Kinematics on oblique axes

Size: px

Start display at page:

Download "Kinematics on oblique axes"

Roger Paul
5 years ago
Views:

1 Bolina 1 Kinematics on oblique axes Oscar Bolina Departamento de Física-Matemática Uniersidade de São Paulo Caixa Postal São Paulo Brasil ; bolina@if.usp.br Abstract We sole a difficult problem inoling elocity acceleration components along oblique axes, propose two problems of central force motion to be soled using oblique axes. Key Words: Oblique axes, plane motion. PACS numbers: d, j Copyright c 2001 by the author. Reproduction of this article in its entirety is permitted for non-commercial purposes.

2 Bolina 2 1 Introduction Any ector quantity can be resoled into components according to the parallelogram law, both by rectangular oblique resolution. In rectangular coordinates, one component is perpendicular to the other. In oblique coordinates, one component has a projection on the other. We hae to take this projection into account when finding elocity acceleration components along these axes. Because of this, problems inoling oblique axes are ery difficult. Howeer, once one realizes that a problem requires oblique axes, the solution is not in general that hard, although the deriation of the kinematics on oblique axes is somewhat disgusting. 2 The elocity components Consider the motion of a particle in a plane. Suppose that the geometry of the motion is such that the elocity of the particle is more coneniently referred to two oblique axes Oξ Oη which make angles respectiely with a fixed direction Ox in the plane, as shown in Fig. 1. Theses angles may ary arbitrarily with time as the particle moes. Suppose that at the time t the components of the elocity in the directions Oξ Oη are u, respectiely. The perpendicular projections of these components along Oξ Oη are, respectiely, u + cos( ) + u cos( ), (2.1) as shown in the figure for the projection on the Oξ axis only. At the time t + t the axes Oξ Oη take the positions ξ η, asshown in Fig. 2, with ξoξ = ηoη =. Let the components of the elocity along these axes at this time be u + u +. The perpendicular projections of these components along the axes Oξ Oη are, respectiely, (u + u)cos +( + )cos( + ) ( + )cos +(u + u)cos( ). (2.2)

3 Bolina 3 Οη u Οξ cos ( ) O x Figure 1: Oblique coordinate system the components of the elocity along oblique axes. Each component has a projection on the other axis. By taking the difference between the projections (2.2) (2.1) of the elocities along the axes Oξ Oη at the corresponding times t + t t, diiding the result by t, letting t go to zero, we obtain the projections of the acceleration along those axes at the time t: u + cos( ) sin( ) (2.3) + u cos( )+u sin( ), (2.4) where u is the limiting alue of u/ t, when t approaches zero. 3 The acceleration components Now let a ξ a η represent the components of the acceleration of the particle along Oξ Oη at the time t. The same relationship (2.1) for elocities hold for accelerations. Thus, the perpendicular projections of the components

4 Bolina 4 η Οη u+ u u ξ Οξ Figure 2: The component of the elocity u + u along the new axis ξ is projected on the old axis Oξ. along the axes Oξ Oη are a ξ + a η cos( ) a η + a ξ cos( ). (3.5) On equating (2.2) (3.5) we obtain a ξ + a η cos( ) = u + cos( ) sin( ) a η + a ξ cos( ) = + u cos( )+u sin( ), (3.6) from which we can sole for a ξ a η. The first two terms in equations (3.6) are the rates of change of the projections (2.1) along fixed axes. The last terms are the consequence of the motion of the axes themseles. (Reference [2] suggests an alternatie approach to obtaining these equations.)

5 Bolina 5 Οη line circle Οξ u point Figure 3: An example of a difficult problem whose solution depends on the kinematics of oblique axes. 4 A difficult problem As an illustration, consider the following mind boggling problem [3]: A circle, a straight line, a point lie in one plane, the position of the point is determined by the lengths of its tangent to the circle p of its perpendicular to the line. Proe that, if the elocity of the point in made up of components u, in the directions of these lengths, if their mutual inclination is θ, the component accelerations will be u u cos θ, u +. Solution. Take the axis Oξ to be the tangent to the circle, Oη to be the axis perpendicular to the line. Set θ = note that does not ary with time. It is then easy to check that equations (3.6), with due change in notation, reduce to the following set of equations a t + a p cos θ = u + cos θ

6 Bolina 6 line circle O R t Q P Figure 4: The geometry of the illustratie problem. The particle moes from the point O to the point P. Its position is determined by the tangent line to the circle also by a perpendicular to a gien line in the plane. The tangent lines at two different times meet at Q make an angle. a p + a t cos θ = + u cos θ + u sin θ. (4.7) If we sole (4.7) for a t a p we obtain a t = u cos θ sin θ u a p = + u sin θ. (4.8) To eliminate the ariable we need to consider the (messy) geometry of the problem. Let the two tangent lines to the circle, drawn from the two positions of the particle at O P, meet at a point Q, asshown in Fig. 4. Note that the lines OQ PQ form an angle with each other at Q. Next, draw from the point P a perpendicular to the gien line. Let this perpendicular meet the line OQ at a point R. The perpendicular PR makes an angle with OQ. For small t, P is near O, we hae, approximately, PQ = PR = t. Thelaw

7 Bolina 7 of sines, applied to the triangle PQR,gies t = sin θ or sin θ =. (4.9) Substituting gien aboe in (4.8), we obtain the desired result. 4.1 Further examples The equations for elocity acceleration components on oblique axes can be used to proide solution to problems of motion in a central force field when these problems are phrased as follows. 1. A particle P moes in a plane in such a way that its elocity has two constant components u, with u parallel to a fixed direction in the plane while is normal to a straight line from the particle to a fixed point O in the plane. Show that the acceleration of the particle is directed along the line OP. (In fact, the particle moes in a ellipse of eccentricity u/, haing O as a focus.) 2. A boat crosses a rier with elocity of constant magnitude u always aimed toward a point S on the opposite shore directly across its starting position. The riers also runs with uniform elocity u. Compare this problem with the preceding one. (How far downstream from S does the boat reach the opposite shore?) Acknowlgement. I was supported by Fapesp under grant 01/ References [1] R.L. Halfman, Dynamics, Addison-Wesley, 1962, p.19 [2] A.S. Ramsey, Dynamics, Part II, Cambridge Uniersity Press, (Chapter 3, p.75, example 15.) [3] E.T. Whittaker, A Treatise on the Analytical Dynamics of Particles Rigid Bodies, Fourth Edition, Doer Publication, N.Y (Chapter 1, p.24, problem 13.)

JUST THE MATHS SLIDES NUMBER 5.2. GEOMETRY 2 (The straight line) A.J.Hobson

JUST THE MATHS SLIDES NUMBER 5.2 GEOMETRY 2 (The straight line) by A.J.Hobson 5.2.1 Preamble 5.2.2 Standard equations of a straight line 5.2.3 Perpendicular straight lines 5.2.4 Change of origin UNIT 5.2