Is the motion of a pendulum restricted by a cycloid surface cyclical?

[alejandrito29]

the coordinates of cycloide are

$x= a (\theta- sin \theta)$
$y= a(cos \theta -1)$

If i use $\theta =\omega t$ this is a example of cycloid

but, if i use $\theta=\cos (\omega t)$, ¿this is a cycloid?

My teacher says that in a cycloid pendulum $\theta$ must be oscillatory, but i think that if $\theta=\cos \omega t$ then, this is not cycloid.

Also the dynamics of the pendulum on the cycloid is oscillatory, but, the $x= a (\theta- sin \theta)$ is not oscillatory.

 

[Simon Bridge]

Have you tried plotting the different cases out and seeing for yourself if it's a cycloid and of what kind?
Also - what is the definition of a cycloid? (How would you know one if you saw it?)

 

[alejandrito29]

sss

Have you tried plotting the different cases out and seeing for yourself if it's a cycloid and of what kind?
Also - what is the definition of a cycloid? (How would you know one if you saw it?)

this is my graph, but, at the lagrangian the solution is $\theta=\omega t$ but, my teacher says that the solution must be $\theta$ oscillatory.

I Says that the length of the proyection cycloid is oscillatory $ds/d\theta = \sqrt{(dx/d\theta)^2+(dy/d\theta)^2}$ but, the x coordinate is not oscillatory.

Clearly in the graph, x, and y are oscillatory, but this is not the solution of the lagrangian

Pd: the originall problem is the huygens pendulum

http://books.google.cl/books?id=E64...nepage&q=huygens pendulum lagrangian&f=false

 

[Simon Bridge]

The domain of your graph is not big enough to see the cycloidal behavior of the equations.
Did you try plotting the cyclical angle version of the formula from before?

What is the angle $\theta$ the angle of? What is the physical behavior of the angle going to be?

Perhaps you should be asking about the problem you have to solve instead of (as well as) the particular solutions you have?

The diagram in post #1 looks like you are supposed to solve for the motion of a pendulum whose path is restricted by a cycloidal surface. In which case, I'd have expected the angle to be measured between the line joining the pivot to the bob and the vertical line through the pivot. This angle would, indeed, be oscillatory, but it has nothing to do with the form of the cycloid restriction.

In order to solve the Lagrangian for that problem, you need to express the restriction in terms of y(x)

parameterizing the cylcoid by "A" (so we don't get confused with the time parameter)

$x(A)= R(A-\sin k A),\; y(A)=-R(1-\cos kA)$ where R and k are constants.