16.07 Dynamics Fall 2004 Version 1.1 Lecture D4- Intrinsic Coordinates In lecture D2 we introduced the position, velocity and acceleration vectors and referred them to a fixed cartesian coordinate system. While it is clear that the choice of coordinate system does not affect the final nswer, we shall see that, in practical problems, the choice of a specific system may simplify the calculations considerably. In previous lectures, all the vectors at all points in the trajectory were expressed in the same frame. When using intrinsic coordinates(as well as the other coordinate systems presented in the next lecture), we shall see that the reference frame is a function of the current position of the particle. This means hat, the position, velocity and acceleration vectors at different points are expressed in different frame Intrinsic coordinates: Tangential. Normal and binormal compo- nents From the previous lecture, we know that the velocity vector, u, can be expressed as et where r(t) is the ve s is the speed, e, is the unit tangent vector to the trajectory, and s is he path coordinate along the trajector s(t) Also, recall that the unit tangent vector can be written as dr The acceleration vector is the derivative of the velocity vector with respect to time, and therefore we can write, from(1)
J. Peraire 16.07 Dynamics Fall 2004 Version 1.1 Lecture D4 - Intrinsic Coordinates In lecture D2 we introduced the position, velocity and acceleration vectors and referred them to a fixed cartesian coordinate system. 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 considerably. In previous lectures, all the vectors at all points in the trajectory were expressed in the same frame. When using intrinsic coordinates (as well as the other coordinate systems presented in the next lecture), we shall see that the reference frame is a function of the current position of the particle. This means that, the position, velocity and acceleration vectors at different points are expressed in different frames. Intrinsic coordinates: Tangential, Normal and Binormal components. From the previous lecture, we know that the velocity vector, v, can be expressed as v = dr dt = vet , (1) 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. Also, recall that the unit tangent vector can be written as, et = dr ds . (2) The acceleration vector is the derivative of the velocity vector with respect to time, and therefore we can write, from (1), a = dv dt = dv dt et + v det dt . (3) 1
The vector et is the local unit tangent vector to the curve which changes from point to point. Consequently, the time derivative of et w al. be nonzero The time derivative of et can be written as det det ds det dt ds dt d (4) 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 O et p +dev d > et+det 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 dB 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 dB. Thus, if en is a unit normal vector in the direction of det, we can write det= dBe, dB (6) Here iB/ds is aa local property of the curve, called the curvature, and p=1/k is called the radius of Note that in the picture, the sizes of det, ds, and dB 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 here two tangent vectors et and e +det, separated by a small ds and having an angle between them 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, dB. The point O is called the center of curvature, and the istance, P, between O and A is the radius of curvature. Thus, from the sketch, we have that ds= pdB, or 1/p
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 dt = det ds ds dt = det ds v . (4) 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 dβ. 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 dβ. Thus, if en is a unit normal vector in the direction of det, we can write det = dβen . (5) Dividing by ds yields, det ds = dβ ds en = κen = 1 ρ en . (6) Here, κ = dβ/ds is a a local property of the curve, called the curvature, and ρ = 1/κ is called the radius of curvature. Note that in the picture, the sizes of det, ds, and dβ 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 dβ. 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, dβ. The point O′ is called the center of curvature, and the distance, ρ, between O′ and A is the radius of curvature. Thus, from the sketch, we have that ds = ρ dβ, or dβ/ds = κ = 1/ρ. 2
P Intuitively, we can see that each infinitesimal arc, ds, can be represented by a circle segment of radius p having its center at the center of curvature. It is clear that both the radius of curvature and the center of urvature are functions of s, and consequently they change 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 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(4), we have that Finally, we have that the acceleration, can be written as du v2 a=-et + -en= atet+ anen ere, at = i, is the tangential component of the acceleration, and an =12/p, is the normal component of e acceleration. Since an is the component of the acceleration pointing towards the center of curvature, it is ometimes referred to as centripetal acceleration. When at is nonzero, the velocity vector changes magnitude
Intuitively, we can see that each infinitesimal arc, ds, can be represented by a circle segment of radius ρ 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 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 (4), we have that det dt = dβ ds v en = β˙ en = v ρ en . (7) Finally, we have that the acceleration, can be written as a = dv dt et + v 2 ρ en = atet + anen . (8) Here, at = ˙v, is the tangential component of the acceleration, and an = v 2/ρ, 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, 3
or stretches. When an is nonzero, the velocity vector changes direction, or swings. The modulus of the total acceleration can be calculated as a= vaf+a? Calculation of the radius of curvature for a trajectory In some situations the trajectory will be known as a curve of the form y= f(r). The radius of curvature in this case can be computed according to the expression, 3 P day/dri This expression is not hard to derive. Try it! 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 can be computed as 3/2 P )(F·)-( where i=dr/dt, and r=dr/dt Example ball is ejected horizontally from the tube with a speed vo. 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 of the acceleration will be in the normal direction, i. e. an=g. Thus, from an=o/p, we have that Alternatively, we can obtain an equation for the trajectory of the form y= f(a) and use expression(9)to calculate the curvature. The trajectory is given as Thus, elin g t, we have At r=0, dy/dx=0, d2y/dx2=-g/u2, and the above expression gives, p=v2/9, as expected
or stretches. When an is nonzero, the velocity vector changes direction, or swings. The modulus of the total acceleration can be calculated as a = p a 2 t + a 2 n . Note Calculation of the radius of curvature for a trajectory 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, ρ = [1 + (dy/dx) 2 ] 3/2 |d 2y/dx2| . (9) This expression is not hard to derive. Try it! 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 can be computed as ρ = (r˙ · r˙) 3/2 p (r˙ · r˙)(r¨ · r¨) − (r˙ · r¨) 2 , where r˙ = dr/dt, and r¨ = d 2r/dt2 . 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 of the acceleration will be in the normal direction, i.e. an = g. Thus, from an = v 2 0/ρ, we have that, ρ = v 2 0 g . Alternatively, we can obtain an equation for the trajectory of the form y = f(x) and use expression (9) to calculate the curvature. The trajectory is given as, x = v0t y = − 1 2 gt2 . Thus, eliminating t, we have y = − g 2v 2 0 x 2 . At x = 0, dy/dx = 0, d 2y/dx2 = −g/v2 0 , and the above expression gives, ρ = v 2 0 /g, as expected. 4
Note Relationship between s, v and at The quantities s, v and at are related in the same manner as the quantities s, u and a for rectilinear motion. In particular we have that v=s, at= i, and at ds= v du. This means that if we have a way of knowing at ve 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 right handed 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 x 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 Uet+—e 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 follows dt where ro =r(o) is given by the initial condition 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 hen 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
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 right handed 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 a = ˙vet + v 2 ρ en . 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 follows, r = r0 + Z t 0 v dt , where r0 = r(0) is given by the initial condition. 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 5
rith s is called the torsion, T. Thus, eb We see that whenever the torsion is zero, the trajectory is planar, and whenever the curvature is zero, the Equations of Motion Newtons 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= Ftet+ Fnen and a observe that the positive direction of the normal coordinate is that pointing towards the center of curvature. Thus in component fo m at =U=ns Fn= man=m P 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 he 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 he tangent and the normal. Thus, the binormal is always orthogonal to the external force. trinsic coordinates are sometimes useful when we are dealing with problems in which the motion is con- strained, 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, a-posteriori, the reaction force using the normal component of the equation of motion Note(optional) frenet formulae The Frenet fromulae give us the variations of the unit vectors et, en and eb with respect to the path coordinate s. The first formula 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 d
with s is called the torsion, τ. Thus, deb ds = −τ en or, deb dt = −τv en. We see that whenever the torsion is zero, the trajectory is planar, and whenever the curvature is zero, the trajectory is linear. Equations of Motion 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 = Ftet + Fnen and a = atet + anen. 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 = m at = m v˙ = m s¨ Fn = m an = m v 2 ρ 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, a-posteriori, the reaction force using the normal component of the equation of motion. Note (optional) Frenet formulae The Frenet fromulae give us the variations of the unit vectors et, en and eb with respect to the path coordinate s. The first formula det ds = 1 ρ en , 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 ds = btet + bnen . 6
If we perform the dot product of this expression with e,, we obtai deb ds‘eb=--en·eb=0 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 =-T, we obtain the third frenet formula Ten Finally, the second formula can be obtained in a similar manner (we leave the details as an exercise)and + TE or, multiplying by u, --et Tveb (10) 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 en if the rotation is in the direction of a right hand screw, and this is taken as the positive direction for the 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 ng en. The roll rate of th e aircraft can be interpreted as TU, and the pitch rate as u/p. 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 equati (10), the direction of the lift vector is changed by rolling TU, as well as pitching u/p. Consider the following example. An aircraft follows a spiral path in the sky while doing a barrel roll. The coordinates are given below, where vo= 194 ft/s, w=0.4 rad/s, and h= 125 ft are constants. h cos wt h sint
If we perform the dot product of this expression with et, we obtain bt = deb ds · et = − det ds · eb = − 1 ρ en · eb = 0 . 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 = −τ, we obtain the third Frenet formula deb ds = −τen . Finally, the second formula can be obtained in a similar manner (we leave the details as an exercise) and gives, den ds = − 1 ρ et + τeb , or, multiplying by v, den dt = − v ρ et + τveb . (10) 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 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 τv, and the pitch rate as v/ρ. 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 (10), the direction of the lift vector is changed by rolling τv, as well as pitching v/ρ. 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, ω = 0.4 rad/s, and h = 125 ft are constants. x = v0t y = h cos ωt z = h sin ωt 7
We have i- hw sin uti hw coswtk U=Vt6+b22=2001/s≈mo -hw-cos uti- wtk Since i=0, a=(0/pen, or, U2 U- P 加2=2000f coswti-sin wtk sIn w costs cos wt,+-sin wtk which corresponds to a roll rate of 0.grad/s ADDITIONAL READING J. L. Meriam and L G. Kraige, Engineering Mechanics, DYNAMICS, 5th Edition 2/5, 3/5(normal and tangential coordinates only)
We have, v = v0i − hω sin ωtj + hω cos ωtk v = q v 2 0 + h 2ω2 = 200ft/s ≈ v0 a = −hω2 cos ωtj − hω2 sin ωtk a = hω2 = 20ft/s2 Since ˙v = 0, a = (v 2/ρ)en, or, hω2 = v 2 ρ , ρ = v 2 hω2 = 2000ft , et = v0 v i − hω v sin ωtj + hω v cos ωtk en = − cosωtj − sin ωtk eb = hω v i + v0 v sin ωtj − v0 v cos ωtk Finally, deb dt = v0ω v cos ωtj + v0ω v sin ωtk which corresponds to a roll rate of τv = v0ω v ≈ ω = 0.4rad/s ADDITIONAL READING J.L. Meriam and L.G. Kraige, Engineering Mechanics, DYNAMICS, 5th Edition 2/5, 3/5 (normal and tangential coordinates only) 8
