Chapter 9: Problem 9
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
Expert verified
The Lagrangian for the system is \[ L = \frac{1}{2} m (2 \dot{r}^2 + r^2 \dot{\theta}^2 ) - m g r \] Lagrange's equations are \[ \ddot{r} = \frac{1}{2} (r \dot{\theta}^2 - g) \] and \[ \ddot{\theta} + 2 \frac{\dot{r}}{r} \dot{\theta} = 0 \].
Step by step solution
01
Identify the coordinates
Given the cone's equation is given by the surface relation: \[ r = z \] The cylindrical coordinates are used where \( r \) represents the radial distance and \( \theta \) the angular coordinate, with gravity acting in the -z direction.
02
Express z in terms of r
Given that \( r = z \), it follows that \( z = r \). This substitution will help in formulating the Lagrangian solely in terms of \( r \) and \( \theta \).
03
Write the kinetic energy
The kinetic energy (T) in cylindrical coordinates is given by: \[ T = \frac{1}{2} m ( \dot{r}^2 + r^2 \dot{\theta}^2 + \dot{z}^2 ) \]Since \( \dot{z} = \dot{r} \) (from \( z = r \)), this simplifies to: \[ T = \frac{1}{2} m ( \dot{r}^2 + r^2 \dot{\theta}^2 + \dot{r}^2 ) = \frac{1}{2} m (2 \dot{r}^2 + r^2 \dot{\theta}^2 ) \]
04
Write the potential energy
The potential energy (V) due to gravity is: \[ V = m g z = m g r \]Here, since \( z = r \), the potential energy remains in terms of \( r \).
05
Construct the Lagrangian
The Lagrangian (L) is the difference between the kinetic and potential energies: \[ L = T - V = \frac{1}{2} m (2 \dot{r}^2 + r^2 \dot{\theta}^2 ) - m g r \]
06
Derive Lagrange's equations
For the coordinate \( r \): \[ \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{r}} \right) - \frac{\partial L}{\partial r} = 0 \] First, calculate \( \frac{\partial L}{\partial \dot{r}} \): \[ \frac{\partial L}{\partial \dot{r}} = 2m \dot{r} \] Next, calculate \( \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{r}} \right) \): \[ \frac{d}{dt} \left( 2m \dot{r} \right) = 2m \ddot{r} \] Finally, \( \frac{\partial L}{\partial r} \): \[ \frac{\partial L}{\partial r} = m r \dot{\theta}^2 - mg \]
07
Solve the Euler-Lagrange equation for r
Combining the Lagrange terms for \( r \): \[ 2m \ddot{r} - (m r \dot{\theta}^2 - mg) = 0 \] Simplify to obtain: \[ 2 \ddot{r} = r \dot{\theta}^2 - g \] Hence, \[ \ddot{r} = \frac{1}{2} (r \dot{\theta}^2 - g) \]
08
Solve Euler-Lagrange equation for θ
For the coordinate \(\theta\): \[ \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{\theta}} \right) - \frac{\partial L}{\partial \theta} = 0 \]First, calculate \( \frac{\partial L}{\partial \dot{\theta}} \): \[ \frac{\partial L}{\partial \dot{\theta}} = mr^2 \dot{\theta} \]Next, simplify \( \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{\theta}} \right) \):\[ \frac{d}{dt} \left( mr^2 \dot{\theta} \right) = mr^2 \ddot{\theta} + 2mr\dot{r} \dot{\theta} \] Since \( \frac{\partial L}{\partial \theta} = 0 \), we get:\[ mr^2 \ddot{\theta} + 2mr\dot{r} \dot{\theta} = 0 \] Thus:\[ \ddot{\theta} + 2 \frac{\dot{r}}{r} \dot{\theta} = 0 \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Lagrange's equations
Lagrange's equations play a crucial role in mechanics. They are used to describe the motion of a system without directly dealing with forces. Instead, they use energy functions like kinetic and potential energy. These equations stem from the principle of least action. This principle states that the path taken by a system between two states is the one for which the action integral (which involves the Lagrangian) is stationary. Thus, this offers a powerful way to analyze complex systems beyond Newton's laws.
kinetic energy
Kinetic energy is the energy possessed by an object due to its motion. In mechanics, it's often denoted as T. The expression for kinetic energy depends on the coordinates used to describe the system.
For a mass m moving in cylindrical coordinates (r, θ, z), the kinetic energy is expressed as: \
\[ T = \frac{1}{2} m ( \ldot{r}^2 + r^2 \ldot{θ}^2 + \ldot{z}^2) \]
In our exercise, because the motion lies on the surface of a cone given by r = z, we can simplify the kinetic energy to: \
\[ T = \frac{1}{2} m (2 \ldot{r}^2 + r^2 \ldot{θ}^2 ) \]
Understanding this concept is vital as it influences the formulation of the Lagrangian.
For a mass m moving in cylindrical coordinates (r, θ, z), the kinetic energy is expressed as: \
\[ T = \frac{1}{2} m ( \ldot{r}^2 + r^2 \ldot{θ}^2 + \ldot{z}^2) \]
In our exercise, because the motion lies on the surface of a cone given by r = z, we can simplify the kinetic energy to: \
\[ T = \frac{1}{2} m (2 \ldot{r}^2 + r^2 \ldot{θ}^2 ) \]
Understanding this concept is vital as it influences the formulation of the Lagrangian.
potential energy
Potential energy represents the energy stored within a system due to its position or configuration. In our cone exercise, the potential energy is derived from gravity, which acts downwards along the z-axis.
For a mass m in a gravitational field, the potential energy V is expressed as:
\[ V = m g z \]
Given the constraint of the cone (r = z), we can substitute z with r, simplifying the potential energy expression to:
\[ V = mg r \]
Grasping the potential energy concept is crucial for constructing the overall Lagrangian of the system, which combines both kinetic and potential energies.
For a mass m in a gravitational field, the potential energy V is expressed as:
\[ V = m g z \]
Given the constraint of the cone (r = z), we can substitute z with r, simplifying the potential energy expression to:
\[ V = mg r \]
Grasping the potential energy concept is crucial for constructing the overall Lagrangian of the system, which combines both kinetic and potential energies.
cylindrical coordinates
Cylindrical coordinates are a three-dimensional coordinate system that extends polar coordinates by adding a height dimension (z). In this system, any point in space is defined by:
The coordinates are especially useful for problems exhibiting cylindrical symmetry, as seen in our cone exercise. By employing cylindrical coordinates, complex three-dimensional motion can often be simplified into more manageable equations.
This facilitates the construction of the Lagrangian and the subsequent application of Lagrange’s equations to describe the system’s dynamics.
- Radial distance (r)
- Angular coordinate (θ)
- Height (z)
The coordinates are especially useful for problems exhibiting cylindrical symmetry, as seen in our cone exercise. By employing cylindrical coordinates, complex three-dimensional motion can often be simplified into more manageable equations.
This facilitates the construction of the Lagrangian and the subsequent application of Lagrange’s equations to describe the system’s dynamics.
Euler-Lagrange equation
The Euler-Lagrange equation is derived from the Lagrangian's principle of least action. It provides the condition that must hold for the system’s actual path:
\[ \frac{d}{dt} \left( \frac{\partial L}{\partial \ldot{q}} \right) - \frac{\partial L}{\partial q} = 0 \]
Here, q represents the generalized coordinates (such as r and θ), and L is the Lagrangian.
For our specific problem, we apply these steps to both coordinates r and θ. This yields differential equations governing the system's motion.
By solving these equations, we find the evolution of the coordinates over time, encapsulating the dynamic behavior of the mass on the cone's surface.
\[ \frac{d}{dt} \left( \frac{\partial L}{\partial \ldot{q}} \right) - \frac{\partial L}{\partial q} = 0 \]
Here, q represents the generalized coordinates (such as r and θ), and L is the Lagrangian.
For our specific problem, we apply these steps to both coordinates r and θ. This yields differential equations governing the system's motion.
By solving these equations, we find the evolution of the coordinates over time, encapsulating the dynamic behavior of the mass on the cone's surface.