Velocities due to angular rates

[steenaap]

Dear all, I am missing some insight concerning dynamic mechanics:

1. Problem statement
A angular rate sensor is attached to a massive body. I would like to calculate the velocity of a certain point on that body due to rotational motion of the body itself.

Homework Equations

The whole problem, equations and question have been described in detail in a http://members.lycos.nl/eyefragm/question.pdf" file

The Attempt at a Solution

I proposed two approaches. Who can tell me which one is wrong and why. I cannot figure it out myselve.

Many thanks!

 

[D H]

They are both correct. The cross product of two vectors in represented in some frame A is another vector in that frame. Transforming this into another frame B,

(\mathbf u \times \mathbf v)_B =<br /> \boldsymbol T_{A\to B} \, (\mathbf u_A \times \mathbf v_A)

Reference frames are just a representation technique. The position vector from point a to point b is the same vector in all reference frames; it just has different representations in difference frames. he cross product is a pseudo vector rather than a true vector. Pseudo transforms from one frame to another given a proper transformation. (They do not transform under reflection, which is why they are called pseudo vectors). Assuming a transformation from one right-handed system to another, we must have

<br /> \mathbf u_B \times \mathbf v_B<br /> = (\boldsymbol T_{A\to B}\,\mathbf u_A) \times (\boldsymbol T_{A\to B}\,\mathbf v_A)<br /> = \boldsymbol T_{A\to B}(\mathbf u_A \times \mathbf v_A)<br />One way to look at the cross product is to view \mathbf u \times as an operator that acts on some vector \mathbf v that produces the cross product of the two vectors:

(\mathbf u \times) \mathbf v = \mathbf u \times \mathbf v

This in turns suggests determining if a corresponding matrix operator exists that yields the exact same vector for all vectors v as does the cross product operator. The cross product matrix generated from some vector

\mathbf u = \bmatrix u_x \\ u_y \\u_z\endbmatrix

is

\boldsymbol X(\mathbf u) \equiv<br /> \bmatrix<br /> 0 &amp; -u_z &amp; \phantom{-}u_y \\<br /> \phantom{-}u_z &amp; 0 &amp; -u_x \\<br /> -u_y &amp; \phantom{-}u_x &amp; 0\endbmatrix

does exactly that. This matrix transforms the vector cross product into a matrix-vector product:

\mathbf u \times \mathbf v<br /> = (\mathbf u \times) \mathbf v<br /> = \boldsymbol X(\mathbf u)\,\mathbf v

A column vector transforms from frame A to frame B via

\mathbf u_B = \boldsymbol T_{A\to B} \, \mathbf u_A

The cross product matrix, on the other hand, transforms via the matrix transformation rule

\boldsymbol X(\mathbf u_B) =<br /> \boldsymbol T_{A\to B} \,<br /> \boldsymbol X(\mathbf u_A) \,<br /> \boldsymbol T_{A\to B}^{\;\;\;T}<br /> = \boldsymbol X(\boldsymbol T_{A\to B}\,\mathbf u_A)

The proof of the above is left as an exercise to the reader. With this one can examine whether

<br /> (\boldsymbol T_{A\to B}\,\mathbf u_A) \times (\boldsymbol T_{A\to B}\,\mathbf v_A)<br /> = \boldsymbol T_{A\to B}(\mathbf u_A \times \mathbf v_A)<br />

Step by step,

<br /> \begin{aligned}<br /> (\boldsymbol T_{A\to B}\,\mathbf u_A) \times<br /> (\boldsymbol T_{A\to B}\,\mathbf v_A)<br /> &amp;=<br /> ((\boldsymbol T_{A\to B}\,\mathbf u_A) \times)<br /> (\boldsymbol T_{A\to B}\,\mathbf v_A) \\<br /> &amp;=<br /> (\boldsymbol X(\boldsymbol T_{A\to B}\,\mathbf u_A))\,<br /> (\boldsymbol T_{A\to B}\,\mathbf v_A) \\<br /> &amp;=<br /> (\boldsymbol T_{A\to B} \,<br /> \boldsymbol X(\mathbf u_A) \,<br /> \boldsymbol T_{A\to B}^{\;\;\;T}) \,<br /> (\boldsymbol T_{A\to B}\,\mathbf v_A) \\<br /> &amp;=<br /> \boldsymbol T_{A\to B} \,<br /> \boldsymbol X(\mathbf u_A) \,<br /> \boldsymbol T_{A\to B}^{\;\;\;T} \,<br /> \boldsymbol T_{A\to B}\,\mathbf v_A \\<br /> &amp;=<br /> \boldsymbol T_{A\to B} \,<br /> \boldsymbol X(\mathbf u_A) \, \mathbf v_A \\<br /> &amp;= \boldsymbol T_{A\to B} (\mathbf u_A \times \mathbf v_A)<br /> \end{aligned}

 

[steenaap]

Thanks.

However, in your derivation, you say nothing about the necessary orthogonal property of the transformation matrix.

Just try in matlab:

C(AxB)
(CA)x(CB)

Once with C a non-orthogonal matrix and once with C a orthogonal matrix. In first case both results will be different, in second case both results will be equal.

[edit]: from the third last line till the second last line of previous reply, the orthogonal property of the transformation matrix is supposed. Conclusion of the story in a pure algebraic way:

C(AxB)==(CA)x(CB) only if C is orthogonal