Bead on a vertical ring centripetal acceleration

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SUMMARY

The discussion focuses on the dynamics of a bead on a vertical ring spinning at 4 revolutions per second, with a radius of 0.100 m. The key question is determining the angle at which the bead achieves equilibrium and whether it can reach the elevation of the ring's center. The radial acceleration formula, defined as radial acceleration = v²/R = 4(π)²R/(T)², is essential for calculating the bead's motion. The inquiry also addresses the scenario when the ring spins at one revolution per second, indicating a change in the bead's behavior.

PREREQUISITES
  • Understanding of centripetal acceleration and its formula
  • Knowledge of free body diagrams and forces acting on objects
  • Familiarity with angular velocity and its relation to linear velocity
  • Basic principles of equilibrium in physics
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  • Explore the concept of radial acceleration in circular motion
  • Learn how to draw and analyze free body diagrams for dynamic systems
  • Investigate the effects of varying angular velocities on equilibrium positions
  • Study the principles of frictionless motion in circular paths
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Homework Statement


There is a vertical ring with a bead strung on it. The vertical ring spins at 4 revolutions per second, and the the bead moves up the ring at what angle from the bottom of the ring will the bead be at equilibrium? Is it possible for the bead to make it all the way up to the same elevation as the center of the ring? What will happen if the ring only spins at one revolution per second? The ring itself has a radius of 0.100 m and the bead has no friction with the ring.


Homework Equations


radial acceleration= v^2/R = 4(pi)^2*R/(T)^2
v=velocity
R=radius
T=time of revolution

The Attempt at a Solution



I tried to use the acceleration equations above to calculate the radial acceleration of the bead, but that doesn't explain why or how much the bead moves up the ring?
 
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Draw a free body diagram for the bead.
 

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