A bead slides on a circular wire

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SUMMARY

The discussion focuses on calculating the time it takes for a bead to slide to the bottom of a smooth circular wire of radius b in a vertical plane. Initially, the bead has potential energy (U = mgb) and zero kinetic energy (KE = 0). At the bottom, the potential energy is zero, and the kinetic energy is expressed as KE = mv²/2, where v is derived as v = √(2bg). The key challenge highlighted is determining the time of descent from the starting point to the bottom of the hoop.

PREREQUISITES
  • Understanding of gravitational potential energy and kinetic energy concepts.
  • Familiarity with basic mechanics and motion equations.
  • Knowledge of calculus for solving differential equations.
  • Experience with circular motion dynamics.
NEXT STEPS
  • Research the equations of motion for objects in circular paths.
  • Study the principles of energy conservation in mechanical systems.
  • Learn about the application of calculus in deriving motion equations.
  • Explore the effects of friction and air resistance on bead motion.
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and dynamics, as well as educators looking for practical examples of energy conservation and motion in circular paths.

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Homework Statement


A bead slides on a smooth circular wire of radius b which lies in a vertical plane. The bead starts from rest at a point that is level with the center of the hoop.
Find the time it takes for the bead to slide to the bottom of the hoop.

The attempt at a solution
at t=0
U=mgb, KE=0

at the bottom
U=0, KE=mv2/2, v=(2bg)1/2
 
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