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
01
- 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 \(V=mgz\).
02
- Write the Kinetic Energy
The kinetic energy in cylindrical coordinates (r, \theta, z) is given by \(T = \frac{1}{2}m(\dot{r}^2 + r^2\dot{\theta}^2 + \dot{z}^2)\).
03
- 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 \[L = \frac{1}{2}m(\dot{r}^2 + r^2\dot{\theta}^2 + \dot{z}^2) - mgz\].
04
- 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 \(r, \theta, z\).
05
- 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 \(\ddot{r} = r\dot{\theta}^2\).
06
- 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, \(r^2\dot{\theta} = \text{constant}\).
07
- 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:
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.
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.
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.
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.
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.
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.