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- Thread starter Fusiontron
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What is your evidence that this is so? I don't know one way or the other but I find it unlikely. I can certainly agree that it would be hard for a person to do it, but imagine a machine that spins the book about a line that give you the rotation that you want, and then it shoots the book straight up into the air. Why would that not work?

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BruceW

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nasu

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Try to throw the book when spinning around a different axis.

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http://www.aerostudents.com/files/dynamicsAndStability/rigidBodies.pdf

(section 3.3)

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http://www.aerostudents.com/files/dynamicsAndStability/rigidBodies.pdf

(section 3.3)

Thank you. I'll look through this document.

EDIT: Is there a more rigorous derivation of this? I was told it involves complex variables.

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BruceW

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Thank you. I'll look through this document.

EDIT: Is there a more rigorous derivation of this? I was told it involves complex variables.

I'll look around and see what I can find.

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AlephZero

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Is there a more rigorous derivation of this? I was told it involves complex variables.

See http://farside.ph.utexas.edu/teachin...on/node71.html [Broken]

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Matterwave

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This is sometimes called the "tennis racket theorem". It can be derived from a perturbative solution to Euler's equations (torque free).

If you have rotation around any of the 3 principle axes, perturbations about 2 of the axes will lead to nice oscillations in the rotation (I forget if it's called nutation or w/e), but perturbations about the intermediate axis will lead to exponential growth in the perturbation (and therefore violating the validity of your perturbative solution).

Euler's equations (torque free):

[tex]I_1\frac{d}{dt}\omega_1=(I_2-I_3)\omega_2\omega_3[/tex]

[tex]I_2\frac{d}{dt}\omega_2=(I_3-I_1)\omega_3\omega_1[/tex]

[tex]I_3\frac{d}{dt}\omega_3=(I_1-I_2)\omega_1\omega_2[/tex]

So, if we take the rotation to be almost all in the 1-axis (e.g. omega2 and omega 3 are small), and work to first order

We find that omega1 is roughly constant because we neglect the second order in smallness for the omega2*omega3 term; however, for omega 2 term (for example):

[tex]\frac{d^2}{dt^2}\omega_2=\left[\frac{(I_3-I_1)(I_1-I_2)}{I_3 I_2}\omega_1^2\right]\omega_2[/tex] (hopefully I did the algebra right).

We see then that for the omega2 term to remain small (and oscillate), then I1 must either be the largest or the smallest moment of inertia (thereby making the coefficient on the right hand side negative). If I1 is an intermediate moment of inertia, the coefficient on the right hand side is positive, and that means the solution for omega2 is exponential growth and not oscillatory.

One should note, though, that if you can PERFECTLY make the rotation ONLY on the 1-axis (with omega2 and omega3 being identically 0), then even if the 1-axis is the unstable axis, you won't get any wobbling. But this only works if you can make the rotation ONLY around the 1-axis.

If you have rotation around any of the 3 principle axes, perturbations about 2 of the axes will lead to nice oscillations in the rotation (I forget if it's called nutation or w/e), but perturbations about the intermediate axis will lead to exponential growth in the perturbation (and therefore violating the validity of your perturbative solution).

Euler's equations (torque free):

[tex]I_1\frac{d}{dt}\omega_1=(I_2-I_3)\omega_2\omega_3[/tex]

[tex]I_2\frac{d}{dt}\omega_2=(I_3-I_1)\omega_3\omega_1[/tex]

[tex]I_3\frac{d}{dt}\omega_3=(I_1-I_2)\omega_1\omega_2[/tex]

So, if we take the rotation to be almost all in the 1-axis (e.g. omega2 and omega 3 are small), and work to first order

We find that omega1 is roughly constant because we neglect the second order in smallness for the omega2*omega3 term; however, for omega 2 term (for example):

[tex]\frac{d^2}{dt^2}\omega_2=\left[\frac{(I_3-I_1)(I_1-I_2)}{I_3 I_2}\omega_1^2\right]\omega_2[/tex] (hopefully I did the algebra right).

We see then that for the omega2 term to remain small (and oscillate), then I1 must either be the largest or the smallest moment of inertia (thereby making the coefficient on the right hand side negative). If I1 is an intermediate moment of inertia, the coefficient on the right hand side is positive, and that means the solution for omega2 is exponential growth and not oscillatory.

One should note, though, that if you can PERFECTLY make the rotation ONLY on the 1-axis (with omega2 and omega3 being identically 0), then even if the 1-axis is the unstable axis, you won't get any wobbling. But this only works if you can make the rotation ONLY around the 1-axis.

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