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Angular velocity of a moving frame and the rotating-frame acceleration identity

Statement

Use one-based labels zi:=z(i−1) and Aij:=A(i−1,j−1) for 1≤i,j≤3, so the displayed components correspond in order to the imported coordinates 0,1,2. Matrix-vector products mean (Az)(k):=∑j<3A(k,j)z(j), obtained from the rectangular product by representing z as the column Cz(j,0)=z(j) (Rectangular matrix multiplication and the identity matrix In, including zero-sized shapes).

Let I⊆R be a nondegenerate interval, let R:I→SO(3) and c:I→R3 be twice continuously differentiable in the sense that all entries of R and all components of c have continuous second derivatives, and put

Ω(t):=R(t)TR′(t)(t∈I).

  1. Ω(t) is skew-symmetric for every t∈I, that is Ω(t)T=−Ω(t).
  2. For every t∈I there is exactly one vector ω(t)∈R3 such that

Ω(t)z=ω(t)×z(z∈R3),

the cross product being that of The cross product in R3. Its components in the standard basis are ω(t)=(−Ω23(t),Ω13(t),−Ω12(t)). Since R is C2, Ω=RTR′ and hence ω are C1; thus ω′(t) in clause 3 is defined. 3. If y:I→R3 is twice continuously differentiable and

x(t):=R(t)y(t)+c(t)(t∈I),

then for every t∈I

R(t)Tx′′(t)=y′′(t)+2 ω(t)×y′(t)+ω′(t)×y(t)+ω(t)×(ω(t)×y(t))+R(t)Tc′′(t).

  1. For every t∈I and z∈R3, ∥R(t)z∥2=∥z∥2: the rotation part of the change of coordinates preserves Euclidean norms (The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn).

Facts & Assumptions

Given: A nondegenerate interval I, twice continuously differentiable maps R:I→SO(3) and c:I→R3, and a twice continuously differentiable y:I→R3; write Ω=RTR′ and x=Ry+c.

[L1]

R(t)TR(t)=I3 for every t by The orthogonal group O(3) and the rotation group SO(3). The matrix entries of R are twice continuously differentiable by the given hypothesis.

[L3]

The cross product is bilinear and alternating, and Ωz=ω×z for all z holds exactly for the matrix Ω with entries Ω12=−ω3, Ω13=ω2, Ω23=−ω1 and the opposite entries fixed by skew-symmetry (The cross product in R3, The cross product is bilinear, alternating, and orthogonal to both factors).

[L4]

The Euclidean norm satisfies ∥z∥22=⟨z,z⟩2=zTz, and ⟨z,z⟩2=0 only for z=0 (The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn).

[L5]

If R∈SO(3), then det⁡R=1, so R is invertible by the square-matrix determinant criterion. Since RTR=I3, its inverse is RT and RRT=I3 (The orthogonal group O(3) and the rotation group SO(3), A finite square real matrix is invertible if and only if its determinant is nonzero).

Proof

technique · direct
1.1L1L2algebra

Differentiate the identity RTR=I3 of [L1] componentwise using [L2]: 0=(RTR)′=R′TR+RTR′, so RTR′=−R′TR. Transposing this last identity with [L2] gives ΩT=(RTR′)T=R′T(RT)T=R′TR=−Ω. This proves clause 1.

1.2L1L2algebra

Differentiate x=Ry+c twice using [L2]: x′=R′y+Ry′+c′ and x′′=R′′y+2R′y′+Ry′′+c′′. Multiplying on the left by RT gives RTx′′=RTR′′y+2RTR′y′+y′′+RTc′′.

2.1step 1.1L3algebra

Write the entries of the skew matrix Ω(t) as Ωij; by step 1.1, Ωji=−Ωij and Ωii=0, so Ω is determined by the three entries Ω12,Ω13,Ω23. Define ω:=(ω1,ω2,ω3) by ω1:=−Ω23, ω2:=Ω13, ω3:=−Ω12, and substitute into the defining formula of [L3]: for every z, the i-th component of Ωz and of ω×z agree by the displayed finite computation of the cross product, so Ωz=ω×z.

2.2step 1.1step 1.2L1L2L5algebra

By definition of Ω and [L2], Ω′=(RTR′)′=R′TR′+RTR′′, hence RTR′′=Ω′−R′TR′=Ω′+Ω2. Indeed, R′=RΩ follows from RTR′=Ω after multiplying on the left by R and using RRT=I3 from [L5]; then R′TR′=(RΩ)T(RΩ)=ΩTRTRΩ=ΩTΩ=−Ω2 by clause 1. Also RTR′=Ω. Substituting into step 1.2, RTx′′=(Ω′+Ω2)y+2Ωy′+y′′+RTc′′.

3.1step 2.1givenL2L3algebra

If another vector η∈R3 satisfies Ωz=η×z for every z, then (ω−η)×z=0 for every z; taking z to be each standard basis vector in the formula of [L3] gives ω−η=0, proving uniqueness. Each entry of Ω=RTR′ is C1 by the given C2 regularity and the product rule [L2]; the three component formulas of step 2.1 therefore make ω a C1 curve. This completes clause 2.

4.1step 2.1step 2.2step 3.1L2L3algebra

Differentiating the three component formulas in step 2.1, justified by the C1 regularity proved in step 3.1, gives Ω′z=ω′×z for each fixed z. Thus Ω′y=ω′×y, Ω2y=Ω(ω×y)=ω×(ω×y) and Ωy′=ω×y′. Substituting these into step 2.2 gives the displayed identity of clause 3.

5.1L1L2L4algebra∎

For z∈R3, ∥Rz∥22=(Rz)T(Rz)=zTRTRz=zTz=∥z∥22 by [L1], [L2] and [L4]; taking square roots proves clause 4.

Remarks

  • The vector ω(t) is the angular velocity of the frame in the body-fixed convention Ω=RTR′. The vector attached to the other convention R′RT is the spatial angular velocity, and the two differ by a rotation by R; this item records only the convention used in clause 3, and its consumers must state which convention they use.

  • The identity of clause 3 is a coordinate identity for twice differentiable curves. It asserts nothing about forces: reading the terms 2ω×y′ and ω×(ω×y) as "Coriolis" and "centrifugal" accelerations is an interpretation made by the physical items, under their own mass and force hypotheses.

Depends on

Dependency tree · 2 levels

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