Free rotation of a rigid body 1 D. E. Soper 2 University of Oregon Physics 611, Theoretical Mechanics 5 November 2012

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1 Free rotation of a rigi boy 1 D. E. Soper 2 University of Oregon Physics 611, Theoretical Mechanics 5 November Introuction In this section, we escribe the motion of a rigi boy that is free to rotate about its center of mass. If the boy is just floating in space with no forces acting on it, then the center of mass moves with constant velocity. If the boy is in free fall in the gravitational fiel of the earth, then the center of mass moves along a parabolic trajectory. In both cases, the motion of the center of mass is not very interesting, so we just eliminate it from consieration by supposing that the center of mass is fixe. What the boy can o is rotate, so we have to figure out how to escribe an arbitrary rotation. 2 Rotations Let x I be the coorinates of a point in the inertial frame in which we suppose that the center of mass of the boy is fixe. We will call this the lab frame. Let x i be the coorinates of the same point in a reference frame fixe on the boy. (I use capital letters for components in the lab frame, lower case letters for components in the boy frame.) Then the two coorinates are relate by a rotation matrix, x I = R Ij x j. (1) The matrix R tells the rotation that the boy has unergone. Thus the ynamical question is to etermine how R changes with time. But we have a problem: a 3 3 matrix has nine components, but there are only three egrees of freeom. Thus we nee to write R in terms of three variables. 1 Copyright, 2012, D. E. Soper 2 soper@uoregon.eu 1

2 One straightforwar way to o this (not mentione in your book) is to say that R is a rotation through angle ψ about an axis n whose components are (n 1, n 2, n 3 ) = ( n 1, n 2, n 3 ) = (sin θ cos φ, sin θ sin φ, cos θ). (2) (To think about: why oes (n 1, n 2, n 3 ) = ( n 1, n 2, n 3 )?). The rotation matrix is R Ij = cos ψ (δ Ij n I n j ) sin ψ ɛ Ijk n k + n I n j (3) Where i this come from? It is the stanar result in the case that n i is along one of the coorinate axes an it is covariant uner rotations of the coorinate axes in the lab an boy system mae before we rotate one of these systems with respect to the other. This establishes that R can be written in terms of three coorinates ψ, θ, an φ. However, people usually use three ifferent coorinates, calle the Euler angles. The Euler angles are efine by writing R as a prouct of three simple matrices: R = R (1) R (2) R (3). (4) The matrix R (1) is The matrix R (2) is The matrix R (3) is The prouct is cos φ sin φ 0 R (1) = sin φ cos φ 0. (5) R (2) = 0 cos θ sin θ. (6) 0 sin θ cos θ cos ψ sin ψ 0 R (3) = sin ψ cos ψ 0. (7) R = (8) cosφ cosψ cosθ sinψ sinφ sinψ cosφ cosθ sinφ cosψ sinθ sinφ cosψ sinφ + cosθ cosφ sinψ sinψ sinφ + cosθ cosφ cosψ sinθ cosφ. sinθ sinψ sinθ cosψ cosθ 2

3 To escribe this in physical terms, think of the relation giving the coorinates x i in the boy system of a point with coorinates x i in the lab system: x i = [R T (3)R T (2)R T (1)] ii x I. (9) We begin with a rotation of coorinates about the boy 3-axis (which at the start is the same as the lab 3-axis) through angle φ: cos φ sin φ 0 R(1) T = sin φ cos φ 0. (10) (That is, the coorinate axes attache to the boy rotate though angle +φ about the boy 3-axis.) Then we perform a rotation of coorinates about the boy 1-axis through angle θ: R(2) T = 0 cos θ sin θ. (11) 0 sin θ cos θ (That is, the coorinate axes attache to the boy rotate though angle +θ about the boy 1-axis.) Finally, we rotate the coorinates about the boy 3-axis through angle ψ: cos ψ sin ψ 0 R(3) T = sin ψ cos ψ 0. (12) (That is, the coorinate axes attache to the boy rotate though angle +ψ about the boy 3-axis.) Exercise 2.1: Multiply the three matrices together to show that this result hols. You might want to use Mathematica to o this. Exercise 2.2: What if we efine the Euler angles as follows. We subject the boy to three successive rotations. We begin with a rotation about the lab 3-axis through angle φ. Then we perform a rotation about the lab 1- axis through angle θ. Finally, we rotate about the lab 3-axis through angle 3

4 ψ. The boy coorinates woul then be relate to the lab coorinates by x I = R Ij x j, but with a ifferent matrix R. Please write R as a matrix prouct of three matrices, but o not perform the matrix multiplication. Exercise 2.3: What if we efine the Euler angles as follows. We subject the boy to three successive rotations. We begin with a rotation about the lab 3-axis through angle φ. Then we perform a rotation about the boy 1-axis through angle θ. Finally, we rotate about the lab 1-axis through angle ψ. The boy coorinates woul then be relate to the lab coorinates by x I = R Ij x j, but with a ifferent matrix R. Please write R as a matrix prouct of three matrices, but o not perform the matrix multiplication. 3 The lagrangian Now that we know how to escribe a rotation, we can write own the lagrangian. It is L = 1 m 2 p v (p) I v (p) I (13) p where we have a sum over an inex p that labels the particles in the rigi boy an where v (p) I is the Ith component of the velocity of particle p in the lab system. There is an implicit sum over the inices I. We have x (p) I = R Ij x (p) j (14) where R Ij epens on the time t but x (p) j oes not. Thus Thus L = p v (p) I = ṘIj x (p) j (15) 1 m 2 pṙij x (p) j Ṙ Ik x (p) k (16) We efine the components of the inertia tensor in the boy frame by I jk = [ ] m p (x (p) l x (p) l )δ jk x (p) j x (p) k. (17) p 4

5 Note that the components of this tensor o not change as the boy moves because we are taking its components along the boy fixe axes. We actually on t have quite this tensor in the lagrangian. What we have is Ĩ jk = p m p x (p) j x (p) k. (18) We can relate these. We have so Thus This gives us the lagrangian I jk = Ĩll δ jk Ĩjk (19) I jj = Ĩll 3 Ĩjj = 2 Ĩjj (20) Ĩ jk = 1 2 I ll δ jk I jk. (21) L = 1 2 ṘIj ṘIk ( 1 2 I ll δ jk I jk ). (22) Here R is a function of the three Euler angles, so L is a function of the three Euler angles an their time erivatives. We simply have to multiply everything together to express L as a simple function of the three Euler angles an their time erivatives, then ifferentiate to obtain the equations of motion. However, a little bit of thought may be better than blin algebra. We unertake this in the next section. 4 Analysis of the lagrangian In orer to unerstan the lagrangian better, let us write ṘIj in the form Ṙ Ij = R Ik Ω kj (23) where this equation efines the matrix Ω: Ω = R 1 Ṙ. The matrix Ω has a special structure. To see what it is, recall that because R is a rotation matrix we have R T R = 1 (24) Thus R 1 = R T, so Ω = R T Ṙ. (25) 5

6 Differentiating (24) with respect to t we have or Ṙ T R + R T Ṙ = 0 (26) Ω T + Ω = 0. (27) This says that Ω is antisymmetric, so we can write it as Ω ij = ɛ ijk ω k. (28) Let s see how to interpret ω l. To o this, consier a fixe point with coorinates x I in the lab. The boy coorinates of this point are x i = R T ij x J (29) Of course, the boy coorinates of the point are changing because R is always changing. Let s see how fast the coorinates x i are changing. We use a matrix notation in orer to keep the number of inices own: That is ẋ = ṘT x = ṘT R R T x = Ω T x = Ω x. (30) ẋ i = ɛ ijk ω k x j = ɛ ikj ω k x j (31) or, in a vector notation, x = ω x. (32) t That is, an observer on the boy sees a point in the lab rotating with angular velocity ω. Thus the components ω i are the components in the boy fixe coorinates of the instantaneous angular velocity of the boy. Since Ω is so nice, let s write the Lagrangian in terms of it. We have L = 1 2 R Im Ω mj R In Ω nk ( 1 2 I ll δ jk I jk ) = 1 2 δ mn Ω mj Ω nk ( 1 2 I ll δ jk I jk ) = 1 2 Ω nj Ω nk ( 1 2 I ll δ jk I jk ) = 1 2 ɛ nji ω i ɛ nkm ω m ( 1 2 I ll δ jk I jk ) = 1 2 (δ jk δ im δ jm δ ik ) ω i ω m ( 1 2 I ll δ jk I jk ) = 1 2 ω i ω i ( 1 2 I ll δ jj I jj ) ω k ω j ( 1 2 I ll δ jk I jk ) = 1 ω 2 i ω i ( 3I 1 2 ll I ll ) ω i ω i 2 ll + ω k ω j I jk = 1 ω 2 j ω k I jk. (33) 6

7 We may as well simplify things a little more. We i not specify how to choose the boy axes. Since I is a real symmetric matrix, we can choose axes such that it is iagonal: I I = 0 I 2 0. (34) 0 0 I 3 The values I 1, I 2, I 3 are known as the principle moments of inertia. Using this choices of axes in the boy, we have L = 1 2 (I 1ω I 2 ω I 3 ω 2 3). (35) Of course, we knew how L ha to turn out. The lagrangian is the kinetic energy an the kinetic energy is 1 2 Iω2 with suitable tensor inices. So what we have that is really new is the relation of ω to the angles that efine the orientation of the boy in the lab. Using the efinitions in Eqs. (25) an (28) one fins ω 1 = φ sin θ sin ψ + θ cos ψ ω 2 = φ sin θ cos ψ θ sin ψ ω 3 = φ cos θ + ψ. (36) Exercise 4.1: Use Eqs. (35) an (36) to fin the equations of motion for the three Euler angles. 5 The angular velocity an the Euler angles Actually, Eq. (36) is amazingly simple. In principle, we ifferentiate R with respect to time, then form Ω = R T Ṙ. Finally ω 1 is Ω 23, etc.. But that s pretty messy. Couln t we be more clever? Lanau gives a nice geometrical argument. But we can o it with algebra, too, an that woul teach us something. We have Ω = R T (3)R T (2)R T (1)[R (1) R (2) Ṙ (3) + R (1) Ṙ (2) R (3) + Ṙ(1)R (2) R (3) ]. (37) 7

8 Let s consier the three terms separately. Begin with the first. Using R T R = 1 we have R T (3)R T (2)R T (1)R (1) R (2) Ṙ (3) = R T (3)Ṙ(3). (38) That s pretty easy to compute: cos ψ sin ψ 0 sin ψ cos ψ 0 R(3)Ṙ(3) T = sin ψ cos ψ 0 cos ψ sin ψ 0 ψ (39) Multiplying these an using sin 2 ψ + cos 2 ψ = 1, we get R(3)Ṙ(3) T = ψ (40) That is [ ] R(3)Ṙ(3) T = ψ ɛ ijk ẑ k (41) ij where ẑ is a unit vector in the 3-irection: ẑ k = δ k3. We now have one thir of our answer. Now, try the secon term: R T (3)R T (2)R T (1)R (1) Ṙ (2) R (3) = R T (3)R T (2)Ṙ(2)R (3). (42) As in the previous analysis, we get [ ] R(2)Ṙ(2) T = θ ɛ ijk ˆx k (43) ij where ˆx k = δ k1 represents a unit vector in the 1-irection. But now we have this matrix wege in between R T (3) an R (3). We nee a eep truth. Here is a eep truth. The epsilon tensor is invariant uner rotations: R ia R jb R kc ɛ abc = ɛ ijk (44) for any rotation matrix R. To see this, note that the left han sie of the equation is ɛ ijk times the eterminant of R. But if R T R = 1 we have (et R T )(et R) = 1 or (et R) 2 = 1 so et R = ±1. In fact mirror reflections have eterminant 1, but any rotation that you can reach by a continuous path from the unit matrix must have eterminant +1, since the 8

9 unit matrix has eterminant +1 an if we change R continuously from the unit matrix, its eterminant can t jump to 1. We can apply this result by inserting one R T R, which is the unit matrix, into our previous result, then using the invariance of the epsilon tensor uner rotations: [R T(3)RT(2)Ṙ(2)R ] (3) = θ R (3) ai R(3) bj ɛ abk ˆx k The vector u l = R (3) kl through angle ψ: ij = θ R (3) ai R(3) bj ɛ abcr (3) cl R (3) kl ˆx k = θ ɛ ijl R (3) kl ˆx k (45) ˆx k is what you get if you rotate ˆx about the 3-axis R (3) kl ˆx k = cos ψ ˆx k sin ψ ŷ k. (46) (That s most easily seen with a picture, but you coul just multiply the vector by the matrix to verify it.) Thus [R T(3)RT(2)Ṙ(2)R ] (3) = θ ɛ ijk [cos ψ ˆx k sin ψ ŷ k ]. (47) Now we are reay for the last term. We fin Then where ij R T (3)R T (2)R T (1)Ṙ(1)R (2) R (3). (48) [R T (1)Ṙ(1)] ij = φ ɛ ijk ẑ k (49) R T (3)R T (2)R T (1)Ṙ(1)R (2) R (3) = φ ɛ ijk u k (50) u k = R (2) jl R(3) lk ẑj = R (3) 1 kl R (2) 1 lj ẑ j. (51) Thus we take a unit vector in the 3-irection an rotate it about the the 1-axis through angle θ then about the 3-axis through angle ψ. The first step gives the vector cos θ ẑ + sin θ ŷ. (52) Then the secon step gives cos θ ẑ + sin θ sin ψ ˆx + sin θ cos ψ ŷ. (53) 9

10 Thus R T (3)R T (2)R T (1)Ṙ(1)R (2) R (3) = φ ɛ ijk [cos θ ẑ k + sin θ sin ψ ˆx k + sin θ cos ψ ŷ k ]. (54) We can now put this together where Ω ij = ɛ ijk ω k, (55) ω = ψ ẑ k + θ [cos ψ ˆx k sin ψ ŷ k ] This is the result shown in Eq. (36). + φ [cos θ ẑ k + sin θ sin ψ ˆx k + sin θ cos ψ ŷ k ]. (56) 6 The moment of inertia tensor an angular momentum Let s look at the relation between the angular momentum of the boy, its angular velocity, an the moment of inertia tensor. Working, to start with, in the lab system, we have L I = p m p ɛ IJK x J p x K p = p m p ɛ IJK x J p ɛ KLM ω L x M p = (δ IL δ JM δ IM δ JL ) p m p x J p ω L x M p (57) = p m p ( x J p x J p δ IL x I p x L p ) ω L This is where Ī IL = p L I = ĪIL ω L, (58) m p ( x J p x J p δ IL x I p x L p ). (59) 10

11 This equation is covariant uner rotations, so we can write it also in the boy frame: L i = I il ω l, (60) where I il = p This equation is the same as our efinition (17). m p (x j px j pδ il x i px l p). (61) 7 Euler s equations We can write simple equations, known as Euler s equations, for the rate of change of ω in the boy coorinate system. Knowing ω as a function of time oes not irectly tell you the Euler angles as a function of time because the components of ω are relate to not just the Euler angles but also their time erivatives. Nevertheless, Euler s equations are quite useful. We begin by recalling Eq. (32) for the rate of change of the boy frame components of a vector A that is fixe in the lab system. t A i = ɛ ijk ω j A k. (62) The lab frame components of the angular angular momentum vector L are fixe, so we can apply this to the angular momentum. Using Eq. (60), we have t I ij ω j = ɛ ijk ω j I kl ω l. (63) Since the components of the inertia tensor are fixe, this is I ij ω j = ɛ ijk ω j I kl ω l. (64) We have chosen our boy coorinates such that I is iagonal. Thus we have three equations, the first of which is I 1 ω 1 = ω 2 I 3 ω 3 + ω 3 I 2 ω 2, (65) or ω 1 = ω 2 ω 3 [I 2 I 3 ]/I 1. (66) 11

12 Thus the three equations are ω 1 = ω 2 ω 3 [I 2 I 3 ]/I 1 8 The tennis racket theorem ω 2 = ω 3 ω 1 [I 3 I 1 ]/I 2 ω 3 = ω 1 ω 2 [I 1 I 2 ]/I 3. (67) One simple consequence of Euler s equations is that if the boy is rotating about one of the principle axes of the inertia tensor, then it continues to o so an the angular velocity stays constant. That is, ω 2 = ω 3 = 0 at one time, then ω 2 an ω 3 stay zero an ω 1 is constant. Is this stable against small perturbations? Suppose that at time zero, ω 1 > 0 while ω 2 an ω 3 are small. Then we can write the equations to first orer in the small quantities: ω 1 = 0 ω 2 = ω 3 ω 1 [I 3 I 1 ]/I 2 ω 3 = ω 1 ω 2 [I 1 I 2 ]/I 3. (68) The first equation says that ω 1 is constant (in this approximation, an as long as ω 2 an ω 3 stay small). We can ifferentiate the other equations again an substitute for ω 2 an ω 3 on the right han sies to get ω 2 = ω1 2 [I 1 I 2 ][I 1 I 3 ] ω 2 I 2 I 3 ω 3 = ω1 2 [I 1 I 2 ][I 1 I 3 ] ω 3. I 2 I 3 (69) Thus we have two ientical equations that escribe small oscillations if [I 1 I 2 ][I 1 I 3 ] > 0 or a combination of exponential growth an ecay if [I 1 I 2 ][I 1 I 3 ] < 0. If I 1 is the largest eigenvalue we have a positive sign an thus small oscillations. If I 1 is the smallest eigenvalue we have a positive sign an thus small oscillations. But if I 1 is in-between I 2 an I 3, then we have exponential growth an ecay. Of course, unless the initial conitions are just right, there will be a small amplitue for exponential ecay an after a little while it is the growing part that you will see. The conclusion is that rotation about the intermeiate axis is unstable. 12

13 9 Conserve quantities We can gain some insight into how the boy moves by making use of conserve quantities. Begin by consiering the angular momentum, L i = I ij ω j. The components in the boy frame, with our choice of axes such that the inertia tensor is iagonal, are L = (I 1 ω 1, I 2 ω 2, I 3 ω 3 ). (70) We know that the laboratory components of the angular momentum remain constant, but these boy-frame components are continually changing since the boy is rotating. We can get ω from L using this equation, but it is perhaps more useful to consier how the boy components of L change because the lab components of this vector o not change. (In contrast, the components of ω change in both frames.) Although the boy components of L change, the sum of their squares equals the sum of the squares of the components in the lab frame. Thus L 2 = L L L 2 3 (71) is a constant. Thus L remains on a sphere. We also know that the kinetic energy E is conserve, with 2E = I 1 ω I 2 ω I 3 ω 2 3. (72) The surface of constant E is an ellipsoi. In terms of L, this is 2E = L2 1 I 1 + L2 2 I 2 + L2 3 I 3. (73) Thus L must stay on this ellipsoial surface. Combining these results, we see that L remains on the curve that is the intersection of these two surfaces. Eviently, the motion of L is perioic. Let us choose the labelling so that I 1 < I 2 < I 3. Then suppose that E is about as big as it can be for a given L 2. That is, E L 2 /I 1. Then we must have L 2 1 L 2 an L an L That is, L moves along a little path aroun the 1-axis. On the other han, suppose that E is about as small as it coul be, E L 2 /I 3. Then we must have L 2 3 L 2 an L an L That is, L moves along a little path aroun the 3-axis. We see that there are motions in which the boy is approximately rotating about either the 13

14 axis with the smallest moment of inertia or about the axis with the largest moment of inertia. On the other han, with E L 2 /I 2, the curve that L follows oes not stay near the 2-axis. Motion about the 2-axis is not stable. That is what we saw in the previous section with a ifferent style of analysis. 10 Symmetrical rigi boy An interesting special case occurs when two of the principal moments of inertia are equal, say I 1 = I 2 I T. (We coul have either I 3 > I T or I 3 < I T.) In this case the intersection of the sphere L 2 = const. with the ellipsoi 2E = const. is a circle about the symmetry axis, which is the 3-axis. We instantly conclue that L executes a uniform motion along this circle. The same statement hols for ω, with (ω 1, ω 2, ω 3 ) = (L 1 /I T, L 2 /I T, L 3 /I 3 ). (74) We can see this by solving the Euler equations, which become ω 1 = ω 2 ω 3 [I T I 3 ]/I T ω 2 = ω 3 ω 1 [I T I 3 ]/I T ω 3 = 0. (75) The thir equation says that ω 3 remains constant. The first two equations give ω 1 = [ω 3 (I T I 3 )/I T ] 2 ω 1. (76) The solution of this is with Then ω 1 = A sin( ω t + δ). (77) ω = ω 3 (I T I 3 )/I T. (78) ω 2 = A cos( ω t + δ). (79) That is, ω an L move in circles with a precession velocity ω. In the lab frame, the symmetry axis moves in circles about the (fixe) angular momentum vector with this precession velocity. 14

15 Let s see what this looks like in the lab frame. As a preliminary step, we write ( L1 (ω 1, ω 2, ω 3 ) =, L 2, L ( ) ) L 3. (80) I T I T I T I 3 I T That is ω = ω L u + ω n n (81) where n is a unit vector along the boy 3-axis, u is a unit vector along the irection of L, an ω L = L I T ω n = I T I 3 I 3 I T L 3. (82) This representation will be useful to us because u is fixe in the lab an n is fixe in the boy. How o the lab components of n change with time? Since n is fixe in the boy system, we have t n I = ɛ IJK ω J n K. (83) (Recall our notation that vector names with bars an uppercase subscripts refer to lab coorinates.) We can insert our ecomposition of ω into this. The secon term, proportional to n oes not contribute since ɛ IJK n J n K = 0. Thus t n I = ω L ɛ IJK ū J n K. (84) This says that n rotates about u with a uniform angular velocity ω L. Now while the boy 3-axis is rotating about u, the boy as a whole is rotating somehow. To escribe this motion, let s look at it in a reference frame that rotates about u with a uniform angular velocity ω L. In this frame u is still fixe an n is also fixe. Consier a fixe point on the boy with lab coorinates x I. Let its coorinates in the new frame be x a. We have x a = R(t) ai x I (85) where R is the rotation matrix relating the two frames. The time erivative of this is t x a = R ai t x I ω L ɛ abc ũ b x c. (86) 15

16 We know what the rate of change of x I is: We can use Thus That is Recall that t x I = ɛ IJK ω J x K. (87) R ai ɛ IJK ω J x K = ɛ abc R bj R ck ω J x K = ɛ abc ω b x c. (88) t x a = ɛ abc ω b x c ω L ɛ abc ũ b x c. (89) t x a = ɛ abc ( ω b ω L ũ b ) x c. (90) ω ω L u = ω n n. (91) Thus t x a = ω n ɛ abc ñ b x c. (92) This says that (in this frame) the boy rotates about the boy 3-axis with a uniform angular velocity ω n. Thus the motion of the boy is really pretty simple. It is a combination of two uniform rotations. The ratio of the two angular velocities is ω n ω u = I T I 3 I 3 L 3 L T. (93) Thus it epens on the shape of the boy an on the initial conitions. Exercise 10.1: Analyze the motion of a boy with principle moments of inertia At time zero let I 1 = 1 kg m 2 I 2 = 2 kg m 2 I 3 = 3 kg m 2 (94) φ = π/4 θ = π/4 ψ = π/4 (95) 16

17 an φ = 1 s 1 θ = 0 ψ = 0. (96) Fin the location of the points with boy coorinates (1 m, 0, 0), (0, 1 m, 0) an (0, 0, 1 m) at times t = 0, t = 1 sec, an t = 10 sec. Optionally, you can a graphs of where these three points go. If you use Mathematica, I recommen trying ListPointPlot3D[points]. 17