By Euler rotation Equation, torque $$\tau (t)=I \omega '(t) + \omega(t) \times I \omega(t)$$

where $I$ is the inertia tensor in rotating frame and is constant; $\omega(t)$ is the Angular velocity in the rotating frame, change with time; $\tau (t)$ is the torque in the rotating frame;

but we know that $$\tau (t) = L'(t) = (I \omega (t))' = I \omega '(t);$$ as $I$ is constant and where $L$ is the Angular momentum in the rotating frame;

$W(t)\times IW(t)=0$ always???


The Euler rotation equation, $$ \vec{\tau} = \mathbf{I} \dot{\vec{\omega}} + \vec{\omega} \times \left(\mathbf{I} \vec{\omega} \right), $$ implicitly refers to the description of the motion in a rotating reference frame, namely the reference frame rotating with the rigid body. In such a reference frame, it is no longer true that $d\vec{L}/dt = \vec{\tau}$.

  • $\begingroup$ Can you prove why? $\endgroup$ – Guy Ab Jul 29 at 15:30
  • $\begingroup$ @GuyAb: I'm not sure which aspect you want proved, but the derivations of this are in most intermediate-level classical mechanics texts. See §11.8 of Thornton & Marion; §5.5 of Goldstein; or §10.7 of Taylor. $\endgroup$ – Michael Seifert Jul 29 at 15:42
  • $\begingroup$ I am working on the solution of: physics.stackexchange.com/questions/493028/… $\endgroup$ – Guy Ab Jul 29 at 15:52
  • $\begingroup$ he use a rotation matrix U(t) to rotate I,W from rotating frame to world frame, can use it to rotate L and Torque also? $\endgroup$ – Guy Ab Jul 29 at 15:54
  • $\begingroup$ question 2: are all I,W,Torque are vectors taking along the rotating frame? $\endgroup$ – Guy Ab Jul 29 at 15:56

Your second equation is a formula which is obtained for a certain reference frame.

On the other hand, your first one is relating two different frames.

  • $\begingroup$ what you mean? is any of I, W or Torque is not in the rotating frame that is fixed to the body? $\endgroup$ – Guy Ab Jul 29 at 15:48
  • $\begingroup$ You tell me, haha. But it looks so. Otherwise, why are you adding the $\omega \ \times$ term? $\endgroup$ – FGSUZ Jul 29 at 19:33

The Euler rotation equation

$$ \vec{\tau} = \mathbf{I} \dot{\vec{\omega}} + \vec{\omega} \times \left(\mathbf{I} \vec{\omega} \right), \tag 1$$

with $\frac{d}{dt}L=\vec{\tau}$ and $L=\mathbf{I}\,\vec{\omega}$

but if $\vec{\tau}=0$ then $L=\mathbf{I}\,\vec{\omega}=const.\quad $

thus equation (1):

$$ \vec{0} = \vec{0} + \vec{\omega} \times \left(\mathbf{I} \vec{\omega} \right), \quad \surd$$

if the components of the angular momentum $L_b$,are given in body fixed coordinate system index $b$ then the time derivative of the angular momentum components $L_o$ in space system (index o) is:

$$\frac{d}{dt} L_o=\frac{d}{dt} L_b+\vec{\omega}\times L_b$$

now $\frac{d}{dt}L_o=\vec{\tau}=\vec{0}$ doesn't obtain that $\frac{d}{dt}L_b=\vec{0}$ thus $L_b$ is not conserved


the external torque $\vec{\tau}=\vec{r}\times \vec{F}$ is equal to zero if the force $\vec{F}$ is parallel to the radius of the rotation point $\vec{r}$, this force is call central force.

  • $\begingroup$ I don't understand, assuming all vector/tensor in the right side of your equations (L,W,I) are taking along the rotating frame, does the left side (torque) is taking along the rotating frame or the world frame? $\endgroup$ – Guy Ab Jul 30 at 3:34
  • $\begingroup$ What is wrong in this $\overrightarrow {0}=\dfrac {d}{dt}\overrightarrow {L}_{b}+\overrightarrow {\omega }\times \overrightarrow {L}_{b}$ equation? $\endgroup$ – Eli Jul 30 at 6:57
  • $\begingroup$ When it is = 0 it does not matter, but what if not? just want to be sure, see other comment and your from physics.stackexchange.com/questions/494086/… $\endgroup$ – Guy Ab Jul 30 at 7:02
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    $\begingroup$ Of course all vector components must be taken in the same frame. $\endgroup$ – Eli Jul 30 at 7:05

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