Intermediate axis theorem (Tennis racket theorem)

In summary, the conversation revolved around the Euler equations for a special case where certain variables were constant. The speaker attempted to solve for the angular velocities using approximations and substitutions, but faced difficulties due to not knowing certain values. They also made a mistake in their derivation, but ultimately came to a solution involving imaginary quantities. The other person in the conversation suggested rewriting the solution without imaginary quantities or redefining one of the variables to avoid them altogether.
  • #1
Lambda96
157
59
Homework Statement
see screenshot
Relevant Equations
Euler equation
Hi,

unfortunately, I am not getting anywhere with task b

Bildschirmfoto 2022-12-17 um 20.39.23.png


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##.
 
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  • #2
Lambda96 said:
After that I got the following equation$$\dot{\omega}_2+\frac{(I_1-I_3)\omega_1}{I_2}\omega_3=0$$$$\dot{\omega}_3-\frac{(I_2-I_1)\omega_1}{I_3}\omega_2=0$$
Shouldn't the second equation have a plus sign for the second term on the left side?

Lambda96 said:
With ##\frac{(I_1-I_3)\omega_1}{I_2}=\Omega_1## and ##\frac{(I_2-I_1)\omega_1}{I_3}=\Omega_2##,
With these definitions, note that ##\Omega_2## is a negative number. It might be less confusing if you define ##\Omega_2 = \frac{(I_1-I_2)\omega_1}{I_3}## so that both ##\Omega_1## and ##\Omega_2## are positive.

Lambda96 said:
$$\dot{\omega}_2+\Omega_1 \omega_3=0$$
$$\dot{\omega}_3- \Omega_2 \omega_2=0$$

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##.

Take the time derivative of the first equation and then use the second equation to substitute for ##\dot \omega_3##.
 
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  • #3
Thank you TSny for your help 👍

Regarding the minus sign, I had made a mistake, I actually meant that ##\Omega_2## is negative, since ##I_1>I_2## and thus becomes negative. But I'll leave this interpretation out of the derivation for now.

The equation ##\dot{\omega_2}+\Omega_1 \omega_3=0## derived in time is as follows, actually I would have to use the product rule, but since ##\dot{\omega_1}=0## I can leave this term out directly and get

$$\ddot{\omega_2}+\Omega_1 \dot{\omega_3}=0$$

The second equation then solved for ##\dot{\omega_3}## is ##\dot{\omega_3}=-\Omega_2 \omega_2##.

Then substituting back into the first equation gives

$$\ddot{\omega_2}-\Omega_1 \Omega_2 \omega_2=0$$

Solving this differential equation then gives:

$$\omega_2=c_1e^{\sqrt{\Omega_1 \Omega_2} t}+c_2e^{-\sqrt{\Omega_1 \Omega_2} t}$$
 
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  • #4
Lambda96 said:
The equation ##\dot{\omega_2}+\Omega_1 \omega_3=0## derived in time is as follows, actually I would have to use the product rule, but since ##\dot{\omega_1}=0## I can leave this term out directly and get

$$\ddot{\omega_2}+\Omega_1 \dot{\omega_3}=0$$
ok

Lambda96 said:
The second equation then solved for ##\dot{\omega_3}## is ##\dot{\omega_3}=-\Omega_2 \omega_2##.
ok. It appears to me that you are still defining ##\Omega_2## as ##\frac{(I_2-I_1)\omega_1}{I_3}##. So, ##\Omega_2## is a negative number. That's alright, but then you need to keep in mind that ##\sqrt{\Omega_1 \Omega_2}## is imaginary.

Lambda96 said:
Then substituting back into the first equation gives

$$\ddot{\omega_2}-\Omega_1 \Omega_2 \omega_2=0$$

Solving this differential equation then gives:

$$\omega_2=c_1e^{\sqrt{\Omega_1 \Omega_2} t}+c_2e^{-\sqrt{\Omega_1 \Omega_2} t}$$
ok. The exponents on ##e## are imaginary. I think it would be nicer to rewrite this without imaginary quantities. You could use Euler's formula. Or, you could go back and define ##\Omega_2## such that it is positive and rederive the differential equation for ##\ddot \omega_2##. Then, you don't run into imaginary quantities.
 

1. What is the Intermediate Axis Theorem (Tennis Racket Theorem)?

The Intermediate Axis Theorem, also known as the Tennis Racket Theorem, is a principle in physics that describes the motion of a rotating object around its intermediate axis of rotation. It states that an object with three distinct axes of rotation will exhibit stable rotation around its intermediate axis, but unstable rotation around its other two axes.

2. How does the Intermediate Axis Theorem apply to a tennis racket?

In the case of a tennis racket, the handle is the intermediate axis, while the other two axes are the vertical and horizontal axes. This means that when a tennis racket is spinning, it will remain stable around the handle, but will be less stable around the other two axes. This is why a tennis player can easily spin the racket around the handle, but not around the other two axes.

3. What are some real-life applications of the Intermediate Axis Theorem?

The Intermediate Axis Theorem has many applications in engineering and physics, such as in the design of gyroscopes and spacecrafts. It is also used in sports equipment, such as tennis rackets, baseball bats, and golf clubs, to ensure stability during rotation.

4. How is the Intermediate Axis Theorem related to the conservation of angular momentum?

The Intermediate Axis Theorem is closely related to the conservation of angular momentum, which states that the total angular momentum of a system remains constant unless acted upon by an external torque. In the case of the tennis racket, the rotation around the intermediate axis is stable due to the conservation of angular momentum.

5. Can the Intermediate Axis Theorem be applied to other objects besides a tennis racket?

Yes, the Intermediate Axis Theorem can be applied to any object with three distinct axes of rotation, such as a spinning top, a frisbee, or a spinning coin. It is a fundamental principle in physics that helps explain the behavior of rotating objects in various contexts.

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