Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Consider two particles moving unconstrained in three dimensions, with potential energy \(U\left(\mathbf{r}_{1}, \mathbf{r}_{2}\right) .\) (a) Write down the six equations of motion obtained by applying Newton's second law to each particle. (b) Write down the Lagrangian \(\mathcal{L}\left(\mathbf{r}_{1}, \mathbf{r}_{2}, \dot{\mathbf{r}}_{1}, \dot{\mathbf{r}}_{2}\right)=T-U\) and show that the six Lagrange equations are the same as the six Newtonian equations of part (a). This establishes the validity of Lagrange's equations in rectangular coordinates, which in turn establishes Hamilton's principle. since the latter is independent of coordinates, this proves Lagrange's equations in any coordinate system.

Short Answer

Expert verified
The Newtonian and Lagrangian equations for two particles are equivalent.

Step by step solution

01

Identify the position and forces

Consider the two particles with positions \(\mathbf{r}_1\) and \(\mathbf{r}_2\) in three-dimensional space. We need to consider the force expressions for each particle using Newton's second law. Let \(\mathbf{F}_1\) and \(\mathbf{F}_2\) represent the net forces on particles 1 and 2, respectively.
02

Apply Newton's Second Law

Newton's second law states that \(\mathbf{F} = m \mathbf{a}\), where \(m\) is the mass of the particle and \(\mathbf{a}\) is its acceleration. For particle 1 and particle 2, their equations of motions are:\[\begin{align*}m_1 \ddot{\mathbf{r}}_1 &= -abla_{\mathbf{r}_1} U(\mathbf{r}_1, \mathbf{r}_2), \m_2 \ddot{\mathbf{r}}_2 &= -abla_{\mathbf{r}_2} U(\mathbf{r}_1, \mathbf{r}_2).\end{align*}\]These give us three equations for each particle component (x, y, z), resulting in a total of six equations.
03

Write the Kinetic Energy expression

The kinetic energy \(T\) for two particles is:\[T = \frac{1}{2} m_1 (\dot{x}_1^2 + \dot{y}_1^2 + \dot{z}_1^2) + \frac{1}{2} m_2 (\dot{x}_2^2 + \dot{y}_2^2 + \dot{z}_2^2)\]
04

Write the Lagrangian

The Lagrangian \(\mathcal{L}\) is the difference between the kinetic and potential energy:\[\mathcal{L}(\mathbf{r}_1, \mathbf{r}_2, \dot{\mathbf{r}}_1, \dot{\mathbf{r}}_2) = T - U(\mathbf{r}_1, \mathbf{r}_2).\]
05

Derive Lagrange's Equations

Lagrange's equations are derived as:\[\frac{d}{dt}\left(\frac{\partial \mathcal{L}}{\partial \dot{\mathbf{r}}_{i}}\right) - \frac{\partial \mathcal{L}}{\partial \mathbf{r}_{i}} = 0\] for each component of each particle. Calculating these for \(\mathbf{r}_1\) and \(\mathbf{r}_2\), we obtain:\[m_1 \ddot{\mathbf{r}}_1 + abla_{\mathbf{r}_1} U(\mathbf{r}_1, \mathbf{r}_2) = 0\] \[m_2 \ddot{\mathbf{r}}_2 + abla_{\mathbf{r}_2} U(\mathbf{r}_1, \mathbf{r}_2) = 0\]These match the Newtonian equations derived earlier.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Newton's Second Law
Newton's second law of motion is one of the fundamental principles in classical mechanics. It states that the force \( \mathbf{F} \) acting on an object is equal to the mass \( m \) of the object multiplied by its acceleration \( \mathbf{a} \). This can be expressed mathematically as \( \mathbf{F} = m \mathbf{a} \). In simple terms, it means that if you apply a force to an object, it will accelerate in the direction of the force, and the acceleration is directly proportional to the force and inversely proportional to the mass of the object.
In the context of particles moving in three dimensions, this law helps us to determine the motion of each particle by considering the forces acting on them. By applying Newton's second law to each particle, we derive a set of six equations—three for each particle—representing their motion in the x, y, and z directions. These form the basis for understanding how forces impact the movement of particles on a basic level.
  • Force \( (\mathbf{F}) \) is a vector, meaning it has both magnitude and direction.
  • Inertia, the resistance of an object to change in motion, is characterized by its mass \( (m) \).
  • The acceleration \( (\mathbf{a}) \) is also a vector, which describes how quickly the velocity of the particle is changing.
Potential Energy
Potential energy is the stored energy in an object due to its position or configuration. It represents the capability of the object to do work due to its position in a force field such as gravitational or electrostatic fields. For our particles, the potential energy \( U(\mathbf{r}_1, \mathbf{r}_2) \) depends on their positions in space.
Potential energy is crucial in analyzing the system of particles because it contributes to the Lagrangian. It accounts for the potential interactions between the particles based on their relative positions. For example, if two particles are close together, the potential energy could increase if there is a repulsive force between them or decrease if there is an attractive force.
  • The potential energy function \( U \) typically varies with the position \( (\mathbf{r}_1, \mathbf{r}_2) \) of particles.
  • A change in potential energy indicates that work will be performed either by or upon the system.
  • In conservative force fields, potential energy can be transformed into kinetic energy and vice versa, while the total energy remains constant.
Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. For a particle with mass \( m \) moving at velocity \( \mathbf{v} \), the kinetic energy \( T \) is given by \( T = \frac{1}{2} m v^2 \). In the context of two particles moving in a three-dimensional space, their kinetic energy depends on their velocities in the x, y, and z directions.
The kinetic energy is a crucial part of the Lagrangian, reflecting how the velocity of particles contributes to the system's overall energy. Changes in kinetic energy occur when the speed of the particles changes, such as when they accelerate due to forces.
  • Kinetic energy increases as the velocity of the particle increases.
  • It's always a positive quantity, as it's proportional to the square of the velocity.
  • Kinetic energy converts into potential energy when a moving object is acted upon by a force that changes its motion path.
Lagrange's Equations
Lagrange's equations form the core of Lagrangian mechanics, providing a powerful method to derive the equations of motion for a system. They state that for a system described by the Lagrangian \( \mathcal{L} = T - U \), the path taken by the system is such that the Lagrangian is stationary.
Lagrange's equations are derived from the principle of stationary action, and in their general form, they provide a set of second-order differential equations. Given a system, you can derive its Lagrange's equations using: \[ \ \frac{d}{dt}\left(\frac{\partial \mathcal{L}}{\partial \dot{\mathbf{r}}_i}\right) - \frac{\partial \mathcal{L}}{\partial \mathbf{r}_i} = 0 \ \] for each coordinate. In practice, they contribute to a more straightforward method to solve dynamics problems compared to directly using Newton's equations, especially when dealing with complex systems.
  • Lagrangian \( \mathcal{L} \) is the difference between total kinetic and potential energies.
  • They can be applied in any coordinate system, not just rectangular coordinates.
  • Simplifying the motion equations as it removes the need to analyze constraint forces explicitly.
Hamilton's Principle
Hamilton's Principle, or the principle of stationary action, posits that the path taken by a physical system between two states is the one for which the action integral is stationary (no change). This principle forms the basis of deriving Lagrange's equations, emphasizing the motion of particles in terms of action rather than force.
Hamilton's principle is independent of the choice of coordinates, which means it provides a unifying perspective over mechanical systems and can be applied universally. It shows that the dynamics of a system can be derived not just by forces, as in Newtonian mechanics, but by optimizing an action. This broader viewpoint provides deeper insights into the nature of physical laws across different areas of physics.
  • The action integral is obtained by integrating the Lagrangian over time between two states.
  • The principle provides a bridge to quantum mechanics and field theories by extending the concept of stationarity to other areas.
  • It demonstrates that physical systems follow paths that optimize their dynamic variables.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A pendulum is made from a massless spring (force constant \(k\) and unstretched length \(l_{\mathrm{o}}\)) that is suspended at one end from a fixed pivot \(O\) and has a mass \(m\) attached to its other end. The spring can stretch and compress but cannot bend, and the whole system is confined to a single vertical plane. (a) Write down the Lagrangian for the pendulum, using as generalized coordinates the usual angle \(\phi\) and the length \(r\) of the spring. (b) Find the two Lagrange equations of the system and interpret them in terms of Newton's second law, as given in Equation (1.48). (c) The equations of part (b) cannot be solved analytically in general. However, they can be solved for small oscillations. Do this and describe the motion. [Hint: Let \(l\) denote the equilibrium length of the spring with the mass hanging from it and write \(r=l+\epsilon .\) "Small oscillations" involve only small values of \(\epsilon\) and \(\phi,\) so you can use the small-angle approximations and drop from your equations all terms that involve powers of \(\epsilon\) or \(\phi\) (or their derivatives) higher than the first power (also products of \(\epsilon\) and \(\phi\) or their derivatives). This dramatically simplifies and uncouples the equations.]

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.

A particle is confined to move on the surface of a circular cone with its axis on the \(z\) axis, vertex at the origin (pointing down), and half-angle \(\alpha\). The particle's position can be specified by two generalized coordinates, which you can choose to be the coordinates \((\rho, \phi)\) of cylindrical polar coordinates. Write down the equations that give the three Cartesian coordinates of the particle in terms of the generalized coordinates ( \(\rho, \phi\) ) and vice versa.

Consider a particle of mass \(m\) and charge \(q\) moving in a uniform constant magnetic field \(\mathbf{B}\) in the \(z\) direction. (a) Prove that \(\mathbf{B}\) can be written as \(\mathbf{B}=\nabla \times \mathbf{A}\) with \(\mathbf{A}=\frac{1}{2} \mathbf{B} \times \mathbf{r} .\) Prove equivalently that in cylindrical polar coordinates, \(\mathbf{A}=\frac{1}{2} B \rho \hat{\phi}\). (b) Write the Lagrangian (7.103) in cylindrical polar coordinates and find the three corresponding Lagrange equations. (c) Describe in detail those solutions of the Lagrange equations in which \(\rho\) is a constant.

A mass \(m\) is suspended from a massless string, the other end of which is wrapped several times around a horizontal cylinder of radius \(R\) and moment of inertia \(I\), which is free to rotate about a fixed horizontal axle. Using a suitable coordinate, set up the Lagrangian and the Lagrange equation of motion, and find the acceleration of the mass \(m\). [The kinetic energy of the rotating cylinder is \(\frac{1}{2} I \omega^{2} .\)]

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free