- #1

Lambda96

- 63

- 21

- Homework Statement:
- see screenshot

- Relevant Equations:
- Euler equation

Hi,

unfortunately, I am not getting anywhere with task b

In the lecture we had the special case that ##\vec{M}=0## , ##I_x=I_y=I , I \neq I_z## and ##\omega_z=const.##

Then the Euler equation looks like this.

$$I_x\dot{\omega_x}+\omega_y \omega_z(I_z-i_y)=0$$

$$I_y\dot{\omega_y}+\omega_z \omega_x(I_x-i_z)=0$$

$$I_z\dot{\omega_z}=0$$

With this, we then set up the following equations, where ##\Omega=\frac{I_z-I}{I}##.

$$\dot{\omega_x}+\Omega \omega_y=0$$

$$\dot{\omega_y}-\Omega \omega_x=0$$

The solution for ##\dot{\omega_x}## and ##\dot{\omega_y}## are then as follows

$$\dot{\omega_x}(t)=Acos{\Omega t + \alpha}$$

$$\dot{\omega_y}(t)=Asin{\Omega t + \alpha}$$

I then wanted to use the approximation ##\omega_1 \gg \omega_2## , ##\omega_1 \gg \omega_3## and ##\omega_3 \omega_2=0## to solve the Euler equation also according to the same recipe

The Euler equation looks like this:

$$I_1\dot{\omega_1}=0$$

$$I_2\dot{\omega_2}+\omega_1 \omega_3(I_1-I_3)=0$$

$$I_3\dot{\omega_3}+\omega_2 \omega_1(I_2-I_1)=0$$

After that I got the following equation

$$\dot{\omega}_2+\frac{(I_1-I_3)\omega_1}{I_2}\omega_3=0$$

$$\dot{\omega}_3-\frac{(I_2-I_1)\omega_1}{I_3}\omega_2=0$$

With ##\frac{(I_1-I_3)\omega_1}{I_2}=\Omega_1## and ##\frac{(I_2-I_1)\omega_1}{I_3}=\Omega_2##, the above equation is

$$\dot{\omega}_2+\Omega_1 \omega_3=0$$

$$\dot{\omega}_3- \Omega_2 \omega_2=0$$

Unfortunately, I'm not getting anywhere now, because I don't know what ##\omega_1## and ##\omega_2## have to look like to satisfy the two equations, because unfortunately I don't have ##\Omega##, but two different ones with ##\Omega_1## and ##\Omega_2##.

unfortunately, I am not getting anywhere with task b

In the lecture we had the special case that ##\vec{M}=0## , ##I_x=I_y=I , I \neq I_z## and ##\omega_z=const.##

Then the Euler equation looks like this.

$$I_x\dot{\omega_x}+\omega_y \omega_z(I_z-i_y)=0$$

$$I_y\dot{\omega_y}+\omega_z \omega_x(I_x-i_z)=0$$

$$I_z\dot{\omega_z}=0$$

With this, we then set up the following equations, where ##\Omega=\frac{I_z-I}{I}##.

$$\dot{\omega_x}+\Omega \omega_y=0$$

$$\dot{\omega_y}-\Omega \omega_x=0$$

The solution for ##\dot{\omega_x}## and ##\dot{\omega_y}## are then as follows

$$\dot{\omega_x}(t)=Acos{\Omega t + \alpha}$$

$$\dot{\omega_y}(t)=Asin{\Omega t + \alpha}$$

I then wanted to use the approximation ##\omega_1 \gg \omega_2## , ##\omega_1 \gg \omega_3## and ##\omega_3 \omega_2=0## to solve the Euler equation also according to the same recipe

The Euler equation looks like this:

$$I_1\dot{\omega_1}=0$$

$$I_2\dot{\omega_2}+\omega_1 \omega_3(I_1-I_3)=0$$

$$I_3\dot{\omega_3}+\omega_2 \omega_1(I_2-I_1)=0$$

After that I got the following equation

$$\dot{\omega}_2+\frac{(I_1-I_3)\omega_1}{I_2}\omega_3=0$$

$$\dot{\omega}_3-\frac{(I_2-I_1)\omega_1}{I_3}\omega_2=0$$

With ##\frac{(I_1-I_3)\omega_1}{I_2}=\Omega_1## and ##\frac{(I_2-I_1)\omega_1}{I_3}=\Omega_2##, the above equation is

$$\dot{\omega}_2+\Omega_1 \omega_3=0$$

$$\dot{\omega}_3- \Omega_2 \omega_2=0$$

Unfortunately, I'm not getting anywhere now, because I don't know what ##\omega_1## and ##\omega_2## have to look like to satisfy the two equations, because unfortunately I don't have ##\Omega##, but two different ones with ##\Omega_1## and ##\Omega_2##.