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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

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

Step by step solution

01

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 \( L = T - V = \frac{1}{2}m_{1}\dot{x}^2 + \frac{1}{2}m_{2}\dot{y}^2 - m_{2}gy \).
02

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: \( x + y = \text{const} \).
03

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 \( \ddot{y} = -\ddot{x} = \frac{m_{2} g}{m_{1} + m_{2}} \).
04

Solve for Lagrange Multiplier \( \lambda \)

Using \( \lambda = m_{1} \ddot{x} \), substitute in the value of \( \ddot{x} \): \[ \lambda = m_{1} \frac{-m_{2} g}{m_{1} + m_{2}} = \frac{-m_{1} m_{2} g}{m_{1} + m_{2}} \]
05

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.
06

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

Write down the Lagrangian for a cylinder (mass \(m\), radius \(R\), and moment of inertia \(I\) ) that rolls without slipping straight down an inclined plane which is at an angle \(\alpha\) from the horizontal. Use as your generalized coordinate the cylinder's distance \(x\) measured down the plane from its starting point. Write down the Lagrange equation and solve it for the cylinder's acceleration \(\ddot{x}\). Remember that \(T=\frac{1}{2} m v^{2}+\frac{1}{2} I \omega^{2},\) where \(v\) is the velocity of the center of mass and \(\omega\) is the angular velocity.

The center of a long frictionless rod is pivoted at the origin, and the rod is forced to rotate in a horizontal plane with constant angular velocity \(\omega\). Write down the Lagrangian for a bead threaded on the rod, using \(r\) as your generalized coordinate, where \(r, \phi\) are the polar coordinates of the bead. (Notice that \(\phi\) is not an independent variable since it is fixed by the rotation of the rod to be \(\phi=\omega t\).) Solve Lagrange's equation for \(r(t) .\) What happens if the bead is initially at rest at the origin? If it is released from any point \(r_{\mathrm{o}}>0,\) show that \(r(t)\) eventually grows exponentially. Explain your results in terms of the centrifugal force \(m \omega^{2} r\).

Write down the Lagrangian for a projectile (subject to no air resistance) in terms of its Cartesian coordinates \((x, y, z),\) with \(z\) measured vertically upward. Find the three Lagrange equations and show that they are exactly what you would expect for the equations of motion.

Noether's theorem asserts a connection between invariance principles and conservation laws. In Section 7.8 we saw that translational invariance of the Lagrangian implies conservation of total linear momentum. Here you will prove that rotational invariance of \(\mathcal{L}\) implies conservation of total angular momentum. Suppose that the Lagrangian of an \(N\) -particle system is unchanged by rotations about a certain symmetry axis. (a) Without loss of generality, take this axis to be the \(z\) axis, and show that the Lagrangian is unchanged when all of the particles are simultaneously moved from \(\left(r_{\alpha}, \theta_{\alpha}, \phi_{\alpha}\right)\) to \(\left(r_{\alpha}, \theta_{\alpha}, \phi_{\alpha}+\epsilon\right)\) (same \(\epsilon\) for all particles). Hence show that $$\sum_{\alpha=1}^{N} \frac{\partial \mathcal{L}}{\partial \phi_{\alpha}}=0.$$ (b) Use Lagrange's equations to show that this implies that the total angular momentum \(L_{z}\) about the symmetry axis is constant. In particular, if the Lagrangian is invariant under rotations about all axes, then all components of \(\mathbf{L}\) are conserved.

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.

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