Angular Acceleration: Min Speed for Cord to Not Slack

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SUMMARY

The discussion focuses on calculating the minimum speed required for a ball of mass 22.8 g, attached to a cord of length 0.739 m, to prevent the cord from becoming slack while rotating in a vertical circle. The acceleration due to gravity is 9.8 m/s². Participants emphasize the importance of applying energy conservation principles alongside the centripetal force equation, F = mv²/r, to derive the solution. The key insight is that at the top of the circle, the tension in the cord can be considered zero for minimum speed calculations.

PREREQUISITES
  • Understanding of centripetal force (F = mv²/r)
  • Basic principles of energy conservation in physics
  • Knowledge of gravitational acceleration (9.8 m/s²)
  • Familiarity with kinetic and potential energy concepts
NEXT STEPS
  • Study the conservation of mechanical energy in rotational motion
  • Learn how to apply centripetal force equations in circular motion problems
  • Explore the relationship between tension and speed in vertical circular motion
  • Practice solving similar problems involving mass, radius, and gravitational forces
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and circular motion, as well as educators seeking to enhance their teaching methods in these topics.

notsam
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Homework Statement

Hello every one :) I do NOT understand rotational speed ect. SO
A ball of mass 22.8 g is attached to a cord of
length 0.739 m and rotates in a vertical circle.
What is the minimum speed the ball must
have at the top of the circle so the cord does
not become slack? The acceleration of gravity
is 9.8 m/s2 .
Answer in units of m/s.



Homework Equations

F= ma, F=mv^2/r



The Attempt at a Solution

Ok So this seems fairly simple, I'm pretty sure that I am supposed to set ma=mv^2/r? Is that correct?
 
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notsam said:
Ok So this seems fairly simple, I'm pretty sure that I am supposed to set ma=mv^2/r? Is that correct?

Yes but using only this will not work :redface:

Try energy conservation

kinetic energy at lower most point provides potential energy to reach top + kinetic energy at top.
This gives you 1 eqn

Now use mv2/r for speed at top point ... which is provided by weight and tension
and for minimum speed, tension = 0​
:biggrin:
 
Thank you, I never thought about using the conservation of energy to work it out. Then using each end of the "radius" as a refrense point for 0 meters. That's smart!
 

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