Moment of linear momentum

Conservation of Moment of Linear Momentum and Symmetry of Stress... 

Let Af be a position vector in the deformed body then the total moment of linear momentum of the body with respect to the origin O is calculated by... 

The conservation law for the moment of linear momentum states that H = T therefore we have... 

That is, if we have the conservation law of moment of linear momentum and assume no point-wise source term of moment, the Cauchy stress is symmetric. A further exposition of the symmetry property of the Cauchy stress is given by Selvadurai... 

The governing equations that control material responses are given by the mass conservation law (2.97) and the equation of motion (2.104) if no energy conservation is considered. Note that the Cauchy stress is symmetric under the conservation law of moment of linear momentum. Furthermore, if the change of mass density is small (or it may be constant), the equation to be solved is given by (2.104). The unknowns in this equation are the velocity v (or displacement u in the small strain theory) and the stress a, i.e. giving a total of nine, that is, three for v (or u) and six for three components, therefore it cannot be solved, suggesting that we must introduce a relationship between v (or u) and [Pg.40]

The conservation of moment of linear momentum for a mixture is introduced in the same way as in Sect. 2.5, and we conclude that the averaged stress [Pg.130]

It is, perhaps, well to pause for a moment to take stock of our developments to this point. We have successfully derived DEs that must be satisfied by any velocity field that is consistent with conservation of mass and Newton s second law of mechanics (or conservation of linear momentum). However, a closer look at the results, (2-5) or (2-20) and (2 32), reveals the fact that we have far more unknowns than we have relationships between them. Let us consider the simplest situation in which the fluid is isothermal and approximated as incompressible. In this case, the density is a constant property of the material, which we may assume to be known, and the continuity equation, (2-20), provides one relationship among the three unknown scalar components of the velocity u. When Newton s second law is added, we do generate three additional equations involving the components of u, but only at the cost of nine additional unknowns at each point the nine independent components of T. It is clear that more equations are needed. 

If we denote by T, Cauchy s stress tensor of our material and by b, the density of body forces, then by Truesdell s third principle the balances of linear momentum and of moment of momentum for the whole mixture in local form turn out to be... 

Angular Momentum (Moment of Momentum). Angular momentum is linear momentum (kg-m/s) times moment arm (m). Its SI unit is kg-m /s. For a rotating body the total angular momentum is equal to the moment of inertia I (kg-m ) times the angular velocity CO (rad/s or 1/s). 

Let us consider systems which consist of a mixture of spherical atoms and rigid rotators, i.e., linear N2 molecules and spherical Ar atoms. We denote the position (in D dimensions) and momentum of the (point) particles i with mass m (modeling an Ar atom) by r, and p, and the center-of-mass position and momentum of the linear molecule / with mass M and moment of inertia I (modeling the N2 molecule) by R/ and P/, the normalized director of the linear molecule by n/, and the angular momentum by L/. 

Bahn, /. way, road, track path orbit trajectory railway breadth (of cloth), bahnbrechend, p.a. pioneer, epoch-making. Bahn-brecher, m. pioneer, -durchmesser, m. orbital diameter, -ebene, /. orbital plane, -elektron, n. orbital electron, bahnen, v.t. beat, smooth, clear (a way). Bahn-hof, m. (railway) station, -impuls, m. linear momentum orbital moment, -sdileife, -schlinge, /., orbital loop, -spur, /, track, -tibergang, m. orbital transition, -zug, m. railway train. 

Without essential limitation of generality it may be assumed that the orientation of the molecule and its angular momentum are changed by collision independently, therefore F(JU Ji+, gt) = f (Jt, Ji+i)ip(gi). At the same time the functions /(/ , Ji+ ) and xp(gi) have common variables. There are two reasons for this. First, it may be due to the fact that the angle between / and u must be conserved for linear rotators for any transformation. Second, a transformation T includes rotation of the reference system by an angle sufficient to combine axis z with vector /. After substitution of (A7.16) and (A7.14) into (A7.13), one has to integrate over those variables from the set g , which are not common with the arguments of the function / (/ , /j+i). As a result, in the MF operator T becomes the same for all i and depends on the moments of tp as parameters. 

This is the classical-mechanical Hamiltonian for the rotation of a rigid body. [Note the similarity of (5.20), (5.22), and (5.23) to the corresponding equations for linear motion we get the equations for rotational motion by replacing the velocity v with the angular velocity to, the linear momentum p with the angular momentum P, and the mass with the principal moments of inertia.)... 

Balance equations for angular momentum, or moment of momentum, may also be written. They are used less frequently than the linear momentum equations. See Whitaker (Introduction to Fluid Mechanics, Prentice-Hall, Englewood Cliffs, N.J., 1968, Krieger, Huntington, N.Y., 1981) or Shames (Mechanics of Fluids, 3d ed., McGraw-Hill, New York, 1992). 

In the impulse model, the excess energy Ek is transferred to an NO molecule as the momentum p0 given only to an N atom. Here, p0 is normal to the surface and Ek = p /2m, where m is mass of the N atom. Recoil of substrate Pt atoms can be ignored, because the mass of a Pt atom is much larger than that of an N atom. After desorption the momentump0 is converted to the linear momentum of the center of mass, P, and the linear momentum of the internal coordinate, p. A relationship p0 = m dri/df is satisfied in the impulse model and it can be approximated to dr2/df = 0 at the moment of the Pt-N bond breaking, where and r2 are the position vectors of N and O atoms, respectively, in an adsorbed NO molecule. 

Moment of Inertia. We next discuss momentum, rotation, torque, moment of inertia, and angular momentum. A body with velocity v and mass m is defined to have (linear) momentum p ... 

All the correlation functions above are normalized, therefore equations (4 and 5) are identical to correlation functions over linear momentum p = mv and angular momentum J — lu, respectively. Note that, in this context I is the moment of inertia tensor The correlation function in equation (6) is calculated over the spherical harmonics. If m = 0, this reduces to time correlation function over Legendre polynomials ... 

From the above derivation it is seen that after the series expansion of the exponential in the space part of the vector potential, the transition moment operator involves the linear momentum operator fij or the gradient operator Equation (1.32) is obtained from Equation (1.30) in the following way From the commutation relation [A, rj = Htj - rfi = fij, we have... 

In linear motion, we are concerned with the momentum p = mv of an object as it heads toward a particular point the linear momentum measures the impact that the object can transfer in a collision as it arrives at the point. To extend this concept to circular motion, we define the angular momentum of an object as it revolves around a point as L = mvr. This is in effect the moment of the linear momentum over the distance r, and it is a measure of the torque felt by the object as it executes angular motion. The angular momentum of an electron around a nucleus is a crucial feature of atomic structure, which is discussed in Chapter 5. 

The electric and magnetic moments, <01// a> and , are based on the linear and angular momentum of electrons involved in the transition. The angular momentum corresponds to the rotational motion of an electron, while the linear momentum corresponds to the linear motion of the electron. If the electric and magnetic moment vectors, <01// a> and , are parallel to each other in one enantiomer, they should be antiparallel in the other enantiomer. Therefore, the rotational strength R, OR [a]A, and CD spectra of enantiomers are opposite in sign but of equal intensity. The problem of how to determine the ACs of... 

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