Chapter 9 Rigid Body Motion in 3D

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1 Chapter 9 Rigid Body Motion in 3D Rigid body rotation in 3D is a complicated problem requiring the introduction of tensors. Upon completion of this chapter we will be able to describe such things as the motion of a top, or of a bicycle. 1 Rotation of body about an arbitrary axis Moments and Products of Inertia, Inertia Tensor, Angular Momentum and Kinetic Energy Consider an arbitrary rigid object and a set of xyz axes. Our first task will be to find the moment of inertia about an axis passing through the origin but oriented in an arbitrary direction, not one of the coordinate axes. The line is bidirectional, but once we set the object into rotation the vector ω will define a unique direction. In Chapter 1 we discussed direction cosines. Angles α, β, γ are the angles between a vector and the xyz axes (pages 16, 17 of the text) so that A x = A cos α etc. Refer to Figure that shows the object, the z-axis of the coordinate system, a rotation axis with ω, and a small piece of the object labeled m i a distance r from the origin, with moment arm r i from the axis. It has a velocity v i that is perpendicular to both r i and ω. The scalar moment of inertia of the object about the rotation axis is by definition I = m i r 2 i (1) where we have used a summation, but we could easily move to continuous mass distributions and do an integral. Now we want to express this in terms of the cartesian coordinates x i, y i, z i and the direction cosines. 1

2 Figure 1: Rigid body rotating about a general axis, not one of the coordinate axes. Unit vector for axis set by angular velocity is ˆn One infinitesimal piece of object is shown. Define the unit vector ˆn as ω = ωˆn. We can recognize that r i = r i sin θ i = r i ˆn (2) where θ i is the angle between the axis of rotation ˆn and the radius vector r i. The position of the infinitesimal mass is r i = x i î+y i ĵ +z iˆk. The unit vector can be written in terms of the cartesian unit vectors and the direction cosines and using this and some algebra r 2 i = r i ˆn 2 ˆn = î cos α + ĵ cos β + ˆk cos γ (3) = (y i cos γ z i cos β) 2 + (z i cos α x i cos γ) 2 + (x i cos β y i cos α) 2 = (y 2 i + z 2 i ) cos 2 α + (z 2 i + x 2 i ) cos 2 β + (x 2 i + y 2 i ) cos 2 γ 2y i z i cos β cos γ 2z i x i cos γ cos α 2x i y i cos α cos β (4) Hence the moment of inertia relative to the axis of rotation is I = m i (y 2 i + z 2 i ) cos 2 α + m i (z 2 i + x 2 i ) cos 2 β + m i (x 2 i + y 2 i ) cos 2 γ 2 m i y i z i cos β cos γ 2 z i x i cos γ cos α 2 x i y i cos α cos β (5) Wow. A seemingly small change results in a really messy expression. We can recognize the terms in the first line as containing moments of inertia about the coordinate axes, and modify our notation to have double subscripts: I xx = m i (y 2 i + z 2 i ) moment of inertia about x axis (6) I yy = m i (z 2 i + x 2 i ) moment of inertia about y axis (7) I zz = m i (x 2 i + y 2 i ) moment of inertia about z axis (8) 2

3 The sums involving products will be called the products of inertia. Our text includes the negative sign in the definition other texts define the products as positive numbers. Hence we can write I xy = I yx = m i x i y i (9) I yz = I zy = m i y i z i (10) I xz = I zx = m i x i z i (11) I = I xx cos 2 α + I yy cos 2 β + I zz cos 2 γ +2I xy cos α cos β + 2I yz cos β cos γ + 2I xz cos α cos γ (12) It is much more compact to cast this into tensor/vector notation. Represent the unit vector ˆn as a column vector, cos α ˆn = cos β cos γ (13) The transpose of this vector is a row vector For the moment of inertia tensor use ˆn = (cos α cos β cos γ) (14) I = I xx I xy I xz I yx I yy I yz I zx I zy I zz (15) where we know relations like I xy = I yx. The scalar moment of inertia can be found by simple matrix multiplication I = ˆnIˆn (16) E.g. Moment of inertia tensor of a rectangular lamina Consider a rectangular lamina lying in the x-y plane with length a in the x and 2a in the y. Find the moment of inertia tensor relative to these coordinates. We can write the surface density ρ = m/2a 2. The moments of inertia about the coordinate axes can be written from previous work, I xx = (1/3)m(2a) 2 = (4/3)ma 2, I yy = (1/3)ma 2 and from the perpendicular axis theorem, I zz = I xx + I yy = (5/3)ma 2. 3

4 Figure 2: A uniform rectangular lamina of dimensions a by 2a. Since z = 0 for the lamina, two products of inertia I xz = I yz = 0. So now we just need to find I xy. So the tensor is I xy = I = a 2a 0 = m 2a 2 0 a 0 ρxy dx dy (17) x dx 2a 0 y dy (18) = 1 2 ma2 (19) 4 3 ma2 1 2 ma ma2 1 3 ma ma2 (20) Find the scalar moment of inertia for the rectangular lamina with an axis at 37 to the x-axis The direction cosines are cos α = cos 37 = 0.80, cos β = cos 53 = 0.60, cos γ = 0 so need to evaluate I = ( ) ma = ma 2 (21) Angular Momentum Vector and Dyad Products We now introduce a new vector product. Recall that the dot product of two vectors gave a scalar and the cross product gave a vector. The dyad product 1 of two vectors will give a tensor via A x B x A x B y A x B z A B A y B x A y B y A y B z (22) A z B x A z B y A z B z 1 See Wikipedia, for more discussion. 4

5 A unit tensor (be sure you can distinguish one 1 and I) will be = = îî + ĵĵ + ˆkˆk (23) The product of a unit tensor with a vector returns the vector: 1 ω = ω where putting the tensor next to the vector impies the usual matrix-vector multiplication. Note that the text often (but not always I think) puts a between the tensor and vector. I will usually reserve the dot for dot products of vectors. Some dyad properties these are shown in the text. Dyad product with vector (yields a vector): ( a b) c = a ( b c) Commutation rule: a b = ( b a) Dyad between vectors (yields a scalar): d( a b) c = ( d a)( b c) = ( d a)( b c) Now we want to write the relation between angular momentum vector and the angular velocity vector. In Chapter 8, with a fixed axis and generally symmetrical objects, this was just L = I ω. The general case is considerably more complicated. We will do it with summations. We know that v i = ω r i. The angular momentum is L = r i m i v i = [m i r i ( ω r i )] (24) and expanding the triple cross product as discussed in Chapter 1 (BAC - CAB) L = m i ri 2 ω m i r i ( r i ω) (25) [( = mi ri 1) 2 ] (m i r i r i ) ω (26) This takes a bit to process, but with a little work you should see that it works. Now we write the relation between angular momentum and angular velocity in tensor form: L = x 2 i + y2 i + z2 i 0 0 m i 0 x 2 i + y2 i + z2 i 0 ω x 2 i x i y i x i z i m i x i y i y 0 0 x 2 i + y2 i + i 2 y i z i ω z2 i x i z i y i z i zi 2 = yi 2 + z2 i x i y i x i z i m i y i x i x 2 i + z2 i y i z i ω = I ω (27) x i z i y i z i x 2 i + y2 i This should tell you that The direction of the angular velocity and the direction of the angular momentum are not, in general, the same! 5

6 E.g. Find angle between angular momentum and angular velocity for the rectangular lamina rotating about an axis at 37 to the x-axis. First compute L: L = ma ω = 0.20 ma 2 ω (28) The magnitude of this ( L L) is 0.793ma 2 ω and since L ω = Lω cos φ we can find φ = 51. Be sure to look at the examples in the text, and 9.1.2, for a square lamina and rotation about (a) the x-axis and (b) the diagonal. In the case of the diagonal, the angle is just 0, and we will shortly describe this as one of the principal axes of the object. 1.2 Rotational Kinetic Energy Using Tensors First consider an object in pure rotation, and use the relation v i = ω r i T rot = 1 2 m i v i v i = 1 ( ω mi r i ) v i (29) 2 But ( A B) C = A ( B C) from chapter 1. Hence T rot = 1 2 ω (mi r i v i ) (30) = 1 2 ω (m i r i v i ) (31) = 1 2 ω L (32) = 1 2 ωiω (33) E.g. Kinetic Energy of Rectangular Plate about 37 Evaluating this gives T rot = 1 2 (0.493)ma2 ω 2 (34) and this is just what we get using the scalar moment of inertia. 2 Principal Axes of a Rigid Body The origin of our coordinates may be fixed in a problem, but the orientation of the axes is usually our choice. We can always find some orientation in which the products of inertia 6

7 are all zero and thus the moment of inertia tensor is diagonal. In this case the diagonal elements are called the principal moments, and the axes are called the principal axes. We will use the following notation when we have the axes oriented to be principal axes: and I xx I 1 ω x ω 1 î ê 1 I yy I 2 ω y ω 2 ĵ ê 2 (35) I zz I 3 ω z ω 3 ˆk ê3 I = I I I 3 (36) In some cases one or more of the principal axes may be found by inspection. In more general cases we must diagonalize the inertia tensor, thereby determining the principal moments and the principal axes. The rigid object can still be rotated about any axis, not just the principle axes. Once we have found the principal axes and moments, it is easy to determine the scalar moment of inertia about an axis other than the principal axes, and to find the angular momentum and rotational kinetic energy about this new axis. Suppose we rotate abut an axis with normal ˆn = (cos α cos β cos γ). Then the scalar moment of inertia is I cos α I = ˆnIˆn = (cos α cos β cos γ) 0 I 2 0 cos β 0 0 I 3 cos γ = I 1 cos 2 α + I 2 cos 2 β + I 3 cos 2 γ (37) The angular velocity is in the same direction as ˆn and so the angular momentum is I ω 1 L = I ω = 0 I 2 0 ω I 3 ω 3 = I 1 ω 1 I 2 ω 2 I 3 ω 3 Finally the rotational kinetic energy is T rot = 1 2 ωi ω = 1 2 (ω 1 ω 2 ω 3 ) = 1 2 = ê 1 I 1 ω 1 + ê 2 I 2 ω 2 + ê 3 I 3 ω 3 (38) ( I1 ω I 2 ω I 3 ω 2 3) I I I 3 ω 1 ω 2 ω 3 (39) 7

8 2.1 Symmetry Often we can see the principal axes by inspection. Here we assume the origin is at the center of mass. Rectangular Block about the cm We expect the principal axes to be parallel to the sides of the block. If the sides are a, b and c, from Chapter 8 we have the principal moments I 1 = m 12 (b2 + c 2 ) I 2 = m 12 (a2 + c 2 ) I 3 = m 12 (a2 + b 2 ) (40) Notice that for a cube each moment is ma 2 /6. Example shows that the scalar moment of inertia of the cube about any other axis through the center of mass of the cube is also ma 2 /6. Laminar Objects If the laminar object lies in the x y plane, then the z axis must be a principal axis since I xz = I yz = 0. If there is also an axis of symmetry, such as for a ping-pong paddle or tennis racquet, then that is another principal axis. The third principal axis is orthogonal to these two. Figure 3: A model ping-pong paddle. Handle has mass m/2 and length 2a, head has mass m/2 and radius a. The paddle is laminar. Consider a model paddle consisting of circular lamina of mass m/2 and radius a attached to a thin rod of mass m/2 and length 2a. The center of mass is at the point where the rod meets the circle (see Figure 9.2.1). Call axis 1 the axis of symmetry and axis 3 normal to the lamina. About axis 1 we can write the total inertia as I 1 = I rod + I circle = m 4 2 a2 = 1 8 ma2 (41) About axis 2, I 2 = I rod + I circle = 1 m 3 2 (2a)2 + 8 ( 1 m 4 2 a2 + m ) 2 a2 = ma2 (42)

9 and from the perpendicular axis theorem, I 3 = ma2 (43) 2.2 Dynamic Balancing Suppose an object is rotating about one of its principal axes, say axis 1. Then ω 2 = ω 3 = 0 and the angular momentum is just L = ê 1 I 1 ω 1 = I 1 ω (44) This tells us that for rotation about a principal axis, the angular momentum and angular velocity are parallel, while for rotation about an axis that is not a principal axis, angular momentum and angular velocity are NOT parallel. This has direct consequence in balancing rotating systems such as automobile tires. Static balancing occurs when the center of mass lies on the axis of rotation. Thus the gravitational force will not cause the object to rotate. However if we set the object into rotation it may wobble terribly unless it is dynamically balanced so that the axis of rotation is along a principal axis. When the wheel is NOT dynamically balanced we can view the situation as having ω along the axle, but L in some other direction, so that as the wheel rotates the angular momentum changes direction, tracing out a cone. Whenever the angular momentum is changing we know that there must be a net torque and it will be at right angles to the axis of rotation. 2.3 Finding Principal Axes and Principal Moments I: One Axis Known For a general inertia tensor we can find the principal moments and axes by diagonalizing the tensor. We start with a simpler case where one of the principal axes is known, call it I 3. The tensor looks like I xx I xy 0 I xy I yy 0 (45) 0 0 I 3 What we want to find is the orientation of the other principal axes, 1 and 2, relative to the starting x y axes. The angle sought is just tan θ = ω y ω x (46) 9

10 For rotation about principal axis 1, the angular momentum and angular velocity will be parallel and related by the principal moment, so ω x I xx I xy 0 ω x L = I 1 ω = I 1 ω y = I ω = I xy I yy 0 ω y (47) ω z 0 0 I 3 ω z Doing the matrix multiplication and equating terms we get Eliminate ω y by introducing θ to get and eliminating I 1 we get I xx ω x + I xy ω y = I 1 ω x (48) I xy ω x + I yy ω y = I 1 ω y (49) I xx + I xy tan θ = I 1 (50) I xy + I yy tan θ = I 1 tan θ (51) (I xx I yy ) tan θ = I xy (1 tan 2 θ) (52) Using the trig identity tan 2θ = 2 tan θ/(1 tan 2 θ) results in tan 2θ = 2I xy I xx I yy (53) Find the principal axes for our rectangle relative to coordinate axes at the corner. We have I xx = 4ma 2 /3, I yy = ma 2 /3, I xy = ma 2 /2. Putting these into our expression we get tan 2θ = 1 (54) 2θ = 45, 135 (55) θ = 22.5 (or ), 67.5 (56) So the 1-axis is at -22 and the 2-axis is at 67.5 relative to the original x-axis. Notice that the 2-axis is not the body diagonal, at Balancing a tire Suppose that we have a solid disk of radius a and mass m. It has a symmetry axis that we will call the x-axis. We know that this is a principal axis for the disk with I xx = ma 2 /2, I yy = ma 2 /4, I xy = 0. The axle for the disk is mounted at a slight angle θ away from the symmetry axis. What we will do is show that by adding two small masses m at distances b from the disk we can move the principal axis to align with the axle. Refer to Figure for the visual. 10

11 Figure 4: A wheel is made from a solid disk of radius a and mass m and lies in the yz plane. It s axle is crooked, tilted an angle θ to the x-axis. By adding symmetrical placed masses m a distance b from the wheel, the principle axis can be made to coincide with the axle, leading to a wheel that is balanced both statically and dynamically. For the combined system (disk plus masses) we can write I xx = 1 2 ma2 + 2m a 2 (57) I yy = 1 4 ma2 + 2m b 2 (58) I xy = [( b)m a + (b)m ( a)] = 2abm (59) tan 2θ = For small angles and m m, 2I xy I xx I yy = 4abm ma 2 /4 + 2m (a 2 b 2 ) θ 8 bm (61) am For values of m = 10 kg, a = 18 cm, b = 5 cm and θ = 1, we find the required balancing masses to be m = 76 grams. (60) 2.4 Principal Axes by Diagonalizing a Matrix Suppose we have an inertial tensor that has all non-zero elements relative to an xyz set of coordinates. We want to find the principal axes (and their orientations relative to xyz) and the principal moments of inertia. This is an example of an eigenfunction problem that will arise extensively in quantum mechanics. 11

12 When we have rotation about a principal axis ê i, angular momentum and angular velocity are parallel, I ω and ω are parallel. We can write the requirement for component i as Iê i = λ i ê i (62) (I λ i 1)ê i = 0 (63) where the λ i are the principal moments. For this to be true, the determinant of the matrix in parentheses must be zero I xx λ I xy I xz I λ i 1 = I xy I yy λ I yz I xz I yz I zz λ = 0 (64) This is a cubic equation in λ whose solutions are the principal moments. The next task is to find the unit vectors that describe the principal axes relative to the original xyz coordinates. If ê 1 is the unit vector of the axis that has a principal moment λ 1, we can write I xx λ 1 I xy I xz I xy I yy λ 1 I yz I xz I yz I zz λ 1 (I λ 1 1)ê 1 = 0 (65) cos α cos β = 0 (66) cos γ and the three component equations can be solved for the direction cosines. Like most matrix efforts, if the matrix is given in numbers it is easier to let a computer do the work. An online resource is matcalc.html or search on (ubc java matrix ). If we use the rectangle and diagonalize we get principal moments I 1 = ma 2, I 2 = ma 2, I 3 = ma 2 with corresponding unit vectors ê 1 = ( ) (67) ê 2 = ( ) (68) ê 3 = (0 0 1) (69) From the unit vectors we can get the angles of the principal axes relative to the original axes. Remember that inverse cosines are double valued, ±θ, so some adjusting and choosing is necessary. I get α = 112.5, β = γ = 90 for the first vector, α = 22.5, β = γ = 90 for the second, and α = 90, β = 90 γ = 0 for the third. The origin of the coordinate system is determined by us. If we change the coordinate system we will change the inertia tensor, the principal moments and the principal axes. 12

13 For example, if we consider the center of the rectangle as the origin, 1/3 0 0 I = 0 1/12 0 ma 2 (70) 0 0 5/12 and the xyz axes are the principal axes. 3 Euler s Equations For Rigid Body Motion Newton s Second Law for rotation is, in an inertial frame, N = d L dt (71) For a rigid body L can be easily expressed if we use principal axes. We then imagine rotation about an axis that is NOT a principal axis. When the object rotates about this axis by some amount, and if the axes are inertial (fixed), then the inertia tensor will change. To make the mathematics tractable, we use two sets of coordinates. One is an inertial or fixed set, the other rotates with the body and we choose the principal axes for the body set. The body coordinates are a non-inertial reference frame like those discussed in Chapter 5. In that chapter we determined the following relation for any vector. The subscript fixed means evaluate the derivative for variables referred to a fixed inertial reference system. The subscript rotating means evaluate the derivative relative to variables measured in the rotating reference system. ( d L dt ) fixed = ( dl ) + ω L dt = Lrot + ω L (72) rotating By choosing the body coordinates to be the principal axes we can use L = I ω, with a diagonal inertia tensor, hence Lrot = I ω and ω L = ω I ω. In tensor form this is N 1 I 1 ω 1 ω 2 ω 3 (I 3 I 2 ) N 2 = I 2 ω 2 + ω 3 ω 1 (I 1 I 3 ) (73) N 3 I 3 ω 3 ω 1 ω 2 (I 2 I 1 ) The second term on the right is obtained by the usual method of cross products using L rot = (I 1 ω 1 I 2 ω 2 I 3 ω 3 ). 13

14 3.1 Rigid Body About Fixed Axis in Uniform Rotation Neither the size nor direction of ω change, so the time derivatives are zero. Euler s equations become N 1 = ω 2 ω 3 (I 3 I 2 ) (74) N 2 = ω 3 ω 1 (I 1 I 3 ) (75) N 3 = ω 1 ω 2 (I 2 I 1 ) (76) These are the components of torque that the axle must supply to the system to maintain the uniform rotation. Example. What torque must be supplied to rotate the rectangle about an axis through the short side? We use the principal axes, and note that the angles between the axis of rotation and the principal axes are 112.5, 22.5, 90 so that the angular velocity in the frame of the principal axes is ω = ω (77) Recalling that I 1 = ma 2 and I 2 = ma 2, we find N 1 = N 2 = 0, N 3 = ma 2 ω 2. If we want to think in terms of force, the r is in the 1-2 plane, the torque is along the 3 axis, so the required force must be in the 1-2 plane. 4 Rigid Body in Free Rotation and Poinsot s Ellipsoids In Chapter 7 we showed that the motion of a system can be decomposed into translational motion of the center of mass plus rotational motion about the center of mass. We are going to apply Euler s equations to a rigid body on which no torques act. An example of such a situation is throwing an object in the air. Neglecting air resistance, the only force acting on the object is the weight and it acts at the center of mass 2.If we use the center of mass as the origin of our body coordinate system, there is no torque due to this force. Since there is no torque, in the fixed reference frame the angular momentum vector L is constant both in direction and magnitude. On the rotating body frame, however, the angular momentum is fixed in magnitude but not direction. We can write this (body frame) as L L = L L L 2 3 = constant (78) 2 Well technically at the center of gravity, but in a uniform gravitational field this is the same thing. 14

15 In the body frame the tip of the angular momentum vector lies on the surface of a sphere. If we look at the angular velocity in the body frame, I 2 1ω I 2 2ω I 2 3ω 2 3 = L 2 = constant (79) The angular velocity vector thus lies on the surface of an ellipsoid. The lengths of the semimajor axes in the three principal axes are inversely proportional to the inertia component, in the ratio 1 I 1 : 1 I 2 : 1 I 3 (80) Since there is no torque, the rotational kinetic energy in the body frame is also constant, and we can write ω L = 2T rot = const or in terms of the angular velocity I 1 ω I 2 ω I 3 ω 2 3 = 2T rot = constant (81) This shows that the angular velocity also lies on the surface an ellipsoid, the Poinsot Ellipsoid or Inertial Ellipsoid, with semi-major axes in the ratio 1 I1 : 1 I2 : 1 I3 (82) The angular momentum vector must therefore lie on two different ellipsoids, one determined by angular momentum, one determined by kinetic energy. Since there is an angular momentum, the ellipsoids must intersect at at least one point. In general the angular velocity vector lies on a path that is the intersection of the two ellipsoids. One example is shown in Figure in the text, where the angular velocity follows a circular path about axis 3. The path is given the rather obscure name polhode. A special case is when the object rotates about a principal axis, say ω = ω 1 ê 1. In this case the two ellipsoids intersect at a point on the 1-axis. 4.1 The Tennis Racquet Theorem Consider a tennis racquet. The principal axes are easily seen to be along the handle, through the center of mass normal to the racquet, and through the center of mass in the plane of the racquet. See Figure If the racquet is spun about one of the principal axes and launched into a torque-free environment, the rotation will be stable providing the rotation axis is along a principal axis associated with either the maximum or the minimum principal moment. The axis with the intermediate moment will be unstable. This is seen very clearly if you spin a racquet and launch it into the air. 15

16 The text discusses this intersections of the ellipsoids qualitatively. First I will try to justify the text figure Suppose I 1 : I 2 : I : 3 = 5 : 4 : 3. Then the ratio of ellipsoid axes from angular momentum is (0.20:0.25:0.33) and the ratio from kinetic energy is (0.44:0.50:0.58). If rotation is about the 1-axis, maximum moment, I will rescale the angular momentum ratio so that the first term is equal to the first term in the kinetic energy ratio, resulting is (0.44:0.55:0.73). The ellipsoids intersect at a single point since on ellipsoid lies completely outside the other. Doing this for the 3-axis, the axis with minimum moment, the angular momentum ratio becomes (0.35:0.43:0.58), and again the ellipsoids intersect at a single point. However for the 2-axis, intermediate moment, the ratio becomes (0.40:0.50:0.66), and now the ellipsoids must intersect along a curve rather than just a point. For the axes with minimum and maximum moments, a slight disturbance from rotation around the axis will result in a restoring movement, with oscillations about the stable point. For the axis with intermediate moment, a slight disturbance from rotation about the axis will result in unstable equilibrium. For more details, consider the model of the tennis racquet, from Classical Mechanics, Barger and Olsson. Figure 5: Model of a tennis racquet. The head is a ring, the handle a thin rod. The left vertical axis is used to find the center of mass. The right axis passes through the center of mass and is used for moments of inertia. Consider a circular head that is a ring of mass m a = 0.15 kg and radius a = 0.13 m. Attached is a handle that is a thin rod of mass m l = 0.18 kg and length l = 0.38 m. First find the center of mass. Measured from the center of the head this is R = m l(a + l/2) m a + m l = meters (83) Now find the principal moments of inertia relative to the center-of-mass origin. Call the 16

17 axis through the handle the 1-axis. I 1 = 1 2 m aa 2 = kg m 2 (84) Call the 2-axis the one in the plane of the racquet. The parallel axis theorem must be used ( ) ( ) 1 1 I 2 = 2 m aa 2 + m a R m ll 2 + m l [a + l/2 R] = kg m 2 (85) and from the perpendicular axis theorem, I 3 = I 1 + I 2 = kg m 2 Now we can write Euler s equations, one from each component. Remember that the torque is zero. I 1 ω 1 + ω 3 ω 2 (I 3 I 2 ) = 0 (86) I 2 ω 2 + ω 1 ω 3 (I 1 I 3 ) = 0 (87) I 3 ω 3 + ω 2 ω 1 (I 2 I 1 ) = 0 (88) Since the racquet is a laminar object with I 3 = I 1 + I 2 these can be simplified to ω 1 + ω 3 ω 2 = 0 (89) ω 2 ω 1 ω 3 = 0 (90) ( ) I2 I 1 ω 3 + ω 1 ω 2 = 0 (91) I 2 + I 1 For the tennis racquet model I 2 I 1, so the last equation is approximately ω 3 + ω 1 ω 2 = 0 (92) At this point we have 3 coupled differential equations for the angular velocity of the tennis racquet Axis 1 or Axis 3 Suppose that the racquet is launched in perfect alignment with one of the principal axes, say 1. Then ω 2 = ω 3 = 0 and Euler s equations say that ω 1 = constant. This is true for all three principal axes. Now let us suppose that there is a slight disturbance from the perfect alignment. Since ω 2 ω 3 is small the first Euler equation is still ω 1 0, ω 1 constant. The other two equations can be combined by defining a complex quantity for the angular velocity transverse to axis 1 (I will denote complex with ă) ω T = ω 3 + iω 2 (93) 17

18 Using this the other two Euler equations can be combined into ω T iω 1 ω T = 0 (94) with solution Hence ω T = Ae i(ω 1t+α) (95) ω 2 = A sin(ω 1 t + α) (96) ω 3 = A cos(ω 1 t + α) (97) Since by our initial assumption the disturbance is small, A is small and the transverse angular velocity components trace out a circle around the main angular velocity ω 1, or in other words the situation is stable, slight disturbances do not lead to radically different orientations. Looking at the Euler equations for this situation, notice that 1 and 3 are in similar positions with similar signs. Thus rotation about the principal axes with either maximum or minimum moment is shown to be stable Axis 2 For rotation about the principal axis with intermediate moment of inertia, we combine the other two Euler equations into with solutions ( ω 1 + ω 3 ) + (ω 1 + ω 3 )ω 2 = 0 (98) ( ω 1 ω 3 ) (ω 1 ω 3 )ω 2 = 0 (99) (ω 1 + ω 3 ) = Ae ω 2t (ω 1 ω 3 ) = Be +ω 2t (100) (101) or ω 1 = 1 ( Ae ω 2 t + Be +ω 2t ) 2 (102) ω 3 = 1 ( Ae ω 2 t Be +ω 2t ) 2 (103) These solutions show that the disturbances grow in size with time. The solutions shown are only valid for small transverse angular velocities, but it is clear that about this axis that the racquet will tumble. 18

19 5 Analysis of a football The text does the analysis of the football, using the symmetry axis as the 3-axis and defining Euler s equations become I s = I 3 (104) I = I 1 = I 2 (105) I ω 1 + ω 3 ω 2 (I s I) = 0 (106) I ω 2 + ω 1 ω 3 (I I s ) = 0 (107) I s ω 3 = 0 (108) The latter equation tells us that ω 3 = constant. Define a new constant Ω = ω 3 I s I I so that the first two Euler equations can be written (109) ω 1 + Ωω 2 = 0 (110) ω 2 Ωω 1 = 0 (111) Differentiate the first of these and use the second to yield which is simple harmonic motion ω 1 + Ω ω 2 = 0 (112) ω 1 + Ω 2 ω 1 = 0 (113) Solving for the other angular velocity ω 1 = ω 0 cos(ωt + α) (114) ω 2 = ω 0 sin(ωt + α) (115) As discussed in the text we can view the football as having an angular velocity ω that is constant in magnitude. It has a component ω 3 = ω cos α along the symmetry axis, where cos α is the direction cosine for the angular velocity and the 3-axis. The projection onto the 1-2 plane is ω 0 = ω sin α which has a constant magnitude but rotates around the 3-axis with angular velocity Ω. Figure shows this. The circular path that the angular velocity traces out represents the intersection of the two ellipsoids discussed earlier. 19

20 5.1 General 3-D Object Solving Euler s equations for a more complicated object can be done also as is illustrated on pages of the text. Specifically, the text looks at an ellipsoid Using Mathematica to evaluate the integrals, Euler s equations become x y2 4 + z2 1 = 1 (116) I 1 = π I 2 = 2π I 3 = (117) A 1 ω 2 ω 3 = ω 1 (118) A 2 ω 1 ω 3 = ω 2 (119) A 3 ω 2 ω 1 = ω 3 (120) Using initial conditions of ω 1 (0) = ω 2 (0) = ω 3 (0) = 1 rad/s and numerically solving the differential equations results in the phase plot shown in Figure Refer to that section for more details. 6 More General Rotation and Eulerian Angles We have looked at two cases so far of applications of Euler s Equations: rotation about a fixed axis at constant angular velocity ( ω = 0) and torque free motions ( N = 0). To most easily deal with the more general case we need to discuss the Eulerian angles. The Eulerian angles provide a way to transform from the fixed frame to the body frame via three rotations (recall the rotation matrices from Chapter 1). Call the fixed frame Oxyz and the rotating body frame O123. A third frame, Ox y z, also rotating, will be the intermediary. The z axis will coincide with the 3-axis, and the x axis will lie in the xy plane and will define the line of nodes. Starting from the fixed axis, rotate around the z axis by an angle φ to get to the x axis. Next rotate about the x axis through an angle θ to bring the z axis to z. Finally rotate about the z axis through an angle ψ to get to the 123 set of axes. The angles φ, θ, ψ are called the Eulerian angles. There are two planes to consider: The xy plane in which x and y axes lie, and the x y plane in which axes 1 and 2 also lie. The x axis lies at the intersection of the two planes, and is called the line of nodes. 20

21 Next we look at how to transform from the Oxyz frame to the O123 frame using the Eulerian angles. Recall the rotation matrices from Chapter 1. We can write cos φ sin φ cos ψ sin ψ 0 λ φ = sin φ cos φ 0 λ θ = 0 cos θ sin θ λ ψ = sin ψ cos ψ sin θ cos θ (121) The three rotations can then be written as the product λ = λ ψ λ θ λ φ which is cos φ cos ψ sin φ cos θ sin ψ sin φ cos ψ + cos φ cos θ sin ψ sin θ sin ψ cos φ sin ψ sin φ cos θ cos ψ sin φ sin ψ + cos φ cos θ cos ψ sin θ cos ψ sin φ sin θ cos φ sin θ cos θ (122) Now suppose that the body is rotating about some arbitrary axis with angular velocity ω NOT one of the principal axes. We set about determining the angular velocity components in the body frame using the Eulerian angles. In a small time dt the rigid body rotates through an angle and thus the angular velocity can be written as The meaning of the time derivatives is φ θ ψ d β = ωdt = d φ + d θ + d ψ (123) ω = φ + θ + ψ (124) Rotation about the z axis of the fixed (inertial) system Rotation about the line of nodes, the x axis (intermediate rotating system) Rotation about the 3 axis of the body (rotating) system So to get the components of ω we need the components of the time derivatives of the vector Eulerian angles. These directions of these derivatives are shown on Figure We would like to get the derivatives in all three reference frames, Oxyz, Ox y z, and O123. Careful inspection of Figure lets us write the derivatives in terms of the Ox y z coordinates (note error in text Equation 9.6.3) φ x = 0 θx = θ ψ x = 0 (125) φ y = φ sin θ θy = 0 ψ y = 0 (126) φ z = φ cos θ θz = 0 ψ z = ψ (127) 21

22 Thus the components of angular velocity in the intermediate system are Now for the components in the body, O123 system ω x = θ (128) ω y = φ sin θ (129) ω z = φ cos θ + ψ (130) φ 1 = φ sin θ sin ψ θ1 = θ cos ψ ψ 1 = 0 (131) φ 2 = φ sin θ cos ψ θ2 = θ sin ψ ψ 2 = 0 (132) φ 3 = φ cos θ θ3 = 0 ψ 3 = ψ (133) making the angular velocity components in the body system ω 1 = φ sin θ sin ψ + θ cos ψ (134) ω 2 = φ sin θ cos ψ θ sin ψ (135) ω 3 = φ cos θ + ψ (136) Similarly we could get the components relative to the fixed reference system, and if I did this right the result is ω x = θ cos ψ ψ sin θ cos φ (137) ω y = θ sin ψ ψ sin θ sin φ (138) ω z = φ + ψ cos θ (139) 6.1 Free Rotation (No Torques) Of a Symmetric Object Let s restate the meaning of the angles: θ is the angle between the z axis and both the z and 3 axes; φ is the angle between the x and x axes; ψ is the angle between the line of nodes and the 1 axis measured in the 12 plane. Note that x is perpendicular to the 3, z and z axes! For free rotation, the angular momentum in the fixed coordinates is constant in magnitude and direction. For convenience, orient the fixed coordinates so that L is along the z axis. This will be called the invariable line. Referring to Figure we can write the components in the intermediate system L x = 0 L y = L sin θ L z = L cos θ (140) We restrict ourselves to a body with rotational symmetry about the 3-axis, I 3 = I s, I 1 = I 2 = I as for the football. For such symmetric objects, the 1 and 2 axes can be rotated 22

23 around the 3 axis and still result in a diagonal inertia tensor. We will choose the 1 axis to be coincident with the x axis, meaning ψ = 0. Hence the body and intermediate frames are identical, and we can write L x = Iω x L y = Iω y L z = I s ω z (141) Since ω x = θ and L x = 0, we have ω x = 0, θ = 0. Thus we see that ω lies in the y z plane with an angle between it and the z axis being called α. Thus L (along z axis), ω and the 3-axis lie in the same plane. A useful visualization is the Java applet at Precession.html (programmed initially in Easy Java Simulations). If you open this, initially turn off the Show Point Trace. Start the simulation and you will see two red vectors on a dark background (or two blue vectors on a white background.). The vertical one is the angular velocity of the rotation of the symmetry axis (the precession) while the inclined one is the angular velocity of the object in the body frame. The vector sum, in yellow on the dark background (red on the white background), is the combined angular velocity, and is the angular velocity of the object in the space frame. Note that it is NOT constant in direction, but only in magnitude. The angle between the red (blue) vectors, i.e. between the z and 3-axes, is θ. The angle between vertical and the yellow(red) vector, i.e. between z and ω, is α Here is Marion s description (Classical Dynamics of Particles and Systems). In addition to constant angular momentum in the fixed frame, the rotational kinetic energy T rot = ω L/2 is constant. The dot product can be interpreted as proportional to the projection of the angular velocity along the direction of the angular momentum. So the projection of ω onto the z axis is constant, and the angular velocity ω precesses around the z-axis making a constant angle α with the angular momentum. This is the same thing that we got before. View the situation from the fixed frame. We can visualize a space cone traced out by the angular velocity as it precesses around the z-axis. This is the fixed cone shown on the right of the simulation. In the body frame, the inertia tensor is diagonalized, and a body cone is traced out by the angular velocity rotating about the 3-axis. You can look at the simulation and consider the view from a frame fixed in the body to see this. Since ω lies on both cones, it must lie on the intersection of the cones, and we visualize the body cone rolling around the space cone as is suggested in Figure and shown in the simulation. Back to Fowles and Cassiday. We have chosen our coordinates so that ω lies in the y z and 23 planes. The inertia tensor is diagonal in both these frames. Thus ω x = 0 ω y = ω sin α ω z = ω cos α (142) 23

24 L x = 0 L y = L sin θ = Iω sin α L z = L cos θ = I s ω cos α (143) and we can get a relation between angles θ the angle between L and 3 and α the angle between ω and 3. tan θ = I I s tan α (144) For prolate objects like a rod, α < θ while for oblate objects like a coin, α > θ. This means that the space cone can engulf the body cone (prolate case) or the space cone can be outside the body cone (oblate case). See Figure You can adjust the aspect ratio in the simulation between 0.5 (oblate) and 5 (prolate). Prolateness of 1 refers to a sphere. We end up with three useful angular velocities: ω of the body about the rotation axis (yellow line in the simulation), Ω of the angular velocity vector about the body symmetry axis 3, and φ of the symmetry axis (3) about the space axis z, the invariable line. When we did the football we determined ( ) Is Ω = I 1 ω cos α (145) The φ for this symmetric situation is the angular rate of precession of both the body symmetry axis, 3, and the angular velocity, ω, about the invariable line (z axis or L.) This appears as a wobble of an imperfectly thrown frisbee or football. Earlier we had φ = ω y / sin θ and ω y = ω sin α, so we can combine these to get φ = ω sin α sin θ With some trig identities and algebra we get [ ( I 2 φ = ω 1 + s I 2 1 ) ] cos 2 α (146) (147) giving the wobble rate in terms of the angular speed ω of the body and the inclination α between the body 3-axis and the angular velocity. E.g. Precession of a Frisbee A highly oblate object like a Frisbee is approximately a laminar object, so I s = 2I, I s /I = 2. If the disk is tossed into the air with an angular velocity inclined to the body symmetry axis, 3, by an angle α, then Ω = ω cos α (148) and φ = ω cos 2 α 2ω (149) for small angles. The wobble rate is twice the angular speed of rotation. 24

25 This had an interesting historical impact on Richard Feynman 3. I was in the cafeteria and some guy, fooling around, throws a plate in the air. As the plate went up in the air I saw it wobble, and I noticed the red medallion of Cornell on the plate going around. It was pretty obvious to me that the medallion went around faster than the wobbling. I had nothing to do, so I start figuring out the motion of the rotating plate. I discovered that when the angle is very slight, the medallion rotates twice as fast as the wobble rate two to one. It came out of a complicated equation! I went on to work out equations for wobbles. Then I thought about how the electron orbits start to move in relativity. Then there s the Dirac equation in electrodynamics. And then quantum electrodynamics. And before I knew it... the whole business that I got the Nobel prize for came from that piddling around with the wobbling plate. E.g. Free Precession of Earth The earth is a slightly oblate spheroid, I s /I The axis of rotation of the earth is inclined by α 0.2 = rad to the symmetry axis. We then have Ω = ω. We know that ω = 2π/(1 day) so the period of the precession of the the earth s axis of rotation about the pole is predicted to be 2π/Ω = 305 days. In fact the observed value is 440 days, attributed to the fact that the Earth is not a perfect oblate spheroid, and that is is not a rigid object. The wobble of the Earth s body axis is φ = ω yielding a period of days. 7 Mr. Euler, Meet Mr. Lagrange 7.1 Free Rotation of a General Rigid Body Consider the O123 coordinate system in which the inertia tensor is diagonal. Then T = 3 I i ωi 2 V = constant (150) 1 In Chapter 10 we used L = T V for the Lagrangian. In this chapter L represents angular momentum, so I will use Λ = T V. In the torque free case, the Lagrangian Λ = T. Let s 3 Feynman R P 1985 Surely You Are Joking, Mr. Feynman! (New York: W W Norton) see pp for a discussion of the rotating plate motion 25

26 choose the Euler angles as the generalized coordinates. Then for the Euler angle ψ, T ψ = d T dt ψ (151) The ω i can be expressed as functions of the Euler angles as we did earlier. Hence we can write for the ψ coordinate, 3 1 T ω i ω i ψ = d dt From the expression for kinetic energy we have 3 1 T ω i ω i ψ (152) T ω i = I i ω i (153) Now let s evaluate the other partials using the previous expressions for the angular velocities in terms of the Euler angles, Equations??,??,??. ω 1 ψ = φ sin θ cos ψ θ ω 2 sin ψ = ω 2 = 0 ψ (154) ω 2 ψ = φ sin θ sin ψ θ ω 2 cos ψ = ω 1 = 0 ψ (155) ω 3 ψ = 0 ω 3 = 1 ψ (156) Putting these values into Equation?? we get or I 1 ω 1 ω 2 + I 2 ω 2 ( ω 1 ) = d dt I 3ω 3 (157) (I 1 I 2 )ω 1 ω 2 I 3 ω 3 = 0 (158) Since the designation of the 3-axis is arbitrary, similar relations hold for the other Euler angle Lagrangians, (I i I j )ω i ω j I k ω k ɛ ijk = 0 (159) k where the permutation symbol ɛ ijk is ɛ ijk = 0 any two subscripts equal (160) = +1 any even permutation 123, 231, 312 (161) = 1 any odd permutation, 132, 213, 321 (162) These are just the Euler equations for a torque free environment. 26

27 7.2 Rotation with Applied Torque In the case of applied torque we start with ( dl N = ) ( d = ) L + ω L dt dt (163) fixed body and take components along the 3-axis N 3 = L 3 + ω 1 L 2 ω 2 L 1 (164) In the body frame the inertia is diagonalized so L i = I i ω i and we get the general Euler equations that we can write in the somewhat obscure form (I i I j )ω i ω j k ( I k ω k N k ) ɛ ijk = 0 (165) If i = j this is 0=0. For i j, If k equals either i or j, so the summation is non-zero only for a different value of k and the sign of the last terms depends on the perturbation symbol. So in reality Equation?? can be written as (I i I j )ω i ω j ( I k ω k N k ) ɛ ijk = 0 (166) 8 Motion of a Symmetrical Top We now turn to the motion of a rotationally symmetric top, I 1 = I 2 = I, I 3 = I s, that rotates about a tip fixed in location, but with a uniform gravitational field. In Chapter 8 we discussed possible choices of an axis of rotation. The center of mass is always a good choice, and is what we have used thus far. When an object rotates about a fixed point, that point is also a good choice of origin, and is what we will use here. Figure shows the top in a tilted position. The inertial system has a vertical z-axis, and the body 3-axis is tilted by an angle θ. As before the x axis is perpendicular to z, z and 3 axes. Also due to the symmetry of the top the inertia tensor is diagonal in both the body and intermediate frames of reference. Call the distance from the tip of the top to the center of mass l, measured along the 3-axis, hence we can write the components of the gravitational torque. N x = mgl sin θ N y = N z = 0. (167) 27

28 Figure 6: Earlier we expressed the angular velocities in the primed frame in terms of the Euler angles, and using this we get L x = I θ (168) L y = I φ sin θ (169) L z = I s ( φ cos θ + ψ) I s S (170) where we have introduced a quantity S = φ cos θ + ψ that is the spin of the top about the z = 3 axis. Using Lagrangian formulation, we can write T = 1 2 I(ω2 1 + ω 2 2) I sω 2 3 (171) Using the expressions for the angular velocity components in terms of the Euler angles, Equations??,??,??, and doing the algebra we see this becomes T = 1 2 I( φ 2 sin 2 θ + θ 2 ) I s( φ cos θ + ψ) 2 (172) The potential energy is so the Lagrangian is V = mgl cos θ (173) Λ = 1 2 I( φ 2 sin 2 θ + θ 2 ) I s( φ cos θ + ψ) 2 mgl cos θ (174) Notice that φ and ψ are ignorable coordinates meaning that the conjugate angular momenta are p φ = Λ φ = (I sin2 θ + I s cos 2 θ) φ + I s ψ cos θ = constant (175) p ψ = Λ ψ = I s( ψ + φ cos θ) = I s S I s ω 3 = constant (176) 28

29 Thus the spin, as defined by Fowles and Cassiday, is constant. The spin is the angular velocity about the body symmetry axis. Now referring to Figure we can see that p φ L z, the angular momentum about the fixed z-axis and a little more algebra shows that this can be written L z p φ = I φ sin 2 θ + I s S cos θ (177) Likewise we can recognize that p ψ = I s S L z, the angular momentum component about the body symmetry axis. Since the momenta are constants of the motion, we can use the above equations to solve for ψ and φ. Use Equation?? and solve for Put this into Equation?? and solve for ψ = L z I s φ cos θ I s (178) Then put this into Equation?? to get ψ = L z I s For the θ generalized coordinate we write φ = L z L z cos θ I sin 2 θ (L z L z cos θ) cos θ I sin 2 θ (179) (180) L θ = I θ (181) L θ = I φ 2 sin θ cos θ I s ( φ cos θ + ψ) φ sin θ + mgl sin θ (182) = I φ 2 sin θ cos θ I s S φ sin θ + mgl sin θ (183) So the equation of motion is I θ = I φ 2 sin θ cos θ I s S φ sin θ + mgl sin θ (184) The expression for φ can be inserted to give a differential equation in one variable. Once θ(t) is found, the other Euler angle functions can be found. We will be able to deduce much of the top s motion without solving this equation. 29

30 8.1 Steady Precession of A Top Consider first a horizontal top, θ = 90 = constant. The equation of motion then reduces to mgl = I s S φ (185) So φ = constant. The precession rate, φ should decrease if the spin is increased, but increase if the mass is increased. We ll see if we can demo this. If we consider the case of arbitrary angle, but still want steady precession, θ = 0, then the equation of motion becomes mgl = I s S φ S φ 2 cos θ (186) This is a quadratic in φ so there are two possible rates of steady precession. Usually (Fowles and Cassaday say) the slower will occur, but with the correct initial conditions, the fast one may occur. The precession rate is φ = I ss ± I 2 s S 2 4mglI cos θ 2I cos θ For precession to occur, this must be real hence (187) I 2 s S 2 4mglI cos θ (188) If a there is friction in the bearings so that the top s spin slows, once it reaches this threshold the top will begin to fall, and eventually topple over. E.g. Rotating Top Consider a simple top made from a spindle of negligible mass with a solid disk of mass 2.00 kg and radius 5.0 cm mounted 4.0 cm above the tip of the top. (a) Find the minimum spin of the top so that it can remain in its vertical position without toppling. I s = 0.5(2.00)(0.05) 2 = kg m 2, using the parallel axis theorem I = (0.25(2.00)(0.05) (0.04) 2 ) = kg m 2, l = 0.04 m. Hence the required spin is 4mglI S Is 2 = 47.3 rad/s = 7.52 rev/sec = 451 rpm (189) This is a period of s (b) If the axis is set horizontally and the spin is 500 rpm, find the precession rate and period. S = 52.4 rad/s The period is then 1.05 s. φ = mgl I s S = 5.98 rad/s = 57.1 rpm (190) 30

31 (c) If the top has spin of 500 rpm and is released at an initial angle of 30, find the two rates of precession and the two periods. The precession rates are 7.75 rad/s with period 0.81 s and 26.2 rad/s with period 0.24 s. 9 Nutation and the Energy Equation The normal force at the tip of the top is assumed to do no work (no friction) so energy is also a constant of motion E = 1 2 I( φ 2 sin 2 θ + θ 2 ) I s( φ cos θ + ψ) 2 + mgl cos θ = constant (191) = 1 2 I( φ 2 sin 2 θ + θ 2 ) I ss 2 + mgl cos θ = constant (192) where the second equation has inserted the spin. However we already know that the spin S is a constant of motion, so we can look at the quantity E = E 1 2 I ss 2 = 1 2 I( φ 2 sin 2 θ + θ 2 ) + mgl cos θ = constant (193) From Equation?? we can rewrite the equation as where E = 1 2 I θ 2 + V (θ) (194) V (θ) = (L z L z cos θ) 2 2I sin 2 + mgl cos θ (195) θ What we have done is to reduce the three dimensional problem to an equivalent onedimensional problem in θ. This is the same treatment that we used in the central-force problem, where we looked at radial variations by introducing an effective potential. Figure shows the general shape of the effective potential. Note that it has a minimum and is concave up. This implies some sort of a restoring torque will cause the top to bob up and down (in θ) in a pattern called nutation. The minimum of the potential is the situation of steady precession. 10 And then... We have discussed only a small portion of rotational problems. Even for the symmetrical top there are other things that we could discuss such as a top whose point of contact with a table makes a circle as the top spins, or a Tippy Top, v=xu_dp9ifgsu. 31

Let s first see how precession works in quantitative detail. The system is illustrated below: ...

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