J. Peraire 16.07 Dynamics Fall 2004 Version 1.2 Lecture D5 - Other Coordinates Systems In this lecture we will look at someother common systems of coordinates. We will present polar coordinates in two dimensions and cylindrical and spherical coordinates in three dimensions. We shall see that these systems are particularly useful for certain classes of problems. Like in the case of intrinsic coordinates presented in the previous lecture, the reference frame changes from point to point. However, for the coordinate systems to be presented below, the reference frame depends only on the position of the particle. This is in contrast with the intrinsic coordinates, where the reference frame is a function of the position, as well as the path. Polar Coordinates (r ?θ) In polar coordinates, the position of a particle A, is determined by the value of the radial distance to the origin, r, and the angle that the radial line makes with an arbitrary fixed line, such as the x axis. Thus, the trajectory of a particle will be determined if we know r and θ as a function of t, i.e. r(t),θ(t). The directions of increasing r and θ are defined by the orthogonal unit vectors er and eθ. The position vector of a particle has a magnitude equal to the radial distance, and a direction determined by er. Thus, r = rer . (1) Since the vectors er and eθ are clearly different from point to point, their variation will have to be considered when calculating the velocity and acceleration. Over an infinitesimal interval of time dt, the coordinates of point A will change from (r,θ), to (r+dr, θ+dθ) as shown in the diagram. 1 We note that the vectors er and eθ do not change when the coordinate r changes. Thus, der/dr = 0 and deθ/dr = 0. On the other hand, when θ changes to θ + dθ, the vectors er and eθ are rotated by an angle dθ. From the diagram, we see that der = dθeθ, and that deθ = ?dθer. This is because their magnitudes in the limit are equal to the unit vector as radius times dθ in radians. Dividing through by dθ, we have, der dθ = eθ, and deθ dθ = ?er . Multiplying these expressions by dθ/dt ≡ ˙θ, we obtain, der dθ dθ dt ≡ der dt = ˙θeθ, and deθ dt = ? ˙θer . (2) Note Alternative calculation of the unit vector derivatives An alternative, more mathematical, approach to obtaining the derivatives of the unit vectors is to express er and eθ in terms of their cartesian components along i and j. We have that er = cosθi+ sinθj eθ = ?sinθi + cosθj . Therefore, when we differentiate we obtain, der dr = 0, der dθ = ?sinθi+ cosθj ≡ eθ deθ dr = 0, deθ dθ = ?cosθi?sinθj ≡?er . Velocity vector We can now derive expression (1) with respect to time and write v = ˙r = ˙rer + r ˙er , or, using expression (2), we have v = ˙rer + r ˙θeθ . (3) 2 Here, vr = ˙r is the radial velocity component, and vθ = r ˙θ is the circumferential velocity component. We also have that v =radicalbigv2r + v2θ. The radial component is the rate at which r changes magnitude, or stretches, and the circumferential component, is the rate at which r changes direction, or swings. Acceleration vector Differentiating again with respect to time, we obtain the acceleration a = ˙v = ¨rer + ˙r ˙er + ˙r ˙θeθ + r¨θeθ + r ˙θ ˙eθ Using the expressions (2), we obtain, a = (¨r?r ˙θ2)er + (r¨θ + 2˙r˙θ)eθ , (4) where ar = (¨r ? r ˙θ2) is the radial acceleration component, and aθ = (r¨θ + 2˙r˙θ) is the circumferential acceleration component. Also, we have that a =radicalbiga2r + a2θ. Change of basis In many practical situations, it will be necessary to transform the vectors expressed in polar coordinates to cartesian coordinates and vice versa. Since we are dealing with free vectors, we can translate the polar reference frame for a given point (r,θ), to the origin, and apply a standard change of basis procedure. This will give, for a generic vector A, ? ? Ar Aθ ? ?= ? ? cosθ sinθ ?sinθ cosθ ? ? ? ? Ax Ay ? ? and ? ? Ax Ay ? ?= ? ?cosθ ?sinθ sinθ cosθ ? ? ? ? Ar Aθ ? ? . Example Circular motion Consider as an illustration, the motion of a particle in a circular trajectory having angular velocity ω = ˙θ, and angular acceleration α = ˙ω. We see that, for this problem, the circumferential and radial directions are very similar to the intrinsic tangential and normal directions. The only difference is that in polar coordinates, the radial direction points outwards, whereas, in intrinsic coordinates, the normal direction always points towards the center of curvature O. 3 In polar coordinates, the equation of the trajectory is r = R = constant, θ = ωt+ 12αt2. The velocity components are vr = ˙r = 0, and vθ = r ˙θ = R(ω + αt) = v , and the acceleration components are, ar = ¨r?r ˙θ2 = ?R(ω + αt)2 = ?v 2 R , and aθ = r ¨θ + 2˙r˙θ = Rα = at , where we clearly see that, ar ≡?an, and that aθ ≡ at. In cartesian coordinates, we have for the trajectory, x = Rcos(ωt+ 12αt2), x = Rsin(ωt+ 12αt2) . For the velocity, vx = ?R(ω + αt)sin(ωt+ 12αt2), vy = R(ω + αt)cos(ωt+ 12αt2) , and, for the acceleration, ax = ?R(ω+αt)2 cos(ωt+12αt2)?Rαsin(ωt+12αt2), ay = ?R(ω+αt)2 sin(ωt+12αt2)+Rαcos(ωt+12αt2) . We observe that, for this problem, the result is much simpler when expressed in polar (or intrinsic) coordi- nates. Example Motion on a straight line Here we consider the problem of a particle moving with constant velocity v0, along a horizontal line y = y0. 4 Assuming that at t = 0 the particle is at x = 0, the trajectory and velocity components in cartesian coordinates are simply, x = v0t y = y0 vx = v0 vy = 0 ax = 0 ay = 0 . Note that, for this problem, cartesian and intrinsic coordinates are virtually identical (for a straight line the normal direction is not defined, in which case we can arbitrarily choose any direction perpendicular to the tangent direction). In polar coordinates, on the other hand, we have, r = radicalBig v20t2 + y20 θ = tan?1( y0v 0t ) vr = ˙r = v0 cosθ vθ = r˙θ = ?v0 sinθ ar = ¨r ?r˙θ2 = 0 aθ = r¨θ + 2˙r˙θ = 0 . Here, we see that the expressions obtained in cartesian coordinates are simpler than those obtained using polar coordinates. Example Spiral motion (Kelppner/Kolenkow) A particle moves with ˙θ = ω = constant and r = r0eβt, where r0 and β are constants. 5 We shall show that for certain values of β, the particle moves with ar = 0. a = (¨r?r ˙θ2)er + (r¨θ + 2˙r˙θ)eθ = (β2r0eβt ?r0eβtω2)er + 2βr0ωeβteθ If β = ±ω, the radial part of a vanishes. It seems quite surprising that when r = r0eβt, the particle moves with zero radial acceleration. The error is in thinking that ¨r makes the only contribution to ar; the term ?r˙θ2 is also part of the radial acceleration, and cannot be neglected. The paradox is that even though ar = 0, the radial velocity vr = ˙r = r0βeβt is increasing rapidly in time. In polar coordinates vr negationslash= integraldisplay ar(t)dt , because this integral does not take into account the fact that er and eθ are functions of time. Equations of Motion In two dimensional polar rθ coordinates, the force and acceleration vectors are F = Frer + Fθeθ and a = arer + aθeθ. Thus, in component form, we have, Fr = mar = m(¨r?r ˙θ2) Fθ = maθ = m(r¨θ + 2˙r˙θ) . Cylindrical Coordinates (r ?θ ?z) Polar coordinates can be extended to three dimensions in a very straightforward manner. We simply add the z coordinate, which is then treated in a cartesian like manner. Every point in space is determined by the r and θ coordinates of its projection in the xy plane, and its z coordinate. The unit vectors er, eθ and k, expressed in cartesian coordinates, are, er = cosθi+ sinθj eθ = ?sinθi + cosθj , 6 and their derivatives, ˙er = ˙θeθ, ˙eθ = ?˙θer, ˙k = 0 . The kinematic vectors can now be expressed relative to the unit vectors er, eθ and k. Thus, the position vector is r = rer + zk , and the velocity, v = ˙rer + r ˙θeθ + ˙zk , where vr = ˙r, vθ = r ˙θ, vz = ˙z, and v =radicalbigv2r + v2θ + v2z. Finally, the acceleration becomes a = (¨r?r˙θ2)er + (r¨θ + 2˙r˙θ)eθ + ¨zk , where ar = ¨r ?r˙θ2, aθ = r¨θ + 2˙r ˙θ, az = ¨z, and a =radicalbiga2r + a2θ + a2z. Note that when using cylindrical coordinates, r is not the modulus of r. This is somewhat confusing, but it is consistent with the notation used by most books. Whenever we use cylindrical coordinates, we will write |r| explicitly, to indicate the modulus of r, i.e. |r| = √r2 + z2. Equations of Motion In cylindrical rθz coordinates, the force and acceleration vectors are F = Frer + Fθeθ + Fzez and a = arer + aθeθ + azez. Thus, in component form we have, Fr = mar = m(¨r?r ˙θ2) Fθ = maθ = m(r¨θ + 2˙r˙θ) Fz = maz = m ¨z . Spherical Coordinates (r ?θ ?φ) In spherical coordinates, we utilize two angles and a distance to specify the position of a particle, as in the case of radar measurements, for example. 7 The unit vectors written in cartesian coordinates are, er = cosθ cosφi + sinθ cosφj + sinφk eθ = ?sinθi+ cosθj eφ = ?cosθ sinφi?sinθ sinφj + cosφk The derivation of expressions for the velocity and acceleration follow easily once the derivatives of the unit vectors are known. In three dimensions, the geometry is somewhat more involved, but the ideas are the same. Here, we give the results for the derivatives of the unit vectors, ˙er = ˙θcosφ eθ + ˙φ eφ , ˙eθ = ?˙θcosφ er + ˙θsinφ eφ , ˙eφ = ?˙φ er ? ˙θsinφ eθ , and for the kinematic vectors r = rer v = ˙rer + r ˙θcosφeθ + r ˙φeφ a = (¨r ?r ˙θ2 cos2 φ?r ˙φ2)er + (2˙r ˙θcosφ + r¨θcosφ?2r˙θ ˙φsinφ)eθ + (2˙r ˙φ + r ˙φ2 sinφ cosφ + r¨φ) eφ . Equations of Motion Finally, in spherical rθφ coordinates, we write F = Frer +Fθeθ +Fφeφ and a = arer +aθeθ +aφeφ. Thus, Fr = mar = m(¨r?r ˙θ2 cos2 φ?r ˙φ2) Fθ = maθ = m(2˙r˙θcosφ + r¨θcosφ?2r˙θ ˙φsinφ) Fφ = maφ = m(2˙r ˙φ+ r ˙φ2 sinφcosφ + r¨φ) . ADDITIONAL READING J.L. Meriam and L.G. Kraige, Engineering Mechanics, DYNAMICS, 5th Edition 2/6, 2/7, 3/5 8