[paul.fr]

I have my work access pass on a retractable string keyring ( like a yoyo ). In a moment of boredom, i was spinnig a weight on one end and the string extended out of the keychain. When i stopped spinning it, it obviously kept spinning and slowly retracting in to the keychain due to the momentum created. My question is: why did the pull on the string get greater the shorter the string got.

[vdurbha]

I am not very sure if I understand the question completely. But here's what I think might be happening.

The keychain probably has a coiled spring, just like the one that we have in our watches. The weight at the end of the spring is doing some work on the spring, and hence some amount of potential energy (elastic potential energy to be precise) is stored in the string. When you take off your hand on the weight, the spring does work on the weight by exerting a force on the weight. Since any body that has a net force acting on it accelerates, the weight seems to retract quicker and quicker as the string coils inside the keychain, and hence it has the maximum speed when the string is shortest.

Hope this answers your question. Sorry if I misunderstood the question in the first place.

Thanks,
Viswanath

[another_someone]

My suspicion is that as the string unwinds, it uncoils, and the remaining string is closer to the core or the yo-yo/keyring.  As the string winds up upon itself, it coils upon itself, so the newer coils being added are being piled on the earlier coils, and are getting closer to the rim of the yo-yo, and thus ever increasing in radius.  At a constant angular speed (and ofcourse, you will not actually have a constant angular speed, as the complicating factor is that the angular speed will be greatest when the string is longest), then the wider the radius of the coil, then the faster the yo-yo will climb for each turn, and so the greater the pull on the string as the yo-yo climbs up the string.

I suspect that the difference is more noticeable as the yo-yo starts its climb, where is still has a lot of rotational energy, and probably at the initial stages you will feel an increase in pull as the yo-yo climbs, but I suspect that as it gets to the top of its climb, and has very little rotational energy left, the effect will probably be slight.

The other complicating factor is how much pull you actually exert yourself on the string.  When the yo-yo is at the topmost position, the yo-yo will have zero rotational energy, and you will want to impart rotation energy to it quickly at that point, so may actually be giving a small yank to the string yourself; whereas as the yo-yo reaches the bottom of its travels, you will not need to give it more rotational energy, and you probably will not be choosing to pull very much on the string, but simply allow the rotational energy already imparted to the yo-yo to do its job.

[paul.fr]

I am not very sure if I understand the question completely. But here's what I think might be happening.

Me neither and i wrote it! Another lesson learned, don't post when you are too tired to think straight....

[lyner]

This is a bit of 'bookwork' physics.
There is a quantity called angular momentum (symbol L) which all rotating bodies (or systems) posses.
For a simple mass on a string it is given by 
L=mass X velocity X length of string.

The total angular momentum of a system stays the same unless there is an outside force applied.  This is a very well known and well tested law of nature. It works all over the place - for instance, planetary orbits  obey the rule.
The force, from your re-coil spring, is inside the system, so L stays the same in your experiment.
As the string gets shorter, the velocity will increase (see formula) to keep L the same. Half the length will double the speed.  This explains its speed increase.

There is also a formula to tell you the force on the string (F):
F = mass X velocity squared  ⁄  length of string

Looking at this formula, if the length gets less and the velocity gets, proportionally, more,  (as a result of the first formula) then the velocity squared wins - so the force goes up. Half the length will double the force on the string.

Sorry about the Maths but Physics is a bit like that - a 'simple' explanation  in mechanics nearly always does involve some! 

[daveshorts]

I am not actually convinced because the angular momentum isn't conserved (as friction will be acting upon your hand or the air), actually the thing that is limiting the force is the strength of the spring.
If the centrifugal force (mv²/r) is larger than the spring force at that extension then the string will pay out, if it is smaller it will come back in. So the centrifugal force you experience as a function of length should be the same as the restoring force of the spring as a function of length, and if it gets stiffer when there isn't much string payed out the centrifugal force will be larger then.

[lyner]

Yes; BUT angular momentum absolutely has to be conserved. If it seems not to be, then there's ALWAYS an explanation. As I understand, the rules apply even with quantum mechanics.

Friction takes some of the KE from the rotation but the angular momentum you originally imparted to the whirring mass was, of course, balanced by corresponding (negative) angular momentum given to you and the Earth (not measurable, of course). Friction in the string and through the air just reduces the angular momentum  of the mass and reduces the corresponding (negative) momentum of the Earth until they both go to zero again, which is what we started with.

My simple sums only apply  once you have  actually stretched the string out to its max and the system is in a stable condition.

I think your point about the tensioning of the string by the spring can be answered by the way you actually get it to stretch in the beginning.  The string can only act in tension, so, if you want to get the system going, you have to pull the string in a non-radial direction, 'ahead of the mass', if you like, so as to accelerate it.  (Imparting angular momentum to you etc.)
At this stage, your hand is no longer at the centre of the rotation but follows a circular path just ahead of the mass. The faster you move your hand the greater the force and the more in advance of the mass you are pulling. This will stretch the spring more and more and, I think this will be a 'runaway' situation which will cause the mass, rapidly to go to the end. I think if you draw a force vector diagram this will show what I mean.
Once your efforts cease, my described motion will be followed. If you introduce friction   into the problem there is a modification due to energy loss but, unless the friction is very high (too high to allow the recoil mechanism to work, I suspect)  the speed would expect to increase.

You could have a similar sort of situation, using a  control line powered model aircraft-once the fuel ran out.

[lyner]

I have just thought of another point.
As the spring coils up the string, a small amount of energy is transferred into the  rotating system (work done is half k xsquared for a spring).
The only way this can turn up is a KE - i.e. faster motion. To explain this and keep Angular Momentum  conserved, there must be a small but finite amount of circular movement  of the inner end of the string, I guess.

[daveshorts]

Angular momentum has to be conserved in the universe, but not always in a non-isolated system. I would have thought that any processes slowing down the weight will affect the angular momentum as well as the energy it has. Either by affecting the air, or your finger. The fact that the weight will eventually stop indicates this also.

I think the system can be modelled as a mass on a spring orbiting something with friction:
[diagram=166_0]

If everything is in the horizontal plane - as otherwise it gets difficult and non-circular, then I think what I was saying earlier still holds.

[lyner]

I think this has lost direction a bit.
If we look for the simplest analysis of the simplest system first we have the best chance of understanding what's going on. Good science tends to be reductionist.
Let's, initially, ignore friction or the changing diameter of the reel holding the string.  The only force is along the string. If we can find a good reason for 1. a speed increase and ,2. an increase in tension, then we have made a valid start.
I think that was what the original question was really about.
We know that, as the radius of rotation of a rotating object decreases, its angular velocity goes up - this  can be explained in terms of conservation of angular momentum.  This principle yields Keppler's laws for  planetry orbits and relies on a simple, central, attractive force.  It explains skaters spinning fast as they bring their legs closer inwards  and many other phenomena.
The simple sums show that, for the same basic reason, the tension increases in our string.
Just looking at the 'before and after' conditions and using angular momentum gives the right answer.
If you don't like my momentum ideas for this particular system then you could replace the central pivot with a balanced system of two equal masses, rotating round their common centre of mass, like a  bola, used for hunting. The string  could be wound in by each mass, at the same rate, to keep things balanced and we would have a completely isolated system, eliminating any possible apparent violation of conservation.   The same result would be produced and  could be a bit more acceptable?
Putting energy in or taking it out should still not violate the momentum conservation. Friction will, eventually, slow things down, in this case, but is it relevant to the basic system?
Of course, this is not the same situation as a satellite in orbit, where the central force follows the inverse square law. The loss in energy caused by friction reduces the orbital height and, as the satellite falls further and further into the potential well, there is more and more (gravitational potential) energy available; the satellite speeds up. Angular momentum is transferred to the atmosphere (and speeds up the world a bit), so the  model, in that case relies on the satellite losing some momentum - due to a non-radial force.  Without the friction, the satellite would just follow a Newtonian orbit, not spiraling in at all.
If the arrangement was like the diagram in Daveshorts' diagram and if the spring was  'open' and the coils never touched each other, following Hooke's Law, there would be no loss of energy and the mass would follow some sort of oscillating path around the centre, with varying radius - either spiraling in and out or like flower petals  /a fried egg, depending on the rate of rotation and the stiffness of the spring. (I tried, briefly, to draw it but failed).

These discussions certainly make you think, don't they?

[diagram=170_0]

[acsinuk]

The force inside the spring holding the mass inwards is not dependent on the length of the string.  It is the quadrature velocity that creates the momentum So Force = M x L/T per each second is a better way of looking at the formula.

Force= Mass x Velocity per second in Newtons

[acsinuk]

OR is it

Force = mass x velocity squared/dr
where dr is the quadrature change in the length of string holding the mass inwards which is nominally zero Sin90.    We need to do an experiment to prove whether force is proportional to velocity or velocity squared as the dimensions ML/T^2 can be correct either way???

[lyner]

How can you possibly say that all this needs to be verified by experiment? It's all been measured years and years ago. Every design of turbine and high speed electric motor has relied on this theory for decades. The theory is copper bottomed.
If you need to prove it for yourself, that is another matter but you can believe the theory 100% for every situation at sub relativistic speeds.