Unformatted text preview: S. Widnall, J. Peraire 16.07 Dynamics Fall 2009 Version 2.0 Lecture L6  Intrinsic Coordinates
In lecture L4, we introduced the position, velocity and acceleration vectors and referred them to a fixed cartesian coordinate system . Then we showed how they could be expressed in polar coordinates. While it is clear that the choice of coordinate system does not affect the final answer, we shall see that, in practical problems, the choice of a specific system may simplify the calculations and/or improve the understanding considerably. In many problems, it is useful to use a coordinate system that aligns with both the velocity vector of the particlewhich is of course tangent to the particle's trajectory and the normal to this trajectory, forming a pair of orthogonal unit vectors. The unit vectors aligned with these two directions also define a third direction, call the binormal which is normal to both the velocity vector and the normal vector. This can be obtained from these two unit vectors by taking their cross product. Such a coordinate system is called an Intrinsic Coordinate System. Intrinsic coordinates: Tangential, Normal and Binormal components.
We follow the motion of a point using a vector r(t) whose position along a known curved path is given by the scalar function s(t) where s(t) is the arc length along the curve. We obtain the velocity v from the time rate of change of the vector r(t) following the particle v= dr dr ds dr = =s dt ds dt ds
dr ds (1) as the unit vector tangent to the curve (2) We identify the scalar s as the magnitude of the velocity v and at the point s(t). Therefore we have v = vet , where r(t) is the position vector, v = s is the speed, et is the unit tangent vector to the trajectory, and s is the path coordinate along the trajectory. 1 The unit tangent vector can be written as, et = dr . ds
dr ds (3) depends only The acceleration vector is the derivative of the velocity vector with respect to time. Since on s, using the chain rule we can write dv dr d a= =s +s dt ds dt curvature. dr ds =s dr d2 r dr d2 r + s2 2 = v + v2 2 . ds ds ds ds (4) The second derivative d2 r/ds2 is another property of the path. We shall see that it is relate to the radius of Taking the time derivative of Equation (2), an alternate expression can be written in terms of the unit vector et as a= dv det et + v . dt dt (5) The vector et is the local unit tangent vector to the curve which changes from point to point. Consequently, the time derivative of et will, in general, be nonzero. The time derivative of et can be written as, det det ds det = = v. dt ds dt ds (6) In order to calculate the derivative of et , we note that, since the magnitude of et is constant and equal to one, the only changes that et can have are due to rotation, or swinging. When we move from s to s + ds, the tangent vector changes from et to et + det . The change in direction can be related to the angle d. 2 The direction of det , which is perpendicular to et , is called the normal direction. On the other hand, the magnitude of det will be equal to the length of et (which is one), times d. Thus, if en is a unit normal vector in the direction of det , we can write det = den . Dividing by ds yields, (7) det d 1 = en = en = en . ds ds (8) Here, = d/ds is a local property of the curve, called the curvature, and = 1/ is called the radius of curvature. Note that in the picture, the sizes of det , ds, and d are exaggerated for illustration purposes and actually represent the changes in the limit as ds (and also dt) approach zero. Note Curvature and radius of curvature We consider here two tangent vectors et and e + det , separated by a small ds and having an angle between them of d. If we draw perpendiculars to these two vectors, they will intersect at a point, say, O . Because the two lines meeting at O are perpendicular to each of the tangent vectors, the angle between them will be the same as the angle between et and e + det , d. The point O is called the center of curvature, and the distance, , between O and A is the radius of curvature. Thus, from the sketch, we have that ds = d, or d/ds = = 1/. Intuitively, we can see that each infinitesimal arc, ds, can be represented by a circle segment of radius having its center at the center of curvature. It is clear that both the radius of curvature and the center of curvature are functions of s, and consequently they change from point to point. There are two limiting cases which are of interest. When the trajectory is a circle, the center of curvature does not change and coincides with the center of the circle, and the radius of curvature is equal to the radius 3 of the circle. On the other hand, when the trajectory is a straight line, the curvature is zero, and the radius of curvature is infinite. Note also, that, in this case, the derivative of et is always zero, and the normal direction is not defined. Going back to expression (6), we have that det d v = v en = en = en . dt ds Finally, we have that the acceleration, can be written as a= dv v2 et + en = at et + an en . dt (10) (9) Here, at = v is the tangential component of the acceleration, and an = v 2 /, is the normal component of , the acceleration. Since an is the component of the acceleration pointing towards the center of curvature, it is sometimes referred to as centripetal acceleration. When at is nonzero, the velocity vector changes magnitude, or stretches. When an is nonzero, the velocity vector changes direction, or swings. The modulus of the total acceleration can be calculated as a = a2 + a2 . t n Example A ball is ejected horizontally from the tube with a speed v0 . The only acceleration on the ball is due to gravity. We want to determine the radius of curvature of the trajectory just after the ball is released. The simplest way to determine the radius of curvature is to note that, initially, the only nonzero component
2 of the acceleration will be in the normal direction, i.e. an = g. Thus, from an = v0 /, we have that, = 2 v0 . g 4 Alternatively, we can obtain an equation for the trajectory of the form y = f (x) and use expression (11) to calculate the curvature. The trajectory is given as, x = v0 t 1 y =  gt2 . 2 Thus, eliminating t, we have y= g 2 2x . 2v0 2 2 At x = 0, dy/dx = 0, d2 y/dx2 = g/v0 , and the above expression gives, = v0 /g, as expected. Note Relationship between s, v and at The quantities s, v and at are related in the same manner as the quantities s, v and a for rectilinear motion. , In particular we have that v = s at = v and at ds = v dv. This means that if we have a way of knowing at , , we may be able to integrate the tangential component of the motion independently. We will be exploiting these relations in the future. The vectors et and en , and their respective coordinates t and n, define two orthogonal directions. The plane defined by these two directions, is called the osculating plane. This plane changes from point to point, and can be thought of as the plane that locally contains the trajectory (Note that the tangent is the current direction of the velocity, and the normal is the direction into which the velocity is changing). In order to define a righthanded set of axes we need to introduce an additional unit vector which is orthogonal to et and en . This vector is called the binormal, and is defined as eb = et en . At any point in the trajectory, the position vector, the velocity and acceleration can be referred to these axes. In particular, the velocity and acceleration take very simple forms, v = vet v2 en . a = vet + The difficulty of working with this reference frame stems from the fact that the orientation of the axis depends on the trajectory itself. The position vector, r, needs to be found by integrating the relation dr/dt = v as 5 follows, r = r0 + where r 0 = r(0) is given by the initial condition. t v dt , 0 We note that, by construction, the component of the acceleration along the binormal is always zero. When the trajectory is planar, the binormal stays constant (orthogonal to the plane of motion). However, when the trajectory is a space curve, the binormal changes with s. It can be shown (see note below) that the derivative of the binormal is always along the direction of the normal. The rate of change of the binormal with s is called the torsion, . Thus, deb =  en ds trajectory is linear. Note Calculation of the radius of curvature and torsion for a trajectory or, deb =  v en . dt We see that whenever the torsion is zero, the trajectory is planar, and whenever the curvature is zero, the In some situations the trajectory will be known as a curve of the form y = f (x). The radius of curvature in this case can be computed according to the expression, = [1 + (dy/dx)2 ]3/2 . d2 y/dx2  (11) This expression is not hard to derive. Try it! Since y = f (x) defines a planar curve, the torsion is zero. On the other hand, if the trajectory is known in parametric form as a curve of the form r(t), where t can be time, but also any other parameter, then the radius of curvature and the torsion can be computed as = (r r )3/2 , (r r )(r r )  (r r )2 = ... where r = d3 r/dt3 . ... (r r ) r r r 2 (12) where r = dr/dt, and r = d2 r/dt2 and Equations of Motion in Intrinsic Coordinates
Newton's second law is a vector equation, F = ma, which can now be written in intrinsic coordinates. In tangent, normal and binormal components, tnb, we write F = Ft et + Fn en and a = at et + an en . We observe that the positive direction of the normal coordinate is that pointing towards the center of curvature. Thus, in component form, we have Ft Fn = m at = m v = m s v2 = m an = m 6 Note that, by definition, the component of the acceleration along the binormal direction, eb , is always zero, and consequently the binormal component of the force must also be zero. This may seem surprising, at first, but recall that the tangent and normal directions are determined by the motion, and, hence, we can say that the motion "chooses" the binormal direction to be always orthogonal to the applied force. In other words, if we apply a force to a particle, the particle will experience an acceleration which is parallel to the force. The normal direction is chosen so that the acceleration vector is always contained in the plane defined by the tangent and the normal. Thus, the binormal is always orthogonal to the external force. Intrinsic coordinates are sometimes useful when we are dealing with problems in which the motion is constrained, such as a car on a roller coaster. The geometry of the trajectory is known, and, therefore, the directions of the tangent, normal and binormal vectors are also known. In these cases it may be possible to integrate the component of the equation of motion along the tangential direction (especially if there is no friction), and then calculate, aposteriori, the reaction force using the normal component of the equation of motion. Note (optional) Frenet formulae The Frenet formulae give us the variations of the unit vectors et , en and eb with respect to the path coordinate s. The first formula det 1 = en , ds has already been defined. Now, since eb is a unit vector, deb /ds will be orthogonal to eb . Hence, it will be of the form, deb = bt et + bn en . ds If we perform the dot product of this expression with et , we obtain bt = deb det 1 et =  eb =  en eb = 0 . ds ds The second equality follows from the fact that the derivative of et eb is zero, i.e. det eb + et deb = 0. Therefore, only bn is nonzero. Defining bn =  , we obtain the third Frenet formula deb =  en . ds Finally, the second formula can be obtained in a similar manner (we leave the details as an exercise) and gives, den 1 =  et + eb , ds den v =  et + veb . dt (13) or, multiplying by v, As we move along s, the osculating plane (and hence eb ) may rotate, making the curve non planar. As an example, an aeroplane may be rolling as it flies along et . The derivative of eb is in the direction opposite to 7 en if the rotation is in the direction of a right hand screw, and this is taken as the positive direction for the torsion. Example "Simplified" Aircraft Kinematics (W. M. Hollister) The flight of an aircraft through the sky is an example of curvilinear motion. Think of et as the roll axis aligned with the velocity vector of the aircraft. Think of eb as being the pitch axis. The lift is then directed along en . The roll rate of the aircraft can be interpreted as v, and the pitch rate as v/. In order to turn the aircraft out of the vertical plane, it is necessary to rotate the direction of the lift en so that there is a component of acceleration out of the vertical plane. Neglecting gravity, the velocity vector along et determines where the aircraft is going, and the lift along en determines where the velocity vector is going. The roll rate determines how the lift vector will be rotated out of the osculating plane. As shown by equation (13), the direction of the lift vector is changed by rolling v, as well as pitching v/. Consider the following example. An aircraft follows a spiral path in the sky while doing a barrel roll. The coordinates are given below, where v0 = 194 ft/s, = 0.4 rad/s, and h = 125 ft are constants. x = v0 t y z = h cos t = h sin t We have, v v a = v0 i  h sin tj + h cos tk 2 = v0 + h2 2 = 200ft/s v0
2 = h 2 cos tj  h 2 sin tk a = h 2 = 20ft/s Since v = 0, a = (v 2 /)en , or, h 2 = v2 , = v2 = 2000ft h 2 8 , et en eb Finally, deb v0 v0 = cos tj + sin tk dt v v which corresponds to a roll rate of v = v0 = 0.4rad/s v v0 h i sin tj + v v =  cos tj  sin tk h v0 = i+ sin tj  v v = h cos tk v v0 cos tk v Example Aircraft flying an "imperfect" loop (Hollister) It is more common to fly a loop keeping the normal acceleration, an , approximately constant at, say, ng Let be the flight path angle which the velocity vector (or tangent) makes with the horizontal, and let be the radius of curvature of the path. Then, an at From, v dv = at ds, with ds = d, we have, v dv = at d = (g sin )( or, integrating, n dv =  sin d , v and = n ln vv0 = cos 0 ,
2 v0  2 (1cos ) e n . ng v (n 3  4). v2 = v = ng = g sin . = v2 )d , ng
1 v = v0 e n (1cos ) , (14) A sketch of this path is shown in the figure below. PULL UP RECOVERY 9 In going over the top of the imperfect loop, the aircraft does not go as high or loose as much velocity as it does going over the perfect loop. Unlike the perfect loop case however, the aircraft does need to pull up before, and recover after, the point of maximum altitude (see diagram). Note that the solution of this example and the previous one would have been rather difficult using rectangular coordinates. Note also that the form of the solution given by equation (14) is rather unusual, e.g. the radius of curvature is given as a function of the attitude angle. A possible way to plot the trajectory starting from an initial position and velocity would be to first determine , and then draw a small circle segment with the appropriate . After calculating the new position, the process would be repeated to draw the entire trajectory. Example Pursuit Problem A classic problem in particle kinematics is the pursuit problem. This has an elegant history in mathematics and practical applications in air combat. At its most basic, we have a "target" A traveling on a known curve at constant speed. The target is pursued by B whose strategy is to turn such that her velocity vector is always pointed at the target. . We take the simplest version of this problem where the velocities of the two vehicles are equal: vA = vB and the evading vehicle travels on a straight line in the positive x direction. Analyzing this problem using intrinsic coordinates can give us some insight into the motion, into the formalism of tangential and normal unit vectors and the role of radius of curvature of the path. We examine this problem at the instant where the distance between A and B is given by R and the angle between their directions of travel is . (i.e. if the particles are moving in the same direction, = 0.) We allow a time dt to elapse and examine the resulting geometry. Of course to allow us to see the motion, we take dt to be rather large. Obviously we are talking about a limiting process as dt > 0. From the figure, we see that in a time dt, the vector R between the two points A and B changes direction by an angle d = vA sin /Rdt. Therefore the direction of the tangential vector also changes by d. From 10 our earlier analysis the change of the tangential vector due to a change in ds is given by det d 1 = en = en ds ds Therefore, using the chain rule, the magnitude of radius of curvature at the point r(t) is given by 1 d = = ds angle changes as a result of the motion of point A.
ds dt d dt d dt ds dt (15) (16) Now, ds/dt is the scalar velocity of the point B along the pursuit curve, while d/dt is the rate at which the = vB
vA sin R (17) (18) = so that the radius of curvature at the point A is given by 1/ = vA sin vB R (19) 2 Given this, the acceleration of particle B is easily found as a = vB /. ADDITIONAL READING J.L. Meriam and L.G. Kraige, Engineering Mechanics, DYNAMICS, 5th Edition 2/5, 3/5 (normal and tangential coordinates only) 11 MIT OpenCourseWare http://ocw.mit.edu 16.07 Dynamics
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 Derivative, Acceleration, Force, dR ds

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