Oh, it’s your standard “boy meets girl, boy loses girl, boy invents a new deposition technique for ultra-thin film semiconductors, boy gets girl back” story!

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 1st term in (1) depends only on (X,Y,Z) &

the 2nd 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)MV2 + (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)MV2 + (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 1st term in (2) depends only on (X,Y,Z), 2nd term depends on

(,θ,ψ): T = (½)MV2 + 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 2nd 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(riω)] (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

Ijk ∫Vρ(r)[r2δjk - xjxk]dV

Or: Ixx = ∫Vρ(r)[r2 - x2]dV, Iyy = ∫Vρ(r)[r2 - y2]dV

Izz = ∫Vρ(r)[r2 - z2]dV

Ixy = Iyx = ∫Vρ(r)[r2 - xy]dV, Ixz = Izx = ∫Vρ(r)[r2 - xz]dV

Iyz = Izy = ∫Vρ(r)[r2 - yz]dV

• 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 2nd rank. Now, a general discussion.• In Cartesian 3d-space, DEFINE a tensor of the Nth

rank T a quantity with 3N 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 0th rank: 1 element scalar. Invariant under an

orthogonal transformation A. • Tensor of 1st rank: 3 elements ordinary vector. Transforms under

orthogonal transformation A as:

T´i(x´) = aij Tj(x)

• Tensor of 2nd rank: 9 elements. Transforms under orthogonal transformation A as:

T´ij(x´) = aik ajl Tkl(x)

• Rigorously, we have a distinction between the 2nd 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 = 2nd 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 = Iω – 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)mivivi

• 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)ωL

• Using L = Iω this becomes:

T = (½)ωIω – 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)ωIω (1)• n unit vector along rotation axis, so that ω = ωn

Can write (1) as:

T = (½)ω2nIn (½)Iω2 (2)

I nIn Moment of Inertia (about rotation axis)

• Using forms of I from before, can write:

I = nIn = mi[(ri)2 - (rin)2] (3) • 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!

T = (½)ωIω (1)• n unit vector along rotation axis:

T = (½)ω2nIn (½)Iω2 (2)

I nIn 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 nIn 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 3rd 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 +

MR2sin2θ

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)ωIω (1)• Can rewrite in terms of tensor elements as

(summation convention):

T = (½)ωjIjkωk (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)[r2δjk - xjxk]dV