Question

Find the Lagrangian and Lagrange's equations for a simple pendulum (Problem 4) if the cord is replaced by a spring with spring constant \(k\). \(Hint\): If the unstretched spring length is \(r_{0},\) and the polar coordinates of the mass \(m\) are \((r, \theta),\) the potential energy of the spring is \(\frac{1}{2} k\left(r-r_{0}\right)^{2}\).

Short answer

The Lagrangian: \[L = \frac{1}{2} m (\dot{r}^2 + r^2 \dot{\theta}^2) - \left( \frac{1}{2} k (r - r_0)^2 + m g r \cos(\theta) \right)\] Equations of motion: \[m \ddot{r} - m r \dot{\theta}^2 - k (r - r_0) + m g \cos(\theta) = 0\] \[m r^2 \ddot{\theta} + 2m r \dot{r} \dot{\theta} + m g r \sin(\theta) = 0\]

Step-by-step solution

Define the generalized coordinates

For the system, use polar coordinates \((r, \theta)\) to describe the position of the mass attached to the spring.

Write the kinetic energy (T)

The kinetic energy for the mass moving in polar coordinates is given by: \[T = \frac{1}{2} m \left( \dot{r}^2 + r^2 \dot{\theta}^2 \right) \]

Write the potential energy (V)

The potential energy of the system includes the potential energy of the spring and the gravitational potential energy. The spring potential energy is given in the hint: \[V_{\text{spring}} = \frac{1}{2} k \left(r - r_{0}\right)^2 \] \The gravitational potential energy is: \[V_{\text{gravity}} = m g r \cos(\theta) \] \Therefore, the total potential energy is: \[V = \frac{1}{2} k \left(r - r_{0}\right)^2 + m g r \cos(\theta) \]

Write the Lagrangian (L)

The Lagrangian is defined as the difference between the kinetic and potential energies: \[L = T - V \] \Substitute the expressions for T and V: \[L = \frac{1}{2} m \left( \dot{r}^2 + r^2 \dot{\theta}^2 \right) - \left( \frac{1}{2} k \left(r - r_{0}\right)^2 + m g r \cos(\theta) \right) \]

Apply the Euler-Lagrange equations

The Euler-Lagrange equations are given by: \[\frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q_i}} \right) - \frac{\partial L}{\partial q_i} = 0 \] \where \(q_i\) represents the generalized coordinates \(r, \theta\).

Derive the equation of motion for r

For the coordinate \(r\), the Euler-Lagrange equation is: \[\frac{d}{dt} \left( \frac{\partial L}{\partial \dot{r}} \right) - \frac{\partial L}{\partial r} = 0 \] \Compute \(\frac{\partial L}{\partial \dot{r}} = m \dot{r}\) and \(\frac{\partial L}{\partial r} = m r \dot{\theta}^2 + k (r - r_0) - m g \cos(\theta) \). \Thus the equation of motion for \(r\) becomes: \[m \ddot{r} - m r \dot{\theta}^2 - k (r - r_0) + m g \cos(\theta) = 0\]

Derive the equation of motion for \(\theta\)

For the coordinate \(\theta\), the Euler-Lagrange equation is: \[\frac{d}{dt} \left( \frac{\partial L}{\partial \dot{\theta}} \right) - \frac{\partial L}{\partial \theta} = 0 \] \Compute \(\frac{\partial L}{\partial \dot{\theta}} = m r^2 \dot{\theta}\) and \(\frac{\partial L}{\partial \theta} = m g r \sin(\theta) \). \Thus the equation of motion for \(\theta\) becomes: \[\frac{d}{dt} (m r^2 \dot{\theta}) + m g r \sin(\theta) = 0 \] \[m r^2 \ddot{\theta} + 2m r \dot{r} \dot{\theta} + m g r \sin(\theta) = 0\]

Key concepts

Euler-Lagrange Equations

The Euler-Lagrange equations are a fundamental part of Lagrangian mechanics and provide a powerful method for deriving the equations of motion for a system. These equations arise from the principle of least action, which states that the actual path taken by a system is the one that minimizes the action.
The action is defined as the integral of the Lagrangian over time:\[ S = \int_{t_1}^{t_2} L \, dt \]In this context, it means we can determine the system's behavior by finding the path that makes this integral stationary. The Euler-Lagrange equations are formulated as:\[ \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q_i}} \right) - \frac{\partial L}{\partial q_i} = 0 \]where \( q_i \) represents the generalized coordinates. This equation tells us how to find the equations of motion for our system.
By applying these equations to our system with the Lagrangian \( L \), we derive the equations of motion for each coordinate step by step. This process converts the problem into a set of differential equations that describe the dynamics of the system.

Potential Energy

Potential energy represents the stored energy of the system due to its position or configuration. In our exercise, the total potential energy includes contributions from both the spring and gravity.
Spring Potential Energy:
The potential energy stored in the spring is due to its deformation (stretching or compression). If the unstretched length is \( r_0 \) and the spring constant is \( k \), it is given by:\[ V_{\text{spring}} = \frac{1}{2} k (r - r_0)^2 \]
Gravitational Potential Energy:
The gravitational potential energy depends on the height of the mass in the gravitational field, which can be expressed in polar coordinates as:\[ V_{\text{gravity}} = m g r \cos(\theta) \]
Total Potential Energy:
Combining these, we get the total potential energy of the system:\[ V = \frac{1}{2} k (r - r_0)^2 + m g r \cos(\theta) \]
This formula captures how both the spring's deformation and the height of the mass contribute to the energy stored in the system.

Kinetic Energy

Kinetic energy is the energy possessed by an object due to its motion. For our system with polar coordinates \( (r, \theta) \), the kinetic energy consists of the radial motion and the angular motion.
The kinetic energy for a mass moving in polar coordinates is given by:\[ T = \frac{1}{2} m ( \dot{r}^2 + r^2 \dot{\theta}^2 ) \]
This formula indicates:

• \( \dot{r} \) is the radial velocity, describing how fast the distance \( r \) is changing.
• \( r \dot{\theta} \) is the tangential velocity, describing how fast the angle \( \theta \) is changing.

The first term \( \frac{1}{2} m \dot{r}^2 \) represents the kinetic energy due to the radial motion, while the second term \( \frac{1}{2} m r^2 \dot{\theta}^2 \) represents the kinetic energy due to the angular motion.
Combining these components gives us the total kinetic energy of the system.

Generalized Coordinates

Generalized coordinates are variables that describe the configuration of a system with respect to some reference. They are essential in Lagrangian mechanics because they simplify the description of motion, especially for systems with constraints.
In our exercise, we use the polar coordinates \( (r, \theta) \) as generalized coordinates:

• \( r \): The radial distance from the origin to the mass.
• \( \theta \): The angular position around the origin.

These coordinates help describe the motion of the mass attached to the spring in a two-dimensional plane, making the problem more manageable.
Generalized coordinates can vary depending on the system and constraints. Unlike Cartesian coordinates (x, y, z), generalized coordinates allow us to account for complex motions and constraints (like pendulums, springs, or rotating systems) more naturally.
Using these coordinates, we derive expressions for kinetic and potential energy, which ultimately help us construct the Lagrangian \( L \) and apply the Euler-Lagrange equations to find the equations of motion.