Rotational Motion Chapter 4 P. J. Grandinetti Chem. 4300 Sep. 1, 2017 P. J. Grandinetti (Chem. 4300) Rotational Motion Sep. 1, 2017 1 / 76
Angular Momentum The angular momentum of a particle with respect to the origin, O, is given by l = r p x O z r p m y The rate of change of angular momentum is given by the cross product of r with the applied force. d l dt = r F = τ This cross product is defined as the applied torque, τ. Unlike linear momentum, angular momentum depends on choice of origin. P. J. Grandinetti (Chem. 4300) Rotational Motion Sep. 1, 2017 2 / 76
Conservation of Angular Momentum Total angular momentum is Consider a system of N Particles z L = N l i = i=1 N r i p i i=1 m 2 m 3 m 5 Rate of change of angular momentum is m 1 y d L N dt = i=1 d l i dt = N i=1 r i d p i dt x m 4 which becomes d L N dt = r i F net i i=1 P. J. Grandinetti (Chem. 4300) Rotational Motion Sep. 1, 2017 3 / 76
Conservation of Angular Momentum d L N dt = r i F net i Taking an earlier expression for a system of particles from chapter 1 we obtain we obtain i=1 F net i i=1 = F ext i + d L N dt = r i F ext i + N j=1 N f ij N i=1 j=1,i j r i f ij d L N dt = r i F ext i + N N 0 r i f ij i=1 i=1 j=1,i j Again, the double sum disappears from Newton s third law ( f 12 = f 21 ). P. J. Grandinetti (Chem. 4300) Rotational Motion Sep. 1, 2017 4 / 76
Conservation of Angular Momentum d L N dt = r i F ext i = i=1 N τ i = τ total i=1 If there is no net external torque on our system of particles then the systems total angular momentum, L, is constant, if d L dt = 0, then L = constant. This is the principle of conservation of angular momentum. It is true in quantum mechanics as well as in classical mechanics. P. J. Grandinetti (Chem. 4300) Rotational Motion Sep. 1, 2017 5 / 76
Orbital and Spin Angular Momentum P. J. Grandinetti (Chem. 4300) Rotational Motion Sep. 1, 2017 6 / 76
Orbital and Spin Angular Momentum Consider a system of particles again. The total angular momentum relative to the origin is L = N l i = i=1 N i=1 r i m i d r i dt Defining the position of each particle relative to the center of mass, r i = R + r i r i is the position of the particle position relative to the origin, R is the center of mass relative to the origin, r i is the particle position relative to the center of mass. Expression for L becomes (see notes for derivation) L = R p total relative to origin + N d r i r i m i dt i=1 relative to CM P. J. Grandinetti (Chem. 4300) Rotational Motion Sep. 1, 2017 7 / 76
Orbital and Spin Angular Momentum L = R p total relative to origin + N d r i r i m i dt i=1 relative to CM Identify total angular momentum as split into orbital and spin angular momentum terms: L = L orbital + L spin Imagine L orbital as angular momentum of the the earth as it moves around sun as the origin, and L spin as angular momentum of earth as it spins about its center of mass. It is often true, to a good approximation, that the orbital and spin parts are separately conserved. P. J. Grandinetti (Chem. 4300) Rotational Motion Sep. 1, 2017 8 / 76
Rotational energy Total kinetic energy of our system of particles is given by K = N i=1 1 2 m i ( ) 2 d ri dt Written in terms of R and r i gives K = 1 ( )2 d 2 M R + 1 dt 2 N i=1 ( ) 2 d r i m i dt First term is energy associated with motion of the center of mass of system. Second term is energy associated with rotational motion of system about center of mass. P. J. Grandinetti (Chem. 4300) Rotational Motion Sep. 1, 2017 9 / 76
Rigid Bodies P. J. Grandinetti (Chem. 4300) Rotational Motion Sep. 1, 2017 10 / 76
Rigid Bodies Definition When all particles in the system are rigidly connected we have a rigid body. This ideal model assumes no relative movement of composite particles. m 2 z m 5 m 3 m 1 y x m 4 P. J. Grandinetti (Chem. 4300) Rotational Motion Sep. 1, 2017 11 / 76
Euler s Rotation Theorem Definition Euler s rotation theorem states that the general displacement of a rigid body about a fixed point can be described by a rotation through an angle about a single axis. Since it takes only two angles to define the orientation of the rotation axis we find that the orientation of any rigid body about a fixed point can be described by just three parameters: e.g., the polar, θ, and azimuthal, φ, angles and the angle of rotation, χ, as shown below. The set of φ, θ, and χ are called Euler angles. P. J. Grandinetti (Chem. 4300) Rotational Motion Sep. 1, 2017 12 / 76
Euler angles Another example of a set of Euler angles is the convention for describing the orientation of an airplane by three parameters called yaw, pitch, and roll, as illustrated below. Pitch Axis (y) Roll Axis (x) Yaw Axis (z) Yaw, pitch, and roll are just one of many ways of defining the three angles implied by Euler s rotation theorem. P. J. Grandinetti (Chem. 4300) Rotational Motion Sep. 1, 2017 13 / 76
Angular velocity vector, ω Definition The angular velocity vector, ω, is a vector passing through the origin along the axis of rotation and whose magnitude equals the magnitude of the angular velocity, ω = ω e r, where ω = dχ dt and e r is a unit vector defined by e r = sin θ cos φ e x + sin θ sin φ e y + cos θ e z The direction of ω is determined by the right-hand rule: if you curl your right-hand fingers in the direction of the rotation, then your thumb points in the direction of ω. Keep in mind when describing the motion of a rigid body that the magnitude and orientation of ω can change with time. P. J. Grandinetti (Chem. 4300) Rotational Motion Sep. 1, 2017 14 / 76
Angular velocity vector, ω Choice of origin through which ω passes depends on specifics of the rigid body motion being described. For molecular rotations the natural choice for the origin is the center of mass of the molecule. For a top spinning on a table surface the origin is better located at a fixed point where the tip of the top meets the table surface. y z x y x P. J. Grandinetti (Chem. 4300) Rotational Motion Sep. 1, 2017 15 / 76
Moment of Inertia Tensor The total angular momentum of the rigid body relative to the origin is J = N ( r i m i v i ) i=1 Note: notation change, J instead of L for rigid body. In rotating rigid body the linear velocity vector, v i, of ith particle is related to its angular velocity, ω, by v i = ω r i r i is the particle position. ω, is identical for all particles since all inter-particle distances are constant. We can combining these two expressions... P. J. Grandinetti (Chem. 4300) Rotational Motion Sep. 1, 2017 16 / 76
Moment of Inertia Tensor Total angular momentum of rigid rotor relative to the origin is J = N ( r i m i v i ) = i=1 N i=1 m i [ ri ( ω r i ) ] vector triple product Vector triple product has a well known expansion ( ) a b c = ( a c) b ( a b) c With triple product expansion we get J = N [( ) ( m i ri r i ω ri ω ) ] r i i=1 P. J. Grandinetti (Chem. 4300) Rotational Motion Sep. 1, 2017 17 / 76
Moment of Inertia Tensor J = N [( ) ( m i ri r i ω ri ω ) ] r i i=1 can be rewritten as a matrix equation J = J x J y J z = I xx I xy I xz I yx I yy I yz I zx I zy I zz I ω x ω y = I ω ω z where matrix I is called the moment of inertia tensor. J = I ω When a rigid body rotates about the origin with angular velocity ω the size and direction of the angular momentum, J, is determined by its moment of inertia tensor, I. P. J. Grandinetti (Chem. 4300) Rotational Motion Sep. 1, 2017 18 / 76 ω
Moment of Inertia Tensor of a rigid body Elements of Moment of Inertia Tensor about the center of mass If the rigid body rotates about its center of mass then the relevant moment of inertia tensor relative to the center of mass is calculated according to I xx = I yy = I zz = N i=1 N i=1 N i=1 m i ( y 2 i + z 2 i ) M(Y 2 + Z 2 ) m i ( x 2 i + z 2 i ) M(X 2 + Z 2 ) m i ( x 2 i + y 2 i ) M(X 2 + Y 2 ) M is total mass X, Y, Z are coordinates of the center of mass. I xy = I yx = I yz = I zy = I xz = I zx = N m i x i y i + MXY i=1 N m i y i z i + MYZ i=1 N m i x i z i + MXZ Notice that off-diagonal elements are symmetric about diagonal, I xy = I yx, I xz = I zx, etc. Tensors with this property are called symmetric tensors. P. J. Grandinetti (Chem. 4300) Rotational Motion Sep. 1, 2017 19 / 76 i=1
The Moment of Inertia Tensor is a symmetric matrix Off-diagonal elements are symmetric about the diagonal I = I xx I xy I xz I xy I yy I yz I xz I yz I zz For any symmetric tensor one can always find an axis system in which it is diagonal. I PAS = I a 0 0 0 I b 0 0 0 I c = R(φ, θ, χ) I xx I xy I xz I xy I yy I yz I xz I yz I zz R(φ, θ, χ) is a rotation matrix and R T (φ, θ, χ) is its transpose. Definition R T (φ, θ, χ) Principal Axis System (PAS) is the coordinate system in which I is diagonal. P. J. Grandinetti (Chem. 4300) Rotational Motion Sep. 1, 2017 20 / 76
Moment of Inertia Tensor Principal Axis System When the moment of inertia tensor is calculated in the PAS we have I PAS = I a 0 0 0 I b 0 0 0 I c Diagonal elements, I a, I b, and I c, called the principal moments of inertia. Axes of body-fixed frame, a, b, and c, are assigned such that I c I b I a. A principal axis is axis for which I ω i = I i ω i, where ω i is an angular velocity vector along one of the axes in the PAS, i.e., ω a, ω b, or ω c. If ω points along one of the principal axis directions then so does J (recall J = I ω). Symmetry can help identify the principal axis system. A two-fold or higher axis of symmetry in a molecule is a principal axis. P. J. Grandinetti (Chem. 4300) Rotational Motion Sep. 1, 2017 21 / 76
Moment of Inertia Tensor Principal Axes If you see an object freely rotating (no external torque applied) about a fixed axis with a constant angular velocity then this axis is a principal axis. The object s angular momentum is conserved when freely rotating, d J dt = 0. If ω is not aligned with J then J rotates away from its original direction violating the principle of conservation of angular momentum. If all three principal moments of inertia are equal then any axis in space is a principal axis. If only two three principal moments are equal then any axis in the plane of the two equal principal moments is a principal axis. P. J. Grandinetti (Chem. 4300) Rotational Motion Sep. 1, 2017 22 / 76
Intermediate Axis Theorem aka Tennis Racket Theorem aka the Dzhanibekov 1 effect Definition Intermediate axis theorem states that the rotation of an object with I a I b I c around its I a and I c principal axes is stable, while rotation around the I b axis (or intermediate axis) is not. Dzhanibekov effect Intermediate Axis Theorem 1 Soviet Cosmonaut P. J. Grandinetti (Chem. 4300) Rotational Motion Sep. 1, 2017 23 / 76
Moment of Inertia Tensor Example Calculate the moment of inertia tensor of a water molecule. Take the OH bond length as d = 0.957 Å, the H-O-H angle as Ω = 105, and the mass of oxygen and hydrogen as 16.0 u and 1.0 u. (1 u = 1.66053904 10 27 kg). We know M = 18 u, and X = 0.06473 Å, and Y = Z = 0. Simplified expressions for the moment of inertia components give I xx = m H z 2 1 + m Hz 2 2 = (1 u)(0.7592å)2 + (1 u)( 0.7592Å) 2 = 1.914 10 47 m 2 kg ( I yy = m H x 2 1 + ) ( z2 1 + mh x 2 2 + 2) z2 MX 2, I yy = (1 u) ( (0.5826 Å) 2 + (0.7592Å) 2) + (1 u) ( (0.5826 Å) 2 + ( 0.7592Å) 2) (18 u)(0.06473å) 2 = 2.916 10 47 m 2 kg, I zz = m H x 2 1 + m H x2 2 MX2, I zz = (1 u)(0.5826 Å) 2 + (1 u)(0.5826 Å) 2 (18 u)(0.06473å) 2, = 1.002 10 47 m 2 kg P. J. Grandinetti (Chem. 4300) Rotational Motion Sep. 1, 2017 24 / 76
Moment of Inertia Tensor Example Calculate the moment of inertia tensor of a water molecule. Take the OH bond length as d = 0.957 Å, the H-O-H angle as Ω = 105, and the mass of oxygen and hydrogen as 16.0 u and 1.0 u. And the rest I xy = I yx = 0 I yz = I zy = 0 I xz = I zx = m H x 1 z 1 m H x 2 z 2 I xz = I zx = (1 u)(0.5826 Å)(0.7592Å) (1 u)(0.5826 Å)( 0.7592Å) = 0 P. J. Grandinetti (Chem. 4300) Rotational Motion Sep. 1, 2017 25 / 76
Moment of Inertia Tensor Example Calculate the moment of inertia tensor of a water molecule. Take the OH bond length as d = 0.957 Å, the H-O-H angle as Ω = 105, and the mass of oxygen and hydrogen as 16.0 u and 1.0 u. In the coordinate system for this calculation we obtain a diagonal moment of inertia tensor given by I 10 47 m 2 kg = 1.914 0 0 0 2.916 0 0 0 1.002 But we re not done yet. We must follow the convention for labeling the axes in the principal axis system. In this example we identify (I yy = I c ) > (I xx = I b ) > (I zz = I a ). P. J. Grandinetti (Chem. 4300) Rotational Motion Sep. 1, 2017 26 / 76
Moment of Inertia Tensor Example Calculate the moment of inertia tensor of a water molecule. Final answer Water molecule has moment of inertia tensor in the PAS given by I PAS 10 47 m 2 kg = 1.002 0 0 0 1.914 0 0 0 2.916 105 b a When labeling the axes we need to maintain a right-handed coordinate system where a b = c. P. J. Grandinetti (Chem. 4300) Rotational Motion Sep. 1, 2017 27 / 76
Rotational energy of a rigid rotor In the PAS the kinetic energy only comes from rotational motion K = 1 2 N i=1 m i v 2 i Note that v i is the velocity of the ith particle relative to the center of mass. Since v i = ω r i we can write K = 1 2 N m i v i ( ω r i ) = 1 N 2 ω m i ( r i v i ) = 1 N 2 ω r i p i i=1 i=1 i=1 Recognizing the final sum as the total angular momentum we have K = 1 2 ω J P. J. Grandinetti (Chem. 4300) Rotational Motion Sep. 1, 2017 28 / 76
Rotational energy of a rigid rotor K = 1 2 ω J Since J = I ω we obtain our expression for the kinetic energy of the rotational motion of a rigid rotor, K = 1 ω I ω 2 In terms of the principal moments the kinetic energy reduces to a convenient form: K = 1 ( ω 2 2 a I a + ω 2 b I b + ω 2 c I ) c More convenient to calculate in body-fixed frame than space-fixed frame. P. J. Grandinetti (Chem. 4300) Rotational Motion Sep. 1, 2017 29 / 76
Rotational energy of a rigid rotor We can also show that K = 1 2 ( ω 2 a I a + ω2 b I b + ω2 c I ) c J J = ω I I ω = I 2 a ω2 a + I2 b ω2 b + I2 c ω2 c Since J J is also equal to J J = J 2 a + J2 b + J2 c we can identify J a = I a ω a J b = I b ω b and J c = I c ω c and rewrite the kinetic energy as K = J2 a 2I a + J2 b 2I b + J2 c 2I c P. J. Grandinetti (Chem. 4300) Rotational Motion Sep. 1, 2017 30 / 76
Classification of molecules Molecules grouped into 5 classes based on principal components. Name Diagonal values Examples Spherical I a = I b = I c = I CH 4 Prolate Symmetric I = I a < I b = I c = I CH 3 F Oblate Symmetric I = I a = I b < I c = I CHF 3 Asymmetric I a < I b < I c CH 2 Cl 2, CH 2 CHCl Linear I a = 0, I b = I c = I OCS, CO 2 All molecules with one three-fold or higher rotational symmetry axis are symmetric tops because the principal moments about two axes normal to the n-fold rotational symmetry axis (n 3) are equal. Molecules with two or more three-fold or higher rotational symmetry axes are spherical tops. P. J. Grandinetti (Chem. 4300) Rotational Motion Sep. 1, 2017 31 / 76
Spherical: I a = I b = I c = I Methane is an example of a molecule with this rotational symmetry. H H C H I a = I b = I c = 5.3 10 40 g cm 2 H Spherical molecules have more than one three-fold axis of symmetry and all three principal moments of inertia are equal. The principal moments of inertia of spherical tetrahedral and octahedral molecules: In spherical symmetry case kinetic energy takes the form K = J2 a + J2 b + J2 c 2I = J2 2I P. J. Grandinetti (Chem. 4300) Rotational Motion Sep. 1, 2017 32 / 76
Symmetric molecules: two principal moments are equal Symmetric molecules have a three-fold or higher axis of symmetry. Presence of this symmetry axis leads to two of the principal moments of inertia being equal. The third moment is associated with rotation about the axis with highest rotational symmetry, called the figure axis. In prolate and oblate symmetric tops the figure axis is associated with the smallest and largest principal moment of inertia, respectively. P. J. Grandinetti (Chem. 4300) Rotational Motion Sep. 1, 2017 33 / 76
Prolate Symmetric: I = I a < I b = I c = I think cigar-shaped Here we label the perpendicular moment of inertia component I for I b = I c and the parallel component I for I a. Fluoromethane is an example of a molecule with prolate symmetry. F C H H H a I a = 5.3 10 40 g cm 2 I b = I c = 32.9 10 40 g cm 2 In the prolate symmetric case the kinetic energy expression takes a form given by K = J2 a + J2 b + ( J2 c = J2 1 + 1 ) J 2 2I 2I 2I 2I 2I a P. J. Grandinetti (Chem. 4300) Rotational Motion Sep. 1, 2017 34 / 76
Oblate Symmetric: I = I a = I b < I c = I think frisbee-shaped Here we label the perpendicular moment of inertia component I for I a = I b and the parallel component I for I c. Trifluoromethane is an example of a molecule with oblate symmetry. F H C F c I a = I b = 81.1 10 40 g cm 2 I c = 149.1 10 40 g cm 2 F In the oblate symmetric case the kinetic energy expression takes a form given by K = J2 a + ( J2 b + J2 c = J2 1 + 1 ) J 2 2I 2I 2I 2I 2I c P. J. Grandinetti (Chem. 4300) Rotational Motion Sep. 1, 2017 35 / 76
Symmetric Rotor Generic Cases P. J. Grandinetti (Chem. 4300) Rotational Motion Sep. 1, 2017 36 / 76
Asymmetric: I a < I b < I c Molecules with I a < I b < I c have the lowest symmetry. Dichloromethane is an example of a molecule with asymmetric symmetry. H H C Cl Cl a b I a = 26.41 10 40 g cm 2 I b = 254.51 10 40 g cm 2 I c = 275.71 10 40 g cm 2 An asymmetric molecule can possess two-fold axes and planes of symmetry but cannot have any three-fold or higher symmetry axes. In the asymmetric case the kinetic energy expression cannot be simplified from the general form K = J2 a 2I a + J2 b 2I b + J2 c 2I c P. J. Grandinetti (Chem. 4300) Rotational Motion Sep. 1, 2017 37 / 76
Linear: I a = 0, I b = I c = I Linear molecules have an axis with infinite rotational symmetry (C ) since all the atoms lie on one axis. Carbon dioxide and OCS are two examples of a molecule with linear symmetry. O C S a O C O a I a = 0, I b = I c = I = 138.0 10 40 g cm 2 I a = 0, I b = I c = I = 71.3 10 40 g cm 2 P. J. Grandinetti (Chem. 4300) Rotational Motion Sep. 1, 2017 38 / 76
Linear Molecule: General Expressions P. J. Grandinetti (Chem. 4300) Rotational Motion Sep. 1, 2017 39 / 76
Linear Molecule: Kinetic Energy To obtain the kinetic energy expression in the linear case we need to reconsider the derivation. Since I a = 0 and I b = I c = I we modify K = 1 2 ( ω 2 a I a + ω2 b I b + ω2 c I ) c to be K = 1 2 ( ω 2 b + ) 1 ω2 c I = 2 ω2 I Similarly we see that J 2 = J J = I 2 b ω2 b + I2 c ω2 c = I2 ω 2 and we have K = J2 2I as the kinetic energy of the linear rigid rotor. Warning: looks the same as spherical case, but not the same. Linear case derived with I a = 0. P. J. Grandinetti (Chem. 4300) Rotational Motion Sep. 1, 2017 40 / 76
Diagonalizing a moment of inertia tensor P. J. Grandinetti (Chem. 4300) Rotational Motion Sep. 1, 2017 41 / 76
Diagonalizing a moment of inertia tensor So far, we have worked in body-fixed coordinate systems cleverly choosing axes aligned with molecular axes of highest rotational symmetry making our moment of inertia tensors diagonal. If we had chosen this coordinate system for the water molecule, z 105 x we would have calculated I 10 47 m 2 kg = 1.576 0 0.441 0 2.916 0 0.441 0 1.340 How do we find the principal components of I and the transformation back to a coordinate system where I is diagonal? P. J. Grandinetti (Chem. 4300) Rotational Motion Sep. 1, 2017 42 / 76
Diagonalizing a moment of inertia tensor Recall that when the angular velocity vector, ω, is aligned with one of the principal axes we get a result of the form J = I ω i = I i ω i ω i is vector aligned with a principal axis with principal component value I i. This can be rearranged to ( I Ii 1 ) ω i = 0 1 is identity matrix. If we further divide both sides by the magnitude ω i, we obtain ( I Ii 1 ) e i = 0 e i is a unit vector along the direction of ω i. This equation is a set of three simultaneous linear equations. P. J. Grandinetti (Chem. 4300) Rotational Motion Sep. 1, 2017 43 / 76
Diagonalizing a moment of inertia tensor ( I Ii 1 ) e i = 0 When the determinant det ( I I i 1 ) = I xx I i I xy I xz I yx I yy I i I yz I zx I zy I zz I i = 0, is expanded we get a 3rd-order polynomial equation in I i. Roots of this equation are the three principal components I a, I b, I c. By substituting each principal component back into the top equation we can solve for a corresponding e i whose direction tells us the direction of that principal component in the original coordinate system where I was non-diagonal. P. J. Grandinetti (Chem. 4300) Rotational Motion Sep. 1, 2017 44 / 76
Diagonalizing a moment of inertia tensor Example Given the inertia tensor calculated in an arbitrary body-fixed frame shown below 1.576 0 0.441 I 10 47 m 2 kg = 0 2.916 0 0.441 0 1.340 determine (a) its principal components of the inertia tensor and (b) the orientation of the PAS. (a) For the general form of this tensor we can create the determinant, I xx I i 0 I xz 0 I yy I i 0 = 0 I xz 0 I zz I i and expand to obtain the polynomial (I 2 i (I xx + I zz )I i + I xx I zz I 2 xz ) (I yy I i ) = 0. P. J. Grandinetti (Chem. 4300) Rotational Motion Sep. 1, 2017 45 / 76
Diagonalizing a moment of inertia tensor Finding the principal components (I 2 i (I xx + I zz )I i + I xx I zz I 2 xz )(I yy I i ) = 0. As expected, one root is just I 1 = I yy = 2.916 10 47 m 2 kg, leaving us to find the other two roots from I 2 i (I xx + I zz )I i + I xx I zz I 2 xz = 0 Applying the quadratic formula we find two solutions I 2 = 1 [ ] I 2 xx + I zz + (I xx I zz ) 2 + 4Ixz 2 = 1.914 10 47 m 2 kg, and I 3 = 1 2 [ ] I xx + I zz (I xx I zz ) 2 + 4Ixz 2 = 1.002 10 47 m 2 kg. Following convention we assign I a = I 3, < I b = I 2, < I c = I 1. P. J. Grandinetti (Chem. 4300) Rotational Motion Sep. 1, 2017 46 / 76
Diagonalizing a moment of inertia tensor Finding the principal components Following convention we assign I a = I 3, < I b = I 2, < I c = I 1, and obtain I 10 47 m 2 kg = 1.002 0 0 0 1.914 0 0 0 2.916 P. J. Grandinetti (Chem. 4300) Rotational Motion Sep. 1, 2017 47 / 76
Diagonalizing a moment of inertia tensor Finding the principal axis system (b) To find the orientation of each principal axis we substitute the principal component values back into ( I Ii 1 ) e i = 0 Simultaneous equations obtained in this manner are usually redundant and only relationships among the x i, y i, and z i can be obtained, rather than unique values for them. Normalization of unit vectors gives additional constraint to pin down the values for x i, y i, and z i. P. J. Grandinetti (Chem. 4300) Rotational Motion Sep. 1, 2017 48 / 76
Diagonalizing a moment of inertia tensor Finding the principal axis system Starting with I 1, also defined as I c, we obtain = 1.576 2.916 0 0.441 0 2.916 2.916 0 0.441 0 1.340 2.916 1.34 0 0.441 0 0 0 0.441 0 1.576 x 1 y 1 z 1 = 0 x 1 y 1 z 1 Here y 1 can have any value since it is multiplied by zero in all three cases. Setting x 1 = z 1 = 0 gives a solution, leading to the normalized vector of e c = e y P. J. Grandinetti (Chem. 4300) Rotational Motion Sep. 1, 2017 49 / 76
Diagonalizing a moment of inertia tensor Finding the principal axis system Switching to I 2, also defined as I b, we obtain Here we set y 2 = 0 and have 0.338 0 0.441 0 1.002 0 0.441 0 0.574 x 2 y 2 z 2 = 0 0.338x 2 + 0.441z 2 = 0 0.441x 2 0.574z 2 = 0 These two equations are equivalent. Setting x 2 = 1 gives z 2 = 0.76644. Using x 2 2 + z2 = 1.256 we obtain the normalized vector 2 e b = 1 ( ) ex + 0.766 e 1.256 z = 0.7937 ex + 0.608 e z P. J. Grandinetti (Chem. 4300) Rotational Motion Sep. 1, 2017 50 / 76
Diagonalizing a moment of inertia tensor Finding the principal axis system Finally for I 3, also defined as I a, we obtain 0.574 0 0.441 0 1.914 0 0.441 0 0.338 Once again we set y 3 = 0 and have x 3 y 3 z 3 = 0 0.574x 3 + 0.441z 3 = 0 0.441x 3 + 0.338z 3 = 0 Again, both equations are equivalent. Setting x 3 = 1 gives z 3 = 1.30159. Using x 2 2 + z2 = 1.6414 we obtain the normalized vector 2 e a = 1 ( ) ex 1.30159 e 1.6414 z = 0.609 ex 0.793 e z P. J. Grandinetti (Chem. 4300) Rotational Motion Sep. 1, 2017 51 / 76
Diagonalizing a moment of inertia tensor Finding the principal axis system The directions of the e a and e b vectors are draw in the figure below. z 105 x The direction of the e c vector is directed up and perpendicular to the plane of the page. All this is consistent with our earlier moment of inertia calculation on water. P. J. Grandinetti (Chem. 4300) Rotational Motion Sep. 1, 2017 52 / 76
Rotating frame transformation P. J. Grandinetti (Chem. 4300) Rotational Motion Sep. 1, 2017 53 / 76
Rotating frame transformation When rigid body rotates the body-fixed frame becomes a non-inertial frame. Newton s laws suggest that they would no longer apply in such a non-inertial (i.e., accelerating or rotating) frame. There are approaches to obtain the correct equations of motion in a rotating frame which can be related back to the some inertial space-fixed frame. P. J. Grandinetti (Chem. 4300) Rotational Motion Sep. 1, 2017 54 / 76
Rotating frame transformation Arbitrary Q in the inertial space-fixed frame Consider some vector Q. In inertial space-fixed frame it can be decomposed into its projections onto unit vectors Q = Q x e x + Q y e y + Q z e z Rate of change of Q in inertial space-fixed frame is ( ) d Q = dq x dt dt e x + dq y dt e y + dq z dt e z space P. J. Grandinetti (Chem. 4300) Rotational Motion Sep. 1, 2017 55 / 76
Rotating frame transformation Arbitrary Q in the non-inertial body-fixed frame Same vector can also be can be decomposed into its projections onto unit vectors in the rotating body-fixed frame of PAS, Q = Q a e a + Q b e b + Q c e c Rate of change of Q in rotating body-fixed frame is ( ) d Q = dq a dt dt e a + dq b dt e b + dq c dt e c How are body ( ( ) d Q dt )space and d Q dt related? body P. J. Grandinetti (Chem. 4300) Rotational Motion Sep. 1, 2017 56 / 76
Rotating frame transformation Calculate rate of change of Q in space-fixed frame in terms of its decomposition in body-fixed frame. ( ) d Q = dq a dt dt e a + dq b dt e b + dq c dt e d e c +Q a d e a + Q b d e dt b + Q c dt c dt space ( ) d Q dt body We identify first 3 terms on right as ( ) d Q dt body P. J. Grandinetti (Chem. 4300) Rotational Motion Sep. 1, 2017 57 / 76
Rotating frame transformation Calculate rate of change of Q in space-fixed frame in terms of its decomposition in body-fixed frame. ( ) d Q = dq a dt dt e a + dq b dt e b + dq c dt e d e c +Q a d e a + Q b d e dt b + Q c dt c dt space ( ) d Q dt body Unit vectors in body-fixed frame, e a, e b, and e c, are time dependent in space-fixed frame from rotation, so last 3 derivatives are non-zero. Linear velocity of vector as it rotates with angular velocity about a given direction is given by cross product ( ) d ei v i = = Ω e dt i space P. J. Grandinetti (Chem. 4300) Rotational Motion Sep. 1, 2017 58 / 76
Rotating frame transformation Last term can be rewritten Q a d e a dt + Q b d e b dt + Q c d e c dt = Q a Ω e a + Q b Ω e b + Q c Ω e c = Ω Q Leaving us with ( ) ( ) d Q d Q = + Ω Q dt dt space body This relates rate of change of Q in space-fixed frame to its rate of change in frame rotating with Ω. Generally, operator relating time derivatives between inertial and rotating frames is ( ) ( ) d d = + Ω dt dt space body P. J. Grandinetti (Chem. 4300) Rotational Motion Sep. 1, 2017 59 / 76
Newton s 2nd law in the rotating frame P. J. Grandinetti (Chem. 4300) Rotational Motion Sep. 1, 2017 60 / 76
Newton s 2nd law in the rotating frame To obtain equations of motion in rotating frame we go back to space-fixed inertial frame where Newton s 2nd law applies, F = m a = m d2 r dt 2 Take rate of change of r or velocity in space-fixed frame as ( ) ( ) d r d r = + Ω r dt dt space body Calculate acceleration in space-fixed frame as ( ) [ (d r d 2 r = d ) dt 2 space dt dt body + Ω r ] P. J. Grandinetti (Chem. 4300) Rotational Motion Sep. 1, 2017 61 / 76
Newton s 2nd law in the rotating frame ( ) d 2 r = d dt 2 space dt [ (d r ) dt body + Ω r Applying derivative operator to expression above one gives ( ) d 2 r = dt 2 space ( ) d 2 r + Ω (Ω r) + 2Ω dt 2 body ] ( ) d r + dω r dt body dt Combining this with Newton s 2nd law, F space = m a space, gives F space = m a space = m a body + m Ω (Ω r body ) + 2mΩ v body + m dω dt r body r body, v body, and a body are position, velocity, and acceleration of body as measured relative to rotating frame. P. J. Grandinetti (Chem. 4300) Rotational Motion Sep. 1, 2017 62 / 76
Newton s 2nd law in the rotating frame In space-fixed inertial frame there is only F space acting on body. In body-fixed rotating frame there appear to be additional forces at play. Equations of motion as seen in rotating frame are F body = m a body = F space m Ω (Ω r body ) 2mΩ v body m dω r dt body centrifugal Coriolis azimuthal Last 3 terms on right are fictitious forces that appear in rotating frames. Centrifugal force is directed away from axis of rotation and acts on all objects when viewed in rotating frame. Coriolis force only acts on objects in motion relative to rotating frame. In clockwise rotating frame, this force acts to left of object s motion; in counter-clockwise frame it acts to right. Azimuthal force acts on all objects when there is a change of magnitude or direction of rotating frame angular velocity. It is the sudden push experienced on a merry-go-round when rotation starts or stops. P. J. Grandinetti (Chem. 4300) Rotational Motion Sep. 1, 2017 63 / 76
Centrifugal and Coriolis force web link: Centrifugal and Coriolis force web link: Coriolis effect web link: Global Wind Patterns for Kids P. J. Grandinetti (Chem. 4300) Rotational Motion Sep. 1, 2017 64 / 76
Euler s equations of motion Relationship between rate of change of J in space- and body-fixed frames is ( ) d J = dt space ( ) d J + ω J dt body In inertial space-fixed frame ( ) d J = τ dt space Combining top 2 equations gives ( ) d J + ω J = τ dt body This is Euler s equation of motion for a rigid body. P. J. Grandinetti (Chem. 4300) Rotational Motion Sep. 1, 2017 65 / 76
Euler s equations of motion ( ) d J + ω J = τ dt body Working in body-fixed frame we relate J and ω by Substituting this into Eq. (66) J = ω a I a e a + ω b I b e b + ω c I c e c ( ) dωa I a dt ( ) dωb I b dt I c ( dωc dt ) body body body (I b I c ) ω b ω c = τ a (I c I a ) ω c ω a = τ b (I a I b ) ω a ω b = τ c What happens if there is no external torques, that is, free rotation? P. J. Grandinetti (Chem. 4300) Rotational Motion Sep. 1, 2017 66 / 76
Euler s equations of motion Free rotation When there is no external torque, that is, τ = 0, Euler s equations describe free rotation of rigid body ( ) dωa = (I b I c ) ω dt I b ω c a ( ) dωb dt ( ) dωc dt body body body = (I c I a ) I b = (I a I b ) I c ω c ω a ω a ω b What happens when rotating about one of the principal axes? P. J. Grandinetti (Chem. 4300) Rotational Motion Sep. 1, 2017 67 / 76
Euler s equations of motion Free rotation about a principal axis If rigid body with I a I b I c is initially rotating about one of the principal axes, say e c, then ω a = ω b = 0 and Euler s equations become ( ) ( ) ( ) dωa dωb dωc = = = 0 dt dt dt body body body Rate of change of all 3 components of ω are zero and magnitude and orientation of ω remains constant in rigid body frame. Knowing J = I c ω c e c we see that magnitude and orientation of J also remains constant in rigid body frame. For freely rotating rigid body the total angular momentum is conserved so J must also remains constant in space-fixed inertial frame. P. J. Grandinetti (Chem. 4300) Rotational Motion Sep. 1, 2017 68 / 76
Euler s equations of motion Free rotation about an arbitrary axis If body with I a I b I c is initially rotating about an arbitrary axis where all 3 components of ω are non-zero in body-fixed frame, then all 3 components of ω will be time dependent in body-fixed frame. For the same reason all 3 components of J will be time dependent in body-fixed frame. In space-fixed frame, however, J must remains constant. But same is not true for ω in space-fixed frame. P. J. Grandinetti (Chem. 4300) Rotational Motion Sep. 1, 2017 69 / 76
Euler s equations of motion Free rotation of oblate symmetric rigid body about an arbitrary axis ( ) dωa dt ( ) dωb dt ( ) dωc dt body body body = (I b I c ) I a = (I c I a ) I b = (I a I b ) I c ω b ω c ω c ω a ω a ω b Consider case of oblate symmetric body, where I a = I b = I < I c = I. P. J. Grandinetti (Chem. 4300) Rotational Motion Sep. 1, 2017 70 / 76
Euler s equations of motion Free rotation of oblate symmetric rigid body about an arbitrary axis Case of oblate symmetric body, where I a = I b = I < I c = I. Euler s equations simplify to ω a = (I I ) ω I c ω b ω b = (I I ) ω I c ω a ω c = 0 Switch to Newton s dot notation for the time derivatives. As ω c remains constant we further simplify Euler s equations to ω a = Ω 0 ω b ω b = Ω 0 ω a where we have defined a new constant angular frequency, Ω 0, as Ω 0 = I I ω I c P. J. Grandinetti (Chem. 4300) Rotational Motion Sep. 1, 2017 71 / 76
Euler s equations of motion Free rotation of oblate symmetric rigid body about an arbitrary axis ω a = Ω 0 ω b ω b = Ω 0 ω a To solve these coupled differential equations we take time derivative again to obtain ω a = Ω 0 ω b and ω b = Ω 0 ω a Substituting first-derivative expressions into above expression we obtain 2 uncoupled second-order homogeneous differential equations: ω a + Ω 2 0 ω a = 0 and ω b + Ω2 0 ω b = 0 For each equation we can propose a solution of the form ω i (t) = A i cos k i t + B i sin k i t P. J. Grandinetti (Chem. 4300) Rotational Motion Sep. 1, 2017 72 / 76
Euler s equations of motion Free rotation of oblate symmetric rigid body about an arbitrary axis Substituting ω i (t) = A i cos k i t + B i sin k i t into gives ω a + Ω 2 0 ω a = 0 and ω b + Ω 2 0 ω b = 0 A i (Ω 2 0 k2 i ) cos k it + B i (Ω 2 0 k2 i ) sin k it = 0 To make this equation true for all values of t we set k i = Ω 0 and obtain ω i (t) = A i cos Ω 0 t + B i sin Ω 0 t Choosing an initial condition where ω a (t = 0) = ω and ω b (t = 0) = 0 leads to ω a (t) = ω cos Ω 0 t and ω b (t) = ω sin Ω 0 t Projection of ω onto a-b plane has constant length, ω, and rotates in a-b plane with frequency of Ω 0. P. J. Grandinetti (Chem. 4300) Rotational Motion Sep. 1, 2017 73 / 76
Euler s equations of motion Free rotation of oblate symmetric rigid body about an arbitrary axis Since ω c is also constant then total length of ω is constant, ω = ω 2 a + ω2 b + ω2 c Define angle ω makes with e c with tan α = ω ω c and get ω(t) = ω sin α cos Ω 0 t e a ω sin α sin Ω 0 t e b + ω cos α e c In body-fixed frame ω precesses e c with angle of α at frequency Ω 0. P. J. Grandinetti (Chem. 4300) Rotational Motion Sep. 1, 2017 74 / 76
Euler s equations of motion Free rotation of oblate symmetric rigid body about an arbitrary axis The motion of J in body-fixed frame is given by J = I ω = ωi sin α cos Ω 0 t e a ωi sin α sin Ω 0 t e b + ωi cos α e c J has constant length and precesses around e c at frequency Ω 0. Angle between J and e c is tan θ = J a J c = I I tan α body-fixed frame In oblate case I < I and θ is smaller than α As both ω and J rotate around e c all 3 vectors remain in same plane. P. J. Grandinetti (Chem. 4300) Rotational Motion Sep. 1, 2017 75 / 76
Euler s equations of motion Free rotation of oblate symmetric rigid body about an arbitrary axis In space-fixed frame we know that J remains unchanged, thus ω and e c must precess about J. body-fixed frame space-fixed frame P. J. Grandinetti (Chem. 4300) Rotational Motion Sep. 1, 2017 76 / 76