# Expressing centripetal action with variables

1. Jul 11, 2008

### kylera

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
a(c) = v^2 / r

3. 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?

2. Jul 11, 2008

### LowlyPion

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. Jul 11, 2008

### kylera

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. Jul 12, 2008

### LowlyPion

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. Jul 12, 2008

### kylera

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. Jul 12, 2008

### cryptoguy

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 $$F_{net} = 0$$. So what is the centripetal force equal to?

7. Jul 12, 2008

### kylera

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. Jul 12, 2008

### cryptoguy

correct. now you can definately solve that equation for r and plug into L = r+d