Lecture L26-3D Rigid Body Dynamics: The Inertia Tensor

Transcription

1 J. Peraire, S. Widnall Dynaics Fall 008 Lecture L6-3D Rigid Body Dynaics: The Inertia Tensor Version.1 In this lecture, we will derive an expression for the angular oentu of a 3D rigid body. We shall see that this introduces the concept of the Inertia Tensor. Angular Moentu We start fro the expression of the angular oentu of a syste of particles about the center of ass, H G, derived in lecture L11, n n H G = (r i i (ω r i)) = i r i ω (1) i=1 i=1 n n H G = (r i i (ω r i)) = i r i ω = r v d () i=1 i=1 H G = r v d. Here, r is the position vector relative to the center of ass, v is the velocity relative to the center of ass. We note that, in the above expression, an integral is used instead of a suation, since we are now dealing with a continuu distribution of ass. For a 3D rigid body, the distance between any particle and the center of ass will reain constant, and the particle velocity, relative to the center of ass, will be given by v = ω r. 1

2 Thus, we have, H G = r (ω r ) d = [(r r )ω (r ω)r ] d. Here, we have used the vector identity A (B C) = (A C)B (A B)C. We note that, for planar bodies undergoing a D otion in its own plane, r is perpendicular to ω, and the ter (r ω) is zero. In this case, the vectors ω and H G are always parallel. In the three-diensional case however, this siplification does not occur, and as a consequence, the angular velocity vector, ω, and the angular oentu vector, H G, are in general, not parallel. In cartesian coordinates, we have, r = x i + y j + z k and ω = ω x i + ω y j + ω z k, and the above expression can be expanded to yield, H G = ω x (x + y + z ) d (ω x x + ω y y + ω z z )x d i + ω y (x + y + z ) d (ω x x + ω y y + ω z z )y d j + ω z (x + y + z ) d (ω x x + ω y y + ω z z )z d k = ( I xx ω x I xy ω y I xz ω z ) i + ( I yx ω x + I yy ω y I yz ω z ) j + ( I zx ω x I zy ω y + I zz ω z ) k. (3) The quantities I xx, I yy, and I zz are called oents of inertia with respect to the x, y and z axis, respectively, and are given by I xx = (y + z ) d, I yy = (x + z ) d, I zz = (x + y ) d. We observe that the quantity in the integrand is precisely the square of the distance to the x, y and z axis, respectively. They are analogous to the oent of inertia used in the two diensional case. It is also clear, fro their expressions, that the oents of inertia are always positive. The quantities I xy, I xz, I yx, I yz, I zx and I zy are called products of inertia. They can be positive, negative, or zero, and are given by, I xy = I yx = x y d, I xz = I zx = x z d, I yz = I zy = y z d. They are a easure of the ibalance in the ass distribution. If we are interested in calculating the angular oentu with respect to a fixed point O then, the resulting expression would be, H O = ( (I xx ) O ω x (I xy ) O ω y (I xz ) O ω z ) i + ( (I yx ) O ω x + (I yy ) O ω y (I yz ) O ω z ) j + ( (I zx ) O ω x (I zy ) O ω y + (I zz ) O ω z ) k. (4)

3 Here, the oents of products of inertia have expressions which are analogous to those given above but with x, y and z replaced by x, y and z. Thus, we have that (I xx ) O = (y + z ) d, (I yy ) O = (x + z ) d, (I zz ) O = (x + y ) d, and, (I xy ) O = (I yx ) O = xy d, (I xz ) O = (I zx ) O = xz d, (I yz ) O = (I zy ) O = yz d. The Tensor of Inertia The expression for angular oentu given by equation (3), can be written in atrix for as, or, H Gx I xx I xy I xz ω x H Gy = I yx I yy I yz ω y. (5) H Gz I zx I zy I zz ω z H G = [I G ]ω, (6) where [I G ] is the tensor of inertia (written in atrix for) about the center of ass G and with respect to the xyz axes. The tensor of inertia gives us an idea about how the ass is distributed in a rigid body. Analogously, we can define the tensor of inertia about point O, by writing equation(4) in atrix for. Thus, we have H O = [I O ] ω, where the coponents of [I O ] are the oents and products of inertia about point O given above. It follows fro the definition of the products of inertia, that the tensors of inertia are always syetric. The iplications of equation (5) are that in any situations of iportance, even for bodes of soe syetry, the angular oentu vector H and the angular velocity vector ω are not parallel. This introduces considerable coplexity into the analysis of the dynaics of rotating bodies in three diensions. 3

4 Principal Axes of Inertia For a general three-diensional body, it is always possible to find 3 utually orthogonal axis (an x, y, z coordinate syste) for which the products of inertia are zero, and the inertia atrix takes a diagonal for. In ost probles, this would be the preferred syste in which to forulate a proble. For a rotation about only one of these axis, the angular oentu vector is parallel to the angular velocity vector. For syetric bodies, it ay be obvious which axis are principle axis. However, for an irregular-shaped body this coordinate syste ay be difficult to deterine by inspection; we will present a general ethod to deterine these axes in the next section. But, if the body has syetries with respect to soe of the axis, then soe of the products of inertia becoe zero and we can identify the principal axes. For instance, if the body is syetric with respect to the plane x = 0 then, we will have I x y = I y x = I x z = I z x = 0 and x will be a principal axis. This can be shown by looking at the definition of the products of inertia. The integral for, say, I x y can be decoposed into two integrals for the two halves of the body at either side of the plane x = 0. The integrand on one half, x y, will be equal in agnitude and opposite in sign to the integrand on the other half (because x will change sign). Therefore, the integrals over the two halves will cancel each other and the product of inertia I x y will be zero. (As will the product of inertia I x z ) Also, if the body is syetric with respect to two planes passing through the center of ass which are orthogonal to the coordinate axis, then the tensor of inertia is diagonal, with I x y = I x z = Iy z = 0. 4

5 Another case of practical iportance is when we consider axisyetric bodies of revolution. In this case, if one of the axis coincides with the axis of syetry, the tensor of inertia has a siple diagonal for. For an axisyetric body, the oents of inertia about the two axis in the plane will be equal. Therefore, the oent about any axis in this plane is equal to one of these. And therefore, any axis in the plane is a principal axis. One can extend this to show that if the oent of inertia is equal about two axis in the plane (I P P = I xx ), whether or not they are orthogonal, then all axes in the plane are principal axes and the oent of inertia is the sae about all of the. In its inertial properties, the body behaves like a circular cylinder. The tensor of inertia will take different fors when expressed in different axes. When the axes are such that the tensor of inertia is diagonal, then these axes are called the principal axes of inertia. The Search for Principal Axes and Moents of Inertia as an Eigenvalue Proble Three orthogonal principal axes of inertia always exist even though in bodies without syetries their directions ay not be obvious. To find the principle axis of a general body consider the body shown in the figure that rotates about an unknown principal axis. The the total angular oentu vector is Iω in the direction of the principle axis. For rotation about the principal axis, the angular oentu and the angular velocity are in the sae direction. 5

6 We seek a coordinate axes x, y and z, about which a rotation ω x, ω y and ω z, which is aligned with this coordinate direction, will be parallel to the angular oentu vector and related by the equation H Gx Iω x I 0 0 ω x H Gy = Iω y = 0 I 0 ω y. (7) H Gz Iω z 0 0 I ω z We then express the general for for angular oentu vector in coponents along the x, y and z axis in ter of the coponents of ω along these axes using the general for of the inertia tensor in the x, y, z syste, we have H Gx I xx I xy I xz ω x H Gy = I yx I yy I yz ω y. (8) H Gz I zx I zy I zz ω z To obtain the special directions of ω that is aligned with a principal axis, we equate these two expressions. H I 0 0 Gx Iω x ω x I xx I xy I xz ω x H 0 I 0 Gy = Iω y = ω y = I yx I yy I yz ω y. (9) H Gz Iω z 0 0 I ω z I zx I zy I zz ω z At this point in the process we know the inertia tensor in an arbitrary x, y, and z syste and are seeking the special orientation of ω which will align the angular oentu H G with the angular velocity ω. Collecting ters fro equation(11) on the left-hand side, we obtain (I xx I) I xy I xz ω 0 x I (I ω y =. (10) yx yy I) I yz 0 I zx I zy (I zz I) ω z 0 resulting in the requireent that I xx I xy I xz ω x 0 I yx I yy I yz I ω y = 0 I zx I zy I zz ω z 0 (11) 6

7 The structure of the solution for finding the principal axes of inertia and their agnitudes is a characteristicvalue proble. The three eigenvalues give the directions of the three principal axis, and the three eigenvectors give the oents of inertia with respect to each of these axis. In principal directions, the inertia tensor has the for I x 0 0 [I G ] = 0 I y I z where we will write I x = I xx, I y = I yy and I z = I zz. Also, in principal axes we will then have H G = I x ω x i + I y ω y j + I z ω z k. Parallel Axis Theore It will often be easier to obtain the tensor of inertia with respect to axis passing through the center of ass. In soe probles however, we will need to calculate the tensor of inertia about different axes. The parallel axis theore introduced in lecture L for the two diensional oents of inertia can be extended and applied to each of the coponents of the tensor of inertia. In particular we can write, (I xx ) O = (y + z ) d = ((y G + y ) + (z G + z ) ) d = (y + z ) + y G y d + z G z d + (y G + z G ) d = I xx + (y G + z G ). Here, we have use the fact that y and z are the coordinates relative to the center of ass and therefore their integrals over the body are equal to zero. Siilarly, we can write, and, (I yy ) O = I yy + (x G + z G ), (I zz ) O = I zz + (x G + y G ), (I xy ) O = (I yx ) O = I xy + x G y G, (I xz ) O = (I zx ) O = I xz + x G z G, (I yz ) O = (I zy ) O = I yz + y G z G. 7

8 Rotation of Axes In soe situations, we will know the tensor of inertia with respect to soe axes xyz and, we will be interested in calculating the tensor of inertia with respect to another set of axis x y z. We denote by i, j and k the unit vectors along the direction of xyz axes, and by i, j and k the unit vectors along the direction of x y z axes. The transforation of the inertia tensor can be accoplished by considering the transforation of the angular oentu vector H and the angular velocity vector ω. We begin with the expression of H and ω in the x 1, x, x 3 syste. H = [I]ω (1) We have not indicated by a subscript where the origin of our coordinates are, i.e. the center of ass G, a fixed point O, or any other point, because as long as we are siply doing a rotational transforation of coordinates about this point, it does not atter, Fro Lecture 3, we have that the transforation of a vector fro a coordinate syste x 1, x, x 3 into a coordinate syste x 1, x x 3 is given by H 1 i 1 i 1 i 1 i i 1 i 3 H 1 H 1 H = i i 1 i i i H = [T ] H. i 3 H 3 i 3 i 1 i 3 i i 3 i 3 H 3 H 3 where we have introduced the sybol [T] for the transforation atrix. 8

9 Likewise, the transforation of ω into the x 1, x, x 3 syste is given by ω 1 i 1 i 1 i 1 i i 1 i 3 ω 1 ω 1 ω = i i 1 i i i i 3 ω = [T ] ω. ω 3 i 3 i 1 i 3 i i 3 i 3 ω 3 ω 3 To deterine H (in the transfored syste) we ultiply both sides of equation (10) by [T ], and obtain H = [T ]H = [T ][I]ω (13) where [I] is the inertia atrix in the original coordinate syste. But this equation is not quite in the proper for; we need to have the for H = [I ]ω to identify [I ], the inertia atrix in the new coordinate syste. We take advantage of the fact that the atrix [T ] T is also the inverse of [T ], [T ] 1, so that [T ] T [T ] = [Iden], where [Iden] is the identity atrix. Now, we can always stick an identity atrix in a atrix equation like ultiplying by 1 so that we ay write H = [T ]H = [T ][I][T ] T [T ]ω = [T ][I][T ] T [T ]ω = [I ]ω (14) were we have grouped the ters so that it is obvious that the inertia tensor in the transfored coordinate syste is [I ] = [T ][I][T ] T (15) Therefore, if we want to calculate the tensor of inertia with respect to axis x y z, we can write in atrix for I i i i j i i j i k i x x I x y I x z k I xx I xy I xz i I y x I y y I y z = j i j j j k I yx I yy I yz i j j i k j. I z x I z y I z z k i k j k k I zx I zy I zz i k j k k k where we have returned to the x, y, z notation. Likewise, expressed in the direction cosines of the figure, the transforation is cos (θ 11 ) cos (θ 1 ) cos (θ 13 ) I xx I xy I xz cos (Θ 11 ) cos (Θ 1 ) cos (Θ 13 ) [I ] = cos (θ 1 ) cos (θ ) cos (θ 3 ) I yx I yy I yz cos (Θ 1 ) cos (Θ ) cos (Θ 3 ).. cos (θ 31 ) cos (θ 3 ) cos (θ 33 ) I zx I zy I zz cos (Θ 31 ) cos (Θ 3 ) cos (Θ 33 ) (16) Inertia Tensor of a Cube We are going to consider and calculate the inertia tensor of a cube of equal length sides b. (This exaple is taken fro Marion and Thorton.) For this choice of coordinates, it is obvious that the center of ass does not lie at the origin. It is also obvious that there is considerable syetry in the geoetry. 9

10 Evaluating the various integrals for the coponents of the inertia tensor, we have 1 1 Mb Mb Mb [I] = Mb Mb Mb Mb Mb Mb (17) This result is perhaps a bit surprising; because of the strong syetry of the cube we ight have expected these coordinates to be principal axes. However, our origin of coordinates is not at the center on ass, and by the parallel axis theore, we incur products of inertia the equal to total ass ties the various distances of center of ass fro the origin of coordinates. There are several options to find a set of principal axis for this proble. One is to use the syetry to identify a coordinate syste in which the products of inertia vanish. The transforation sketched in the figure is a good candidate to obtain the principal axes; that is a rotation of θ = π/4 about the x 3 axis, 1 followed by a rotation of ψ = sin 1 3 about the x axis. This would result in the x 1 axis going through the 1 upper point of the cube and through the center of ass as well. If we were to do this, we would find that indeed this transforation results in coordinate directions which are principal axis directions, and oreover that the inertias are equal about any axis in the plane perpendicular to the diagonal of the cube, so that all these axes are principal with equal oents of inertia. Another ore general ethod is to express the search for principal axes as an eigenvalue proble, as previously outlined. In a proble without strong and obvious syetry, identifying an appropriate transforation is often a atter of guess work; therefore the eigenvalue forulation is ore useful. 10

11 Products of Inertia in a Plane In any situations, we will know one principal axis fro syetry, and will be working with oents of inertia and inertia products in a plane. An exaple would be a flat plate of ass, thickness t, width a and length b. It is clear fro the sketch that the principal axis are the x y coordinates. We ay have reasons for wanting to express the oents of inertia and the products of inertia in the x, y syste (rotated fro the x, y syste by an angle θ) or to ove back and forth between the x, y syste, and the x, y syste. The coordinate transforation that takes a vector in the x, y, z syste into the x, y, z syste is x 0 cosθ sinθ x y = sinθ cosθ 0 y. (18) z z or (x ) = [T ](x); while the coordinate transforation that takes a vector in the x, y, z syste into the x, y, z syste is x cosθ sinθ 0 x y y. (19) = sinθ cosθ 0 z z or (x) = [T ] T where we have taken advantage of the fact that the atrix for the reverse transforation (x to x) is the transpose of the atrix for the forward transforation (x to x ). The inertia atrix for a rectangular plate with its center of ass at the origin in the x, y, z syste, for which the principal axes line up with the coordinate directions is b I x,y,z = M a a +b 1. (0) If we wish to obtain the inertia tensor in the x, y, z syste, we apply the transforation I x,y,z = [T ][I x,y,z][t ] T (1) 11

12 The Syetry of Three An interesting and practical result for inertias in a plane coes fro considering the oent of inertia of a body with tri-syetry, such as a three-bladed propeller or a wind turbine. It is clear that the axis noral to the figure is a principal axis, with asses syetrically balanced about the origin. It is also clear that both the x and y axis are principal axis, with inertia products going to zero by syetry. What is perhaps surprising, is that I xx the oent of inertia about the x axis is equal to I yy the oent of inertia about the y axis. Consider the x axis; for a point r 0 the contribution to the oent of inertia fro all 3 ass points is 0 I (r 0 + (r 0 /) ) = 3r xx =. Consider the y axis; for a point r 0 the contribution to the oent of inertia fro all 3 ass points is I yy = (( 3r 0 /) ) = 3r 0. Therefore, the total oent of inertia about both the x and y axis will be an integral of the ass distribution in d ties 3r. Since the oents of inertia are equal about the principal axis x and y, the oent of inertia about any axis in the x,y plane is also equal to I xx. The three-bladed propeller behaves as a unifor disk. This has iportant iplications for the vibrations of propellers of historically-interesting high perforance propeller fighter aircraft and wind turbines in yawing otion. ADDITIONAL READING J.L. Meria and L.G. Kraige, Engineering Mechanics, DYNAMICS, 5th Edition 7/7, Appendix B S. T. Thorton and J. B. Marion, Classical Dynaics of Particles and Systes, 4th Edition, Chapter

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