Chapter08 - Physics 235 Chapter 8 Central-Force Motion...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Physics 235 Chapter 8 - 1 - Chapter 8 Central-Force Motion In this Chapter we will use the theory we have discussed in Chapter 6 and 7 and apply it to very important problems in physics, in which we study the motion of two-body systems on which central force are acting. We will encounter important examples from astronomy and from nuclear physics. Two-Body Systems with a Central Force Consider the motion of two objects that are effected by a force acting along the line connecting the centers of the objects. To specify the state of the system, we must specify six coordinates (for example, the ( x , y , z ) coordinates of their centers). The Lagrangian for this system is given by L = 1 2 m 1 r 1 2 + 1 2 m 2 r 2 2 U r 1 r 2 ( ) Note: here we have assumed that the potential depends on the position vector between the two objects. This is not the only way to describe the system; we can for example also specify the position of the center-of-mass, R , and the three components of the relative position vector r . In this case, we choose a coordinate system such that the center-of-mass is at rest, and located at the origin. This requires that R = m 1 m 1 + m 2 r 1 + m 2 m 1 + m 2 r 2 = 0 The relative position vector is defined as r = r 1 r 2 The position vectors of the two masses can be expressed in terms of the relative position vector: r 1 = m 2 m 1 + m 2 r r 2 = m 1 m 1 + m 2 r The Lagrangian can now be rewritten as
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Physics 235 Chapter 8 - 2 - L = 1 2 m 1 m 2 m 1 + m 2 2 r 2 + 1 2 m 2 m 1 m 1 + m 2 2 r 2 U r ( ) = 1 2 μ r 2 U r ( ) where is the reduced mass of the system: = m 1 m 2 m 1 + m 2 Two-Body Systems with a Central Force: Conserved Quantities Since we have assumed that the potential U depends only on the relative position between the two objects, the system poses spherical symmetry. As we have seen in Chapter 7, this type of symmetry implies that the angular momentum of the system is conserved. As a result, the momentum and position vector will lay in a plane, perpendicular to the angular momentum vector, which is fixed in space. The three-dimensional problem is thus reduced to a two- dimensional problem. We can express the Lagrangian in terms of the radial distance r and the polar angle θ : L = 1 2 r 2 + r 2 2 ( ) U r ( ) The generalized momenta for this Lagrangian are p r = L r = r p = L = r 2 The Lagrange equations can be used to determine the derivative of these momenta with respect to time: p r = d dt L r = L r = r 2 U r p = d dt L = L = 0 The last equation tells us that the generalized momentum p θ is constant:
Background image of page 2
Physics 235 Chapter 8 - 3 - l = μ r 2 θ = constant The constant l is related to the areal velocity. Consider the situation in Figure 1. During the time interval dt , the radius vector sweeps an area dA where dA = 1 2 r 2 d Figure 1. Calculation of the areal velocity.
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 11/19/2011 for the course PHYSICS 101 taught by Professor Rollino during the Fall '08 term at Rutgers.

Page1 / 15

Chapter08 - Physics 235 Chapter 8 Central-Force Motion...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online