J. Peraire 16.07 Dynamics Fall 2004 Version 1.5 Lecture D12 - Relative Motion using Translating/Rotating Axes In the previous lecture, we related the motion experienced by two observers in relative translational motion with respect to each other. In this lecture, we will extend this relation to our third type of observer. That is, observers who accelerate and rotate with respect to each other. As a matter of illustration, let us consider a very simple situation, in which a particle at rest with respect to the fixed frame xy, i.e. vA = 0, aA = 0, is observed by an observer, B, who is standing at the center of a turn table. The table rotates with a constant angular velocity of ? rad/s. Assuming that the platform only rotates about its center and does not move translationally, the position of B will not change, and therefore vB = 0 and aB = 0. It is clear that relative to B, A will not be at rest. In fact, B will see A rotating about B with a constant angular velocity of ??. Thus, we can easily calculate the motion of A as observed by B. Expressing the velocity and acceleration in local polar coordinates (rotating with x′y′), we will have (vA/B)x′y′ = (˙rA/B)x′y′ = ??rA/B eθ , and (aA/B)x′y′ = (˙vA/B)x′y′ = ?(vA/B) 2 x′y′ rA/B er = ?? 2rA/B er . The notation, ( · )x′y′, is used to indicate the velocity, or acceleration, experienced by an observer that rotates with the axes x′y′. In other words, the time derivatives are taken assuming that the directions x′y′, as seen by a rotating observer, do not change. For illustration purposes, we note that vA/B = ˙rA/B = 0 and aA/B = ˙vA/B = 0. Here, vA/B and aA/B denote the relative velocity and acceleration of A with respect to B, experienced by a non-rotating observer (this was the situation considered in the last lecture). We could, in fact, write (vA/B)xy and (aA/B)xy instead of vA/B and aA/B, but we will often omit the subscripts 1 when the axes considered are fixed. Also, we note that rA/B = (rA/B)x′y′ since this is an instantaneous measurement and is not related to whether the axes rotate or not. At this point, it should be clear that the expressions derived in the last lecture, which includes only trans- lational relative motion, are not applicable when rotating observers are involved. Note Angular velocity and angular acceleration vectors In the two dimensional example presented above, we have dealt with the angular velocity as if it were a scalar quantity. Here, we will introduce the concept of angular velocity and acceleration vectors. We shall see that, especially in three dimensions, it makes a lot of sense to think of these quantities as vectors. Let us consider a rigid body, which is spinning about an axis C ?C with an angular velocity of ? rad/s. We consider a unit vector, eC, along the direction of the axis, and define the angular velocity vector, ?, as a vector having magnitude ? and direction eC. Thus, ? = ?eC . Note that the convention between the direction of rotation and that of ? is determined by the right hand rule. If the body were to rotate in the direction opposite to that shown in the diagram, then we would simply have ? = ??eC. The angular velocity vector is useful to express the velocity, due to rotation, of any point A in the rigid body. Let r be the position vector of A relative to an origin point, O, located on the axis of rotation. It turns out that the velocity of A, v, can be simply expressed as v = ?×r . (1) First, we note that, when the body spins, A describes a circular trajectory around the axis of radius d = rsin?, where ? is the angle between r and the axis of rotation. Since the body is spinning at a rate of ? rad/s, the magnitude of the velocity vector will be v = ?d = ?rsin?. The direction of the velocity vector 2 v will be tangent to the trajectory at A, which means that it will be perpendicular to r and eC. Thus, recalling the definition of the vector product, we can verify that the velocity given by (1) does indeed have the right direction, orientation and magnitude. We also define the angular acceleration vector as, ˙? = d? dt . Since the angular acceleration is the derivative of a vector, it will measure both the changes in magnitude and the changes in direction of the angular velocity vector. Time derivative of a fixed vector in a rotating frame We consider a reference frame x′y′z′ rotating with an angular velocity ? with respect to a fixed frame xyz. Let V be any vector, which is constant relative to the frame x′y′z′. That is, the vector components in the x′y′z′ frame do not change, and, as a consequence, V rotates as if it were rigidly attached to the frame. In the absolute frame, the time derivative will be equal to dV dt = lim?t→0 (V +?V )?V ?t = ?×V , which can be interpreted as the velocity of the tip of vector V . The above expression applies to any vector which is rigidly attached to the frame x′y′z′. In particular, it applies to the unit vectors i′, j′ and k′. Therefore, we have that di′ dt = ?×i ′ dj′ dt = ?×j ′ dk′ dt = ?×k ′ . Time derivative of a vector in a rotating frame: Coriolis’ theorem Now let V be an arbitrary vector (e.g. velocity, magnetic field, force, etc.), which is allowed to change in both the fixed xyz frame and the rotating x′y′z′ frame. The vector V can be expressed, in the x′y′z′ frame, 3 as, V = Vx′i′ + Vy′j′ + Vz′k′. If we now consider the time derivative of V , as seen by the fixed frame, we have ˙V = dV dt = ˙Vx′i′ + ˙Vy′j′ + ˙Vz′k′ + Vx′ ˙i′ + Vy′ ˙j′ + Vz′ ˙k′ = ( ˙V )x′y′z′ +?×V . (2) Here, ˙Vx′i′ + ˙Vy′j′ + ˙Vz′k′ = ( ˙V )x′y′z′ is the time derivative of the vector V as seen by the rotating frame. Hence, for this derivative, the vectors i′, j′ and k′ remain unchanged. The term Vx′ ˙i′ + Vy′ ˙i′ + Vz′ ˙i′ = Vx′(?× i′) + Vy′(?× j′) + Vz′(?× k′) = ?× V is the change in V due to the rotation. The expression (2) above is known as Coriolis’ theorem. Given an arbirary vector, it relates the derivative of that vector as seen by a fixed frame with the derivative of the same vector as seen by a rotating frame. Symbolically, we can write, ˙( )xyz = ˙( )x′y′z′ +?×( ) . We will often omit the notation ( )xyz when the derivative is taken with respect to the fixed frame. Relative Motion using translating/rotating axes Here, we consider the relationship between the motion seen by an observer B that may be accelerating as well as rotating, and the motion seen by a a fixed observer O. Let aB be the acceleration of B with respect to O, and let ? and ˙? denote the angular velocity and angular acceleration, respectively, of the frame x′y′z′ rigidly attached to B. The vectors i, j and k are the unit vectors corresponding to the fixed frame xyz, and i′, j′ and k′ are the unit vectors corresponding to the rotating frame x′y′z′. Position vector The position vector of A with respect to the fixed frame can be written as, rA = rB +rA/B . (3) 4 Note that this is an instantaneous concept, and therefore it is immaterial whether the observer B is rotating or not. In other words, the above expression is valid at any given instant. Of course, if we choose to express rA and rB in xyz components, and rA/B in x′y′z′ components, then we will have to make sure that the proper coordinate transformation is done before the components of the vectors are added. Also, since the orientation of x′y′z′ changes in time, this transformation will depend on the instant considered. Velocity vector Differentiating (3) with respect to time, we have, vA = ˙vA = ˙rB +(˙rA/B)x′y′z′ +?×rA/B = vB +(vA/B)x′y′z′ +?×rA/B . (4) Here, we have used Coriolis’ theorem (2) to calculate the derivative of rA/B, i.e. ˙rA/B = (˙rA/B)x′y′z′ +?× rA/B. In the above expression, vA and vB are the velocities of A and B, respectively, relative to the fixed frame. The term (vA/B)x′y′z′ is the velocity of A measured by the rotating observer, B. Finally, ? is the angular velocity of the rotating frame, and rA/B is the relative position vector of A with respect to B. Acceleration vector Differentiating (4) once again, and making use of Coriolis’ theorem, e.g. d(vA/B)x′y′z′/dt = (˙vA/B)x′y′z′ + ?×(vA/B)x′y′z′, we obtain the following expression for the acceleration, aA = ˙vA = ˙vB +(˙vA/B)x′y′z′ +?×(vA/B)x′y′z′ + ˙?×rA/B +?×(vA/B)x′y′z′ +?×(?×rA/B) = aB +(aA/B)x′y′z′ +2?×(vA/B)x′y′z′ + ˙?×rA/B +?×(?×rA/B) . (5) Here, aA and aB are the accelerationsof A and B, respectively, observedby xyz. The term (aA/B)x′y′z′ is the acceleration of A measured by an observer B that rotates with the axes x′y′z′. The term 2?×(vA/B)x′y′z′ is called Coriolis’ acceleration. The term ˙? × rA/B is due to the change in ?, and, finally, the term ?×(?×rA/B) is called centripetal acceleration. It can be easily seen that this acceleration always points towards the axis of rotation, and is orthogonal to ? (when trying to show this, you may find the vector identity, A× (B × C) = (A · C)B ? (A · B)C, introduced in the Review Notes on Vectors, applied to ?×(?×rA/B) = (?·rA/B)??(?·?)rA/B, useful). 5 Note ˙? vs. ( ˙?)x′y′z′ When writing the angular acceleration vector, ˙?, we did not specify whether the derivative was taken with respect to the fixed observer, O, or with respect to the rotating observer, B. This may seem a little bit sloppy at first, but, in fact, it turns out that it does not really matter. The time derivatives of vectors which are parallel to ? are the same for both observers. This can be easily seen if we go back to Coriolis’ theorem (2) and apply it to ?. That is, ˙? = ( ˙?)x′y′z′ +?×? = ( ˙?)x′y′z′ . Note Non-rotational observers We note that, for the case where ? = ˙? = 0, the x′y′z′ axes only have translational motion relative to the fixed axes xyz, and therefore the equations (3), (4) and (5) reduce to those introduced in the previous lecture. Namely, rA = rB +rA/B vA = vB +vA/B aA = aB +aA/B . Note also that, since the axes x′y′z′ do not rotate, we do not need to use the notation ( · )x′y′z′. We can now go back to our introductory example and verify that expressions (4) and (5) are indeed consistent with the observations made earlier. We have vA = vB = 0, and (vA/B)x′y′z′ = ??rA/Beθ. Thus, equation (4) reduces to 0 = ??rA/Beθ +?k×rA/Ber , which is clearly satisfied since k × er = eθ. For the acceleration equation (5), we have aA = aB = 0, (vA/B)x′y′z′ = ??rA/Beθ, (aA/B)x′y′z′ = ??2rA/Ber, and ˙? = 0. Thus, 0 = ??2rA/Ber +2?k×(??rA/Beθ) +?k×(?k×rA/Ber) , which is also satisfied since k×eθ = ?er and k×(k×er) = ?er. Note Velocity and acceleration of in a rotating rigid body We have seen (see earlier note in this lecture) that the velocity of a point A in a rigid body that is rotating with an angular velocity ? is simply v = vB +?×r , 6 where r is the position vector of A with respect to point B, and vB is the acceleration of point B. In order to determine the acceleration, we simply use expression (5). After setting all the appropriate to terms to zero, we get, a = aB + ˙?×r +?×(?×r) . The second term is due to the changein angular velocity, whereasthe third term is the centripetal acceleration due to the body’s rotation. Example Helicopter Blades We want to determine the instantaneous velocity and acceleration of a point A, which is located at the tip of the blade of the helicopter. The blades are rotating with an angular velocity p, and, at the same time, the helicopter is pitching downwards at an angular velocity q. At the instant considered, the velocity and acceleration of the center of mass of the helicopter, G, are zero. We will solve this problem using two approaches: First approach : In this approach, we consider a fixed set of axes, xyz, and a set of axes attached to the body of the helicopter x′y′z′. The origin of both axes is at the center of mass, G. Taking B ≡ G in (4), we write vA = vG +(vA/G)x′y′z′ +?×rA/G. Here, vG = 0 since the center of mass is at rest. The velocity observed by the x′y′z′ frame is simply (vA/G)x′y′z′ = pk ×rA/G = pk ×(aj + bk) = ?pai. The angular velocity of the frame x′y′z′ is clearly the pitch angular velocity, thus ? = ?qi. Finally, we have vA = ?pai+(?qi)×(aj + bk) = ?pai+ qbj ?qak . 7 For the acceleration, we use expression (5) and write aA = aG +(aA/G)x′y′z′ +2?×(vA/G)x′y′z′ + ˙?×rA/G +?×(?×rA/G) . Here, aG = 0because the center of mass is not accelerating. Relative the the axes x′y′z′, point A describes a circular motion at constant speed. Therefore, the only observed acceleration will be the centripetal acceler- ation. The magnitude of the centripetal acceleration is simply the speed divided by the radius of curvature, and the direction is parallel to the negative y axis. Thus, (aA/G)x′y′z′ = ?p2aj. The angular acceleration of x′y′z′ with respect to xyz is zero because the aircraft is pitching at a constant angular velocity, i.e., ˙? = 0. Finally, we have aA = ?p2aj +2(?qi)×(?pai) +(?qi)×(?qi×(aj + bk)) = ?(p2 + q2)aj ?q2bk . Second approach : Here, we consider the same fixed set of axes xyz, but the moving axes x′y′z′ are now rigidly attached to the blades. Also, the origin of the moving axis is now taken at point B in the diagram. The relative angular velocity of x′y′z′ with respect to xyz is now the sum of the pitch angular velocity ?p = ?qi, and the spin angular velocity ?s = pk. Thus, we have ? = ?p +?s = ?qi+ pk . The expression for the angular acceleration follows from considering the derivative of the above expression. It is clear that the direction of ?p is constant and its magnitude is also constant. Therefore, we have ˙?p = 0. The situation with ?s is different. The magnitude of ?s is constant, but its direction changes. We can use Coriolis’ theorem, (2), to compute its derivative. In order to do that, we consider a set of axes in which ?s remains constant. For instance, we can use the axes x′y′z′ (the axes x′y′z′ of the first approach will also be a good choice). We have, ˙?s = ( ˙?s)x′y′z′ +?×?s . 8 Here, ( ˙?s)x′y′z′ = 0. Therefore, we have ˙?s = (?p +?s)×?s = ?p ×?s = pqj . Now we can apply expressions (4) and (5) to calculate the velocity and acceleration. We note that vB = qbj, and aB = ?q2bk. Also, (vA/B)x′y′z′ = 0 and (aA/B)x′y′z′ = 0 since, in the frame attached to the blades, point A is not moving. Setting rA/B = aj, we have, vA = qbj +(?qi+ pk)×(aj) = ?pai?qbj ?qak , and aA = (?q2bk) +(pqj)×(aj)+(?qi+ pk)×((?qi + pk)×(aj)) = ?(p2 + q2)aj ?q2bk . Multiple observers It should be noted that in all the above derivations we have assumed that O was fixed, and that B was translating and rotating relative to O. It turns out that the above derivations are still valid if O moves, possibly with acceleration and rotation, provided that ? is the difference between the angular velocities of B and O. Realizing this allows us to extend the above expressions to the situation in which we have multiple observers. In particular, if we consider the expression (4) for the velocity, the term (vA/B)x′y′z′ can be computed with the aid of a third observer C, using axes x′′y′′z′′, which has a velocity vC/B and angular velocity ?′ with respect to x′y′z′. Thus, we can write (vA/B)x′y′z′ = (vC/B)x′y′z′ +(vA/C)x′′y′′z′′ +(?′)x′y′z′ ×rA/C , 9 which, combined with (4), gives, vA = vB +(vA/B)x′y′z′ +?×rA/B = vB +(vC/B)x′y′z′ +(vA/C)x′′y′′z′′ +(?′)x′y′z′ ×rA/C +?×rA/B . An analogous expression could be derived for the acceleration involving three observers in relative motion with respect to each other. You may want to try this as an exercise. Example Multiple Observers In this example we illustrate a more systematic procedure for calculating the angular velocities and accel- erations when several reference frames are involved. We want to determine the angular acceleration of the disc D as a function of the angular velocities and accelerations given in the diagram. The angle of ω1 with the horizontal is φ. D We consider three sets of axes. Axes xyz are fixed. Axes x′y′z′ rotate with angular velocity ω3k with respect to xyz. Axes x′′y′′z′′ rotate with angular velocity ω2i′ with respect to x′y′z′. Finally, the disc rotates with angular velocity ω1j′′ with respect to the axes x′′y′′z′′. The angular velocity of the disc with respect to the fixed axes will be simply ? = ω2i′ + ω1j′′ ++ω3k . (6) At the instant considered, i′ = i and j′′ = cosφj +sinφk. Therefore, we can also write, ? = ω2i+ ω1 cosφj +(ω1 sinφ + ω3)k . In order to calculate the angular acceleration of the disc with respect to the fixed axes xyz, we start from (6) and write, parenleftbiggd? dt parenrightbigg xyz = parenleftbigg d dt(ω2i ′ + ω1j′′ + ω3k) parenrightbigg xyz = parenleftbigg d dt(ω2i ′ + ω1j′′) parenrightbigg xyz + ˙ω3k . 10 Here, we have used the fact that k does not change with respect to the xyz axes and therefore only the magnitude of ω3 changes. In order to calculate the time derivative of ω2i′+ω1j′′ with respect to the inertial reference frame, we apply Coriolis’ theorem (2). Since x′y′z′ rotates with angular velocity ω3k with respect to xyz, we write, parenleftbigg d dt(ω2i ′ + ω1j′′) parenrightbigg xyz = parenleftbigg d dt(ω2i ′ + ω1j′′) parenrightbigg x′y′z′ + ω3k×(ω2i′ + ω1j′′) . In the x′y′z′ frame, i′ does not change direction. Therefore, we can write parenleftbigg d dt(ω2i ′ + ω1j′′) parenrightbigg xyz = ˙ω2i′ + parenleftbigg d dt(ω1j ′′) parenrightbigg x′y′z′ + ω3k×(ω2i′ + ω1j′′) . In order to evaluate the derivative of ω1j′′ with respect to the x′y′z′ frame we make use again of Coriolis’ theorem (2), and write parenleftbigg d dt(ω1j ′′) parenrightbigg x′y′z′ = parenleftbigg d dt(ω1j ′′) parenrightbigg x′′y′′z′′ + ω2i′ ×ω1j′′ = ˙ω1j′′ + ω2i′ ×ω1j′′ . Here, we have used the fact that x′′y′′z′′ rotates with angular velocity ω2i′ with respect to x′y′z′, and the derivative of j′′ in the x′′y′′z′′ reference frame is zero. Putting it all together, we have, parenleftbiggd? dt parenrightbigg xyz = ˙ω2i′ + ˙ω1j′′ + ω2i′ ×ω1j′′ + ω3k×(ω2i′ + ω1j′′) + ˙ω3k = ˙ω2i′ + ˙ω1j′′ + ω1ω2k′′ + ω2ω3j ?ω1ω3 cosφ i+ ˙ω3k = (˙ω2 ?ω1ω3 cosφ) i+(˙ω1 cosφ + ω2ω3 ?ω1ω2 sinφ) j +(˙ω3 + ˙ω1 sinφ + ω1ω2 cosφ) k. Example Cylindrical coordinates Cylindrical (and spherical) coordinate systems can be regarded as particular cases of the expressions (3), (4) and (5). Consider the fixed frame (i, j, k) and the rotating frame (er,eθ,k), and consider two observers O and B whose positions coincide, but who refer to the fixed and the rotating frame, respectively. Thus, we have that 11 rB = vB = aB = 0. The angular velocity and acceleration of the rotating frame relative to the fixed frame are ? = ˙θk and ˙? = ¨θk, respectively. Therefore, expressions (3), (4) and (5) become, rA = rA/B = rer + zk vA = (vA/B)rθz +?×rA/B = ˙rer + ˙zk + ˙θk ×(rer + zk) = ˙rer + r ˙θeθ + ˙zk aA = (aA/B)rθz + 2?×(vA/B)rθz + ˙?×rA/B +?×(?×rA/B) = (¨r ?r ˙θ2)er +(r¨θ +2˙r˙θ)eθ + ¨zk , which agree with the previously derived expressions for position, velocity and acceleration in cylindrical coordinates. Here, (aA/B)rθk = ¨rer+¨zk, 2?×(vA/B)rθk = 2˙r˙θeθ, ˙?×rA/B = r¨θeθ, and ?×(?×rA/B) = ?r˙θ2er. ADDITIONAL READING J.L. Meriam and L.G. Kraige, Engineering Mechanics, DYNAMICS, 5th Edition 5/7, 7/6 12