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(a) Write down the Lagrangian for a particle moving in three dimensions under the influence of a conservative central force with potential energy \(U(r),\) using spherical polar coordinates \((r, \theta, \phi)\). (b) Write down the three Lagrange equations and explain their significance in terms of radial acceleration, angular momentum, and so forth. (The \(\theta\) equation is the tricky one, since you will find it implies that the \(\phi\) component of \(\ell\) varies with time, which seems to contradict conservation of angular momentum. Remember, however, that \(\ell_{\phi}\) is the component of \(\ell\) in a variable direction.) (c) Suppose that initially the motion is in the equatorial plane (that is, \(\theta_{0}=\pi / 2\) and \(\dot{\theta}_{0}=0\) ). Describe the subsequent motion. (d) Suppose instead that the initial motion is along a line of longitude (that is, \(\dot{\phi}_{0}=0\) ). Describe the subsequent motion.

Short Answer

Expert verified
The Lagrangian describes the system's dynamics; Lagrange's equations express conservation laws and radial processes. The motion continues in the equatorial plane or precesses along a longitude, depending on initial conditions.

Step by step solution

01

Write the Lagrangian

The Lagrangian for a particle moving in three dimensions with position represented in spherical polar coordinates \((r, \theta, \phi)\) is given by the kinetic energy minus the potential energy. The kinetic energy in spherical coordinates is:\[ T = \frac{1}{2} m \left(\dot{r}^2 + r^2 \dot{\theta}^2 + r^2 \sin^2 \theta \dot{\phi}^2 \right) \]where \(m\) is the mass of the particle and \(\dot{r}, \dot{\theta}, \dot{\phi}\) are the time derivatives of the coordinates. The potential energy due to the conservative central force is \( U(r) \). Therefore, the Lagrangian is:\[ L = T - U = \frac{1}{2} m \left(\dot{r}^2 + r^2 \dot{\theta}^2 + r^2 \sin^2 \theta \dot{\phi}^2 \right) - U(r) \]
02

Derive Lagrange's Equations

Lagrange's equations are derived from the Lagrangian \( L \), using the Euler-Lagrange equation \( \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}_i} \right) - \frac{\partial L}{\partial q_i} = 0 \) for each generalized coordinate \( q_i = r, \theta, \phi \):1. **Radial equation \( r \):**\[ \frac{d}{dt}(m \dot{r}) - m r (\dot{\theta}^2 + \sin^2 \theta \dot{\phi}^2) + \frac{\partial U}{\partial r} = 0 \]2. **Polar angle equation \( \theta \):**\[ \frac{d}{dt}(m r^2 \dot{\theta}) - m r^2 \sin \theta \cos \theta \dot{\phi}^2 = 0 \]3. **Azimuthal angle equation \( \phi \):**\[ \frac{d}{dt}(m r^2 \sin^2 \theta \dot{\phi}) = 0 \]These equations describe the radial acceleration, conservation of angular momentum, and variations in angular dynamics.
03

Analyze Motion in the Equatorial Plane

When the motion is in the equatorial plane, \( \theta_0 = \frac{\pi}{2} \) and \( \dot{\theta}_0 = 0 \), implying \( \theta = \frac{\pi}{2} \) remains constant. In this case:- The radial equation determines radial acceleration based on \( \dot{r} \).- The \( \phi \)-equation \[ \frac{d}{dt}(m r^2 \dot{\phi}) = 0 \] shows that the angular momentum around the \( z \)-axis is conserved. The motion remains purely in the \( \frac{\pi}{2} \) plane.
04

Analyze Motion Along Line of Longitude

If the motion initially is along a line of longitude with \( \dot{\phi}_0 = 0 \):- Initially, there is no azimuthal motion (\( \dot{\phi} = 0 \)).- The \( \theta \)-equation requires \( \theta \) to change as the \( \phi \)-component of angular momentum can vary; this leads to a precessional motion.- Radial and \( \theta \) dynamics depend on the potential shape and the initial values of \( r \) and \( \dot{r} \). The system effectively behaves in two coupled dimensions.

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Key Concepts

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

Understanding Radial Acceleration in Spherical Coordinates
In spherical coordinates, radial acceleration is a crucial concept that describes how the speed of a particle changes as it moves towards or away from the center of a potential field. This concept is essential when dealing with central force problems in Lagrangian Mechanics. To grasp radial acceleration, we must start with the radial component of the force acting on the particle.
  • Radial acceleration is essentially the second derivative of the radial position coordinate, often denoted as \( \ddot{r} \).
  • It is derived from the radial component of Lagrange's equations, which for a particle subject to potential energy \( U(r) \) is expressed as:
    \[m \ddot{r} = m r (\dot{\theta}^2 + \sin^2 \theta \dot{\phi}^2) - \frac{\partial U}{\partial r}\] This equation effectively balances the radial acceleration with centripetal contributions from angular motion, noted by \( m r (\dot{\theta}^2 + \sin^2 \theta \dot{\phi}^2) \).
  • Radial acceleration is influenced by how the potential energy changes with radial distance. If the derivative of the potential with respect to \( r \), denoted \( \frac{\partial U}{\partial r} \) is positive, it suggests a restoring force, akin to a spring trying to return to equilibrium.
Understanding these dynamics is key to predicting a particle’s path in a field governed by a central potential.
Exploring Angular Momentum in Lagrangian Mechanics
Angular momentum is a fundamental concept describing rotational motion, and it's deeply linked to the symmetry of circular or spherical systems. In Lagrangian Mechanics, angular momentum appears naturally in the formulation of the equations of motion and provides insights into the conservation laws.
  • In spherical coordinates, angular momentum has components associated with the angles \( \theta \) (polar angle) and \( \phi \) (azimuthal angle).
  • The azimuthal angle component, often more intuitive, is given by:
    \[m r^2 \sin^2 \theta \dot{\phi}\] The Azimuthal equation \( \frac{d}{dt}(m r^2 \sin^2 \theta \dot{\phi}) = 0 \) asserts that this component is conserved, meaning there's no change over time, frequently leading to circular or elliptical paths.
  • Angular momentum in the \( \theta \) component reveals more complexity. It manifests in the equation
    \[m r^2 \ddot{\theta} + 2m r \dot{r} \dot{\theta} - m r^2 \sin \theta \cos \theta \dot{\phi}^2 = 0\] This implies variations especially when initially \( \theta_0 eq \frac{\pi}{2} \) (not in the equatorial plane), where the angle \( \theta \) changes dynamically, showing potential precession or wobbling in motion.
Angular momentum is thereby much more than a static quantity; it helps poviding a map of dynamic rotational movements in constrained motion scenarios.
Navigating Spherical Coordinates in Lagrangian Mechanics
Spherical coordinates are a natural choice when dealing with problems involving central forces and rotational symmetry, such as planets orbiting a star or electrons around a nucleus. Understanding how to work with spherical coordinates in Lagrangian mechanics can simplify complex three-dimensional problems.
  • The coordinates \( (r, \theta, \phi) \) correspond to the radial distance, polar angle, and azimuthal angle, respectively, changing how we perceive movement in space compared to Cartesian coordinates.
  • The kinetic energy in spherical coordinates is a sum of its radial, polar, and azimuthal components:
    \[T = \frac{1}{2} m (\dot{r}^2 + r^2 \dot{\theta}^2 + r^2 \sin^2 \theta \dot{\phi}^2)\] Each component can independently describe motion influences on the particle, allowing targeted analysis and simplification depending on the initial conditions.
  • Based on how the particle's movement is initiated, different coordinate equations like the one for \( \theta \),\( \phi \), or \( r \), can take precedence, altering the motion's characteristics profoundly.
By breaking down motion into these key components, spherical coordinates help manage the chaos of three-dimensional dynamics, guiding us toward clear solutions in seemingly intractable problems.

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Most popular questions from this chapter

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.

Consider a mass \(m\) moving in a frictionless plane that slopes at an angle \(\alpha\) with the horizontal. Write down the Lagrangian in terms of coordinates \(x,\) measured horizontally across the slope, and \(y\), measured down the slope. (Treat the system as two-dimensional, but include the gravitational potential energy.) Find the two Lagrange equations and show that they are what you should have expected.

A smooth wire is bent into the shape of a helix, with cylindrical polar coordinates \(\rho=R\) and \(z=\lambda \phi,\) where \(R\) and \(\lambda\) are constants and the \(z\) axis is vertically up (and gravity vertically down). Using \(z\) as your generalized coordinate, write down the Lagrangian for a bead of mass \(m\) threaded on the wire. Find the Lagrange equation and hence the bead's vertical acceleration \(\ddot{z}\). In the limit that \(R \rightarrow 0\), what is \(\ddot{z} ?\) Does this make sense?

Consider a double Atwood machine constructed as follows: A mass 4 \(m\) is suspended from a string that passes over a massless pulley on frictionless bearings. The other end of this string supports a second similar pulley, over which passes a second string supporting a mass of \(3 m\) at one end and \(m\) at the other. Using two suitable generalized coordinates, set up the Lagrangian and use the Lagrange equations to find the acceleration of the mass \(4 m\) when the system is released. Explain why the top pulley rotates even though it carries equal weights on each side.

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 \(m \omega^{2} r\).

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