# Homework Help: Wire wrapped around a pipe

1. Sep 25, 2007

### pugfug90

1. The problem statement, all variables and given/known data
Given a pipe with 8cm circumference and length 20cm that is 8 turns of wire wrapped around a pipe, what's the length of the wire?

2. Relevant equations

circumference=2(pi)r=diameter*pi (I think)

3. The attempt at a solution

This is pic with problem. My approach, first, take circumference of 8cm. set to formula to find radius, which can then be used to find diameter.

8cm=2(pi)r..r=4cm/pi

Then pic shows the tube being "divided" into 8 pieces. I then draw a diagram, and measure length as the sum of the diagonal blue parts + the sum of the nonvisible imaginary lines passing behind the tube as it revolves. Tricky because there are 8 diagonal lines and 7 horizontal lines... o.o

So yeah, r=4cm/pi, so d=8cm/pi or 2.55. This is the horizontal part. For the vertical part of the diagram (if you cut the tube into eight thingies), I took the height of 20cm, divided by 8 (you see 8 sections on the pic), and got 2.5. Using pythagorean theorem to find the diagonal, 2.5 squared plus 2.55 squared is 12.7525, so diagonal is 3.57.

Now I have diagonal, add the 8 diagonal parts together, so 8(3.57)=28.56. +7(2.55)=46.41.

But then.. The horizonal part, instead of a straight edged distanced, could also be half of the circumference, or pi*r.. But then that diagonal across the tube isn't flat edged either.. oii

The answer is 67.05 for those that can get it right.

2. Sep 25, 2007

### Dick

If you think hard, you'll realize that if you can 'unwrap' the wire going around the cylinder into a single right triangle and not have to split it into parts. It looks to me like you are calculating the length of the wire projected into the plane of the paper.

3. Sep 26, 2007

### pugfug90

Not sure what you mean...

This is my interpretation of what you just said and I get a mere ~28.

4. Sep 26, 2007

### Dick

I get that the base is 64cm. Eight diameters of the cylinder. Your are projecting again. If you look at a semicircle edge-on it may LOOK like it has length 2R, but that's not it's real length. You are trying to measure distances by essentially putting a ruler on a 3d picture.

5. Sep 26, 2007

### pugfug90

ee

Teacher gave some hints and I got it now. If I were to imagine this problem, but with 1 instead of 8 wraps, it'd be a line, from one end of, say, a paper, to another end.

1------1
1 1
1------1

The horizontal part of the folded out tube, or paper, is the circumference, 8cm. I need to divide the length into 20/8, do pythag with horizontal of 8cm and height of 20/8, multiply that found length by 8 since 8 wraps, voila :D

6. Sep 26, 2007

### Dick

I don't see much point in dividing by 8 and then multiplying by 8, but if it helps you visualize the problem...

7. Sep 27, 2007

### dynamicsolo

As a possible additional source of confusion, does anyone spot something odd about how that winding is drawn?

Do you mean eight circumferences?

As a suggestion to pugfug90 to illustrate what Dick is saying, get the cardboard core from a roll of toilet paper or paper towels. You'll find that the seam running around it forms a spiral or helix like your wire winding. Now take the core apart carefully along the spiral and lay it flat. You should find that this edge is now a straight line (that's how the core is manufactured). How could you find the length of that seam?

Last edited: Sep 27, 2007
8. Sep 27, 2007

### DaveC426913

pug, you're overthinking it.

Examine a simpler case: one winding.

The height is 20/8. The width is not across the tube (the diameter), it's around the tube (the circumference) - which you're given. So what are the xy dimensions of the rectangle that a single winding would span?