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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,θ,ϕ). (b) Write down the three Lagrange equations and explain their significance in terms of radial acceleration, angular momentum, and so forth. (The θ equation is the tricky one, since you will find it implies that the ϕ component of varies with time, which seems to contradict conservation of angular momentum. Remember, however, that ϕ is the component of in a variable direction.) (c) Suppose that initially the motion is in the equatorial plane (that is, θ0=π/2 and θ˙0=0 ). Describe the subsequent motion. (d) Suppose instead that the initial motion is along a line of longitude (that is, ϕ˙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,θ,ϕ) is given by the kinetic energy minus the potential energy. The kinetic energy in spherical coordinates is:T=12m(r˙2+r2θ˙2+r2sin2θϕ˙2)where m is the mass of the particle and r˙,θ˙,ϕ˙ are the time derivatives of the coordinates. The potential energy due to the conservative central force is U(r). Therefore, the Lagrangian is:L=TU=12m(r˙2+r2θ˙2+r2sin2θϕ˙2)U(r)
02

Derive Lagrange's Equations

Lagrange's equations are derived from the Lagrangian L, using the Euler-Lagrange equation ddt(Lq˙i)Lqi=0 for each generalized coordinate qi=r,θ,ϕ:1. **Radial equation r:**ddt(mr˙)mr(θ˙2+sin2θϕ˙2)+Ur=02. **Polar angle equation θ:**ddt(mr2θ˙)mr2sinθcosθϕ˙2=03. **Azimuthal angle equation ϕ:**ddt(mr2sin2θϕ˙)=0These 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, θ0=π2 and θ˙0=0, implying θ=π2 remains constant. In this case:- The radial equation determines radial acceleration based on r˙.- The ϕ-equation ddt(mr2ϕ˙)=0 shows that the angular momentum around the z-axis is conserved. The motion remains purely in the π2 plane.
04

Analyze Motion Along Line of Longitude

If the motion initially is along a line of longitude with ϕ˙0=0:- Initially, there is no azimuthal motion (ϕ˙=0).- The θ-equation requires θ to change as the ϕ-component of angular momentum can vary; this leads to a precessional motion.- Radial and θ dynamics depend on the potential shape and the initial values of r and 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 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:
    mr¨=mr(θ˙2+sin2θϕ˙2)Ur This equation effectively balances the radial acceleration with centripetal contributions from angular motion, noted by mr(θ˙2+sin2θϕ˙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 Ur 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 θ (polar angle) and ϕ (azimuthal angle).
  • The azimuthal angle component, often more intuitive, is given by:
    mr2sin2θϕ˙ The Azimuthal equation ddt(mr2sin2θϕ˙)=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 θ component reveals more complexity. It manifests in the equation
    mr2θ¨+2mrr˙θ˙mr2sinθcosθϕ˙2=0 This implies variations especially when initially θ0eqπ2 (not in the equatorial plane), where the angle θ 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,θ,ϕ) 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=12m(r˙2+r2θ˙2+r2sin2θϕ˙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 θ,ϕ, 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

Using the usual angle ϕ as generalized coordinate, write down the Lagrangian for a simple pendulum of length l suspended from the ceiling of an elevator that is accelerating upward with constant acceleration a. (Be careful when writing T; it is probably safest to write the bob's velocity in component form.) Find the Lagrange equation of motion and show that it is the same as that for a normal, nonaccelerating pendulum, except that g has been replaced by g+a. In particular, the angular frequency of small oscillations is (g+a)/l.

A mass m1 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 m2 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.

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 α 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 x¨. Remember that T=12mv2+12Iω2, where v is the velocity of the center of mass and ω is the angular velocity.

The "spherical pendulum" is just a simple pendulum that is free to move in any sideways direction. (By contrast a "simple pendulum"- unqualified - is confined to a single vertical plane.) The bob of a spherical pendulum moves on a sphere, centered on the point of support with radius r=R the length of the pendulum. A convenient choice of coordinates is spherical polars, r,θ,ϕ, with the origin at the point of support and the polar axis pointing straight down. The two variables θ and ϕ make a good choice of generalized coordinates. (a) Find the Lagrangian and the two Lagrange equations. (b) Explain what the ϕ equation tells us about the z component of angular momentum z (c) For the special case that ϕ= const, describe what the θ equation tells us. (d) Use the ϕ equation to replace ϕ˙ by z in the θ equation and discuss the existence of an angle θo at which θ can remain constant. Why is this motion called a conical pendulum? (e) Show that if θ=θ0+ϵ, with ϵ small, then θ oscillates about θo in harmonic motion. Describe the motion of the pendulum's bob.

Consider the well-known problem of a cart of mass m moving along the x axis attached to a spring (force constant k ), whose other end is held fixed (Figure 5.2 ). If we ignore the mass of the spring (as we almost always do) then we know that the cart executes simple harmonic motion with angular frequency ω=k/m. Using the Lagrangian approach, you can find the effect of the spring's mass M, as follows: (a) Assuming that the spring is uniform and stretches uniformly, show that its kinetic energy is 16Mx˙2. (As usual x is the extension of the spring from its equilibrium length.) Write down the Lagrangian for the system of cart plus spring. (Note: The potential energy is still 12kx2.) (b) Write down the Lagrange equation and show that the cart still executes SHM but with angular frequency ω=k/(m+M/3); that is, the effect of the spring's mass M is just to add M/3 to the mass of the cart.

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