Expressing centripetal action with variables

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Homework Help Overview

The problem involves a string of length L with paper clips attached, where one end is twirled to create circular motion while the other end hangs vertically. The task is to express the distance d as a function of velocity v and length L, and to express gravitational acceleration g in terms of v and the radius r of the circle.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the relationship between the length of the string, the radius of the circular motion, and the distance d. There are attempts to rearrange equations related to centripetal acceleration and velocity, but confusion arises regarding the introduction of variables like acceleration and tension.
  • Some participants question the role of gravitational force and how it relates to the tension in the string and the weights of the paper clips. They explore the implications of equilibrium and the forces acting on the system.
  • Others suggest that the hanging paper clips provide the necessary centripetal force and ponder how to express this relationship mathematically.

Discussion Status

The discussion is ongoing, with participants exploring various interpretations of the forces involved and how they relate to the problem at hand. Some guidance has been offered regarding the relationships between the forces and the variables, but no consensus has been reached on a definitive approach or solution.

Contextual Notes

Participants note the importance of considering the weights of the paper clips and the implications of gravitational force in the context of centripetal motion. There is also mention of the need to clarify the definitions and relationships between the variables involved, particularly in terms of equilibrium and tension.

kylera
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Homework Statement


There is a string of length L dropped through a spool of thread. 10 paper clips are tied to each end and one end is twirled so that one bunch of clips moves in a horizontal circle at speed v while the other hangs vertically. Neglecting friction, write an expression:
a) for the distance d as a function of v and L, where L = r + d (r = radius)
b) for g in terms of v and r, the radius of the circle.


Homework Equations


a(c) = v^2 / r


The Attempt at a Solution


I don't even know where to begin. Since L is already given, d is obviously L - r. I've tried to re-arrange the centripetal acceleration equation to get r = v^2 / a, but that still introduces a. I then took the regular velocity v = 2pi*r / t but rearranging that for r will still give v*t. What am I missing?
 
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When you have "g" involved what do you have to consider?

The 10 paper clips at each end is really saying you have a weight - of 10 paperclips - at each end. Correct?

Is acceleration the only formula you may need to consider at equilibrium?
 
LowlyPion said:
When you have "g" involved what do you have to consider?

The 10 paper clips at each end is really saying you have a weight - of 10 paperclips - at each end. Correct?

Is acceleration the only formula you may need to consider at equilibrium?

If there is g involved, then there's a normal force...but in this case, normal force would equate to tension (as per Newton's Third Law) between the two weights. And since both weights are equal, tension equates to both m x g on the still clips as well as the moving clips. If tension is following the string that's constantly rotating, that means the force is facing out towards the direction of whatever the angle is at the moment. Its tangent force (the equivalent force facing at a 90 degree angle) would be 1/(m * g), no?

However, because this isn't dealing with momentum, length has no relation...argh!
 
kylera said:
If there is g involved, then there's a normal force...but in this case, normal force would equate to tension (as per Newton's Third Law) between the two weights. And since both weights are equal, tension equates to both m x g on the still clips as well as the moving clips. If tension is following the string that's constantly rotating, that means the force is facing out towards the direction of whatever the angle is at the moment. Its tangent force (the equivalent force facing at a 90 degree angle) would be 1/(m * g), no?

However, because this isn't dealing with momentum, length has no relation...argh!

What tension would there be on the string if you were holding the string and the mass was rotating? Have a formula for that?
 
LowlyPion said:
What tension would there be on the string if you were holding the string and the mass was rotating? Have a formula for that?

If it's the string between the rotating clips and the hand, the tension would be solvable via the centripetal acceleration one -- a(c) = v^2 / r

r is given for the radius, but there' still one velocity value. . .god I suck.
 
You have to ask yourself,what is providing the centripetal force? In this case, It's clear that the paper clips that are hanging are providing it (if you took them away, there would not be any centripetal acceleration, would there). Since the string is not accelerating up or down, you know [tex]F_{net} = 0[/tex]. So what is the centripetal force equal to?
 
cryptoguy said:
You have to ask yourself,what is providing the centripetal force? In this case, It's clear that the paper clips that are hanging are providing it (if you took them away, there would not be any centripetal acceleration, would there). Since the string is not accelerating up or down, you know [tex]F_{net} = 0[/tex]. So what is the centripetal force equal to?

Because the net force is zero. . .the centripetal force would equate to the ten clips hanging down -- mg -- that way, the net force is zero. And this is despite whatever value r may be because length is variable!
 
correct. now you can definitely solve that equation for r and plug into L = r+d
 

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