Expressing centripetal action with variables

In summary, the conversation discusses the problem of finding the distance and acceleration of a string with paper clips tied to each end, one end of which is twirled in a horizontal circle while the other hangs vertically. The equation a(c) = v^2 / r is considered for solving the problem, as well as the concept of tension and net force. The conversation also considers the presence of gravity and its impact on the problem.
  • #1
kylera
40
0

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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  • #2
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?
 
  • #3
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!
 
  • #4
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?
 
  • #5
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.
 
  • #6
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?
 
  • #7
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!
 
  • #8
correct. now you can definitely solve that equation for r and plug into L = r+d
 

1. What is centripetal action?

Centripetal action is the force that acts on an object moving in a circular path, constantly pulling it towards the center of the circle.

2. How is centripetal action expressed mathematically?

The mathematical expression for centripetal action is F = mv^2/r, where F is the centripetal force, m is the mass of the object, v is its linear velocity, and r is the radius of the circular path.

3. Can centripetal action be represented by other variables?

Yes, centripetal action can also be expressed as F = mrω^2, where ω is the angular velocity of the object in radians per second.

4. What is the relationship between centripetal action and centripetal acceleration?

Centripetal action and centripetal acceleration are related through the equation a = v^2/r, where a is the centripetal acceleration, v is the linear velocity, and r is the radius of the circular path. Centripetal action is the force that causes centripetal acceleration.

5. How does centripetal action affect the motion of an object?

Centripetal action causes an object to continuously change direction and move in a circular path. It is responsible for keeping the object moving in a curved path and preventing it from moving in a straight line.

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