\theta}^2 \) describes how the radial component changes over time, while \( \frac{d}{dt}(r^2\dot{\theta}) = 0 \) implies the conservation of angular momentum, and \( \ddot{z} = -g \) describes the vertical motion under gravity.