Time derivative of angular velocity in rotating reference frame

Question

I am going through a section in a textbook regarding the Newton Euler equations for a system of rigid bodies (robotics text). There is a particular line in the derivation I don't understand, I've attached the image to make it clear:

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I think the derivative of the second term on the right-hand side of 3.28 is using the product rule to get to 3.32, but I don't understand where the cross-product comes from. Could someone explain this particular part of the derivation? Thanks.

Very funny... I looked it up, it's a reference to a book I don't have and can't seem to download. — JDS, Sep 12 '14 at 17:08
Then I would search for "Newton's second law in a rotating frame." — BMS, Sep 12 '14 at 17:11
it's because in the rotating frame $\frac{dz}{dt} = \omega \times z$ — SuperCiocia, Sep 12 '14 at 17:40
Have you read http://en.wikipedia.org/wiki/Rotating_reference_frame ? — John Alexiou, Sep 13 '14 at 05:54

score 2 · Accepted Answer · edited Apr 13 '17 at 12:40

It is worth it for you reading about differentiating vectors on rotating frames.

Appendix

In addition, to prove equation (3.28) follow this answer https://physics.stackexchange.com/a/65768/392 or https://physics.stackexchange.com/a/105099/392

score 0 · Answer 2 · answered Sep 13 '14 at 04:54

0

You just have to use the corresponding equations from Newton's law for rotating objects. So simply:

Product rule
dz/dt=ω×z

answered Sep 13 '14 at 04:54

Meshal

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Time derivative of angular velocity in rotating reference frame

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