Question

A mass \(m_{1}\) rests on a frictionless horizontal table and is attached to a massless string. The string runs horizontally to the edge of the table, where it passes over a massless, frictionless pulley and then hangs vertically down. A second mass \(m_{2}\) is now attached to the bottom end of the string. Write down the Lagrangian for the system. Find the Lagrange equation of motion, and solve it for the acceleration of the blocks. For your generalized coordinate, use the distance \(x\) of the second mass below the tabletop.

Short answer

The acceleration of the blocks is \(\ddot{x} = \frac{-m_2 g}{m_1 + m_2}\).

Step-by-step solution

Define the system and coordinates

We have two masses, \(m_1\) on a frictionless table and \(m_2\) hanging vertically. The string connects them over a pulley. We choose the distance \(x\) that mass \(m_2\) is below the tabletop as our generalized coordinate. As \(m_2\) moves vertically by \(x\), \(m_1\) moves horizontally by the same distance \(x\).

Write down the Kinetic Energy

The kinetic energy \(T\) of the system is the sum of the kinetic energies of both masses. Since \(m_1\) moves horizontally, its kinetic energy is \(\frac{1}{2} m_1 \dot{x}^2\), and since \(m_2\) moves vertically, its kinetic energy is \(\frac{1}{2} m_2 \dot{x}^2\). Therefore, the total kinetic energy is \(T = \frac{1}{2} m_1 \dot{x}^2 + \frac{1}{2} m_2 \dot{x}^2\).

Write down the Potential Energy

The potential energy \(U\) is due to the gravitational energy of \(m_2\). Since \(m_2\) is at a height \(x\), its potential energy is \(U = m_2 g x\). There is no potential energy for \(m_1\) as it moves horizontally on a frictionless table.

Formulate the Lagrangian

The Lagrangian \(L\) is defined as the difference between the kinetic and potential energies: \(L = T - U\). Substituting the expressions for \(T\) and \(U\), we get \(L = \frac{1}{2} m_1 \dot{x}^2 + \frac{1}{2} m_2 \dot{x}^2 - m_2 g x\).

Derive the Lagrange Equation of Motion

The Lagrange equation is given by \(\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{x}}\right) - \frac{\partial L}{\partial x} = 0\). First, compute \(\frac{\partial L}{\partial \dot{x}} = (m_1 + m_2) \dot{x}\). Then, compute \(\frac{\partial L}{\partial x} = -m_2 g\). Differentiating \(\frac{\partial L}{\partial \dot{x}}\) with respect to time gives \(\frac{d}{dt}\left((m_1 + m_2) \dot{x} \right) = (m_1 + m_2) \ddot{x}\). The Lagrange equation becomes \((m_1 + m_2) \ddot{x} + m_2 g = 0\).

Solve for the acceleration

Rearrange the equation \((m_1 + m_2) \ddot{x} = -m_2 g\) to solve for the acceleration \(\ddot{x}\). This gives \(\ddot{x} = \frac{-m_2 g}{m_1 + m_2}\).

Key concepts

Kinetic Energy

Kinetic energy is the energy that an object possesses due to its motion. In our exercise with two blocks connected by a string over a pulley, both masses have kinetic energy as they move. For the block on the table, mass \(m_1\), which moves horizontally, its kinetic energy is given by the formula:\[T_{1} = \frac{1}{2} m_1 \dot{x}^2\]where \(\dot{x}\) represents the velocity of mass \(m_1\). For the second block, mass \(m_2\), which moves in the vertical direction, its kinetic energy is similarly given by:\[T_{2} = \frac{1}{2} m_2 \dot{x}^2\]Since both blocks move by the same amount due to the string connection, the total kinetic energy of the system is simply the sum of these individual kinetic energies:\[T = \frac{1}{2} m_1 \dot{x}^2 + \frac{1}{2} m_2 \dot{x}^2\]Understanding kinetic energy is important as it helps in formulating the total energy profile of a dynamic system like this one.
It is key to consider both contributions when analyzing the motion of the connected bodies.

Potential Energy

Potential energy relates to the position of an object in a gravitational field. In our problem, only mass \(m_2\) has potential energy because it is elevated above the ground. The formula for its potential energy is:\[U = m_2 g x\]Here, \(g\) is the acceleration due to gravity, which pulls the mass downward, and \(x\) is the height below the tabletop where mass \(m_2\) is located.
Mass \(m_1\), lying on a horizontal frictionless table, doesn't gain or lose height so doesn't possess gravitational potential energy in this setup.
The understanding of potential energy helps in determining how the stored energy can be converted into kinetic energy as the masses move, particularly when mass \(m_2\) descends and mass \(m_1\) is pulled across the tabletop. This concept is crucial for formulating the Lagrangian, where the difference between kinetic and potential energies helps us understand the dynamics of the system.

Equations of Motion

Equations of motion describe how the positions, velocities, or accelerations of a system change over time. Using Lagrangian mechanics, these equations result from finding a path that minimizes the action, a principle central to this formalism.1. **Formulating the Lagrangian:** The Lagrangian \(L\) is defined as the difference between the kinetic energy \(T\) and potential energy \(U\) of the system: \[ L = T - U = \frac{1}{2} m_1 \dot{x}^2 + \frac{1}{2} m_2 \dot{x}^2 - m_2 g x \]2. **Lagrange's Equation of Motion:** Derived using the Lagrangian, it is given by: \[ \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{x}}\right) - \frac{\partial L}{\partial x} = 0 \] Applying this to our formulated Lagrangian gives us: \[ (m_1 + m_2) \ddot{x} + m_2 g = 0 \]3. **Solving for Acceleration:** Rearranging the equation leads to finding the acceleration \(\ddot{x}\): \[ \ddot{x} = \frac{-m_2 g}{m_1 + m_2} \]Understanding these equations allows us to predict how quickly the blocks will accelerate. Such equations are foundational to analyzing and projecting the motion of mechanical systems. They provide a deeper insight into how forces cause changes in motion and are crucial for designing systems in engineering and physics applications.