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Conservative Forces

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Conservative Forces Inertial frames It is useful at this point to introduce the concept of a frame of reference. To specify positions and time, each observer may choose a zero of the time scale, an origin in space, and a set of three Cartesian co-ordinate axes. We shall refer to these collectively as a frame of reference. The position and time of any event may then be specified with respect to this frame by the three Cartesian co-ordinates x, y, z and the time t. If we are located on a solid body, such as the Earth, we may, for example, choose some point of the body as the origin, and take the axes to be rigidly fixed to it (though, this frame is not quite unaccelerated). Formally, an inertial frame may be defined to be one with respect to which any isolated body, far removed from all other matter, would move with uniform velocity. This is of course an idealized definition, since in practice we can never get infinitely far away from other matter. For all practical purposes, however, an inertial frame is one whose orientation is fixed relative to the ‘fixed’ stars, and in which the Sun (or more precisely the centre of mass of the solar system) moves with uniform velocity. Energy; Conservative Forces The kinetic energy of a particle of mass m free to move in three dimensions is defined to be (1) The rate of change of the kinetic energy is, therefore, ˙ (2) by the equation of motion (1). The change in kinetic energy in a time interval dt during which the particle moves a (vector) distance dr is then , (3) with . (4) This is the three-dimensional expression for the work done by the force in the displacement . Note that it is equal to the distance travelled, | |, multiplied by the component of in the direction of the displacement. One might think at first sight that a conservative force in three dimensions should be defined to be a force ( ) depending only on the position r of the particle. However, this is not sufficient to 1 ensure the existence of a law of conservation of energy, which is the essential feature of a conservative force. We require that , (5) where is given by (1), and the potential energy is a function of position, ( ). Now, the rate of change of the potential energy is , or, in terms of the gradient of , , by . (6) Thus, differentiating (5), and using (2) and (6) for and , we obtain . Since this must hold for any value of the velocity of the particle, the quantity in parentheses must vanish, i.e. (7) In terms of components, it reads . (8) For example, suppose that has the form , (9) which describes a three-dimensional harmonic oscillator. Then (8) yields . (10) It is not hard to find the necessary and sufficient condition on the force to ensure the existence of a potential energy function satisfying (7). Any vector function of the form (7) obeys the relation , (11) 2 that is, its curl vanishes. For instance, the z component of (11) is , and this is true because of (8) and the symmetry of the second partial derivative: . Thus (11) is a necessary condition for the force to be conservative. It can be shown that it is also a sufficient condition. More specifically, if the force satisfies this condition, then the work done by the force in a displacement from to is independent of the path chosen between these points. Therefore, we can define the potential energy in terms of the force by a relation analogous to the one dimensional case ( ) . Projectiles As a first example of motion in two dimensions let us consider a projectile moving under gravity. The equation of motion, assuming that atmospheric friction is negligible, is , (12) where is the acceleration due to gravity, a vector of constant magnitude g =9 .81ms−2 directed in the downward vertical direction. (We ignore the small effects due to the Earth’s rotation, as well as the variation of with height.) If we choose the z-axis vertically upwards, then , so that (12) takes the form . (13) If we choose the x- and y-axes so that initially the particle is moving in the xz-plane, then it will do so always. Thus the motion is effectively two-dimensional. The solutions of (13) are , where are constants to be fixed by the initial conditions. Example: Projectile range A projectile is launched from the surface of the Earth (assumed flat) with velocity at an angle to the horizontal. What is its range? We may choose the point of launch as the origin, and the time of launch as . At that instant, . 3 Thus the solution with the required initial conditions is . (14) (The trajectory in the xz-plane is obtained by eliminating t between these equations, and is easily seen to be a parabola.) The projectile will hit the ground when again, i.e., at the time . Substituting into the first of Eqs. (14) yields for the range . (15) It is interesting to note that to maximize the range we should choose ; the maximum value is . Finally, let us consider the much less trivial problem of a projectile subject also to a resistive force, such as atmospheric drag. We assume that the retarding force is proportional to the velocity, so that the equation of motion becomes . (16) The actual dependence of atmospheric drag on velocity is definitely nonlinear, but this equation nevertheless gives a reasonable qualitative picture of the motion. Defining , we may write (16) as . These equations may be integrated by the use of integrating factors. If the initial velocity is , the required solution is . Eliminating between these two equations yields the equation of the trajectory, . (17) A distinctive feature of this equation is that, even if the launch point is far above the ground, there is a maximum value of . The projectile can never reach beyond , and as it approaches that value, . Thus the trajectory ends in a near-vertical drop, when all the forward momentum has been spent. Moments; Angular Momentum The moment about the origin of a force acting on a particle at position is defined to be the vector product 4 . (18) The components of the vector are the moments about the , . The direction of the vector is that of the normal to the plane of and . It may be regarded as defining the axis about which the force tends to rotate the particle. The magnitude of is , where is the angle between and , and is the perpendicular distance from the origin to the line of the force. (See Fig. below) Moments of forces play a particularly important role in the dynamics of rigid bodies. Correspondingly, we define the vector angular momentum (sometimes called moment of momentum) about the origin of a particle at position , and moving with momentum , to be . (19) Its components, the angular momenta about the , are . The momentum is often called linear momentum when it is important to emphasize the distinction between and . The rate of change of angular momentum, obtained by differentiating (19) is . (20) 5 (Recall that the usual rule for differentiating a product applies also to vector products, provided that the order of the two factors is preserved.) Now, the first term in (20) is zero, because it is the vector product of a vector with itself. By Newton’s second law, the second term is simply . Thus we obtain the important result that the rate of change of the angular momentum is equal to the moment of the applied force: . (21) This should be compared with the equation for the rate of change of the linear momentum. Since the definition of the vector product depends on the choice of a right-hand screw convention, the directions of the vectors and also depend on this convention. A vector of this type is known as an axial vector. It is to be contrasted with an ordinary, or polar, vector. Axial vectors are often associated with rotation about an axis. What is specified physically is not the direction along the axis, but the sense of rotation about it (see Fig.above). Central Forces; Conservation of Angular Momentum An external force is said to be central if it is always directed towards or away from a fixed point, called the centre of force. If we choose the origin to be this centre, this means that F is always parallel to the position vector . Since the vector product of two parallel vectors is zero, the condition for a force to be central is that its moment about the centre should vanish: (22) From (21) it follows that if the force is central, the angular momentum is a constant: . (23) This is the law of conservation of angular momentum in its simplest form. It really contains two statements: that the direction of is constant, and that its magnitude is constant. Let us look at these in turn. The direction of is that of the normal to the plane of and . Hence, the statement that this direction is fixed implies that and must always lie in a fixed plane. In other words, the motion of the particle is confined to the plane containing the initial position vector and velocity vector (see Fig. below). This is obvious physically; for, since the force is central, it has no component perpendicular to this plane, and, since the normal component of velocity is initially zero, it must always remain zero. 6 To understand the meaning of the second part of the law, the constancy of the magnitude of , it is convenient to introduce polar co-ordinates and in the plane of the motion.
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