On a UIL Calculator Applications test (A Texas public high school calculator-based math competition), I've come across a problem that I can only remember the problem, not the numbers.

A man is standing on the center of a platform of radius $R$, which makes one full revolution every $P$ (for period) seconds. He begins to walk in a straight line at a velocity of $V$, but his path is curved by the spinning of the platform. What total distance is covered by the man?

The speed of a point on the edge of the platform is given by $2{\pi}R/P$. If the platform speed was $0$, then the distance would simply be $R$, however this is no the case. During the contest (under a very tight time constraint), that his distance covering per second would be what he covers at at distance of $R/2$. The time it takes to cross the platform would be $R/V$, and his speed at $R/2$ would be half of the maximum speed, which results as ${\pi}R/P$. The distance covered due to spinning would be $({\pi}R^2)/(2P)$, added to the original distance gave a final answer of $({\pi}R^2)/(2P) + R/2$.

Did I take the most efficient approach to this problem, if even correct? Maybe there is an application to an Archimedean Spiral? Any information would be greatly appreciated.

  • $\begingroup$ Welcome Math.SE! Take the tour to get familiar with this site. Mathematical expressions and equations can be formatted using $LaTeX$ syntax. . If you receive useful answers, consider accepting one. $\endgroup$ – Shailesh Mar 31 '16 at 3:18

We have $r(t) = Vt, \theta(t) = \omega t$, with $\omega = {2 \pi \over P}$

The position is given by $x(t) = Vt( \cos \omega t, \sin \omega t ) $ and so $\dot{x}(t) = V [ ( \cos \omega t, \sin \omega t ) + \omega t ( -\sin \omega t, \cos \omega t ) ]$ and $\|\dot{x}(t)\| = V \sqrt{1+(\omega t)^2}$ and $L = V \int_0^{R \over V} \sqrt{1+(\omega t)^2} dt $.

This gives $L = {V \over 2\omega} \left[ \operatorname{arcsinh} ({\omega R \over V}) + {\omega R \over V} \sqrt{({\omega R \over V})^2+1} \right] $.

Replacing $\omega$ gives: $L = {VP \over 4 \pi} \left[ \operatorname{arcsinh} ({2 \pi R \over VP}) + {2 \pi R \over VP} \sqrt{({2 \pi R \over VP})^2+1} \right] $.

  • $\begingroup$ That looks familiar to the formula for the length of an Archimedean Spiral. Is that the application? $\endgroup$ – Joe Doe Apr 2 '16 at 0:28
  • $\begingroup$ I don't understand your question. What application are you referring to? We have $\operatorname{arcsinh} x = \log(x+\sqrt{x^2+1})$, so yes it is the same as the length of an Archimedean Spiral. $\endgroup$ – copper.hat Apr 2 '16 at 4:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.