# Confusion regarding derivation of euler's equation

I'm having confusion regarding the derivation of Euler's equation for rigid bodies.

Suppose I have an inertial frame $$O$$, and a rotating frame $$O'$$ fixed to the rotating body and the rotating frame is also a principal axis (the inertia tensor in the rotating frame is diagonal with diagonal entries $$I_{11}, I_{22}, I_{33}$$).

For any vector $$v$$ in the space, we know that $$\left . \frac{dv}{dt} \right \vert_{O} = \left . \frac{dv}{dt} \right \vert_{O'} + (\omega \times v)$$ (where $$\left .\frac{dv}{dt} \right \vert_{O'}$$ is the change of the vector $$v$$ as seen from the frame $$O'$$, and $$\omega$$ is the velocity of the origin of $$O'$$ measured from $$O$$ the cross product is taken in $$L$$).

Suppose $$L(t)$$ is the angular momentum vector at time $$t$$. Then setting $$v = L(t)$$ in the above equation, and using the fact that $$\left . \frac{dL(t)}{dt} \right \vert_{O} = \left . \tau \right \vert_{O}$$, we get $$\tau = \left . \frac{d L(t)}{dt} \right \vert_{O'} + (\omega \times L)$$.

Now here's my main source of confusion: How do you calculate how the vector $$L(t)$$ calculated in $$O$$ looks from $$O'$$ ? Backtracking from what's its supposed to be from Euler's formula, if $$\pmb{\omega} = (\omega_1, \omega_2, \omega_3)$$ as calculated from the inertial frame, since in the rotational frame the angular velocity is zero, and then $$L(t)$$ is supposed to look like $$(I_{11}\omega_1, I_{22} \omega_2, I_{33} \omega_3)$$. Why this should be true ?

I checked up Kleppner Kolenkow and here's what they writes about it (the bold part is the part which I don't understand, italicized is my comments)

Let us introduce an inertial coordinate system which coincides with the instantaneous position of the body's principal axes at time $$t$$.

OK so let $$Q'$$ be $$O'$$ translated such that the origin of $$Q$$ coincides with the origin of $$O$$

We label the axes of the inertial system $$1, 2, 3$$. Let the components of the angular velocity $$\omega$$ at time $$t$$ relative to the $$1, 2, 3$$ system be $$(\omega_1, \omega_2, \omega_3)$$

OK fine till now

At the same instant, the components of $$L$$ are $$L_1 = I_1 \omega_1$$, $$L_2 = I_2 \omega_2$$, $$L_3 = I_3 \omega_3$$ where $$I_1, I_2, I_3$$ are the moments of inertia about the three principal axes.

This is where I'm completely lost. Let $$R$$ be the rotation matrix taking $$Q$$ to $$O$$.

Firstly I don't get what he means exactly by the components of $$L$$, is it in $$O$$ or $$Q$$ ? It can't possible be in $$Q'$$ because the angular momentum in $$Q'$$ has the form $$I_j \omega'_j$$ where $$\pmb{\omega}' = R \pmb{\omega}$$.

So it must be $$O$$. For any vector $$v$$, note that $$\left . v \right \vert_{Q} = R \left . v \right \vert_{O}$$. Note that the vector $$L$$ lies in space independent of the origin, so the coordinates of $$L$$ in $$O$$ must be $$R \left . L \right \vert_{Q} = RI_{Q} \pmb{\omega}' = RI_{Q}R^{-1} \pmb{\omega}$$, which he is claiming is equal to $$I_Q \pmb{\omega}$$. As this is true for all $$\omega$$, this is equivalent to saying $$R$$ commutes with $$I_Q$$ but isn't that false ?

• – John Alexiou Oct 16 '19 at 17:21

The Euler equation

$$\frac{d}{dt}\vec{L}=\vec{\tau}\tag 1$$

Where $$\vec{L}$$ is the angular momentum and $$\vec{\tau}$$ is the external torque .

with $$\vec{L}=(I\,\vec{\omega})$$ you get :

$$\frac{d}{dt}\left(I\,\vec{\omega}\right)=\vec{\tau}\tag 2$$

we transform the components of the vector $$I\,\vec{\omega}$$ from body system to inertial system

$$I\,\vec{\omega}\mapsto R\,(I_B\,\vec{\omega}_B)$$

where $$R$$ is the transformation matrix from $$B$$ system to $$I$$ system

$$\frac{d}{dt}\left[R\,(I_B\,\vec{\omega}_B)\right]=\vec{\tau}_I\tag 3$$

taking time derivative of equation (3) you get:

$$R\,(I_B\,\vec{\dot{\omega}}_B)+\dot{R}\,(I_B\,\vec{\omega}_B)=\vec{\tau}_I\tag 4$$

with:

$$\dot{R}=R\,\tilde{\omega}$$

$$R\,(I_B\,\vec{\dot{\omega}}_B)+R\,\tilde{\omega}\,(I_B\,\vec{\omega}_B)=\vec{\tau}_I\tag 5$$

multiply equation (5) from the left with $$R^T$$ you get the Euler equation in components of the body system.

$$(I_B\,\vec{\dot{\omega}}_B)+\vec{\omega}_B\times \,(I_B\,\vec{\omega}_B)=R^T\vec{\tau}_I=\vec{\tau}_B\tag 6$$

the Euler equation in components of the inertial system is:

$$(I_I\,\vec{\dot{\omega}}_I)+\vec{\omega}_B\times \,(I_I\,\vec{\omega}_I)=\vec{\tau}_I\tag 6$$

where

$$I_I=R^T\,I_B\,R$$

Tilde operator

$$\tilde{\omega}=\begin{bmatrix} 0 & -\omega_z & \omega_y \\ \omega_z & 0 & -\omega_x \\ -\omega_y & \omega_x & 0 \\ \end{bmatrix}$$