Tension and centripital force HELP

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    Force Tension
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Homework Help Overview

The problem involves a 2.0 kg ball attached to a vertical post with two strings of different lengths, set to whirl in a horizontal circle. Participants are exploring the minimum speed required for the lower string to become taut and the tension in each string at a constant speed of 6 m/s.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to understand the relationship between centripetal force and the tension in the strings, expressing confusion about the concept of maximum centripetal force.
  • Some participants suggest visualizing the scenario without the lower string to determine when it becomes taut, indicating a need to consider the angle of the upper string as the radius changes.
  • Others propose drawing a free body diagram to identify the forces acting on the ball, including gravitational force and tensions in both strings.
  • Questions arise regarding the calculations for tension and the angles involved, with participants discussing the components of tension in relation to gravitational force.

Discussion Status

The discussion is active, with participants offering insights into the mechanics of the problem and questioning the assumptions made about the forces involved. Some calculations have been presented, but there is no explicit consensus on the correctness of the approaches taken.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information available and the methods they can use to solve the problem. There is an ongoing exploration of the definitions and relationships between the forces acting on the ball.

warmfire540
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Homework Statement



A 2.0 kg ball is attached to a vertical post with two strings, one 2.0 m long and the other 1.0 m long as shown in the figure. If the ball is set whirling in a horizontal circle, what is the minimum speed necessary for the lower string to be taught? If the ball has a constant speed of 6 ms-1, find the tension on each string.


|\
| \
| \ -2.0m
| \
| \
|----0 -2.0kg
|
| ^1m




The Attempt at a Solution



I don't exactly understand what this means, I've tried tackling it several ways.
Only thing I can figure is that when the string is taught it is at maximum centripetal force. However I'm not sure how to find this.
I have so far
F(centripetal)=mv^2/r
F=2v^2/1
F=2v^2
 
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your equations are correct, but there is no maximum centripetal force in this case.
To find out when the bottom string becomes taught:
image that the bottom string isn't there (when its not taught, it effectively isn't - it doesn't do anything). As the ball speeds up, it will move further out (increasing radius) from the pole (i.e. the angle between the pole (vertical) and the top string will increase). When the radius of rotation reaches the length of the bottom string (1m), the bottom string will be taught.
Its probably easier to find this in terms of the angle afore-mentioned.
 
Draw free body diagram
total 3 forces acting:
1. gravitational
2: Tension 1 (upper)
3 tension 2 (lower)

The resultant of the three forces is providing the necessary centripetal force.
Tension must be greater then zero.
solve accordingly.
 
lzkelley said:
your equations are correct, but there is no maximum centripetal force in this case.
To find out when the bottom string becomes taught:
image that the bottom string isn't there (when its not taught, it effectively isn't - it doesn't do anything). As the ball speeds up, it will move further out (increasing radius) from the pole (i.e. the angle between the pole (vertical) and the top string will increase). When the radius of rotation reaches the length of the bottom string (1m), the bottom string will be taught.
Its probably easier to find this in terms of the angle afore-mentioned.


do these answers comply?

For the first part, figured that the 2m string has two components of tension Tsin60 and Tcos60.
Tsin60=mg=19.6
Tcos60=mv^2/r=2v^2
tan60=19.6/2v^2
v=5.66 m/s

is that right? I calculated the angle that the 2m string made w/ the x-axis and went from there...

Now..for the second part..
i figured that the tension in the 1m string is the centripetal force which is:
T=mv^2/r=2*36/1= 72N
Now for the second string..what would the tension be?
i'm guessing i have to use the Tsin60 and the Tcos60
Tsin60=19.6N
Tcos60=72N
T=sqrt(19.6^2+72^2)
T=74.6N
 

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