What is the speed of a ball released at the bottom of a vertical circular path?

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SUMMARY

The discussion focuses on the mechanics of a 2kg ball released from the bottom of a vertical circular path with a radius of 1.2m, where a 50N force is applied parallel to the motion. The initial speed of the ball at the top of the circle is 20m/s. The relevant equation used is the conservation of mechanical energy, represented as KEi + PEi = KEf + PEf, which allows for the calculation of the speed of the ball upon release at the bottom of the circle.

PREREQUISITES
  • Understanding of kinetic energy (KE) and potential energy (PE)
  • Familiarity with the principles of conservation of energy
  • Knowledge of forces acting on objects in circular motion
  • Basic algebra for solving equations
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  • Explore the calculation of speed and energy transformations in vertical circular paths
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Students studying physics, particularly those focusing on mechanics and energy conservation, as well as educators looking for examples of circular motion problems.

alevis
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Homework Statement


A pitcher rotates a 2kg ball around a vertical circular path of radius 1.2m. he exerts a 50N force directed parallel to the motion of the ball around the completer circular path. The speed of the ball at the top of the circle is 20m/s. If the ball is released at the bottom of the circle, what is the speed upon release?


Homework Equations


KEi+PEi = KEf+PEf


The Attempt at a Solution


wd = Fd
 
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Can you explain what you wrote under "3. The Attempt at a Solution ".
What does that formula give you? Why is it useful?
 

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