1
$\begingroup$

In university I remember learning about a particular curve function with the unusual property that if you were to make a physical curved ramp out of it and roll a ball down it starting from rest, the ball would take the same amount of time to reach the bottom no matter where you started it on the ramp (in a frictionless environment).

I remember that this curve function had a fancy name, but I've long since forgotten it. Can someone tell me what it's called?

$\endgroup$
  • 1
    $\begingroup$ It's a cycloid $\endgroup$ – Omnomnomnom Mar 26 '14 at 12:55
  • 2
    $\begingroup$ Also known as the tautochrone curve $\endgroup$ – user50229 Mar 26 '14 at 12:57
  • $\begingroup$ @Omnomnomnom I think the fancy name the OP was after was "tautochrone" or "isochrone". $\endgroup$ – Daniel Fischer Mar 26 '14 at 12:57
  • $\begingroup$ Actully, it would take the same time sliding. If they were to roll down you would have angular momentum to worry about as well. $\endgroup$ – Arthur Mar 26 '14 at 13:36
  • $\begingroup$ I know that as Brachistocrona $\endgroup$ – user126154 Mar 26 '14 at 14:19
0
$\begingroup$

As already said from other users the name is "tautochrone". I wanted to expand on the last comment from @Arthur and a comment is too small. If I am not mistaken from my quick calculation it shows that the time needed for rolling is not the same as for sliding. Since the conservation of energy contains an additional term. Give a curve $f(x)$ in case you have a particle of mass $m$ sliding your conservation of energy would look like this (potential energy plus kinetic one). Note I am using the notation $x(t)$ for the function that relates the position of the particle at a specific time $t$. The time $t=0$ is the starting moment and the particle is tought to start without velocity. $$ mgf(x(0))=mgf(x)+\frac{1}{2}mv^2 $$ in case you have ball of radius $R$ rolling (without sliding) your conservation of energy contains a term $$ K=\frac{1}{2}I\omega^2 $$ where $\omega$ is the angular velocity. Now is easy to rewrite this term as $$ K=\frac{Mv^2}{5} $$ considering that for a homogenous ball you have $I=2mR^2/5$. So if you write now the conservation of energy you end up with $$ mgf(x(0))=mgf(x)+\frac{7}{10}mv^2 $$ that is similar but with a different coefficient. I hope this helps to clarify the point a bit.

$\endgroup$
  • $\begingroup$ but if you roll, the "isochrony" property is preserved? $\endgroup$ – user126154 Mar 26 '14 at 14:21

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.