Circular Motion: Find Speed at Point A

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SUMMARY

The discussion focuses on calculating the speed of a lead ball at the top of a circular path, specifically point A, using the principles of circular motion. Given a mass of 0.8 kg and a tension of 4.96 N in a 5-meter string, the initial attempt to calculate speed using the formula v = √(F*r)/m was incorrect. The correct approach requires considering both the gravitational force and the tension in the string acting on the ball at point A, leading to the conclusion that a free body diagram is essential for accurate analysis.

PREREQUISITES
  • Understanding of Newton's Second Law (F=ma)
  • Knowledge of centripetal force (F=m(v²/r))
  • Familiarity with free body diagrams
  • Basic principles of circular motion
NEXT STEPS
  • Study the derivation of centripetal force equations in circular motion
  • Learn how to construct and analyze free body diagrams
  • Explore the effects of tension and weight on objects in circular motion
  • Review examples of circular motion problems involving multiple forces
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and circular motion, as well as educators seeking to clarify concepts related to forces in circular paths.

Bryan Tran
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Homework Statement


A .8 kg lead ball is whirled on the end of a string 5 meters long. When the ball passes through point A at the top of the path, the tension in the string is 4.96 N.
What is the speed of the bob at point A?
*NOTE - POINT A is at the top of the circle...

Homework Equations


F=ma
F=m(v2/r)

The Attempt at a Solution


F=m(v2/r)
v = √(F*r)/m

V = √(4.96*5)/.8
V = 5.57 m/s

DOESN'T WORK...!
 
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At the top of the circle, what are the forces acting and in what directions do they act (up or down) ?

The resultant of these forces is equal to the centripetal force mv2/r
 
Umm force of weight and that's all... and it acts down.
But what does that do?
 
Bryan Tran said:
Umm force of weight and that's all... and it acts down.
But what does that do?

Not just weight. Another force acts on it too. You've mentioned it in the problem itself.
 
Bryan Tran said:
Umm force of weight and that's all... and it acts down.
But what does that do?
Have you drawn a free body diagram showing the forces acting on the ball, or do you feel you have advanced beyond the point where you need to use free body diagrams?
 

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