A rope 1 cm wide and 100 m long is wound up. What is the diameter of the wound up rope?

A rope 1 cm wide and 100 m long is wound up. What is the diameter of the wound up rope?

420

sqrt(10000/pi) cm

are you sure

10000 centimeters squared

Derp 10,000 is the area, so divide by 2, get sqrt(10,000/pi) centimeters

How tightly and in what winding pattern is it wound user?

Instead of actually figuring this out, I'm gonna try to get a rough estimate because I'm bored. Maybe someone smarter than me can finish the job. So instead of using a spiral lets use concentric circles, with the center circle being 1 cm in diameter. So this first circle takes off ~ 1 cm off the total, so we are left with 99.99 m. The next circle will have a diameter of 3 cm, and a circumference of 3pi cm. This leaves a total length of 99.98575 m.

If we continue this pattern, we can bring the total length down to 0. So for the nth concentric circle, the diameter will be (2n-1) cm. The circumference of the nth circle is then (2n-1)*pi cm. And so we just have to find the k that satisfies: [eqn] 0=9999- \sum_{n=2} ^{k} (2n-1)*pi [/eqn]
And then plug this k value into the equation for the diameter (2k-1) cm. I went ahead and did this on wolfram alpha and got [math] \sqrt{ \frac{9999+pi}{pi} } [/math] or ~56 cm. So it turns out you can pack 100 meters of rope into a spiral of about a half meter diameter.

Loosely, such that a cylinder containing the wound rope has precisely 50% rope and 50% air by mass. In an even equidistant spiral, somewhat similar to OP pic related.

I guess I understand the answer now

The area of the circle has to be the area of the rope, so the area is 10,000cm^2...

I thought I had come up with a cooler question, but that is just boring

Asking for diameter instead of radius is truly a cruel thing to do.

Yea woops
>sqrt(10000/pi) cm
is the radius

The rope is 1x10000 centimeters you dumbass, that won't change if you wrap it into a circle

No need to call names you piece of shit. I was doing an approximation because spirals cannot have a constant diameter. So instead of reposting what everyone else said in the thread, why don't you figure out the diameter as a function of the distance to the end of the rope for me. Oh wait, the only math you can do is how to find areas of rectangles and circles, you sure are a dumbass. Fuck you.

>being buttblasted because you can't see obvious solutions to problems

Kek stay mad

ok smartie solve this one

couldn't find the diameter as a function of distance to the end of the rope I see. Well I hope you at least tried and failed instead of just being a shitposting faggot.

50% rope and 50% air by mass? Don't you mean by volume?

If by volume, then it would be 2 times the diameter wouldn't it?

Damn, you were very close to what I did, basically the same thing, but with integrals instead of summations.

[eqn]10000 = 2\pi \int_1^x r dr[/eqn]
[eqn]10000 = 2\pi \left(\frac{x^2}{2} - \frac{1}{2}\right)[/eqn]

Solve for x in WolframAlpha, and I got [math]\sqrt{\frac{10000 + \pi}{\pi}}[/math]

btw, x equals the radius, so the diameter would be double of that

Should be from 0 to x, and then you get the normal area of a circle, pi*r^2