How to Find the Shortest Length Covered by a Wound Cable on a Spool?

[Chris L T521]

Sorry about the delay! It slipped my mind last night to post this week's POTW... >_>

Thanks to those who participated in last week's POTW! Here's this week's problem!

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Problem: A cable has radius $r$ and length $L$ and is wound around a spool with radius $R$ without overlapping. What is the shortest length along the spool that is covered by the cable?

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[Chris L T521]

No one answered this week's problem. You can find the solution below.

[sp]We look at points in the center of the bottom of the cable. These points form a helix. The parameterization of this helix is given by the equations\[x = (R+r)\cos t,\quad y=(R+r)\sin t,\quad z=h\left(\frac{t}{2\pi}\right),\]
where $R+r$ comes from the distance from the center of the cable to the point at the center of the bottom of the cable. which is one large radius and one small radius. Here, $h$ is the vertical distance between consecutive loops; as the able makes one loop in $2\pi$ units of $t$, $h$ is the distance traveled upwards in one loop. As soon as we find $h$, we can find the arc length.The width of the small cable is $2r$. The height change of one full rotation is $h$. That same height change can be achieved by moving $2\pi(R+r)$ units ahead on the cable. Drawing this distances as straight lines that create triangles, a similar triangle argument produces
\[\frac{2r}{2\pi(R+r)} = \frac{\sqrt{h^2-4r^2}}{h}\implies h = \frac{2\pi(R+r)r}{\sqrt{\pi^2(R+r)^2-r^2}}.\]Thus, the length of one cycle, $\ell$, around the main axis is
\[\begin{aligned} \ell &= \int_0^{2\pi}\sqrt{x^{\prime}(t)^2 + y^{\prime}(t)^2 + z^{\prime}(t)^2} \,dt\\ &= \int_0^{2\pi} \sqrt{(R+r)^2 + \left(\frac{h}{2\pi}\right)^2 }\,dt \\ &= 2\pi\sqrt{(R+r)^2 + \frac{(R+r)^2r^2}{\pi^2(R+r)^2-r^2} }\\ &= 2\pi(R+r) \sqrt{\frac{\pi^2(R+r)^2-r^2+r^2}{\pi^2(R+r)^2-r^2} } \\ &= \frac{2\pi^2(R+r)^2}{\sqrt{\pi^2(R+r)^2 -r^2}}\end{aligned}\]The number of complete cycles is $n=\lfloor L/\ell \rfloor$. The shortest length then along the length before the final closes, and this length is $hn$, where
\[hn = \frac{2\pi(R+r)r}{\sqrt{\pi^2(R+r)^2-r^2}} \cdot \left\lfloor \frac{L\sqrt{ \pi^2(R+r)^2-r^2}}{ 2\pi^2(R+r)^2}\right\rfloor.\][/sp]