(a) Starting from the rotating-frame derivative identity and , derive Euler's…

PHYSICS · Subjective · Fermi University Physics · Mechanics & Dynamical Systems · Rotation, rigid bodies and accelerated frames · Rotation, rigid bodies and accelerated frames

Action and Symmetry, Ch Euler's equations, free precession and the heavy symmetric top

Problem

(a) Starting from the rotating-frame derivative identity (dL/dt)space=(dL/dt)body+ω×L(\mathrm{d}\mathbf{L}/\mathrm{d}t)_{\rm space} = (\mathrm{d}\mathbf{L}/\mathrm{d}t)_{\rm body} + \boldsymbol\omega\times\mathbf{L} and (dL/dt)space=N(\mathrm{d}\mathbf{L}/\mathrm{d}t)_{\rm space}=\mathbf{N}, derive Euler's equations, in body components, for a rigid body with principal moments I1,I2,I3I_1,I_2,I_3 turning under torque N\mathbf N about its centre of mass or a fixed pivot. (b) A symmetric top (I1=I2I_1=I_2) spins on a frictionless pivot with its symmetry axis exactly vertical, spin angular velocity ω3\omega_3, spin moment of inertia I3I_3 about the pivot, transverse moment I1I_1 about the pivot, mass MM and pivot-to-centre distance dd. Use the third Euler equation to explain why ω3\omega_3 stays constant while the axis is exactly vertical. Then, by expanding the heavy top's effective potential Veff(θ)V_{\rm eff}(\theta) to second order in θ\theta about θ=0\theta=0 with pϕ=pψ=I3ω3p_\phi=p_\psi=I_3\omega_3, derive the sleeping condition I32ω32>4I1MgdI_3^2\omega_3^2 > 4I_1Mgd and evaluate the threshold spin numerically for I1=2.5×10−4 kg m2I_1=2.5\times10^{-4}\,\mathrm{kg\,m^2}, I3=6.0×10−5 kg m2I_3=6.0\times10^{-5}\,\mathrm{kg\,m^2}, M=0.18 kgM=0.18\,\mathrm{kg}, d=3.0 cmd=3.0\,\mathrm{cm}, g=9.8 m s−2g=9.8\,\mathrm{m\,s^{-2}}.

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