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Ch 5: Rigid Body Eqtns of Motion • Ch 4: Introduced all kinematical tools to analyze rigid body motion. • Euler angles: , θ, ψ: A useful set of 3 generalized coordinates for describing the body orientation. • Orthogonal transformation A & associated matrix properties & algebra: A powerful, elegant math formalism for the description of rigid body motion. • Ch 5: Uses these tools & techniques to get general eqtns of motion for rigid body in convenient form The Euler Dynamical Equations. • Also: Applications of Euler Equations.

Sect 5. 1: Angular Momentum & KE • Chasles’ Theorem: The most general displacement of a rigid body is a rotation about some axis plus a translation. It ought to be possible to divide the problem into 2 separate problems: 1. Translation, 2. Rotation. – Obviously if one point is fixed, this separation is trivially true. – Also possible for general problem in some cases. • Ch. 4 discussion: 6 generalized coordinates are needed to describe rigid body motion: 3 Cartesian coordinates of a fixed point in the body + 3 Euler angles describing the body rotation about an axis through the fixed point. In general: Equations of motion for all 6 coordinates do not separate into 3 eqtns for translation + 3 for rotation!!

• Most common special case where fixed point in rigid body Center of Mass (CM): • Can use some Ch. 1 results for angular momentum & kinetic energy which simplify the problem & separate equations of motion for translational & rotational coordinates into 2 sets of 3, so translation & rotation can be treated separately!

• O = arbitrary origin (fixed axes), R = CM position with respect to O. V = CM velocity, M = CM mass. ri´ = position of particle i with respect to CM (body axes) pi´ = mivi´ = momentum of same particle in this system. • A Ch. 1 result for the angular momentum of a system of particles: L = R MV + ∑i[ri´ pi´] (1) The total angular momentum about point O = the angular momentum of the motion of the CM about O + the angular momentum of the motion of particles about the CM

L = R MV + ∑i[ri´ pi´] (1) Total angular momentum about O = angular momentum of the CM about O + angular momentum of particles about the CM • (1) In general, L depends on the origin O, through vector R. Only if the CM is at rest with respect to O, will first term in (1) vanish. Then & only then will L be indep of the point of reference. Then & only then L = angular momentum about the CM. If we choose the origin of the body axes to be the CM, then clearly, as far as L is concerned, we can treat the translation & rotation of the body separately. Then, the 1 st term in (1) depends only on (X, Y, Z) & the 2 nd term depends on ( , θ, ψ) : L = R MV + L´( , θ, ψ)

• O = arbitrary origin (fixed axes), R = CM position with respect to O. V = CM velocity, M = CM mass. ri´ = position of particle i with respect to CM (body axes) vi´ = velocity of same particle in this system. • Ch. 1 result for KE of a system of particles: T = (½)MV 2 + (½)∑imi(vi´)2 (2) The total Kinetic Energy of a many particle system is equal to the Kinetic Energy of the CM plus the Kinetic Energy of motion about the CM.

T = (½)MV 2 + (½)∑i mi(vi´)2 (2) Total KE with respect to O = KE of the CM+ KE of motion about the CM. • (2) In general, T depends on origin O, through vector V. Only if the CM is at rest with respect to O, will first term in (2) vanish. Then & only then will T be indep of point of reference. Then & only then T = KE about the CM. If choose origin of body axes to be the CM, then clearly, as far as T is concerned, we can treat the translation of body & rotation of body the separately. Then, the 1 st term in (2) depends only on (X, Y, Z), 2 nd term depends on ( , θ, ψ): T = (½)MV 2 + T´( , θ, ψ)

• We can make this separation for the angular momentum L & the KE T. To do dynamics, using say, the Lagrangian method, we would like to make this same kind of separation in Lagrangian L = T -V Obviously, we need to be able to separate the PE V in the same way. • We don’t have a theorem which does this in general for V, like we do for L & T! However, from experience, in a large number of problems (special cases) of practical interest, such a separation is possible. – PE in a uniform gravitational field, PE of a magnetic dipole in uniform magnetic field, others, . . . – In such cases, we can write: L = L (X, Y, Z) + L´( , θ, ψ)

• In cases where: L = L (X, Y, Z) + L´( , θ, ψ) Lagrange’s Eqtns of motion for translation (X, Y, Z) obviously separate from Lagrange’s Eqtns of motion for rotation ( , θ, ψ) & we can treat the translation & rotation as independent. • Now, work on expressions for angular momentum L & KE T for motion about some fixed point in a rigid body. Work on 2 nd terms in previous expressions for these. Called L´ & T´ in Eqtns (1) & (2) above. Drop prime notation. • Make repeated use of the Ch. 4 result relating the time derivatives in space & body axes (angular velocity ω): (d/dt)s = (d/dt)r + ω (I)

• Goldstein proves that angular velocity ω is indep of choice of the origin of body axes. See pages 185 -186 for details. Intuitively obvious by definition of rigid body? • Angular momentum L of rigid body about a fixed point in the body. Dropping the primes of before on positions, momenta, & velocities: L = ∑i[ri pi], pi = mivi • Follow text & write using summation convention: L = mi[ri vi] (1) ri = fixed with respect to body axes (definition of rigid!) The only contribution to vi is from the body rotation. Use (I) vi = (dri/dt)s = (dri/dt)r + ω ri = 0 + ω ri (1) becomes: L = mi [ri (ω ri)] (2)

• Use the vector identity for double cross product: L = mi[ω(ri)2 - ri(ri ω)] (3) • Look at the x, y, & z components: Lx = ωxmi[(ri)2 - (xi)2] - ωymixiyi - ωzmixizi Ly = ωymi[(ri)2 - (yi)2] - ωxmiyixi - ωzmiyizi Lx = ωzmi[(ri)2 - (zi)2] - ωymiziyi - ωxmizixi (4) • (4) Each component of L = Linear combination of components of angular velocity ω. Or: The angular momentum vector is related to the angular velocity vector by a linear transformation! Define inertia tensor Ijk (summation convention!) : Lj Ijkωk (5) Ijk also called the moment of inertia coefficients

• Inertia tensor Ijk (summation convention!) : Lj Ijkωk Or: Lx = Ixxωx + Ixyωy + Ixzωz Ly = Iyxωx + Iyyωy + Iyzωz Lz = Izxωx + Izyωy + Izzωz • Diagonal elements Moment of inertia coefficients: Ixx = mi[(ri)2 - (xi)2], Iyy = mi[(ri)2 - (yi)2], Izz = mi[(ri)2 - (zi)2] • Off-diagonal elements Products of inertia: Ixy = Iyx = - mixiyi , Ixz = Izx = - mixizi , Iyz = Izy = - miyizi

• Inertia tensor Ijk (summation convention, i = particle label, j, k = Cartesian x, y, z = 1, 2, 3 labels!) : Lj Ijkωk • Compact form: Ijk mi[(ri)2δjk - (ri)j(ri)k] • All of this is in notation for discrete systems of particles. • Continuous bodies: Sum over particles integral over volume V. Define: ρ(r) Mass density at position r Or: Ijk ∫Vρ(r)[r 2δjk - xjxk]d. V Ixx = ∫Vρ(r)[r 2 - x 2]d. V, Iyy = ∫Vρ(r)[r 2 - y 2]d. V Izz = ∫Vρ(r)[r 2 - z 2]d. V Ixy = Iyx = ∫Vρ(r)[r 2 - xy]d. V, Ixz = Izx = ∫Vρ(r)[r 2 - xz]d. V Iyz = Izy = ∫Vρ(r)[r 2 - yz]d. V

• Recall: The coordinate system is the BODY axis system (the primes from the last chapter are dropped). • Since the body is rigid, all elements Ijk are constants in time. Clearly, they DO depend on the origin of coordinates (different if taken about the CM or elsewhere!). • We can summarize Lj = Ijkωk in an even more compact form: L Iω where I Inertia Matrix • This is in the form of the orthogonal transformation of Ch. 4. Clearly we need to take the interpretation that I acts on the vector ω to produce L, rather than the interpretation of I acting on the coordinate system.

Sect 5. 2: Tensors • More pure math discussion. Brief! Hopefully, a review! • I Tensor of 2 nd rank. Now, a general discussion. • In Cartesian 3 d-space, DEFINE a tensor of the Nth rank T a quantity with 3 N elements or components Tijk. . (N indices) which transform under an orthogonal transformation of coords A (from Ch. 4) as (summation convention!): T´ijk. . (x´) = ail ajm akn … Tlmn. . (x) (N factors of matrix elements aij of A) • Following the discussion of pseudo-vectors in Ch. 4, we also need to define a pseudo-tensor, which transforms as a tensor except under inversion. See footnote, p 189.

• Tensor of 0 th rank: 1 element scalar. Invariant under an orthogonal transformation A. • Tensor of 1 st rank: 3 elements ordinary vector. Transforms under orthogonal transformation A as: T´i(x´) = aij Tj(x) • Tensor of 2 nd rank: 9 elements. Transforms under orthogonal transformation A as: T´ij(x´) = aik ajl Tkl(x) • Rigorously, we have a distinction between the 2 nd rank tensor T & the square matrix formed from its elements: Tensor: Defined only by its transform properties under orthogonal transforms. Matrix: A representation of a tensor in a particular coordinate system. Every tensor equation In some coordinate system, a corresponding matrix equation. Further discussion, p. 189. • Various math properties of tensors, p. 190 -191: Read on your own!

Sect 5. 3: Inertia Tensor & Moment of Inertia • Inertia tensor I = 2 nd rank tensor, transforms as such under orthogonal transformation A. • The angular momentum of a rigid body is: L = Iω. Should more properly be written L = I ω – Where matrix or tensor multiplication. • Now, we’ll derive a form for the KE T for a rigid body in terms of I and ω. • Start with KE of motion about a point in form (summation convention, dropped prime from earlier) : T (½)mi(vi)2

T (½)mi(vi)2 = (½)mivi vi • Rigid body The only contribution to vi is from body rotation. As before use vi = (dri/dt)s = (dri/dt)r + ω ri = 0 + ω ri Convenient to use with only one of the vi in the dot product: T = (½)mivi (ω ri ) • Permuting vectors in triple product: T = (½)ω [mi(ri vi )] (½)ω L • Using L = I ω this becomes: T = (½)ω I ω – Note: Order matters & product with ω when is to left of I & to right of I is clearly different. T is clearly a scalar, as it should be! – Often, I may write it as simply T = (½)ωIω (leaving out )

T = (½)ω I ω (1) • n unit vector along rotation axis, so that ω = ωn Can write (1) as: T = (½)ω2 n I n (½)Iω2 (2) I n I n Moment of Inertia (about rotation axis) • Using forms of I from before, can write: I = n I n = mi[(ri)2 - (ri n)2] • Text proves (p. 192, figure) that definition, (3), of I is consistent with elementary definition as sum over all particles of products of particle masses times square of distances from rotation axis. I = mi[(ri)2] . Read! (3)

T = (½)ω I ω (1) • n unit vector along rotation axis: T = (½)ω2 n I n (½)Iω2 (2) I n I n Moment of Inertia (about rotation axis) • Clearly, moment of inertia I depends on the direction n of the rotation axis. Also, for a rigid body in motion, that direction & thus the direction of ω can be time dependent. In general, I = I(t) • In the (important!) special case where the body is constrained to rotate about a fixed axis, clearly then, I = constant. – When this is the case, T in the form of (1) can be used in the Lagrange formalism if we can write ω as the time derivative of some angle.

I n I n Moment of Inertia (about rotation axis) • Inertia tensor I & moment of inertia I depend on the choice of the body axes origin. However, the moment of inertia about a given axis is simply related to the moment of inertia about a parallel axis through the CM ( Parallel Axis Theorem). • Figure: 2 || rotation axes a & b, direction n. Axis b passes through CM. R, ri = positions of CM & particle i with respect to origin O. ri´ = position of particle i with respect to CM. ri = R + ri´. Moment of inertia about axis a: Ia = ∑imi(ri n)2 = ∑imi[(R + ri´) n]2 = ∑imi(R n)2 + ∑imi(ri´ n)2 + 2∑imi(R n) (ri´ n)

• Moment of inertia about axis a: Ia = ∑imi(R n)2 + ∑imi(ri´ n)2 + 2 ∑imi(R n) (ri´ n) • Moment of inertia about axis b: Ib = ∑imi(ri´ n)2 • Note that: ∑ imi(R n)2 M(R n)2 Ia = Ib + M(R n)2 + 2∑imi(R n) (ri´ n) • Rewrite 3 rd term as: 2(R n) (∑imiri´ n) Note that, by definition of CM, ∑imiri´ = 0 2∑imi(R n) (ri´ n) = 0 Ia = Ib + M(R n)2 For θ = angle between R & rotation axis n, distance between axes a & b = |R n| = Rsinθ = r. Ia = Ib + MR 2 sin 2θ Or: Ia = Ib + M(r )2 Parallel Axis Theorem

Ia = Ib + M(r )2 Parallel Axis Theorem • In words: The moment of inertia about an arbitrary axis is equal to the moment of inertia about a parallel axis passing through the center of mass plus the moment of inertia of the body about the arbitrary axis, taken as if all of the mass M of the body were at the center of mass.

• Summary: Rotational KE in terms of inertia tensor: T = (½)ω I ω (1) • Can rewrite in terms of tensor elements as (summation convention): T = (½)ωj. Ijkωk (2) • Again (summation convention, i = particle label, j, k = Cartesian x, y, z = 1, 2, 3 labels!) : Ijk mi[(ri)2δjk - (ri)j(ri)k] • Continuous bodies: Sum over particles integral over volume V. Define: ρ(r) Mass density at position r Ijk ∫Vρ(r)[r 2δjk - xjxk]d. V