Angular Acceleration: Min Speed for Cord to Not Slack

AI Thread Summary
To determine the minimum speed for a ball attached to a cord rotating in a vertical circle, the conservation of energy principle is essential. At the lowest point, the kinetic energy must equal the potential energy needed to reach the top, plus the kinetic energy at the top. The equation mv^2/r can be applied at the top, where the tension in the cord is zero for minimum speed. This approach effectively combines energy conservation with centripetal force requirements. Understanding these concepts is crucial for solving the problem correctly.
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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