A frictionless pivoted symmetric top has principal moments (transverse

PHYSICS · Multi-part · 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 frictionless pivoted symmetric top has principal moments I1=4.0×10−4 kg m2I_1=4.0\times10^{-4}\,\mathrm{kg\,m^2} (transverse, about the pivot) and I3=1.0×10−4 kg m2I_3=1.0\times10^{-4}\,\mathrm{kg\,m^2} (about its symmetry axis), mass M=0.25 kgM=0.25\,\mathrm{kg} and pivot-to-centre distance d=3.5 cmd=3.5\,\mathrm{cm}. It spins at ω3\omega_3 with its axis exactly vertical. ((a)) Use the third of Euler's equations, together with the torque about the symmetry axis and I1=I2I_1=I_2, to show that ω3\omega_3 stays constant while the axis remains exactly vertical. ((b)) Steady precession at tilt θ\theta requires the quadratic I1cos⁡θ ϕ˙2−I3ω3ϕ˙+Mgd=0I_1\cos\theta\,\dot\phi^2-I_3\omega_3\dot\phi+Mgd=0 for the precession rate ϕ˙\dot\phi to have a real root. By requiring this in the limit θ→0\theta\to0, derive the sleeping condition on ω3\omega_3, and evaluate the threshold spin numerically for the data above, taking g=9.8 m s−2g=9.8\,\mathrm{m\,s^{-2}}.

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