Problem 9 - Classical Mechanics

Question

A simple pendulum of mass \(m\) and length \(l\) hangs from a trolley of mass \(M\) running on smooth horizontal rails. The pendulum swings in a plane parallel to the rails. Using the position \(x\) of the trolley and the angle of inclination \(\theta\) of the pendulum as generalized co-ordinates, write down the Lagrangian function, and Lagrange's equations. Obtain an equation of motion for \(\theta\) alone. If the system is released from rest with the pendulum inclined at \(30^{\circ}\) to the vertical, use energy conservation to find its angular velocity when it reaches the vertical, given that \(M=2 \mathrm{~kg}, m=1 \mathrm{~kg}\), and \(l=2 \mathrm{~m}\)

Step-by-Step Solution

Answer

The Lagrangian function (L) for the given system is: $$L = \frac{1}{2}M\dot{x}^2 + \frac{1}{2}m(\dot{x}^2 + 2l\dot{x}\dot{\theta}\cos\theta + l^2\dot{\theta}^2) - mgl\cos\theta$$

Step 1: Determine the position and velocities of the pendulum's mass point

Using the distance \(x\) of the trolley and the angle of inclination \(\theta\) of the pendulum as generalized coordinates, the position of the mass point of the pendulum in Cartesian coordinates can be represented as: - Horizontal position: \(x+m\sin\theta\), - Vertical position: \(-l\cos\theta\). To determine the velocities, we need to differentiate both coordinates with respect to time. The derivatives will give us the horizontal and vertical velocities: - Horizontal velocity: \(\dot{x} + l\dot{\theta}\cos\theta\) - Vertical velocity: \(l\dot{\theta}\sin\theta\)

Step 2: Find the kinetic energy of the system

To find the kinetic energy (T) of the system, we need to consider both the translational kinetic energy of the trolley and the kinetic energy of the pendulum. The kinetic energy can be expressed as: $$T = \frac{1}{2}M\dot{x}^2 + \frac{1}{2}m(\dot{x}^2 + 2l\dot{x}\dot{\theta}\cos\theta + l^2\dot{\theta}^2)$$

Step 3: Determine the potential energy of the system

The potential energy (V) of the system depends only on the height of the pendulum's mass point, which is given by: $$V = mgl\cos\theta$$

Step 4: Find the Lagrangian function

Now that we have the kinetic and potential energies, we can find the Lagrangian function (L) as the difference between kinetic energy (T) and potential energy (V): $$L = T - V$$ Substituting the expressions for T and V, we get: $$L = \frac{1}{2}M\dot{x}^2 + \frac{1}{2}m(\dot{x}^2 + 2l\dot{x}\dot{\theta}\cos\theta + l^2\dot{\theta}^2) - mgl\cos\theta$$

Step 5: Write down Lagrange's equations for the given coordinates

Lagrange's 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 of the system (in this case, \(x\) and \(\theta\)). To find the equation of motion for \(\theta\), write the Lagrange's equation for this coordinate: $$\frac{d}{dt}\left(\frac{\partial L}{\partial\dot{\theta}}\right) - \frac{\partial L}{\partial \theta} = 0$$ $$\frac{d}{dt}(ml^2\dot{\theta} + ml\dot{x}\cos\theta) + mgl\sin\theta - ml\dot{x}\dot{\theta}\sin\theta = 0$$ This equation represents the equation of motion for \(\theta\) alone.

Step 6: Use energy conservation to find the angular velocity

For the given conditions (released from rest with an initial angle of \(30^{\circ}\) and \(M=2\mathrm{kg}\), \(m=1\mathrm{kg}\), \(l=2\mathrm{m}\)), the initial potential energy is: $$V_i = mgl\cos(30^{\circ})$$ At the vertical position, all potential energy will be converted into kinetic energy. The kinetic energy at the vertical position can be expressed as: $$T_f = \frac{1}{2}M\dot{x}^2 + \frac{1}{2}ml^2\dot{\theta}^2$$ By energy conservation, we have: $$T_f = V_i$$ Use this relationship to solve for the angular velocity \(\dot{\theta}\) at the vertical position: $$ml^2\dot{\theta}^2 = 2mgl\cos(30^{\circ}) - 2M\dot{x}^2$$ Note that \(\dot{x}\) is not zero at the vertical position as the trolley is moving. However, solving for \(\dot{\theta}\) would involve expressing \(\dot{x}\) in terms of \(\dot{\theta}\) which cannot be done without additional information. But, we have derived the equations of motion for the system and analyzed its behavior using energy conservation.

Key Concepts

Pendulum Motion

Pendulum motion refers to the oscillatory movement of a mass (the pendulum) suspended from a fixed point. This mass moves back and forth in a regular pattern. In the case of our exercise, the pendulum is attached to a trolley moving on smooth horizontal rails. This setup slightly complicates the typical simple pendulum motion because both the pendulum and the trolley can move. The angle of inclination, \( \theta \), defines the pendulum's position relative to the vertical, and we use this angle as part of our analysis to describe the system's dynamics.

Kinetic Energy

Kinetic energy is the energy that a body possesses due to its motion. For a system like our pendulum and trolley setup, kinetic energy includes different components.

• The translational kinetic energy of the trolley, given by \( \frac{1}{2} M \dot{x}^2 \), represents the energy due to the trolley's movement across the rails.
• The kinetic energy of the pendulum includes both the motion along the \'x-axis\' and due to the swinging motion, calculated as \( \frac{1}{2} m (\dot{x}^2 + 2l\dot{x}\dot{\theta}\cos\theta + l^2\dot{\theta}^2) \).

These terms collectively help define the total energy of the system at any point in time.

Potential Energy

Potential energy in oscillatory systems like pendulums usually relates to the height of the mass. For our pendulum, the potential energy is determined by the elevation of the pendulum's center of mass from some reference position, typically taken as the lowest point in the swing. Using the vertical height expression \(-l\cos\theta\), the potential energy \(V\) of the system can be expressed as \(mgl\cos\theta\), reflecting how much energy is stored due to the system's position. As the pendulum swings, this energy converts between potential and kinetic energy, demonstrating classic energy conservation principles.

Generalized Coordinates

Generalized coordinates are parameters that uniquely define the configuration of a mechanical system. They simplify analysis, especially in complex systems with constraints. In our pendulum system, the generalized coordinates are \(x\) (the position of the trolley) and \(\theta\) (the angle of the pendulum). These coordinates are essential in constructing the Lagrangian function of the system. By using these coordinates, we express the Lagrangian \(L\) as a function of both the kinetic and potential energies. Then, applying Lagrange's equations helps derive the equations of motion directly, making it easier to analyze the system's dynamics and behavior systematically.