Recall Chapter 4's bead on a wire: a bead of mass slides without friction on a rigid…
PHYSICS · Multi-part · Fermi University Physics · Mechanics & Dynamical Systems · Rotation, rigid bodies and accelerated frames · Lagrangian mechanics
Action and Symmetry, Ch Non-inertial frames from one Lagrangian
Problem
Recall Chapter 4's bead on a wire: a bead of mass m slides without friction on a rigid wire of fixed shape y=w(x) in a vertical plane, and the whole wire is carried along its own horizontal axis at a constant speed u imposed by an external mechanism. With x measured along the wire's axis, the ground coordinates are X=x+ut and Y=w(x), and Chapter 4 found the conserved Jacobi integral h=21mx˙2(1+w′(x)2)−21mu2+mgw(x). For a bead released at rest on the wire at x=0.300 m, where w=0.0900 m, with m=10.0 g and u=2.00 m/s, this gives h=−0.0112 J.
((a)) Write down the Lagrangian for the bead in axes that translate with the wire, at the constant velocity u along its axis, and do not rotate, using the general moving-frame Lagrangian. Show that every fictitious force vanishes for this frame, and that the wire-frame Lagrangian is L′=21mx˙2(1+w′(x)2)−mgw(x). Name the two pieces of Chapter 4's Lagrangian that L′ lacks, and say why dropping them changes no equation of motion.
((b)) Write down the conserved integral h′ of L′ (justify why Lagrange's equations still hold in these coordinates) and evaluate it at release, using g=9.81 m/s2.
((c)) Show that h′−h=21mu2 and use this to write h′ as the sum −0.0112J+0.0200J. State which of h′ and h is the mechanical energy measured by an observer riding on the wire.