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Rotation of a rectangular prism. 
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#1
Oct2711, 10:20 AM

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For some reason it is impossible for a rectangular prism to have pure rotation along one of its axises. For example, if you throw a hardback book in the air and try to spin it about its width then there is some distortion. What is the explanation for this?



#2
Oct2711, 10:28 AM

PF Gold
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#3
Oct2711, 10:36 AM

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The distortion is probably due to air resistance. If you had a really dense rectangular prism (and your arm was strong enough), then I'm sure you could make it work.



#4
Oct2711, 10:42 AM

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Rotation of a rectangular prism.
Rotation about one of the three principal axes is unstable. The "unstable" axis is the one with intermediate moment of inertia. The axes with maximum and minimum moment of inertia are stable. It is a general property, not related to air drag.
Try to throw the book when spinning around a different axis. 


#5
Oct2711, 11:43 AM

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nasu is correct; I do this demonstration in class.
http://www.aerostudents.com/files/dy...igidBodies.pdf (section 3.3) 


#6
Oct2711, 11:48 AM

P: 90

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


#7
Oct2711, 12:02 PM

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Interesting stuff. And thanks Andy for the pdf. Unfortunately, it doesn't give an explanation of why one of the principle axis is unstable, it simply says "It can be derived"



#8
Oct2711, 03:47 PM

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#9
Oct2711, 05:39 PM

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#10
Oct2711, 06:38 PM

Engineering
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#11
Oct2711, 08:27 PM

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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_2I_3)\omega_2\omega_3[/tex] [tex]I_2\frac{d}{dt}\omega_2=(I_3I_1)\omega_3\omega_1[/tex] [tex]I_3\frac{d}{dt}\omega_3=(I_1I_2)\omega_1\omega_2[/tex] So, if we take the rotation to be almost all in the 1axis (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_3I_1)(I_1I_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 1axis (with omega2 and omega3 being identically 0), then even if the 1axis is the unstable axis, you won't get any wobbling. But this only works if you can make the rotation ONLY around the 1axis. 


#12
Oct2811, 08:12 AM

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Thanks, Matterwave. I found derivations in Landau&Lifgarbagez (section 37) and Arnold's "Mathematical Methods of Classical Mechanics" (section 29), but they are both horribly ugly. The derivation in Goldstein (chapter 5.6) is close to what you posted.



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