Length of string around circular rod

**fractal** · December 15th 2008, 04:25 AM

A string is wound symmetrically around a circular rod. The string goes exactly 4 times around the rod. The circumference of the rod is 4 cm. and its length is 12 cm.

Find the length of the string ?

Please see PDF for diagram. Not sure how to simplify this problem to basic geometric assumption & still be correct. Thanks.

**Chop Suey** · December 15th 2008, 04:34 AM

EDIT: Apologies, I've misunderstood the problem.

**fractal** · December 15th 2008, 06:30 AM

Hi Chop Suey:

s=r $\theta$ , where r = C / $(2\pi)$ and $\theta =2\pi$

For 4 revolution, $s=4/(2\pi)*(4*2\pi)=16$

ANSWER = 16, what is your answer?

**Bruce** · December 15th 2008, 07:15 AM

^ I got that answer of 16 by realising that each time it goes around the rod it has gone 4cm. 4 x 4 = 16

Not exactly rocket science. =S

**fractal** · December 15th 2008, 07:28 AM

A google search leads me to believe the answer is not 16.

The Puzzle Instinct: The Meaning of ... - Google Book Search

**earboth** · December 15th 2008, 08:10 AM

Originally Posted by fractal

A string is wound symmetrically around a circular rod. The string goes exactly 4 times around the rod. The circumference of the rod is 4 cm. and its length is 12 cm.

Find the length of the string ?

Please see PDF for diagram. Not sure how to simplify this problem to basic geometric assumption & still be correct. Thanks.

The length of the string is the length of the diagonal of the rectangle with

$l = 4 \cdot 4 cm = 16 cm\ and\ w = 12 cm$

Therefore the length of the string is

$s = \sqrt{16^2 + 12^2} = 20\ cm$

**Soroban** · December 15th 2008, 09:00 AM

Hello, fractal!

There is a wonderful back-door approach to this problem . . .

A string is wound symmetrically around a circular rod.
The string goes exactly 4 times around the rod.
The circumference of the rod is 4 cm and its length is 12 cm.

Find the length of the string.

"Unroll" the cylinderical rod, and we have a 4-by-12 rectangle.
Draw four of them in a row.

Code:

      * - - - - - * - - - - - * - - - - - * - - - - - *
      |           |           |           |     *     |
      |           |           |           *           |
      |           |           |     *     |           |
   12 |           |           *           |           | 12
      |           |     *     |           |           |
      |           *           |           |           |
      |     *     |           |           |           |
      * - - - - - * - - - - - * - - - - - * - - - - - *
            4           4           4           4

Draw the diagonal from one corner to the opposite corner.
This is the path of string as it spirals around the cylinder four times.

Using Pythagorus, we can find the length of that hypotenuse.

**fractal** · December 15th 2008, 09:36 AM

Soroban: your solution is an out of the single box kind.

however, for tutorial to middle & high school students, i think it would be easier to make a single cut through the cylinder, then flatten it, which give you four stripes. each stripe can be treated as a hypotenuse where the height is 4 cm and the width is 3cm. the sum of the four hypotenuse equals 20 cm.

**galactus** · December 15th 2008, 10:13 AM

If I may, here is another way to look at it. Perhaps too complicated, but just for kicks.

I will not go through the derivation of the formula.

The arc length of a circular helix, the string in this case, is given by:

$t_{0}\sqrt{a^{2}+c^{2}}$ if $x=acos(t), \;\ y=bsin(t), \;\ z=ct$

What length of string will make 4 complete revolutions in a length of 12 inches around the cylinder of radius $\frac{2}{\pi}$ ?.

$t_{0}=8{\pi}$ , because we have 4 rev.

a=radius of cylinder = $\frac{2}{\pi}$

$c=\frac{12}{8\pi}=\frac{3}{2\pi}$

$\boxed{8{\pi}\sqrt{\left(\frac{2}{\pi}\right)^{2}+ \left(\frac{3}{2\pi}\right)^{2}}=20}$

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