Lecture35.pdf
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This document discusses principal axes and Eulerian angles for describing the orientation and motion of rigid bodies. It contains the following key points: 1) The principal moments and axes of an inertia tensor can always be found for real, symmetric tensors like the inertia tensor, and they will always be real values. 2) Under translation, the inertia tensor about a point not at the center of mass can be expressed in terms of the inertia tensor at the center of mass using Steiner's theorem. 3) Eulerian angles define a rotation as three successive rotations about fixed axes and can be used to relate an inertial frame to a body-fixed frame for a rigid body. 4) Euler's equations describe
Lecture35.pdf
- 1. Lecture 35: Principal Axes, Translations, and Eulerian Angles
- 2. When Can We Find Principal Axes? • We can always write down the cubic equation that one must solve to determine the principal moments • But if we want to interpret these as physically meaningful quantities, the roots of that cubic have to be real – Recall that in general, cubics can have two complex roots • Fortunately, we’re not in the general case here • The inertia tensor is both real and symmetric – in particular, it satisfies: • Matrices that satisfy this restriction are called Hermitian • For such matrices, the principal moments can always be found, and they are always real (see proof in text) * ij ij I I = This mathematics will come up again in Quantum Mechanics Principal Moments ↔ Eigenvalues Principal Axes ↔ Eigenfunctions
- 3. Inertia Tensor Under Translation • We’ve already seen how the inertia tensor transforms under a rotation of the coordinate system • Now we’ll see how it behaves under translation – that is, keeping the direction of the the coordinate axes fixed, but changing the origin • In particular, we’ll compare I for a coordinate system at the center of mass of an object with that for a different origin: x1 x2 x3 O X1 X2 X3 Q rCM O is at the center of mass; Q can be anywhere
- 4. • In the X reference frame (the one with origin at Q), the position of any point in the rigid body can be written as: which means the inertia tensor becomes: α α = + CM R r r Position in center-of- mass frame ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) 2 , , 2 , , , 2 , , , , , , 2 2 , , , , , , , , , , , , , 2 , , , , 2 2 ij ij k i j k ij k CM k i CM i j CM j k ij k CM k k CM k k i j CM i j i CM j CM i CM j ij CM k k ij k i j k CM k CM k I m X X X m x x x x x x m x x x x x x x x x x x x m x x x m x x x x α α α α α α α α α α α α α α α α α α α α α α α α α δ δ δ δ δ = − = + − + + = + + − − − − = + + − − ( ) , , , , , i j i CM j CM i CM j x x x x x α α α − − Inertia tensor about the center of mass
- 5. • So the inertia tensor about Q is: • Each term in the second line has a factor that looks like: • For a general reference frame, this quantity equals the mass of the system times the position of the ith component of the center of mass – But the xi and in the center of mass frame, so all these terms are zero • Therefore we have the result (Steiner’s Parallel Axis Theorem): ( ) ( ) 2 , , , , , , , , , , 2 ij CM ij ij CM k CM i CM j k ij CM k k CM i j i CM j k I I x x x m m x x x x x x α α α α α α α δ δ = + − + − − ,i m x α α α ( ) 2 , , , , ij CM ij ij CM k CM i CM j k I I M x x x δ = + −
- 6. Eulerian Angles • In analyzing the motion of a rigid body, it is most convenient to use a reference frame that is fixed to the body – For example, a symmetry axis of the body could be the z axis in this “body frame” – But that’s usually not an inertial frame • So we need to define a translation between some inertial “fixed” frame and the body frame • Three angles are needed to specify how one coordinate system is rotated with respect to another (in three dimensions) – These angles determine the elements of the rotation matrix λ λ λ λ
- 7. • There are many ways the three angles can be specified • One convenient choice is the Eulerian Angles, in which the 3D rotation is the result of three 1D rotations: 1. Rotate CCW by φ around axis 2. Rotate CCW by θ around axis 3. Rotate CCW by ψ around axis 3 3 x x ′ ′′ = 1 x′ 2 x′ φ θ 1 x ψ 1 2 3 ( , , ) x x x ′ ′ ′ 1 2 3 ( , , ) x x x ′′ ′′ ′′ 1 2 3 ( , , ) x x x ′′′ ′′′ ′′′ 2 x′′ 1 x′′ 1 x′ 2 x′ 3 3 x x ′ ′′ = 2 x′′′ 3 x′′′ 1 1 x x ′′ ′′′ = 3 3 x x ′′′= 1 x′ 3 x′ 2 x′ 2 x Line of nodes 1 2 3 ( , , ) x x x 3 x′ 1 x′′ 3 x′′′
- 8. • Note that in each rotation, one axis (from the previous rotation) is held fixed. That means the rotation matrices have the form: • The complete rotation is given by: • i.e., the elements of λ λ λ λ are obtained by multiplying the three individual rotation matrices (see Equation 11.99 in the text for their values) cos sin 0 1 0 0 sin cos 0 ; 0 cos sin 0 0 1 0 sin cos cos sin 0 sin cos 0 0 0 1 φ φ φ φ θ θ θ θ ψ ψ ψ ψ = − = − = − ′ ′ = ≡ x x x
- 9. Angular Velocity in the Body Frame • It’s going to be easiest to find the equations of motion for the object in the body frame • In general, we can write: • To write the components of this vector in the body frame we need to do some geometry: = + + 1 x ψ 3 x 1 x′ 3 x′ 2 x′ 2 x φ θ • is along the x3 direction: • is in the x1-x2 plane: • has components in all three directions: ψ = 3 e 2 cos sin θ ψ θ ψ = − 1 e e 2 3 sin sin sin cos cos φ θ ψ φ θ ψ φ θ = + + 1 e e e
- 10. • Adding up these vectors, we find that the components of ω ω ω ω in the body frame are: 1 2 3 sin sin cos sin cos sin cos ω φ θ ψ θ ψ ω φ θ ψ θ ψ ω φ θ ψ = + = − = +
- 11. Euler’s Equations for Rigid Body Motion • Now that we’ve defined the geometry, we can set up the Lagrangian for rigid body motion • To make things easy, we’ll first consider the case where no external forces act on the body, so: • In that case, we can put the origin of both the fixed (meaning inertial) and body frames at the center of mass • Furthermore, we can cleverly choose the body frame to coincide with the principal axes of the body, so that: L T = 2 1 2 i i i I T ω = Principal moment of inertia associated with the ith axis
- 12. • Since the Eulerian angles fully specify the orientation of the rigid body in the inertial frame, we can take them to be our generalized coordinates • The equation of motion for ψ is: which becomes (using the chain rule): • One piece that enters everywhere is which is simply: 0 T d T dt ψ ψ ∂ ∂ − = ∂ ∂ 0 i i i i i i T d T dt ω ω ω ψ ω ψ ∂ ∂ ∂ ∂ − = ∂ ∂ ∂ ∂ , i T ω ∂ ∂ i i i T I ω ω ∂ = ∂
- 13. • The other relations we need are: • Putting these back into the Lagrange equation gives: 1 2 2 1 3 sin cos sin sin sin cos 0 ω φ θ ψ θ ψ ω ψ ω φ θ ψ θ ψ ω ψ ω ψ ∂ = − = ∂ ∂ = − = − ∂ ∂ = ∂ 1 2 3 0 0 1 ω ψ ω ψ ω ψ ∂ = ∂ ∂ = ∂ ∂ = ∂ ( ) ( ) 1 1 2 2 2 1 3 3 1 2 1 2 3 3 0 0 d I I I dt I I I ω ω ω ω ω ω ω ω + − − = − − =
- 14. • The choice of which axis to call x3 was completely arbitrary – That means we can find similar relations for the other components of the angular velocity ( ) ( ) ( ) 1 2 1 2 3 3 3 1 3 1 2 2 2 3 2 3 1 1 0 0 0 I I I I I I I I I ω ω ω ω ω ω ω ω ω − − = − − = − − = Euler’s Equations for force-free motion