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 ?


1 Answer 1


The Euler equation

you start with:

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

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:

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



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

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:

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



Tilde operator

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


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