Sect 5.4: Eigenvalues of I & Principal Axis Transformation

• Definition of inertia tensor (continuous body):

Ijk ∫Vρ(r)[r2δjk - xjxk]dV

– Clearly, Ijk is symmetric: Ijk = Ikj

Out of the 9 elements Ijk only 6 are independent.

Ijk depend on the location of the origin of the body axes set &

on the orientation of the body axes with respect to the body.• Symmetry There exists a set of coordinates

Principal Axes in which the tensor Ijk is diagonal with 3 Principal Values I1, I2, I3. In this system, the angular momentum: L = Iω

becomes:

L1 = I1ω1, L2 = I2ω2, L3 = I3ω3.

the KE T = (½)ωIω becomes:

T = (½)I1(ω1)2 + (½)I2(ω2)2 + (½)I3(ω3)2

• Get principal axes set and principal values of tensor I by diagonalizing I. That is, by finding eigenvalues (principal values I1, I2, I3) & eigenvectors (defining principal axes).

• From Ch. 4, do this by solving determinant eigenvalue problem or by a similarity transformation on I.

• Given inertia matrix I, principal axes & principal values can be found by finding a suitable rotation matrix (finding a proper set of Euler angles

,θ,ψ) A and performing the similarity transformation: ID = AIA-1 = AIÃ such that ID is a diagonal matrix. That is, want ID to have form:

I1 0 0 I1, I2, I3 eigenvalues of I

ID = 0 I2 0 I1, I2, I3 principal Components of I

0 0 I3

Directions x, y, z defined by the eigenvectors principal axes of I

• Once I is diagonalized, Principal Components (I1, I2, I3) & principal Axes (x, y, z) are known.– Then, can get I relative to any other axes set by another

similarity transformation: I = AI(A)-1 = A IÖ Also Parallel Axis Theorem might be used to shift rotation

axis.

• The matrix algebra method to diagonalize I

Solve secular eqtn: (I - I1)R = 0 (1)

The Eigenvalue ProblemValues of I which satisfy (1) Eigenvalues (I1, I2, I3)

Vectors R which satisfy (1) Eigenvectors x, y, z

(I - I1)R = 0 (1) Ixx - I Ixy Ixz

Ixy Iyy - I Iyz = 0 (2)

Ixz Iyz Izz - I

– Have used Ijk = Ikj .

• 3 solutions to (2): Eigenvalues (I1, I2, I3)

• Put these into (1) & get: Eigenvectors

Ri = x, y, z = Principal Axes.• Often, can know principal axes by the object symmetry.

• Some Properties of the Eigenvalues (I1, I2, I3)

1. Can’t be < 0! Ii > 0, (i = 1,2,3)

2. If one Ii = 0, the body is vanishingly small in the

direction given by the corresponding eigenvector.

• Principal axes from geometry:

• Moment of inertia about the rotation axis n: I nIn

• Body Cartesian axes unit vectors: i,j,k Define:

n αi +βj + γk Work out the details of I:

I = Ixxα2 +Iyyβ2 +Izzγ 2 +2Ixyαβ + 2Iyzβγ + 2Ixzαγ

Define the vector ρ n/(I)½ . |ρ| = (I)-½

Write: ρ ρ1i + ρ2j + ρ3k

I = Ixx(ρ1)2 + Iyy(ρ2)2 + Izz(ρ3)2

+ 2Ixyρ1ρ2 + 2Iyzρ2ρ3+ 2Ixzρ1ρ3 (A)

(A) I = I(ρ1,ρ2,ρ3)

Equation of a surface in “ρ” space Inertial Ellipsoid

ρ = n(I)-½ = ρ1i + ρ2j + ρ3k

I = Ixx(ρ1)2 +Iyy(ρ2)2 +Izz(ρ3)2 + 2Ixyρ1 ρ2 + 2Iyzρ2ρ3 + 2Ixzρ1ρ3 (A) Equation of a surface in “ρ” space Inertial Ellipsoid

• Diagonalizing the inertia tensor I Transforms to principal axes where the moment of inertia I has the form:

I = I1(ρ1)2 + I2(ρ2)2 + I3(ρ3)2 (B)

The normal form for an ellipsoid!

A geometric interpretation of the principal moments of inertia (I1,I2,I3): They are exactly the lengths of the axes of the inertial ellipsoid. If 2 Ij’s are equal, this ellipsoid has 2 equal axes & this is an ellipsoid of revolution. If all 3 are equal, this is a sphere!

• Define: The Radius of Gyration R0 in terms of the total mass M & the moment of inertia I: I M(R0)2

I is written as if all mass M were a distance R0 from the rotation axis.

• From the definition of the vector ρ = n(I)-½, we can write:

ρ = n(R0)-1(M)-½

A radius vector to a point on the inertia ellipsoid is inversely proportional to the radius of gyration about the direction of that same vector.

Emphasize: • The inertia tensor I & all associated quantities (principal

axes, principal moments, inertia ellipsoid, moment of inertia, radius of gyration, etc.) are defined relative to some fixed point in the body.

• If we shift this point to somewhere else, all of these in general are changed. The parallel axis theorem can be used.

• The principal axis transformation which diagonalizes I about an axis through CM will not necessarily diagonalize it about another axis! It will be diagonal with respect to both axes only if the shift is along a vector parallel to one of the original principal axes & only if that axis passes through the CM.

Example 1 from Marion• Calculate the inertia tensor of a homogeneous cube of density ρ, mass

M, and side length b. Let one corner be at the origin, and let the 3 adjacent edges lie along the coordinate axes (see figure). (For this choice of coord axes, it should be obvious that the origin does NOT lie at the CM!)

Ijk ∫Vρ(r)[r2δjk - xjxk]dV

β Mb2 I =

Symmetry:

Example 2 from MarionFind the principal moments of inertia &

the principal axes for the same cube:

Solve secular eqtn: (I - I1)R = 0

So:

Row manipulation:

• This results in:

• Or:

• Giving:

Diagonal I =

To get the principal axes, substitute I1, I2, I3 into the secular equation (I - Ij1)Rj = 0 (j = 1,2,3) & solve for Rj. Find: R1 along cube diagonal. R2, R3 each other & R1.

Example 3 from Marion• Calculate the inertia tensor of the

same cube in a coord system with

origin at the CM. (figure).

a = (½)b(1,1,1)

Again: Ijk ∫Vρ(r)[r2δjk - xjxk]dV Student exercise to show:

I11 = I22 = I33 = (1/6)Mb2, I12 = I21 = I13 = I31 = I23 = I32 = 0

1 0 0

ID = = (1/6)Mb2 0 1 0

0 0 1

Or: ID = (1/6)Mb2 1

Sect 5.5: Solving Rigid Body Problems; The Euler Equations

• We now have the tools to solve rigid body dynamics problems. Usually assume holonomic constraints. If not (like rolling friction) need to use special methods.

• Usually start by seeking a reference point in the body such that the problem can be split into pure translational + pure rotational parts.

• If one point in the body is fixed, then obviously all that is needed is to treat the dynamics of the rotation about that point.

• If one point is not fixed, it is most useful to choose the reference point in the body to be the CM.

• We have already seen that, for case where the reference point is the CM, the KE splits into KE of translation of CM + KE of rotation about an axis through the CM:

• We had: T = Ttrans + Trot

– 1st term = Ttrans = (½)Mv2 Translational KE of the CM

– 2nd term = Trot = (½)∑imi(vi)2. Rotational KE about CM

– We’ve written second term as: Trot = (½)ωIω

– For n unit vector along the rotation axis:

Trot = (½)ω2nIn (½) Iω2

• So: T = (½)Mv2 + (½)Iω2 • For all problems considered here, we can make a similar division for the

PE: V = Vtrans + Vrot

Lagrangian similarly divides: L = Ltrans + Lrot

• More generally (Goldstein notation):

L = Ltrans + Lrot = Lc(qc,qc) + Lb(qb,qb)

Lc = CM Lagrangian, qc,qc = generalized coordinates & velocities of the CM.

Lb = Body Lagrangian, qb,qb = generalized coordinates & velocities of body (rotation).

• Can use either Newtonian or Lagrangian methods, of course. In either case, often convenient to work in the principal axes system so that the rotational KE takes the simple form:

Trot = (½)I1(ω1)2 + (½)I2(ω2)2 + (½)I3(ω3)2

• The most convenient generalized coordinates to use are the Euler angles: ,θ,ψ. They are cumbersome, but useable.

• If the motion is confined to 2 dimensions (the rotation axis is fixed in direction), then only 1 angle describes motion!

• Follow Goldstein & start with the Newtonian approach to describe the rotational motion about an axis through a fixed point (like the CM):

• Consider either an inertial frame with the origin at a fixed point in body or a space axes system with CM as the origin.

• In this case, a Ch. 1 result is: (dL/dt)s = N (1)

Newton’s 2nd Law (rotational motion): Time derivative of the total angular momentum L (taken with respect to the space axes) is equal to the total external torque N.

• Make use of the Ch. 4 result relating the time derivatives in the space & body axes (angular velocity ω):

(d/dt)s = (d/dt)b + ω (2)

(dL/dt)s = (dL/dt)b + ω L (3)• Equating (1) & (3) & dropping the “body” subscript (which is

understood in what follows) gives:

(dL/dt) + ω L = N (4)• (4): Newtonian eqtn of motion relative to body axes.• In terms of inertia tensor, I, in general, L = Iω needs to be substituted

into (4).• For ith component, (4) becomes:

(dLi/dt) + ijkωjLk = Ni (4)– Levi-Civita density ijk 0, any 2 indices equal

ijk 1, if i, j, k even permutation of 1,2,3

ijk -1, if i, j, k odd permutation of 1,2,3

122 = 313 = 211 = 0 , etc. 123= 231= 312=1, 132= 213= 321= -1

• Take the body axes to be the principal axes of I, (relative to the reference

point) Li = Iiωi (i = 1,2,3)

• (4) becomes (no sum on i)

Ii(dωi/dt) + ijkωjωkIk = Ni (5)

• Newtonian eqtns of motion relative to the body axes are:

Ii(dωi/dt) + ijkωjωkIk = Ni (5)• Write component by component:

I1(dω1/dt) - ω2ω3(I2 -I3) = N1 (5a)

I2(dω2/dt) - ω3ω1(I3 -I1) = N2 (5b)

I3(dω3/dt) - ω1ω2(I1 -I2) = N3 (5c)

Euler’s Equations of Motion for a Rigid Body with one point fixed.

• Could derive these from Lagrange’s Eqtns, treating torques as generalized forces corresponding to Euler angles as generalized coords.

• Special Case: I1 = I2 I3: In this case, a torque in the “1-2” plane (N3 = 0) causes a change in ω1 & ω2 while leaving ω3 = constant. A very important special case! (Sect. 5.7)