This problem has two independent parts.

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 Non-inertial frames from one Lagrangian

Problem

This problem has two independent parts. ((a)) Tokyo is at latitude λ=35.7∘N\lambda = 35.7^\circ\mathrm N. Using the horizontal-plane result derived from the frame Lagrangian, that a Foucault pendulum's plane of swing turns at rate Ω⊕sin⁡λ\Omega_\oplus\sin\lambda relative to the floor, with Ω⊕=7.292×10−5 rad s−1\Omega_\oplus = 7.292\times10^{-5}\ \mathrm{rad\,s^{-1}}, find this rate in degrees per hour at Tokyo's latitude. ((b)) A bead of mass m=50 gm = 50\ \mathrm g slides without friction along a rigid horizontal rod that rotates at a fixed rate Ω=4.0 rad s−1\Omega = 4.0\ \mathrm{rad\,s^{-1}} about a fixed vertical axis through one end of the rod. The rod is horizontal, so gravity does no work along the bead's motion and may be dropped from the potential energy for this problem. At some instant the bead is at radial distance r′=0.30 mr' = 0.30\ \mathrm m from the axis, sliding outward at r˙′=0.50 m s−1\dot r' = 0.50\ \mathrm{m\,s^{-1}}. Using the frame Lagrangian L=12mr˙′2+12mΩ2r′2L = \tfrac12 m\dot r'^2 + \tfrac12 m\Omega^2 r'^2, compute the Jacobi integral hh at this instant, then compute the energy EE and the angular momentum component LzL_z measured in the inertial (non-rotating) frame at the same instant, and confirm that h=E−ΩLzh = E - \Omega L_z.

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