Why does the spinning book have an unstable axis?

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jpswensen
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I searched through the forums and couldn't find a topic or answer, so I will pose it (possibly again). I can work through deriving the equations of motion (either through Euler-Lagrange methods or Hamilton methods) for the spinning book problem and it is obvious that there *is* an unstable axis, but I am wondering if there is a more fundamental explanation as to *why* this occurs. I guess I can see that it happens in the equations and experiments, but don't have any intuition or understanding as to why this occurs.

I have searched the internet and various mechanics books that show derivations of this problem, but haven't seen an explanation of why. If someone could point me to a good book or article that tries to explain it, I would appreciate it.
 
Because moment of inertia is a tensor, so angular momentum and angular velocity are not co-linear in general. Angular momentum is conserved, which forces angular velocity vector to precess.
 
Will this do?

(after Acheson)
 
Attachments
  • Handwritten derivation numbered 1: a sketch of a rectangular box with x, y and z axes, then Euler's equations of rotational motion: P dω_x/dt = L + (Q - R) ω_y ω_z, Q dω_y/dt = M + (R - P) ω_x ω_z, R dω_z/dt = N + (P - Q) ω_x ω_y, with P, Q, R moments of inertia and L, M, N applied couples.
    box1.jpg
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  • Handwritten derivation on lined paper, page 2, of a small perturbation about a rotation axis: linearised Euler equations for P dωx/dt and R dωz/dt, then eliminating δ3 to give a second order linear differential equation in δx.
    box2.jpg
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  • Handwritten page (3) deriving stability conditions: d2s/dt2 + βs = 0, with solutions for β>0 (bounded cos/sin oscillation, ω = √β) and β<0 (growing exponentials with q = √(-β)), with small sketches of each case.
    box3.jpg
    19.1 KB · Views: 537
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