Question

A mass \(m_{1}\) rests on a frictionless horizontal table. Attached to it is a string which runs horizontally to the edge of the table, where it passes over a frictionless, small pulley and down to where it supports a mass \(m_{2} .\) Use as coordinates \(x\) and \(y\) the distances of \(m_{1}\) and \(m_{2}\) from the pulley. These satisfy the constraint equation \(f(x, y)=x+y=\) const. Write down the two modified Lagrange equations and solve them (together with the constraint equation) for \(\ddot{x}, \ddot{y},\) and the Lagrange multiplier \(\lambda\). Use (7.122) (and the corresponding equation in \(y\) ) to find the tension forces on the two masses. Verify your answers by solving the problem by the elementary Newtonian approach.

Short answer

\( \ddot{x} = \frac{-m_{2} g}{m_{1} + m_{2}} \), \( \ddot{y} = \frac{m_{2} g}{m_{1} + m_{2}} \), \( T = \frac{m_{1} m_{2} g}{m_{1} + m_{2}} \).

Step-by-step solution

Establish the Lagrangian and Constraint

Define the Lagrangian of the system, where the kinetic energy of the masses is given by \( T = \frac{1}{2}m_{1}\dot{x}^2 + \frac{1}{2}m_{2}\dot{y}^2 \) and the potential energy is \( V = m_{2}g y \). The constraint is \( x + y = \text{const} \). So, the Lagrangian is \( L = T - V = \frac{1}{2}m_{1}\dot{x}^2 + \frac{1}{2}m_{2}\dot{y}^2 - m_{2}gy \).

Apply Lagrange's Equation with Multipliers

The Lagrange equations with the multiplier \( \lambda \) are given by \( \frac{d}{dt}(\frac{\partial L}{\partial \dot{q}_i}) - \frac{\partial L}{\partial q_i} + \lambda \frac{\partial f}{\partial q_i} = 0 \) for each coordinate \( q_i \). Applying this, for \( x \): \( m_{1}\ddot{x} - \lambda = 0 \). For \( y \): \( m_{2}\ddot{y} + m_{2}g + \lambda = 0 \). And the constraint: \( x + y = \text{const} \).

Solve the Lagrange Equations

Using the constraint \( \ddot{y} = -\ddot{x} \), plug into the Lagrange equations: 1. \( m_{1}\ddot{x} = \lambda \)2. \( m_{2}(-\ddot{x}) + m_{2}g = -\lambda \)Substitute \( \lambda \) from 1 into 2: \[ m_{1} \ddot{x} = m_{2} \ddot{x} - m_{2} g \]Solve for \( \ddot{x} \): \[ \ddot{x} = \frac{-m_{2} g}{m_{1} + m_{2}} \] Then \( \ddot{y} = -\ddot{x} = \frac{m_{2} g}{m_{1} + m_{2}} \).

Solve for Lagrange Multiplier \( \lambda \)

Using \( \lambda = m_{1} \ddot{x} \), substitute in the value of \( \ddot{x} \): \[ \lambda = m_{1} \frac{-m_{2} g}{m_{1} + m_{2}} = \frac{-m_{1} m_{2} g}{m_{1} + m_{2}} \]

Find Tension Forces Using Equations

Using the relation \( T = -\lambda = \frac{m_{1} m_{2} g}{m_{1} + m_{2}} \), the tension force on both masses is calculated as the Lagrange multiplier with opposite sign. Verify using the Newtonian approach.

Verify with Newtonian Approach

For mass \( m_{1} \), Newton's law gives \( T = m_{1} \ddot{x} \). For \( m_{2} \), it's \( m_{2}g - T = m_{2} \ddot{y} \). Substitute the values to match: \[ T = \frac{m_{1} m_{2} g}{m_{1} + m_{2}} \] This checks out with the equations, confirming our solutions.

Key concepts

Lagrange Equations

In physics, Lagrange's equations revolutionize how we solve dynamics problems, particularly for systems with constraints. These equations emerge from a combination of a system's kinetic and potential energies, forming what we call the Lagrangian, noted as \( L = T - V \), where \( T \) is kinetic energy and \( V \) is potential energy.
Kinetic energy for the two-mass system is given by \( T = \frac{1}{2}m_{1}\dot{x}^2 + \frac{1}{2}m_{2}\dot{y}^2 \) and potential energy only pertains to \( m_{2} \) due to gravity: \( V = m_{2}gy \).
By inserting these into the Lagrangian, we apply Lagrange's equation:

• \( \frac{d}{dt}(\frac{\partial L}{\partial \dot{q}_i}) - \frac{\partial L}{\partial q_i} + \lambda \frac{\partial f}{\partial q_i} = 0 \)

Here, \( \lambda \) is a Lagrange multiplier used for constraints, adding an additional dynamic in analyzing systems such as the one in your exercise, which helps us explore how forces act under restrictions.

Constraint Equations

Constraints in mechanics define the possible movements in a system, linking coordinates of different objects. The constraint equation, \( x + y = \text{const} \), restricts how \( m_1\) and \( m_2\) can move relative to each other. In the exercise, assuming the string is inextensible, the speed \( \dot{x} \) of \( m_1 \) must equal \( -\dot{y} \) of \( m_2 \), linking their accelerations such that \( \ddot{y} = -\ddot{x} \).
With constraints, we introduce a Lagrange multiplier \( \lambda \) into Lagrange’s equations, providing insights into the effect of constraints on the system.
These equations ensure a deeper understanding of how constraints impact motion and introduce important reactions or pseudo-forces within constrained systems like the tension in a string.

Newtonian Mechanics

Newtonian mechanics remains a fundamental principle for understanding motion, focusing on how forces affect objects. It uses Newton's laws of motion as its basis:
1. An object remains at rest or moves uniformly unless acted upon by a force.
2. Force equals mass times acceleration (\( F = ma \)).
3. For every action, there is an equal and opposite reaction.
This framework complemented Lagrangian Mechanics, providing a "hands-on" approach to confirm solutions derived through Lagrange's equations.
In this problem, Newtonian perspectives clarify how forces such as tension and gravitational forces interact with each mass. Applying Newton's second law for each mass enables us to derive the same results obtained via Lagrange’s equations, solidifying that both frameworks, albeit different, cohesively describe the physical system at play.

Tension Forces

Within the context of the exercise, tension forces play a crucial role as they balance the gravitational force acting on \( m_{2} \) and provide the necessary centripetal force for \( m_{1} \).
Using the relations derived from Lagrange’s multipliers, we determined the tension \( T \) in the string as \( T = \frac{m_{1} m_{2} g}{m_{1} + m_{2}} \), reliant on both masses and gravitational acceleration \( g \).
Tension embodies the internal force exerted along the string, facilitating constrained movement between the two masses. With an understanding of tension, students can visualize its function — maintaining equilibrium and transmitting force through the pulley in this mechanical scenario.
This tension is essentially the same as the force required to keep both masses in dynamic equilibrium, resisting gravitational pull yet maintaining the system's integrity.