Solving for Horizontal Displacement: Ball on a String

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SUMMARY

The discussion focuses on calculating the horizontal displacement of a ball on a string after the string breaks at the highest point of its vertical circular motion. The ball moves with a centripetal acceleration 'a' and is released from a height of 2R above the ground. The key equation used is ac = v²/r, leading to the conclusion that the ball initially travels horizontally before descending due to gravity. The participant successfully manipulates the equation to find the solution.

PREREQUISITES
  • Understanding of centripetal acceleration (ac = v²/r)
  • Knowledge of kinematic equations for projectile motion
  • Familiarity with concepts of vertical circular motion
  • Basic principles of gravity (g)
NEXT STEPS
  • Study the principles of projectile motion in physics
  • Learn how to derive equations for motion in vertical circles
  • Explore the effects of centripetal force on objects in circular motion
  • Investigate the relationship between height and potential energy in gravitational fields
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in understanding the dynamics of circular motion and projectile motion in mechanics.

aquapod17
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Homework Statement


A ball on a string moves in a vertical circle of radius R at a constant speed with a centripetal acceleration a. At its
lowest point, the ball is a negligible height above the ground. After several revolutions, the string breaks at the highest
point in the motion. In terms of R, a, and g, find the horizontal displacement of the ball from the time the string breaks.


Homework Equations


ac=v2/r


The Attempt at a Solution


I found v = √acr but I don't know where to go from here.
 
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If it breaks at it's highest point, what direction will the ball be traveling?

How high off the ground was it when it was released?

Doesn't it just become a simple kinematic problem?
 
If it breaks at its highest point it will travel horizontally first, but then reach the ground.

It is 2R off the ground when it was released.

OMG I GOT IT! Thanks so much! I just had trouble manipulating the equation. Embarrassing.
 

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