Lecture L253D Rigid Body Kinematics


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1 J. Peraire, S. Winall Dynamics Fall 2008 Version 2.0 Lecture L253D Rigi Boy Kinematics In this lecture, we consier the motion of a 3D rigi boy. We shall see that in the general threeimensional case, the angular velocity of the boy can change in magnitue as well as in irection, an, as a consequence, the motion is consierably more complicate than that in two imensions. Rotation About a Fixe Point We consier first the simplifie situation in which the 3D boy moves in such a way that there is always a point, O, which is fixe. It is clear that, in this case, the path of any point in the rigi boy which is at a istance r from O will be on a sphere of raius r that is centere at O. We point out that the fixe point O is not necessarily a point in rigi boy (the secon example in this notes illustrates this point). Euler s theorem states that the general isplacement of a rigi boy, with one fixe point is a rotation about some axis. This means that any two rotations of arbitrary magnitue about ifferent axes can always be combine into a single rotation about some axis. At first sight, it seems that we shoul be able to express a rotation as a vector which has a irection along the axis of rotation an a magnitue that is equal to the angle of rotation. Unfortunately, if we consier two such rotation vectors, θ 1 an θ 2, not only woul the combine rotation θ be ifferent from θ 1 + θ 2, but in general θ 1 + θ 2 = θ 2 + θ 1. This situation is illustrate in the figure below, in which we consier a 3D rigi boy unergoing two 90 o rotations about the x an y axis. 1
2 It is clear that the result of applying the rotation in x first an then in y is ifferent from the result obtaine by rotating first in y an then in x. Therefore, it is clear that finite rotations cannot be treate as vectors, since they o not satisfy simple vector operations such as the parallelogram vector aition law. This result can also be unerstoo by consiering the rotation of axes by a coorinate transformation. Consier a transformation (x ) = [T 1 ](x), an a subsequent coorinate transformation (x ) = [T 2 ](x ). The (x ) coorinate obtaine by the transformation (x ) = [T 1 ][T 2 ](x) will not be the same as the coorinates obtaine by the transformation (x ) = [T 2 ][T 1 ](x), in other wors, orer matters. Angular Velocity About a Fixe Point On the other han, if we consier infinitesimal rotations only, it is not ifficult to verify that they o inee behave as vectors. This is illustrate in the figure below, which consiers the effect of two combine infinitesimal rotations, θ 1 an θ 2, on point A. Image by MIT OpenCourseWare. (figure reprouce from J.L. Meriam an K.L. Kraige, Dynamics, 5th eition, Wiley) As a result of θ 1, point A has a isplacement θ 1 r, an, as a result of θ 2, point A has a isplacement θ 2 r. The total isplacement of point A can then be obtaine as θ r, where θ = θ 1 +θ 2. Therefore, it follows that angular velocities ω 1 = θ 1 an ω 2 = θ 2 can be ae vectorially to give ω = ω 1 + ω 2. This means that if at any instant the boy is rotating about a given axis with angular velocity ω 1 an at the same time this axis is rotating about another axis with angular velocity ω 2, the total angular velocity of the boy will be simply ω = ω 1 + ω 2. Therefore, although the finite rotations of a boy about an axis are 2
3 not vectors, the infinitesimal rotations are vectors. The angular velocity is thus a vector an for a complex configuration, the various components can ba vectorially ae to obtain the total angular velocity. Consier the complex rotating configuration shown below. We want to etermine the angular velocity of the isc D. First, we note that the isc is rotating with angular velocity ω 1 about the axis MM. In turn, this axis is rotating with angular velocity ω 2 about the horizontal axis, which is at this instant aligne with the x axis. At the same time, the whole assembly is rotating about the z axis with angular velocity ω 3. Therefore, the total angular velocity of the isc is vector sum of the iniviual angular velocity vectors. The resultant vector is shown in the figure: ω total = ω 1 + ω 2 + ω 3. Expresse in the fixe x, y, z system for which the configuration is instantaneously aligne as shown, we have ω = ω 2 i + ω 1 cos φ j + (ω 1 sin φ + ω 3 ) k. Here, φ is the angle between MM an the y axis. Angular Acceleration In orer to apply the principle of conservation of angular momentum, we nee a general expression for the angular acceleration of the various components of a complex rotating configuration. We first examine this problem with respect to inertial coorinate system, x, y, z. In orer to apply conservation of angular momentum, we must evelop a formula for the time rate of change of angular momentum, BH = (I G ω) = I G ω + ω I G. We will later consier uner what conitions we nee to consier ω I G. For now, we concentrate on etermining ω. We first consier a common situation in the stuy of rotating 3D objects, the rotating wheel with angular velocity ω attache to a central hub which rotates with angular velocity Ω, shown in the figure. The total angular velocity is the vector sum of Ω an ω. 3
4 In this example, we take ω constant in magnitue (but not irection) an Ω as constant. It is clear that the rotation Ω will rotate the vector ω, changing its irection. The magnitue of ω is Ωω, the irection is normal to ω; by Coriolis theorem, the result is ω = Ω ω. It is interesting to note that this result is inepenent of the istance b between the wheel an the axis of rotation for Ω. This is a consequence of our earlier observation that in a rigi boy rotating with angular velocity ω, every point rotates with angular velocity ω. Example Multiple Observers In this example we illustrate a more systematic proceure for calculating the angular velocities an accelerations when several reference frames are involve. We want to etermine the angular acceleration of the isc D as a function of the angular velocities an accelerations given in the iagram. The angle of ω 1 with the horizontal is φ. 4
5 The angular velocity vector for each component is the vector sum of the iniviual angular velocities of the components:ω 3, ω2 total = ω 3 + ω 2 ; ω 1total = ω 1 + ω 2 + ω 3. The angular accelerations of the various components are worke through iniviually. Consier ω 3. Since it is the primary rotor, the rotation ω 2 an ω 1 o not affect its motion. Therefore the angular acceleration is ue solely to ω 3. Consier ω 2. This angular velocity vector will change with time both ue to ω 2 as well as ω 3 ω 2. An finally, for ω 1: this angular velocity vector will change with time both ue to ω 1 as well as (ω 3 + ω 2 ) ω 1. We now resolve these vectors into appropriate coorinate systems. We consier three sets of axes. Axes xyz are fixe. Axes x y z rotate with angular velocity ω 3 k with respect to xyz. Axes x y z rotate with angular velocity ω 2 i with respect to x y z. Finally, the isc rotates with angular velocity ω 1 j with respect to the axes x y z. The angular velocity of the isc with respect to the fixe axes will be simply Ω = ω 2 i + ω 1 j + ω 3 k. (1) At the instant consiere, i = i an j = cos φj + sin φk. Therefore, we can also write, Ω = ω 2 i + ω 1 cos φj + (ω 1 sin φ + ω 3 )k. In orer to calculate the angular acceleration of the isc with respect to the fixe axes xyz, we start from (1) an write, Ω = (ω 2 i + ω 1 j + ω 3 k) = (ω 2 i + ω 1 j ) + ω 3 k. xyz xyz xyz Here, we have use the fact that k oes not change with respect to the xyz axes an therefore only the magnitue of ω 3 changes. In orer to calculate the time erivative of ω 2 i + ω 1 j with respect to the inertial reference frame, we apply Coriolis theorem. Since x y z rotates with angular velocity ω 3 k with respect to 5
6 xyz, we write, (ω 2 i + ω 1 j ) = (ω 2 i + ω 1 j ) + ω 3 k (ω 2 i + ω 1 j ). xyz x y z In the x y z frame, i oes not change irection. Therefore, we can write (ω 2 i + ω 1 j ) = ω 2i + (ω 1 j ) + ω 3 k (ω 2 i + ω 1 j ). xyz In orer to evaluate the erivative of ω 1 j with respect to the x y z frame we make use again of Coriolis x y z theorem, an write (ω 1 j ) = (ω 1 j ) + ω 2 i ω 1 j = ω 1j + ω 2 i ω 1 j. x y z x y z Here, we have use the fact that x y z rotates with angular velocity ω 2 i with respect to x y z, an the erivative of j in the x y z reference frame is zero. Putting it all together, we have, Ω = ω 2i + ω 1 j + ω 2 i ω 1 j + ω 3 k (ω 2 i + ω 1 j ) + ω 3 k xyz = ω 2i + ω 1 j + ω 1 ω 2 k + ω 2 ω 3 j ω 1 ω 3 cos φ i + ω 3 k = ( ω 2 ω 1 ω 3 cos φ) i + ( ω 1 cos φ + ω 2 ω 3 ω 1 ω 2 sin φ) j + ( ω 3 + ω 1 sin φ + ω 1 ω 2 cos φ) k. Instantaneous Axis of Rotation In two imensions, we introuce the concept of instantaneous center of rotation. For a rotating boy with one fixe point, we can exten this concept to an instantaneous axis of rotation. Consier a vertical isc rotating with a constant angular velocity ω, rolling without slip with an angular velocity about the vertical axis with angular velocity Ω. The noslip conition requires that the velocity of the center of mass V G = ωr. The isk rolling aroun a circle of raius b will have an angular velocity Ω = ωr/b. We note that just as in twoimensions, point O is a fixe point, an instantaneous center of rotation. That is, at that instant, the motion of the isc is such that the istance from any point in the isc to the point O remains constant. The total instantaneous angular velocity is the vector sum of Ω + ω. The resultant vector is sketche an as woul be expecte, it is parallel to a line from the center of the hub to the point C. Once the instantaneous angular velocity, ω T = Ω + ω, has been etermine, the velocity of any point in the rigi boy is simply v = ω T r, (2) where r is the position vector of the point consiere with respect to the fixe point O. It follows that for any point which is on the line passing through O an parallel to ω T, the velocity will be zero. This line is therefore the Instantaneous Axis of Rotation. 6
7 We can now efine two space curves, which are instantaneously tangent to the instantaneous axis of rotation. These curves are useful for more complex problems in the general rotation of a boy about a fixe point. This simple example gives us a useful visualization of these space curves. The first is calle the boy cone: it is the locus of points mae by the instantaneous axis of rotation as the boy traces its motion. As can be seen, for this case, it is a cone of raius equal to the raius of the isc, extening to the hub at the z axis. (The term boy cone is a bit of a misnomer; the boy oesn t have to fit entirely insie the boy cone.) The secon space curve is calle the space cone. It is efine as the locus of points trace in space by the instantaneous axis of rotation. At the instantaneous position of the boy, these two curves are both tangent to the instantaneous axis of rotation. In this case, the space cone is the cone boune by the curve trace by the instantaneous contact point an the fixe point, the center of the hub. The boy curve an the space curve for more general threeimensional motions are shown in the figure. For a general motion, as the irection of the instantaneous axis of rotation (or the line passing through O parallel to ω) changes in space, the locus of points efine by the axis generates a fixe Space Cone. If the change in this axis is viewe with respect to the rotating boy, the locus of the axis generates a Boy Cone. 7
8 At any given instant, these two cones are tangent along the instantaneous axis of rotation. When the boy is in motion, the boy cone appears to roll either on the insie or the outsie of the fixe space cone. This situation is illustrate in the figure below. The acceleration of any point in the rigi boy is obtaine by taking the erivative of expression 2. Thus, a = ω T r + ω T ṙ = α r + ω T (ω T r). (3) Here, α is the angular acceleration vector an is locally tangent to both the Space an the Boy Cones. General Motion In the general case, the isplacement of a rigi boy is etermine by a translation plus a rotation about some axis. This result is a generalization of Euler s theorem, which is sometimes known as Chasles theorem. In practice, this means that six parameters are neee to efine the position of a 3D rigi boy. For instance, we coul choose three coorinates to specify the position of the center of mass, two angles to efine the axis of rotation an an aitional angle to etermine the magnitue of the rotation. Unlike the motion about a fixe point, it is not always possible to efine an instantaneous axis of rotation. Consier, for instance, a boy which is rotating with angular velocity ω an, at the same time, has a translational velocity parallel to ω. It is clear that, in this case, all the points in the boy have a nonzero velocity, an therefore an instantaneous center of rotation cannot be efine. It turns out that, in some situations, the motion of the center of mass of a 3D rigi boy can be etermine inepenent of the orientation. Consier, for instance, the motion of an orbiting satellite in free flight. In this situation, the sum of all external forces on the satellite oes not epen on the satellite s attitue, an, therefore, it is possible to etermine the position without knowing the attitue. In more complex situations, however, it may be necessary to solve simultaneously for both the position of the center of mass an the attitue. The velocity, v P, an acceleration, a P, of a point, P, in the rigi boy can be etermine if we know the velocity, v O, an acceleration, a O, of a point in the rigi boy, O, as well as the boy s angular velocity, ω, an acceleration, α. Here, r P v P = v O + ω r (4) P a P = a O + ω r P + ω (ω rp ). (5) is the position vector of the point, P, relative to O. We point out that the angular velocity an angular acceleration are the same for all the points in the rigi boy. ADDITIONAL READING 8
9 J.L. Meriam an L.G. Kraige, Engineering Mechanics, DYNAMICS, 5th Eition 7/1, 7/2, 7/3, 7/4, 7/5, 7/6 (review) 9
10 MIT OpenCourseWare Dynamics Fall 2009 For information about citing these materials or our Terms of Use, visit:
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