Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

My Question was like this and i have realised few things and still have some doubts

I have a book in which a paragraph goes like this

Now, $\dot\phi$, $\dot \theta$, $\dot\psi$ are respectively the angular speeds about the space z-axis, line of node and the body z-axis.

I don't know how Euler's angles of rotation are even connected with the angular velocity. Since it's my assumption that the the space set of axis remains unchanged, and the body set of axes rotate with the the axis of rotation. According to my assumption, the angular velocity vector components should remain constant w.r.t the space set of axes (not equal but unchanged) and the angular velocity vector is zero w.r.t the body set of axes because the body set of axes is rotating with the axis so the angular velocity is zero.

Even if the body set of axes were stationary and the rigid body is rotating, would that mean the the components would be connected to the Euler's angle of rotation anyway? I think that Euler's angle are just angles of rotation that transforms the space set of axes into body set of axes.

And I also don't understand what does this 'line of node mean'.

I have come to realize that in Euler's rotation, The space axis is rotated about space Z axis, new space X-axis and, body Z axis (which is aligned by the new space X axis rotation). Since there is rotation, there is angular speed, and the rotation are $ \phi , \theta ,and \psi $, then obviously the the angular speeds are $\dot\phi$, $\dot \theta$, $\dot\psi$ and the line of node is the new space X-axis from space Z-axis rotation. And there is no rigid body involved. But has this angular velocity got something to do with the rotation of rigid body? Like stability of spinning top?

I don't know but I hope i am right.

share|improve this question
    
en.wikipedia.org/wiki/Euler_angles#Euler_rotations <- does this help? –  Marek Dec 2 '10 at 11:21

1 Answer 1

up vote 1 down vote accepted

If we express the change in anglular velocity as $\Delta\vec\omega$ in local coordinates, with for example the angles $\phi$, $\theta$ and $\psi$ rotations about the $Z$, $X$ and $Z$ respectively then the answer is

$$\Delta\vec{\omega}=\dot{\phi}\hat{k}+\mathrm{{Rot}(\hat{k},\phi})\left(\dot{\theta}\hat{i}+\mathrm{{Rot}(\hat{i},\theta})\left(\dot{\psi}\hat{k}\right)\right)$$

Where Rot(axis,angle) is a 3x3 rotation matrix. We apply the angular rotation components in sequence on the local coordinate axes this way. Also $\hat{i}=(1,0,0)$, $\hat{j}=(0,1,0)$ and $\hat{k}=(0,0,1)$.

Line of node must be the common normal between the two $z$-axes. We typically denote that as the $x$-axis (see Denavit-Hartenberg notation).

share|improve this answer
    
I think my question is wrong please see my edited question again and my apologies for it. And thank you for the answer. –  Santosh Linkha Dec 3 '10 at 4:58

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.