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I do not understand from the line, "Now, in the body frame $T = (T_{x'}, T_{y'}, T_{z'})\ldots$"

How is that in the body frame? That should be in the inertial frame, since it is dL/dt. In the body frame, it is dL'/dt.

How can you treat dl/dt as the torque in the body frame, and then derive the Euler's equation?

Qmechanic
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Black Dagger
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  • Your question is not clear to me. What is the point? It seems that you do not accept the fact that, for instance, $L$ computed with respect to the inertial reference frame can be decomposed with respect to a basis at rest with the non-inertial body. Same for $\omega$. Am I correct? – Valter Moretti Mar 21 '14 at 07:28
  • @ V. Moretti . I do not have a problem with the decomposition. I do not understand the part where dL/dt (where L is computed with respect to inertial frame) is treated as T in the body frame. My question is, will the quantity T remain independent of the frame? – Black Dagger Mar 23 '14 at 09:13
  • When the text says T in the body frame simply means that the components of T w.r.to that frame are considered. T is computed in the inertial frame, in this sense is indipendent from the frame. T includes only true forces and disregards the apparent ones (inertial forces) arising in the body frame. Correspondingly, also the components of L w.r.to the body frame enter the identity, even if L is computed in the inertial frame (i.e., using the velocities of the points of the body referred to the inertial frame). – Valter Moretti Mar 23 '14 at 10:17
  • Can you follow this which explains the derivative of a vector or a rotating frame? – John Alexiou May 22 '14 at 13:09

2 Answers2

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What is done here and maybe not stated so clearly is that the rate of change of angular momentum vector is broken up in two parts.

  1. Change in angular momentum due to angular acceleration $I\dot{\vec{\omega}}$
  2. Change in angular momentum due to inertia tensor rotation $\vec{\omega}\times I \vec{\omega}$

All this is explained in terms of a changing vector $\vec{L}$ riding on a rotating frame. Maybe this a good way to explain the equations, but not such a good way to derive them. To derive the equations follow this logic:

  1. There is a coordinate system attached to the body where the 3×3 inertial tensor is constant $I_{body}$. The 3×3 rotation matrix at some instant is $E(t)$.
  2. To find the 3×3 inertia tensor in world coordinates you need the transformation $$I=E(t)\,I_{body}\,E(t)^\top$$ with $^\top$ the matrix transpose. This is interpreted as a transformation to local coordinates, application of inertia tensor and transformation back to world coordinates.
  3. The body at the same instant has angular velocity $\vec{\omega}(t)$ described in world coordinates.
  4. The columns of the rotation matrix $E(t)$ are the unit vectors $\hat{i}$, $\hat{j}$ and $\hat{k}$ of the coordinate system. Their time derivative equals only to a change in direction, as their magnitude is constant. So $$\frac{{\rm d}}{{\rm d}t} E(t) = \vec{\omega}(t) \times E(t)$$ where $\times$ is the vector cross product.
  5. The rate of change of the 3×3 inertia tensor is derived with the chain rule $$\begin{align}\frac{\rm d}{{\rm d}t} I(t) & = \frac{\rm d}{{\rm d}t} \left( E(t) I_{body} E(t)^\top\right) = \frac{{\rm d}E(t)}{{\rm d}t} I_{body} E(t)^\top + E(t) I_{body} \frac{{\rm d}E(t)}{{\rm d}t}^\top \\ & = (\vec{\omega}\times E(t)) I_{body} E(t)^\top + E(t) I_{body} (\vec{\omega}\times E(t))^\top) \\ & = \vec{\omega} \times I(t) - I(t) \vec{\omega} \times \end{align} $$ Don't worry about the 3×3 matrix cross product operator $[\vec{\omega}\times]$ as it is going be canceled out next.
  6. The angular momentum vector at the same instant is $\vec{L}(t) = I(t) \vec{\omega}(t)$ (by definition of the inertia tensor)
  7. Time derivative is angular momentum vector is derived with the chain rule $$\begin{align} \frac{{\rm d}}{{\rm d}t} \vec{L} &= I(t) \frac{{\rm d} \vec{\omega}(t)}{{\rm d}t} + \frac{{\rm d}I(t) }{{\rm d}t} \vec{\omega}(t) \\ &= I(t) \dot{\vec{\omega}} + \left( \vec{\omega} \times I(t) - I(t) \vec{\omega} \times \right) \vec{\omega} \end{align}$$

$$\boxed{\dot{\vec{L}} = I(t) \dot{\vec{\omega}} + \vec{\omega} \times I(t) \vec{\omega} }$$

The last is Euler's equations of rotational motion.

The typo mentioned in comments is corrected. NOTE: related answer https://physics.stackexchange.com/a/80449/392 for combined rotational and transnational equations of motion.

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John Alexiou
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    There is a thing that I don't understand: I suppose -please correct me if I'm wrong- that $(\vec{\omega}\times E(t))^\top=([\vec{\omega}\times]E(t))^\top= E(t)^\top [\vec{\omega}\times]^\top=- E(t)^\top[\vec{\omega}\times]$ and consequently $E(t)I_{body}\frac{\text{d}E(t)}{\text{d}t}=-I(t)\vec{\omega}\times$, so I don't understand why $\frac{\text{d}}{\text{d}t}I(t)=\vec{\omega}\times I(t)+I(t)\vec{\omega}\times$... where is my error? Thank you very much! – Self-teaching worker May 26 '15 at 16:57
  • You are correct. But the term cancels out because $-\omega \times \omega =0$. I edited the post. – John Alexiou Nov 01 '16 at 23:00
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It clearly states that "...the body frame co-rotates with the body...". So what you are calling $L'$ is in fact $L$ itself. The torque will be the same in both reference frames.