identify appropriate degrees of freedom and coordinates for a rigid body;

Transcription

1 Chapter 5 Rotational motion A rigid body is defined as a body (or collection of particles) where all mass points stay at the same relative distances at all times. This can be a continuous body, or a collection of discrete particles: the same equations of motion hold for both cases. A rigid body will move as a single entity, but it can change its orientation, and this motion can be highly nontrivial as we shall see. The notion of a rigid body is an idealisation, since no real completely rigid bodies exist in the real world. Firstly, real bodies consist of atoms and molecules which always undergo vibrations (and electrons, as quantum particles, are never at rest). These complications can be ignored for macroscopic bodies. Secondly, it is always possible to deform an actual body, and this can happen even in the absence of external forces if for example the internal forces keeping the constituents apart are not strong enough to balance the attractive forces or vice-versa! Still, the idealised description is reasonable for many macroscopic solids, and provide a good description of for example tops, gyroscopes, bicycle wheels, falling sandwiches and tumbling cats. Learning outcomes At the end of this section, you should be able to identify appropriate degrees of freedom and coordinates for a rigid body; describe rotations using rotation matrices, and explain the general properties of rotation matrices; define the inertia tensor and explain the relation between the inertia tensor, rotational kinetic energy and angular momentum; calculate the inertia tensor for simple objects; explain what is meant by principal axes of inertia and how they may be found; use the equations of motion for rotating bodies (Euler equations) to analyse the motion of rotating systems. 69

2 5. How many degrees of freedom do we have? Let us imagine a rigid body consisting of N discrete particles. Altogether this gives us 3N coordinates. The requirement that the body is rigid, ie all the internal distances are fixed, imposes constraints on these coordinates. We denote the distance between particles i and j by r ij. We can work out how many constraints and hence degrees of freedom we have: N = Here we have a single constraint r = r, and hence we have a total of 3 = 5 degrees of freedom. N = 3 We now have 3 internal distances to be fixed: r, r 3 and r 3, giving 3 constraints and 3N 3 = 6 degrees of freedom. N = 4 Here there are 6 internal distances to be fixed, so we have = 6 degrees of freedom. Strictly speaking, there are different configurations which both satisfy these 6 constraints, corresponding to rigid bodies that are mirror images of each other. This ambiguity does not correspond to any physical degree of freedom, however. N = 5 Now it appears there would be constraints, corresponding to the pairs of particles we have. However, if you construct a 5-particle body from a 4-particle one, you will see that once the fifth particle has been positioned relative to 3 of the others, the position relative to the fourth one is also given. (You may try this for yourself!) We therefore only have 9 constraints, and = 6 degrees of freedom. N 6 As was the case for N = 5, we will need to specify 3 relative distances to position the 6th particle relative to the other 5, 3 more for the 7th particle, etc. This cancels out the 3 coordinates that each new particle comes with, so we end up with 6 degrees of freedom in every case. In summary, we find that a rigid body has 6 degrees of freedom, except for N =, which is a special case, and has only 5 degrees of freedom. A more careful analysis will reveal that there are only 5 degrees of freedom whenever the rigid body is -dimensional (all the constituent particles are located on a single line), and 6 otherwise. Three of these degrees of freedom can be taken to represent the position of the body, and it is natural to use the centre of mass coordinates of the body as corresponding generalised coordinates. The three remaining degrees of freedom represent the orientation of the body, and it is natural to choose three angles as coordinates. We will come back to how these angles may be chosen later on. Dynamically, the 3+3 degrees of freedom correspond to two different kinds of motion: the linear motion of the centre of mass, and the rotation of the body about its centre of mass. We can now see why there are 3 such degrees of freedom: they correspond to rotations (changes in orientation) about 3 axes going through the centre of mass. In the case of a -dimensional body, there are only rotational degrees of freedom, since rotation about the line the body is located on does not correspond to any real motion. This discussion can be summarised in Chasles theorem, which states 7

3 Any motion of a rigid body is the sum of a translation and a rotation. From now on we will focus on how we can describe and study the rotational motion and degrees of freedom. 5.. Relative motion as rotation Assume first that one point in the body is fixed. We can choose this point to be the origin of our coordinate system. If we now move a point P relative to this origin, it follows from the definition of a rigid body that although the position r of P changes, the distance r of P from the origin does not change, and the displacement d r r. This defines a rotation of an angle dφ about some axis passing through the origin. Since all the points in the body remain at a fixed distance relative to the origin, they all rotate about an axis through the origin, and since the body retains its shape, they all rotate about the same axis. This can be summarised in Euler s theorem: Any movement of a rigid body with one point fixed is a rotation about some axis. If we say that the vector d φ points along the axis of rotation, and this vector forms and angle θ with the position vector r, we have The velocity of the point P is then d r = r sin θ d φ or d r = d φ r. (5.) v = d r dt = d φ r = ω r. (5.) dt If in addition, the whole body moves with some linear velocity V, the total velocity of the point P is v = V + v rel = V + ω r (5.3) You have in the past studied the case where the axis of rotation, ie the direction of ω is fixed. But in the general case, ω = ω(t) can change both magnitude and direction, and we need to describe this general situation. 5. Rotated coordinate systems and rotation matrices Before we go on to discuss the kinematics and dynamics of rotational motion, let us look at the coordinates we can use to describe the rigid body. It is clearly convenient to describe the shape of the body, or the relative positions of the constituent parts of the body, using a coordinate system that is sitting in the body and moving with it. It will also turn out to be convenient to describe the rotational motion of the body in such a 7

4 system, since it is natural to describe this as the body rotating about its own axes. We call such a coordinate system the body coordinate system. But to describe the motion of the body in space we need a coordinate system fixed in space, not moving with the body. We therefore need to know the relation between the body coordinate system and this fixed coordinate system. Ignoring any linear displacements, these coordinate systems will be rotated relative to each other. We therefore need to know how coordinates change when the coordinate system is rotated. Let us call the original coordinates of a point (eg in the fixed coordinate system) r = (x, y, z) and the coordinates in the rotated (eg the body) coordinate system r = (x, y, z ). We call the angles between the axes in the original and the rotated coordinate systems θ ij, i = x, y, z: for example, θ xz is the angle between the x -axis and the z-axis. We can then write the position vector (or indeed any vector) as r = xˆx + yŷ + zẑ = x ˆx + y ŷ + z ẑ. (5.4) Since ˆx, ŷ, ẑ are orthogonal, we can find the rotated coordinate x by x = r ˆx = xˆx ˆx + yŷ ˆx + zẑ ˆx = x cos θ xx + y cos θ xy + z cos θ xz. (5.5) Similarly we find y == x cos θ yx + y cos θ yy + z cos θ yz, z == x cos θ zx + y cos θ zy + z cos θ zz, (5.6) This can be written as a matrix equation, x cos θ xx cos θ xy cos θ xz x y = cos θ yx cos θ yy cos θ yz y r = A r (5.7) z cos θ zx cos θ zy cos θ zz z The 3 3 matrix A is called the rotation matrix. 5.. Active and passive transformations There are two equivalent ways of thinking about rotations:. You rotate the coordinate system, as described above. This is called a passive rotation, since it is not doing anything to the world, only to the mathematical description of it. The rotation matrix then gives the relation between the old and the new coordinates.. You leave the coordinate system in place, but rotate your points (for example the body you want to describe) in the opposite direction. This is called an active rotation, since you are now doing something to the world. The rotation matrix then gives the positions of the body after it has been rotated, given the position before. The two points of view are mathematically equivalent: the relation between old and new coordinates are exactly the same. 7

5 The two pictures: active and passive, and the equivalence between them, can be extended to other transformations, such as translations, where it is the same whether you move an object from a position x to a position x + a or you shift the coordinate system by a. 5.. Elementary rotation matrices For a rotation about the z-axis, it is clear that the z-coordinates are unchanged. If we rotate an angle θ, then we have that θ xx = θ yy = θ (the angles between the old and new x- and y-axes respectively). The angle θ xy between the old y-axis and the new x-axis is 9 θ, while the angle θ yx between the old x-axis and the new y-axis is 9 + θ. The rotation matrix is therefore A z (θ) = cos θ cos(9 θ) cos(9 + θ) cos θ = cos θ sin θ sin θ cos θ (5.8) Similarly we find for rotations about the x- and y-axes, cos θ sin θ A x (θ) = cos θ sin θ, A y (θ) =. (5.9) sin θ cos θ sin θ cos θ 5..3 General properties of rotation matrices. Any combination of two successive rotations about the same point (albeit about different axes through that point) is a rotation about that point. This is obvious, since the relative distance of each point from the fixed point (origin) remains unchanged throughout. We can therefore describe the combined rotation by a rotation matrix which is the product of the two rotation matrices. If we call the coordinates after the first rotation r and after the second rotation r, and the two rotation matrices A (first rotation) and A (second rotation), we have r = A r = A (A r) = (A A) r = B r with B = A A. (5.) Note that the first rotation matrix is on the right and the second rotation matrix is on the left. Example 5. Find the rotation matrix corresponding to a 3 rotation about the x-axis followed by a 45 rotation about the z-axis. Solution: A rotation of an angle 3 about the x-axis is, according to (5.9), A x (3 ) = cos 3 sin 3 = 3. (5.) sin 3 cos

6 The matrix for a 45 rotation about the z-axis is cos 45 sin 45 A z (45 ) = sin 45 cos 45 =. (5.) The combined rotation is A = A z A x = = (5.3) 3. Rotations (about different axes) are not commutative: the order in which you do them matters. We know that matrix multiplication is not commutative: AB BA if A and B are matrices. You can also show for yourself that if you for example rotate a book 9 about the x-axis followed by 9 about the y-axis, you get something different from doing the two in reverse order. Example 5. The rotation matrix for a 45 rotation about the z-axis followed by a 3 rotation about the x-axis is B = A x A z = 3 3 = B (5.4) All rotation matrices are orthogonal, A T A =. Proof: We know that the distance of any point from the origin is the same before and after the rotation. Therefore we have that x r = ( x y z ) y = r T r = (Ar) T (Ar) = r T A T Ar = r = r T r (5.5) z A T A =. (5.6) We can check explicitly that all the matrices in the examples above are orthogonal. We can also show that if A and B are orthogonal, then their product AB is also 74

7 orthogonal: (AB) T (AB) = (B T A T )(AB) = B T (A T A)B = B T B =. (5.7) 4. For every rotation A there is an inverse rotation given by A T which brings us back to our starting point. It is physically obvious that you can always get back to your starting point by reversing all your rotations in reverse order. It follows from point 3. that the inverse rotation matrix is given by A T. 5. (a) Any (proper) rotation can be expressed as a combination of elementary rotations about coordinate axes. (b) No combination of such rotations can produce a reflection r r. (c) All proper rotations have det A =. Improper rotations (involving an odd number of reflections) have det A =. The proof of (a) is complicated, but we will use this fact later on when we will define coordinates corresponding to the the rotational degrees of freedom. Statement (c) follows from the properties of determinants. It is straightforward to show that the elementary rotation matrices all have det A =. We also have det(ab) = (det A)(det B) (5.8) so any combination of elementary rotations must have determinant. Moreover, any orthogonal matrix A must have det A = ±: det = = det(a T A) = det A T det A = (det A). (5.9) Finally, the reflection matrix, which takes (x, y, z) ( x, y, z) is R =, det R =. (5.) It is clearly impossible to construct this from a product of matrices with determinant. Physically, this means that it is impossible (in 3 dimensions) to rotate a body into its mirror image The rotation group [optional] A group is defined as a set of elements, together with a composition (multiplication) operation, with the following properties:. There exists an identity element (called or e) which is a member of the group.. The combination a b of two elements a and b is also a member of the group. 3. For every member a of the group there is an inverse a which is also a member of the group, such that a a = a a =. 75

8 We see that rotations form a group according to this definition: properties and 4 of Sec correspond to properties and 3 above. The identiy element is the operation of doing nothing, corresponding to the identity matrix. We can use the rotation matrices to define this group, and this gives the rotation group its name: O(3), or the group of real orthogonal 3 3 matrices. The set of all proper rotations also form a (smaller) group, called SO(3), or the group of all special [determinant ] real orthogonal 3 3 matrices. However, this is merely a particular representation of the general operation that we call rotations. We could equally well represent the group elements by actual rotations in space, or by 3 angles (with a suitably defined multiplication operation), or in many other ways. All these different representations would share the same multiplication table, which is what ultimately defines the group. This theory of different representations of the same group is mathematically extremely powerful, and the group-theory properties of rotations are extremely important in modern physics. For example, particles and bodies in general may be classified according to how they behave when rotated, and this turns out to be a fundamental classification. It is possible to express any rotation matrix formally as L A = e T, with T =, T =, T 3 =. (5.) This expression is very useful theoretically, but useless if you want to find an explicit form for A. It does however give a direct connection between rotations and angular momentum. The T matrices obey the commutation relations [T, T ] = T 3, [T, T 3 ] = T, [T 3, T ] = T. (5.) These are essentially the commutation relations of the angular momentum operators in quantum mechanics, and the vector L is proportional to the angular momentum of the particle in question. But it turns out that there is another group with the same multiplication table, namely the group of complex unitary matrices with determinant, called SU(). The operations of this group actually describe the rotations of fermions, while bosons such as photons are described by the usual 3 3 rotation matrices. The SU() matrices can be written in a similar form to (5.), where σ = ( ), σ = A = e i S σ, (5.3) ( ) ( ) i, σ i 3 =, (5.4) are the three Pauli matrices, which obey the same commutation relations as the T - matrices in (5.) (up to a factor i). The vector S is a new form of angular momentum called spin, which corresponds to an internal rotation degree of freedom. A curious result is that we must rotate a fermion by 4π ( full rotations) to get back to where we started! 76

9 5.3 Euler angles Figure 5.: The Euler angles We now go on to discuss which coordinates we can use to describe the orientation of a rigid body, or alternatively, which three parameters can be used to uniquely obtain a rotated coordinate system from an original one. There are several possibilities: The most natural would be to use rotation angles about the x-, y- and z-axes. This leads to the three Tait Bryan angles, which are widely used for aircraft. However, since rotations do not commute, unless supplemented with a prescription for the order of the three rotations, these angles are not unique, except for small rotations. The next most natural parameter would be to find the axis and angle of rotation, ie the vector φ above. This is called the Cayley Klein or Euler parameters. These are mathematically very nice, and can be related to the vector L given above, but are not very practical. The third, most widely used parametrisation in mechanics is in terms of the three Euler angles. Here, a general rotation is constructed from 3 successive elementary rotations:. The body is rotated an angle φ [, π about the z-axis. The x- and y-axes move, while the z-axis is unchanged.. The body is rotated an angle θ [, π] about its x-axis. The z- and y-axes move, but the x -axis is unchanged. 3. The body is rotated an angle ψ [, π about its z-axis. As in the first step, the x- and y-axes move, while the z -axis is unchanged. Note that there is no rotation about any y-axis here; instead there are two rotations about (different) z-axes. These three rotations are summarised in figure 5.. It can be proved that any rotation can be expressed in this way Rotation matrix for Euler angles We can now construct the general rotation matrix explitly as a function of the Euler angles. Calling the vectors in the intermediate coordinate systems ρ, ρ respectively, the 77

10 first rotation gives us ρ = A z (φ) r. The second rotation gives us ρ = A x (θ) ρ, and the final rotation r = A z (ψ) ρ. Putting all this together, we find A = A z (ψ)a x (θ)a z (φ) cos ψ sin ψ cos φ sin φ = sin ψ cos ψ cos θ sin θ sin φ cos φ sin θ cos θ cos ψ cos φ sin ψ cos θ sin φ cos ψ sin φ + sin ψ cos θ cos φ sin ψ sin θ = sin ψ cos φ cos ψ cos θ sin φ sin ψ sin φ + cos ψ cos θ cos φ cos ψ sin θ. sin θ sin φ sin θ cos φ cos θ (5.5) Example 5.3 Find the Euler angles corresponding to a rotation of an angle ϑ about the y-axis. The rotation matrix is cos ϑ sin ϑ A y (ϑ) ==. (5.6) sin ϑ cos ϑ Comparing the bottom right (zz) element of the matrices, we immediately see that θ = ϑ. Comparing the bottom left elements then, we see that sin φ =, so φ = π/. Finally, from the top right element we see that sin ψ =, so ψ = π/. You may then confirm that all the other elements come out as desired. You may also check for yourself that you may indeed achieve a rotation about the y axis by the following combination of rotations:. rotate it by 9 about the z-axis;. rotate it about the x-axis (by your desired angle); 3. rotate it back by 9 about the z-axis Euler angles and angular velocity The total angular velocity can be constructed as the sum of angular velocities that result from the changes in each of the three Euler angles. Note that simply adding these three vectors together is ok, since these correspond to infinitesimal changes in orientation, and for such changes the order does not matter. Hence we can write ω = φ + θ + ψ. (5.7) The magnitude of each of these three vectors is the respective angular velocity: φ, θ, ψ. But in which directions are they pointing? We will in the end want to express the angular velocity in the body coordinate system, since it will turn out that the equations of motion are best expressed in these coordinates. 78

11 φ This vector points along the original z-axis, which is unchanged by the first rotation. The second rotation is about the x-axis, and changes the vector (,, φ) (, φ sin θ, φ cos θ). After the final rotation of an angle ψ about the z-axis, we get φ = ( φ sin θ sin ψ, φ sin θ cos ψ, φ cos θ). (5.8) θ This vector poins along the intermediate x-axis, so after the second rotation we have θ = ( θ,, ). After the final rotation about the z-axis it becomes θ = ( θ cos ψ, θ sin ψ, ). (5.9) ψ This vector points along the body s z-axis, so ψ = (,, ψ). Adding up these, we get φ ω = φ sin θ sin ψ + θ cos ψ + θ + ψ = φ sin θ cos ψ θ sin ψ (5.3) φ cos θ + ψ We will use this to derive the equations of motion. But first we need to determine the kinetic energy. 5.4 The inertia tensor 5.4. Rotational kinetic energy Let us now work out the total kinetic energy of a rigid body. From (5.3) we get that T = m α ( V + ω r α ) = [ m α V + ] V ( ω r α ) + ( ω r α ) α α ( ) = M V + V ω m α r α + (5.3) m α ( ω r α ) α If one point P in the body is fixed, ie the motion is pure rotation then the one can chose the origon of the body coorinate system at P. Then V = so the first two terms vanish, and the third is the rotational energy T rot, where r α is the distance from particle number α to the point P. If we take the origin to be the centre of mass, then we can show that the second term vanishes, since then α m α r α =. The kinetic energy can then be written as the sum of the centre of mass energy and the rotational energy about the centre of mass: R = T CM + T rot = M V CM + m α ( ω r α ), (5.3) α where r α is the distance from particle number α in the body to the centre of mass. 79 α

12 We now use that the angular velocity ω is the same for the whole body. We furthermore use the identity ( A B) = A B sin θ AB = A B ( cos θ AB ) = A B ( A B). (5.33) With this, (5.3) becomes We end up with T rot = [ m α r α ω ( r α ω) ] (5.34) α = [ m α r α ωi (x α,i ω i )(x α,j ω j ) ] (5.35) α i ij = ( m α x α,kδ ij x α,i x α,j )ω i ω j (5.36) = ij ij α k I ij ω i ω j = ω I ω. (5.37) I ij = α T rot = I ij ω i ω j, (5.38) ij ) m α (rαδ ij x α,i x α,j ρ( x)( x δ ij x i x j )d 3 x (5.39) The quantity I = {I ij } is called the inertia tensor of the body. It characterises the rotational properties of the body. If ω is directed purely along one of the coordinate axes, for example ω = (,, ω), we see that T rot reduces to the expression we have encountered before, T rot = I ij ω i ω j = I zzω, (5.4) ij where I zz is the moment of inertia about the z-axis. But the expression (5.38) is much more general, and will hold in any (orthogonal) coordinate system, regardless of which direction the rotation vector is pointing. We can non calculate the inertia tensor once and for all in the body coordinate system (indeed, we will later use it to define this coordinate system). Once we know I in one coordinate system, we shall see that we can relatively easily find it in any other coordinate system. 8

13 Example 5.4 Find the inertia tensor of the dumbbell pictured in fig. 5., and find the kinetic energy if it rotates with angular velocity ω. about the y-axis,. about the z-axis, 3. about its own axis. Answer: Particle is located at r = (r sin θ, r cos θ, ). Particle is located at r = ( r sin θ, r cos θ, ). We divide the computation of I as I = I diag I prod where I diag i,j = δ i,j α r α and I prod i,j = α m αx α,i x α,j. The two terms are computed as y 3m... θ.. r m x Figure 5.: A dumbbell I diag = (m r + m r) m r + m r = m r + m r m r + m r and I prod = r sin θ r sin θ cos θ r m r sin θ cos θ r cos sin θ r sin θ cos θ θ + m r sin θ cos θ r cos θ sin θ sin θ cos θ = (m r + m r) sin θ cos θ cos θ Adding them together we get the full intertial tensor I as cos θ {}}{ sin θ sin θ cos θ I = (m r + m r) sin θ {}}{ sin θ cos θ cos θ. For rotation about the y-axis, ω = (, ω, ), so T rot = I yyω = (m r + m r )ω sin θ. For rotation about the z-axis, ω = (,, ω), so T rot = I zzω = (m r + m r )ω. 8

14 3. For rotation about the body axis, ω = (ω sin θ, ω cos θ, ), so T rot = I xxω x + I xyω x ω y + I yxω y ω x + I yyω y = (m r + m r )ω [cos θ sin θ (sin θ cos θ)(sin θ cos θ) sin θ cos θ] =. It should not be a surprise that we get T rot = here, since nothing is actually moving in this case! Example 5.5 Find the inertia tensor for a homogeneous cube with mass M and length L with the origin at one corner and edges along coordinate axes. Answer: The density of the cube is ρ = M/L 3. We find I xx = I xy = L L L = M L L L L L = M L M L 3 (y + z )dxdydz = M L ( L z L )dz = M ( L 4 L L M L 3 ( xy)dxdydz L ydy = M L L 3 + L3 3 L ) ( ) L = 4 ML. (y + z )dydz = 3 ML, (5.4) (5.4) Since the x, y and z axes are completely symmetric, it is clear that I xx = I yy = I zz and I xy = I xz = I yz, and the inertia tensor is 3 ML 4 ML 4 ML I = 4 ML 3 ML 4 ML = ML (5.43) 4 ML 4 ML ML What is a tensor? Scalars, vectors and tensors. In practice, you can think of a tensor as a kind of matrix, where the rows and columns correspond to directions in space. You may then treat the expression for T rot as a vector matrix vector multiplication. 8

15 In principle, a tensor is a generalisation of a vector, ie a physical (or mathematical) quantity with more than one direction. Tensors (and vectors) are defined by their transformation properties, and in particular how they change when they, or the coordinate system, are rotated. We shall see that in a rotated coordinate system, the inertia tensor is given by I ij = A ik A jl I kl. (5.44) kl This is the defining property of a tensor (to be precise, a rank- tensor, ie a tensor with indices). This mirrors the definition of scalars and vectors, which are defined as follows: A scalar is a quantity that does not change when you rotate your coordinate system. Examples of this is the length of a vector, or the kinetic energy. A vector is a quantity v that transforms in the same way as the position vector, ie v i = j A ij v j. (5.45) There are however some vector and scalar type quantities that transform differently under reflections: An ordinary vector will change its sign when seen in a mirror, but for example the angular momentum vector, L = r p, will keep the same sign (since both r and p change sign). Such vectors are called pseudovectors or axial vectors. Finally, ordinary scalars will be unchanged under reflections, but some quantities change sign. These are called pseudoscalars. For example, v L (the scalar product of the velocity and angular momentum) would be a pseudoscalar. Other examples of tensors y x The stress tensor. The picture on the left depicts a fluid flowing with a velocity u in the x-direction. The fluid flows faster at larger y, so that y u x. In a viscous fluid this will create stress (shear) forces in the y-direction. This can be expressed through the stress tensor σ xy. The diagonal components of this tensor represent the pressure of the fluid, eg σ xx is the pressure in the x-direction. The same picture also governs stress forces in solid materials, if spatially varying forces are applied. The (outer) product of two vectors. If we define T = a b T ij = a i b j, (5.46) then we can easily see that this satisfies the rotation transformation property T ij = kl A ika jl T kl. The vector product c = a b can be obtained as the antisymmetric part of this tensor, c x = T yz T zy, c y = T xz T zx, c z = T xy T yx. (5.47) 83

16 The electromagnetic field. In 4-dimensional spacetime, the electric field E and magnetic field B form the (antisymmetric) field tensor F µν = F νµ, E x E y E y {F µν } = E x B z B y E x B z B x. (5.48) E z B y B x (Note that in 4-dimensional space-time, rotations include Lorentz boosts.) Rotations and the inertia tensor How does the inertia tensor change if we rotate our coordinate system? To answer this question, we can use the fact that the kinetic energy must be the same regardless of how we choose our coordinates (it is a scalar): T rot = ω i I ij ω j = ω ii ijω j. (5.49) ij But ω is a vector, and changes in the same way as the position vector when rotated, ij ω i = k A ik ω k. (5.5) Substituting this into (5.49), we get T rot = A ik ω k I ij A jl ω l = ( ) A ik I ija jl ωl = ω k I kl ω l. (5.5) ij k l So we find that I kl = ij A iki ija jl or I = A T I A. kl ij kl Angular momentum and the inertia tensor We want to find the angular momentum L of a rigid body about some point O. This point can be the centre of mass (for a body tumbling freely in space, for example), or some point fixed in space (for example a spinning top). The momentum of some particle α is p α = m α v α = m α ω r α. Using the vector identity A ( B A) = A B ( A B) A (5.5) Strictly speaking, this is not quite the case: under a reflection, r r, but ω is unchanged. We call a vector which behaves this way a pseudovector or axial-vector. The angular momentum is another example of such a pseudovector. 84

17 we can write the total angular momentum as So we find L = α L i = α = j r α p α = r α ( ω r α ) = ( ) m α rα ω ( r α ω) r α (5.53) α α m α (rαω i x α,j ω j x α,i ) = m α (rαδ ij x α,j x α,i )ω j (5.54) j α j [ ] m α (rαδ ij x α,j x α,i ) ω j = I ij ω j. (5.55) α j L i = j I ij ω j or L = I ω (5.56) 5.5 Principal axes of inertia Life would be a lot simpler if, in some coordinate system, the inertia were diagonal, I I = I (I ij = I i δ ij ). (5.57) I 3 We would then have L i = I i ω i ; T rot = I i ωi. (5.58) In particular, if the body rotates about one of the axes of such a coordinate system, we have L = I ω, T rot = Iω. The good news is that it is always possible to find such a set of (body) coordinate axes. These axes are called principal axes of inertia, and the corresponding I, I, I 3 are the principal moments of inertia. The question then is how we find these axes. There are two methods: Find a rotation matrix A such at AIA T = I is diagonal. The coordinate axes in the rotated system are then the principal axes. Find vectors (directions) ω such that L = I ω = I ω. These vectors form the principal axes of inertia, and the numbers I are the principal moments. In fact, both these methods are mathematically identical. Let us first look at method. If ω points along a principal axis of inertia, we have i L i = j I ij ω j = Iω i, (5.59) 85

18 where I is the corresponding moment of inertia, or L = Iω = I ω + I ω + I 3 ω 3 L = Iω = I ω + I ω + I 3 ω 3 (5.6) L 3 = Iω 3 = I 3 ω + I 3 ω + I 33 ω 3 This is an eigenvalue problem. The condition for a nontrivial solution is det(i I ) =. This gives us a (third-degree) equation for I, called the characteristic equation. The solutions are the principal moments of inertia (or eigenvalues of I). Once we have found one of the solutions (eigenvalues), we can find the corresponding principal axis by substituting the values of I into the equation for L = I ω = I ω. This gives us the direction of ω, or equivalently, the ratios ω : ω : ω 3. The vectors ω are the eigenvectors of I. Example 5.6 Show that the homogeneous cube with the origin at one corner has a principal moment of inertia I = 6 ML, and find the corresponding principal axis of inertia. Answer: We found that the inertia tensor was I = ML (5.6) I is a principal moment of inertia if det(i I ) =. Setting I = 6 ML, we find 3 ML I 4 ML 4 ML 4 ML 3 ML I 4 ML 4 ML 4 ML 3 ML I = ML 4 ML 4 ML 4 ML ML 4 ML 4 ML 4 ML ML = ( 4 ML) 3 = ( 4 ML) 3[ ] ( ) + ( ) ( ( ) The corresponding axis is given by Iω = I ω + I ω + I 3 ω 3 = Iω = I ω + I ω + I 3 ω 3 = Iω 3 = I 3 ω + I 3 ω + I 33 ω 3 = Only two of these equations are independent, =. (5.6) 6 ω = 3 ω 4 ω 4 ω 3 (5.63) 6 ω = 4 ω + 3 ω 4 ω 3 (5.64) 6 ω 3 = 4 ω 4 ω + 3 ω 3 (5.65) since (5.65) is equivalent to 86

19 (5.63)+(5.64). We thus have (5.63) (5.64) : ω 4 ω 4 ω 3 = (5.63) 4 ω ω 4 ω 3 = (5.64) ω ω = ω = ω (5.66) (5.63) = ω 4 ω 4 ω 3 = 4 ω 4 ω 3 = ω 3 = ω (5.67) We thus have ω = ω = ω 3, or the principal axis is along the diagonal (,,). We might have guessed this from the symmetry of the cube, and rotated our system so that the new x-axis is along the diagonal. This can be obtained by a 45 rotation about the x-axis, followed by a rotation of cos /3 about the y-axis. The matrix for this rotation is 3 3 A = = 3 3. (5.68) The rotated inertia tensor is I = AIA T = ML = ML. (5.69) This also gives us the other two principal moments of inertia, which we find to be equal Comments. Finding the principal moments and axes of inertia (diagonalising the inertia tensor) by hand can be a very tedious process. Sometimes it can be simplified by symmetry considerations (see below), but in most cases it is better left to computers.. For the cube with the origin at one corner, we found that two of the principal moments of inertia were equal, I = I 3 = ML. This means that the two corresponding principal axes can be any orthogonal pair of axes perpendicular to the diagonal ie, the moment of inertia is the same about any axis orthogonal to the diagonal. 3. For a body where all three principal moments of inertia are equal, all directions or axes are equivalent. We can see this by noting that in this case the inertia tensor in 87

20 the coordinate system defined by the principal axes is proportional to the identity (matrix), I = I, which commutes with all rotation matrices, so I = IAA T = I for all A O(3). 4. Any body which is rotationally symmetric about some axis has one principal axis of inertia along that axis. The two other axes can be arbitrarily chosen in the plane perpendicular to the first principal axis, ie I = I 3. Definitions A body with I = I = I 3 is called a spherical top. A body with I = I I 3 is called a symmetric top. A body where all principal moments are different is called an asymmetric top. A body with I =, I = I 3 is called a rotor. For example, the cube with the origin at the centre is a spherical top; the cube with the origin at one corner is a symmetric top, and the dumbbell (or a diatomic molecule) is a rotor. 5.6 Equations of motion Having found the principal axes of inertia, we can now use them to define the body coordinate system. This will greatly simplify the equations of motion. The kinetic energy is given by T = I i ωi. (5.7) Using (5.3) we can write this as i T = I ( φ sin θ sin ψ+ θ cos ψ) + I ( φ sin θ cos ψ θ sin ψ) + I 3( φ cos θ+ ψ). (5.7) The equations of motion for generic values of the moments of inertia and a generic potential energy will become very complicated. We will instead focus on two specific cases: a symmetric top (I = I ) with one point fixed and under the influence of constant gravity, and force-free motion of an asymmetric top (all I, I, I 3 are different) The symmetric heavy top For a symmetric top, we can take I = I, and (5.7) becomes T = I ( θ + φ sin θ) + I 3( φ cos θ + ψ). (5.7) We now take the top to be rotating about a fixed point at the bottom of the symmetry axis, under the influence of a constant gravitational field g. We take θ to be the angle 88

21 the symmetry axis forms with the vertical. In that case, the potential energy of the top is V = Mgh cos θ, where M is the mass of the top, and h is the distance from the base of the top to the centre of mass along the symmetry axis. The lagrangian is therefore L = I ( θ + φ sin θ) + I 3( φ cos θ + ψ) Mgh cos θ. (5.73) The meaning of the three Euler angles in this case is: θ denotes the angle the axis of the top makes with the vertical. φ denotes the orientation of the (tilted) axis relative to the fixed reference coordinate system. ψ denotes the orientation of the top relative to its own axis. We will find that in general, θ will oscillate between a minimum and a maximum, while both φ and ψ will be monotonically increasing (or decreasing). The motion in ψ reflects the top spinning around its own axis. The motion in φ corresponds to the orientation of the axis precessing around the vertical axis. Finally, the motion in θ corresponds to periodic wobbles in the tilt of the top, called nutation. We may now derive the Euler Lagrange equations for the top and use them to study this motion in detail. We will not do this here, but instead perform a qualitative analysis of the possible motion. First we determine the canonical momenta, p φ = L φ = I φ sin ω 3 θ + I 3 ω 3 φ = I φ sin θ + I 3 cos θ( φ cos θ + ψ), (5.74) p θ = L θ = I θ (5.75) p ψ = L ψ = I 3ω 3 ω 3 ψ = I 3( φ cos θ + ψ). (5.76) Since L does not depend explicitly on either φ or ψ, L/ φ = L/ ψ =, and we see straightaway that the canonical momenta p φ, p ψ are conserved. In particular, p ψ = I 3 ω 3, so the top spins with a constant angular velocity ω 3 about its own axis. However, the precession rate, governed by the constant p φ is more complicated. We now proceed to derive the hamiltonian of the system. ω 3 = φ cos θ + ψ = p ψ /I 3. We can use this to rewrite (5.74), From (5.76) we see that p φ = I φ sin θ + I 3 cos θω 3 = I φ sin θ + p ψ cos θ (5.77) p φ p ψ cos θ = I φ sin θ φ = p φ p ψ cos θ I sin. (5.78) θ Since the kinetic energy is quadratic in the generalised velocities, and the potential energy does not depend on velocities, the hamiltonian is equal to the total energy, H = T + V = [ (pφ I p ψ cos θ ) sin I sin θ + θ = p θ + p ψ + (p φ p ψ cos θ) I I 3 I sin + Mgh cos θ. θ ( pθ ) ] + ( I I pψ ) 3 + Mgh cos θ I 3 89 (5.79)

22 Since p φ and p ψ are both constant, there is effectively only one degree of freedom θ, and the second and third term in H can be combined with the final term V = Mgh cos θ to form an effective potential V eff (θ). The first term is the usual kinetic energy term. The first term in V eff is just a constant, so it has no effect on the dynamics of the system, other than increasing the minimum energy. Unless p φ and p ψ are both precisely zero or p φ = ±p ψ, the second term is nonzero and diverges to + as θ and θ π. Therefore, the motion in θ is bounded for all but very special values of p φ, p ψ, with θ max in general being smaller the larger the value of p ψ. If θ max < π/ this ensures the top does not fall over Euler s equations for rigid bodies Consider now force-free motion of a rigid body. In this case, we have 3 L = T = i I i ω i. (5.8) The Euler Lagrange equations are given by d dt L L = φ φ ; d dt L L = θ θ ; d L L = dt ψ ψ. (5.8) We will however only consider the third of those, and then derive two additional equations of motion from symmetry considerations. We can write the Euler Lagrange equation for ψ as We first note that d L dt ψ = d ( I3 ( dt φ cos θ + ψ) ) = d dt (I 3ω 3 ) = L ψ = 3 i= I i ω i ω i ψ. (5.8) ω ψ = φ sin θ cos ψ θ sin ψ = ω, (5.83) ω ψ = φ sin θ sin ψ θ cos ψ = ω. (5.84) Since ω 3 does not depend on ψ the last expression in (5.8) becomes L ψ = I ω ω ψ + I ω ω ψ = I ω ω + I 3 ω ( ω ) = (I I )ω ω, (5.85) so (5.8) becomes I 3 dω 3 dt = (I I )ω ω. (5.86) However, the labels, and 3 for the three axes is arbitrary, and we may just as well rename them, as long as we ensure the coordinate system remains right-handed (corresponding to cyclic permutations. This gives us the three equations If p φ = p ψ the second term in V eff diverges as θ π but goes to as θ, and conversely if p φ = p ψ. This, however, requires finely balanced initial conditions. 3 Note that the translational energy can be ignored since R is cyclic. 9

23 These are Euler s equations for force-free motion. I dω dt = (I I 3 )ω ω 3, (5.87) I dω dt = (I 3 I )ω 3 ω, (5.88) I 3 dω 3 dt = (I I )ω ω. (5.89) Stability of rigid-body rotations Let us now look at what happens when we set a body rotating about one of the three principal axes. For example, if it rotates purely about the first axis, we have ω, ω = ω 3 =. In practice, it is not possible to have exactly zero rotation about the other two axes, so what we have is that ω and ω 3 are both much smaller (in magnitude) than ω. We now want to find out how ω, ω and ω 3 each evolve with time. If ω and ω 3 remain small and are either damped to zero or fluctuate around zero, the rotation about the first axis is said to be stable. On the other hand, if the magnitude of ω and/or ω 3 grows with time, they will eventually become as large as ω and the body is no longer rotating about its original axis. In that case, the rotation is unstable. Without any loss of generality, we can take I > I > I 3 since we are allowed to label our axes as we wish. We then have three different cases to deal with:. If ω ω 3 ω the three equations become I ω = (I I )ω ω 3 = ω = constant, (5.9) ( ) dω I dt = (I I3 I 3 I )ω 3 ω = ω = ω ω 3, (5.9) I ( ) dω 3 I 3 dt = (I I I I )ω ω = ω 3 = ω ω, (5.9) I 3 Differentiating (5.9) and using (5.9) we get ω = I 3 I I ω ω 3 = (I 3 I )(I I ) I I 3 ω ω (5.93) ω + Ω ω = with Ω = (I 3 I )(I I ) I I 3 ω >. (5.94) This has the solution ω (t) = A cos Ω t + B sin Ω t (5.95) ω 3 (t) ω (t) = A cos Ω t + B sin Ω t. (5.96) So we see that ω, ω 3 will oscillate about equilibrium values ω = ω 3 =. 9

24 . If ω ω ω 3, by the same procedure as in the first case, we obtain ω 3 = const, ω ω 3, ω + Ω 3 ω =, Ω 3 = (I 3 I )(I 3 I ) I I ω 3 >, which again has solutions ω (t) = A cos Ω 3 t + B sin Ω 3 t, ω (t) = A cos Ω 3 t + B sin Ω 3 t. (5.97) 3. If ω ω 3 ω, then, using the same procedure again, we now find the equations The general solution to (5.) is ω = constant, (5.98) I ω 3 = ω, (I I 3 )ω (5.99) ω = (I 3 I )(I I ) I I 3 ω ω = Ω ω. (5.) ω (t) = Ae Ω t + Be Ω t, (5.) so ω (and ω 3 ) will increase exponentially with time, at least until the approximations ω ω and ω 3 ω are no longer valid. The upshot of this is that rotations about the long and short axes are stable, while those about the intermediate axis are unstable. You can verify this for yourself by tossing a rectangular block (for example a a book held together by an elastic band) up in the air, rotating it about each of its three axes. 9