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As the particle travels around the cylinder (having radius r) and the taught string wraps around the cylinder, the path of the particle will trace out a spiral with radius that decreases as a function of the cylinder radius. In other words, the radius of the spiral (let's call S) equals the string length at any given point. All you need to do is find the ...

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The path will be an Archimedean spiral. I think the key to this is that the speed of the particle does not change. You have an equation for the angular velocity in terms of the radius. You also have an equation for the rate at which r is changing in terms of its angular velocity: r is changing by $$-2 \pi R\omega$$ Which seems to be all you need.

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Yes the energy input into the string comes from the wind. Essentially, the flat cross section acts like a wing producing lift when moving upwards, and downforce when moving downwards. On each cycle there is more and more energy added. This energy travels to the ends and some of it dissipates and some of it bounces back only to add to the motion. When you ...

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Well, if the string was pulled such that it was along a radial arm from the centre of the "earth", in other words, in a vertical, then it wouldn't sag. Any deviation such that it is no longer exactly vertical, then the above answers come into play. I know this isn't what you are really asking, but let's admit, it does apply to the question as stated.

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