Categories
_NOTES_ Analytical Mechanics

Derivation of Euler Dynamic Equation for Free Rigid Body from EL Equation

This article is rearranged based on Qianye YU’s previous work, Physics Reserved Labour is fully authorized to make the reprint.

qyggtql

Lagrangian of free rigid body in its comoving reference frame is given by:

$$
L=\sum_{i} \frac{1}{2} I_{i} \omega_{i}^{2}
$$

where angular velocities can be expressed in Eulerian angles: $\omega_{1}=\dot{\varphi} \sin \theta \sin \psi+$ $\dot{\theta} \cos \psi, \omega_{2}=\dot{\varphi} \sin \theta \cos \psi-\dot{\theta} \sin \psi, \omega_{3}=\dot{\varphi} \cos \theta+\dot{\psi}$.

Substitute into EL equations:

$$
\frac{d}{d t}\left(\frac{\partial \mathcal{L}}{\partial \dot{q}_{i}}\right)=\frac{\partial \mathcal{L}}{\partial q_{i}}
$$

where $q_{i}$ take $q_{1}=\theta, q_{2}=\varphi, q_{3}=\psi$.

Apply chain rule to both sides of (2) :

$$
\begin{aligned}
\frac{\partial \mathcal{L}}{\partial q_{i}} & =\sum_{j=1}^{3} \frac{\partial \mathcal{L}}{\partial \omega_{j}} \frac{\partial \omega_{j}}{\partial q_{i}} \\
\frac{\partial \mathcal{L}}{\partial \dot{q}_{i}} & =\sum_{j=1}^{3} \frac{\partial \mathcal{L}}{\partial \omega_{j}} \frac{\partial \omega_{j}}{\partial \dot{q}_{i}}
\end{aligned}
$$

Substitute $q_{3}$ by $\psi$ in (2), RHS:

$$
\begin{aligned}
\frac{\partial \mathcal{L}}{\partial \psi} & =\sum_{i=1}^{3} \frac{\partial \mathcal{L}}{\partial \omega_{i}} \frac{\partial \omega_{i}}{\partial \psi} \\
& =\sum_{i=1}^{3} I_{i} \omega_{i} \frac{\partial \omega_{i}}{\partial \psi} \\
& =I_{1} \omega_{1}(\dot{\varphi} \sin \theta \cos \psi-\dot{\theta} \sin \psi)+I_{2} \omega_{2}(-\dot{\varphi} \sin \theta \sin \psi-\dot{\theta} \cos \psi)+0 \\
& =I_{1} \omega_{1} \omega_{2}+I_{2} \omega_{2}\left(-\omega_{1}\right) \\
& =\left(I_{1}-I_{2}\right) \omega_{1} \omega_{2}
\end{aligned}
$$

*this note is rearranged based on Qianye YU’s work LHS:

$$
\frac{d}{d t}\left(\frac{\partial \mathcal{L}}{\partial \dot{\psi}}\right)=I_{3} \dot{\omega}_{3}
$$

Combing (2)(4)(5) gives the first Euler dynamic equation:

$$
\left(I_{1}-I_{2}\right) \omega_{1} \omega_{2}-I_{3} \dot{\omega}_{3}=0
$$

Similarly substitute $q_{2}$ by $\varphi$ into equation(2). Since $\frac{\partial \mathcal{L}}{\partial \varphi}=0$, we have:

$$
\frac{d}{d t}\left(\frac{\partial \mathcal{L}}{\partial \dot{\varphi}}\right)=0
$$

Apply chain rule to LHS of (7) gives:

$$
\begin{aligned}
0= & I_{1} \dot{\omega}_{1} \sin \theta \sin \psi+I_{2} \dot{\omega}_{2} \sin \theta \cos \psi+I_{3} \dot{\omega}_{3} \cos \theta+I_{1} \omega_{1}(\cos \theta \sin \psi \dot{\theta} \\
& +\sin \theta \cos \psi \dot{\psi})+I_{2} \omega_{2}(\cos \theta \cos \psi \dot{\theta}-\sin \theta \sin \psi \dot{\psi})-I_{3} \omega_{3} \sin \theta \dot{\theta}
\end{aligned}
$$

We introduce $A=I_{1} \omega_{1}(\cos \theta \sin \psi \dot{\theta}+\sin \theta \cos \psi \dot{\psi})$ and $B=I_{2} \omega_{2}(\cos \theta \cos \psi \dot{\theta}-$ $\sin \theta \sin \psi \dot{\psi})$ to reduce the calculation.

Notice $I_{3} \dot{\omega}_{3}=\left(I_{1}-I_{2}\right) \omega_{1} \omega_{2}$. So (8) is equivalent to:

$$
\begin{aligned}
\sin \theta\left[I_{1} \dot{\omega}_{1} \sin \psi+I_{2} \dot{\omega}_{2} \cos \psi\right] & =\left(I_{2}-I_{1}\right) \omega_{1} \omega_{2} \cos \theta+I_{3} \omega_{3} \sin \theta \dot{\theta}-A-B \\
& =I_{1}\left(-\omega_{1} \omega_{2} \cos \theta-\omega_{1} \cos \theta \sin \psi \dot{\theta}-\omega_{1} \sin \theta \cos \psi \dot{\psi}\right) \\
& +I_{2}\left(\omega_{1} \omega_{2} \cos \theta-\omega_{2} \cos \theta \cos \psi \dot{\theta}+\omega_{2} \sin \theta \sin \psi \dot{\psi}\right) \\
& +I_{3} \omega_{3} \sin \theta \dot{\theta} \\
& =I_{1} \omega_{1}(-\dot{\varphi} \sin \theta \cos \psi \cos \theta+\dot{\theta} \sin \psi \cos \theta-\dot{\theta} \sin \psi \cos \theta-\sin \theta \cos \psi \dot{\psi}) \\
& +I_{2} \omega_{2}(\dot{\varphi} \sin \theta \sin \psi \cos \theta+\dot{\theta} \cos \psi \cos \theta-\dot{\theta} \cos \psi \cos \theta+\sin \theta \sin \psi \dot{\psi}) \\
& +I_{3} \omega_{3} \sin \theta \dot{\theta} \\
& =I_{1} \omega_{1}(-\dot{\varphi} \sin \theta \cos \psi \cos \theta-\sin \theta \cos \psi \dot{\psi}) \\
& +I_{2} \omega_{2}(\dot{\varphi} \sin \theta \sin \psi \cos \theta+\sin \theta \sin \psi \dot{\psi}) \\
& +I_{3} \omega_{3} \sin \theta \dot{\theta}
\end{aligned}
$$

Dividing (9) through $\sin \theta$ gives:

$$
\begin{aligned}
I_{1} \dot{\omega}_{1} \sin \psi+I_{2} \dot{\omega}_{2} \cos \psi & =I_{1} \omega_{1}(-\dot{\varphi} \cos \psi \cos \theta-\cos \psi \dot{\psi}) \\
& +I_{2} \omega_{2}(\dot{\varphi} \sin \psi \cos \theta+\sin \psi \dot{\psi}) \\
& +I_{3} \omega_{3} \dot{\theta}
\end{aligned}
$$

We keep equation (10) for later use

Substituting $q_{3}$ by $\theta$ into equation $(2)$ :

$$
\frac{d}{d t}\left(\frac{\partial \mathcal{L}}{\partial \dot{\theta}}\right)=\frac{\partial \mathcal{L}}{\partial \theta}
$$

So,

$$
\begin{aligned}
\frac{d}{d t}\left(I_{1} \omega_{1} \cos \psi-I_{2} \omega_{2} \sin \psi\right)= & I_{1} \omega_{1} \dot{\varphi} \cos \theta \sin \psi+I_{2} \omega_{2} \dot{\varphi} \cos \theta \cos \psi \\
& +I_{3} \omega_{3} \dot{\varphi}(-\sin \theta)
\end{aligned}
$$

which gives:

$$
\begin{aligned}
I_{1} \dot{\omega}_{1} \cos \psi-I_{2} \dot{\omega}_{2} \sin \psi= & I_{1} \omega_{1}(\dot{\varphi} \cos \theta \sin \psi+\sin \psi \dot{\psi}) \\
& +I_{2} \omega_{2}(\dot{\varphi} \cos \theta \cos \psi+\cos \psi \dot{\psi}) \\
& -I_{3} \omega_{3} \dot{\varphi} \sin \theta
\end{aligned}
$$

Since (10) and (13) are obtained from EL equation of generalized coordinates $\varphi$ and $\theta$ respectively, along with (6), we have used all three independent EL equations, the other two Euler equations must be obtained directly from combinations of (10) and (13):

Firstly, (10) $\cdot \sin \psi+(13) \cdot \cos \psi$ gives:

$$
\begin{aligned}
I_{1} \dot{\omega}_{1} & =I_{2} \omega_{2}(\dot{\varphi} \cos \theta+\dot{\psi})+I_{3} \omega_{3}(\dot{\theta} \sin \psi-\dot{\varphi} \sin \theta \cos \psi) \\
& =I_{2} \omega_{2} \omega_{3}+I_{3} \omega_{3}\left(-\omega_{2}\right) \\
& =\left(I_{2}-I_{3}\right) \omega_{2} \omega_{3}
\end{aligned}
$$

And $(10) \cdot \cos \psi-(13) \cdot \sin \psi$ gives:

$$
\begin{aligned}
I_{2} \dot{\omega}_{2} & =I_{1} \omega_{1}(-\dot{\varphi} \cos \theta-\dot{\psi})+I_{3} \omega_{3}(\dot{\theta} \cos \psi+\dot{\varphi} \sin \theta \sin \psi) \\
& =I_{1} \omega_{1}\left(-\omega_{3}\right)+I_{3} \omega_{3} \omega_{1} \\
& =\left(I_{3}-I_{1}\right) \omega_{3} \omega_{1}
\end{aligned}
$$

Rearranging (6),(14),(15) gives the Eulerian dynamic equations:

$$
\left\{\begin{array}{l}
\left(I_{1}-I_{2}\right) \omega_{1} \omega_{2}-I_{3} \dot{\omega}_{3}=0 \\
\left(I_{2}-I_{3}\right) \omega_{2} \omega_{3}-I_{1} \dot{\omega}_{1}=0 \\
\left(I_{3}-I_{1}\right) \omega_{3} \omega_{1}-I_{2} \dot{\omega}_{2}=0
\end{array}\right.
$$

If you have any question on the article, plz don’t hesitate to contact the website administrator by clicking the email icon below.

Leave a Reply

Your email address will not be published. Required fields are marked *