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Unformatted text preview: S. Widnall, J. Peraire 16.07 Dynamics Fall 2008 Version 2.0 Lecture L5  Other Coordinate Systems
In this lecture, we will look at some other 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. Polar Coordinates (r  )
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 as a function of t, i.e. r(t), (t). The directions of increasing r and are defined by the orthogonal unit vectors er and e . 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 e 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, ), to (r + dr, + d) as shown in the diagram. 1 We note that the vectors er and e do not change when the coordinate r changes. Thus, der /dr = 0 and de /dr = 0. On the other hand, when changes to + d, the vectors er and e are rotated by an angle the limit are equal to the unit vector as radius times d in radians. Dividing through by d, we have, der = e , d and de = er . d d. From the diagram, we see that der = de , and that de = der . This is because their magnitudes in Multiplying these expressions by d/dt , we obtain, der d der = e , d dt dt and de = er . dt (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 e in terms of their cartesian components along i and j. We have that er e Therefore, when we differentiate we obtain, der = 0, dr de = 0, dr der d de d =  sin i + cos j e =  cos i  sin j er . = cos i + sin j =  sin i + cos j . Velocity vector
We can now derive expression (1) with respect to time and write v = r = r er + r er , or, using expression (2), we have v = r er + r e . (3) and the circumferential component, is the rate at which r changes direction, or swings. Here, vr = r is the radial velocity component, and v = r is the circumferential velocity component. We 2 2 also have that v = vr + v . The radial component is the rate at which r changes magnitude, or stretches, 2 Acceleration vector
Differentiating again with respect to time, we obtain the acceleration a = v = r er + r er + r e + r e + r e Using the expressions (2), we obtain, a = (  r2 ) er + (r + 2r) e , r (4) r where ar = (  r2 ) is the radial acceleration component, and a = (r + 2r) is the circumferential acceleration component. Also, we have that a = a2 + a2 . r 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, ), to the origin, and apply a standard change of basis procedure. This will give, for a generic vector A, Ar cos sin Ax cos  sin Ax Ar = = . and A  sin cos Ay Ay sin cos A Example Circular motion Consider as an illustration, the motion of a particle in a circular trajectory having angular velocity = , and angular acceleration = . 3 In polar coordinates, the equation of the trajectory is r = R = constant, The velocity components are vr = r = 0, and the acceleration components are, ar = r  r2 = R( + t)2 =  v2 , R and a = r + 2r = R = at , and v = r = R( + t) = v , 1 = t + t2 . 2 where we clearly see that, ar an , and that a at . In cartesian coordinates, we have for the trajectory, 1 x = R cos(t + t2 ), 2 For the velocity, 1 vx = R( + t) sin(t + t2 ), 2 and, for the acceleration, 1 1 ax = R(+t)2 cos(t+ t2 )R sin(t+ t2 ), 2 2 nates. 1 1 ay = R(+t)2 sin(t+ t2 )+R cos(t+ t2 ) . 2 2 1 vy = R( + t) cos(t + t2 ) , 2 1 y = R sin(t + t2 ) . 2 We observe that, for this problem, the result is much simpler when expressed in polar (or intrinsic) coordi 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 . Assuming that at t = 0 the particle is at x = 0, the trajectory and velocity components in cartesian coordinates are simply, x = v0 t vx = v0 ax = 0 4 y = y0 vy = 0 ay = 0 . In polar coordinates, we have, r= 2 2 v0 t2 + y0 y0 ) v0 t v = r = v0 sin = tan1 ( vr = r = v0 cos ar = r  r2 = 0 a = r + 2r = 0 . Here, we see that the expressions obtained in cartesian coordinates are simpler than those obtained using polar coordinates. It is also reassuring that the acceleration in both the r and direction, calculated from the general twoterm expression in polar coordinates, works out to be zero as it must for constant velocitystraight line motion. Example Spiral motion (Kelppner/Kolenkow) A particle moves with = = constant and r = r0 et , where r0 and are constants. We shall show that for certain values of , the particle moves with ar = 0. a = (  r2 )er + (r + 2r)e r = ( 2 r0 et  r0 et 2 )er + 2r0 et e If = , the radial part of a vanishes. It seems quite surprising that when r = r0 et , the particle moves r2 is also part of the radial acceleration, and cannot be neglected. with zero radial acceleration. The error is in thinking that r makes the only contribution to ar ; the term The paradox is that even though ar = 0, the radial velocity vr = r = r0 et is increasing rapidly in time. In polar coordinates vr = ar (t)dt , because this integral does not take into account the fact that er and e are functions of time. 5 Equations of Motion
In two dimensional polar r coordinates, the force and acceleration vectors are F = Fr er + F e and a = ar er + a e . Thus, in component form, we have, Fr F = m ar = m (  r2 ) r = m a = m (r + 2r) . Cylindrical Coordinates (r   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 coordinates of its projection in the xy plane, and its z coordinate. The unit vectors er , e and k, expressed in cartesian coordinates, are, er e and their derivatives, er = e , e = er , k=0. = cos i + sin j =  sin i + cos j The kinematic vectors can now be expressed relative to the unit vectors er , e and k. Thus, the position vector is r = r er + z k , and the velocity, v = r er + r e + z k , where vr = r v = r, vz = z and v = , , 2 2 2 vr + v + vz . Finally, the acceleration becomes 2 a2 + a2 + az . r 6 a = (  r2 ) er + (r + 2r) e + z k , r where ar = r  r2 , a = r + 2r, az = z , and a = 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 + z 2 . Equations of Motion
In cylindrical rz coordinates, the force and acceleration vectors are F = Fr er + F e + Fz ez and a = ar er + a e + az ez . Thus, in component form we have, Fr F Fz = m ar = m (  r2 ) r = m az = m z . = m a = m (r + 2r) Spherical Coordinates (r   )
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. The unit vectors written in cartesian coordinates are, er e e = cos cos i + sin cos j + sin k =  sin i + cos j =  cos sin i  sin sin j + cos k 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 = cos e + e , e =  cos er + sin e , e =  er  sin e , 7 and for the kinematic vectors r v a = r er = r r + r cos e + r e e = (  r2 cos2  r2 ) er r + (2r cos + r cos  2r sin ) e + (2r + r2 sin cos + r) e . Equations of Motion
Finally, in spherical r coordinates, we write F = Fr er + F e + F e and a = ar er + a e + a e . Thus, Fr F F = m ar = m (  r2 cos2  r2 ) r = m a = m (2r cos + r cos  2r sin ) = m a = m (2r + r2 sin cos + r) . Application Examples
We will look at some applications of Newton's second law, expressed in the different coordinate systems that have been introduced. Recall that Newton's second law F = ma , is a vector equation which is valid for inertial observers. In general, we will be interested in determining the motion of a particle given that we know the external forces. Equation (5), written in terms of either velocity or position, is a differential equation. In order to calculate the velocity and position as a function of time we will need to integrate this equation either analytically or numerically. On the other hand, the reverse problem of computing the forces given motion is much easier and only requires direct evaluation of (5). Is is also common to have mixed type problems, in which we know some components of the force and some components of the acceleration. The goal is then to determine the remaining unknown terms. While no general rules can be given regarding the appropriate choice of a coordinate system, we note that intrinsic coordinates are particularly useful in constrained problems, where the trajectory is known beforehand. Example Aircraft flying on a helix (5) v = 211 ft/s, and = 3o 0.05rad/s. This is standard for gas turbine powered aircraft. We want to know the force on the aircraft and the radius of curvature of the path. 8 A 10, 000 lb aircraft is descending on a cylindrical helix. The rate of descent is z = 10ft/s, the speed is We have, v = r r + re + zez = vet e Since, r = R, r = 0. Therefore, 211 = (0.05R)2 + 102 , or R = 4, 215 ft. For the acceleration, v2 en , a = (  r2 )er + (r + 2r)e + z ez = v t + r e and, considering only the nonzero terms, v2 a = R2 er = en . We see that en = er , and that, a = (0.05)2 4, 215 = 10.54 ft/s = The normal force on the aircraft is Fn = man = and finally, the lift, L, is L = 3, 273 er + 10, 000 ez lb . 10, 000 10.54 = 3, 273 lb , 32
2 v2 , = 211 = 4, 225 ft . 10.54 Here we see that r which means that the helix is very tight. 9 The angle of descent is calculated as sin = z/v, or, = 2.72o . This angle is sometimes called the pitch of the helix. Example Pendulum Now, we consider a simple pendulum consisting of a mass, m, suspended from a string of length l and negligible mass. We can formulate the problem in polar coordinates, and noting that r = l (constant), write for the r and components, mg cos  T mg sin = ml2 = ml , (6) where T is the tension on the string. If we restrict the motion to small oscillations, we can approximate sin , and the equation becomes Integrating we obtain the general solution, g g (t) = C1 cos( t) + C2 sin( t) , l l where the constants C1 and C2 are determined by the initial conditions. Thus, if (0) = max , g (t) = max cos( t) . l 10 g + =0 . l Example Aircraft flying a perfect loop (Hollister) Consider an aircraft flying a perfect loop, i.e. a circle in the vertical plane. Assume that the engine thrust exactly cancels the aerodynamic drag so that the lift and gravity are the only unbalanced forces on the aircraft. This assumption makes the problem into the same dynamical model that we have used in the previous example. Since the lift, L, is perpendicular to the flight path, we have that the force on the aircraft, in normal and tangential components, is F = mg sin et + (L  mg cos ) en . Thus, at an = v = r = g sin 2 v L =  g cos . = R m (7) Since, v dv = at ds = at R d = Rg sin d. Thus, integrating,
2 v 2 = v0 + 2Rg(cos  1) , (8) where v0 is the velocity at the bottom of the loop when = 0. To be able to go over the top we need v > 0 when = . This means that we need v0 > 2 Rg. Note that for v0 < 2 Rg, we can calculate the maximum angle the aircraft can reach, max . If we set v = 0 when = max , we have,
2 v0 ). 2Rg The necessary lift, L, can be calculated as a function of . From (7) and (8), we have max = cos1 (1  L v2 v2 = + g cos = 0 + 3g cos  2g . m R R We have that, in order for to go from 0 to , the aircraft has to have a range of lift capability that extends over 5g. It turns out that most aircraft do not have this capability and consequently do not fly perfect loops. 11 ADDITIONAL READING J.L. Meriam and L.G. Kraige, Engineering Mechanics, DYNAMICS, 5th Edition 2/6, 2/7, 3/5 12 MIT OpenCourseWare http://ocw.mit.edu 16.07 Dynamics
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This note was uploaded on 05/02/2011 for the course DYNAM 101 taught by Professor Matuka during the Spring '11 term at MIT.
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