How Does a Cycloidal Path Ensure Isochronous Oscillations in a Pendulum?

[Vitani11]

Homework Statement

A pendulum is suspended from the cusp of a cycloid cut in a rigid support. The path described by the pendulum bob is cycloidal and is given by x = a (φ − sin φ), y = a (cos φ − 1), where the length of the pendulum is l = 4a, and where φ is the angle of rotation of the circle generating the cycloid. Show that the oscillations are exactly isochronous with a frequency ω_o = sqrt(g/l), independent of the amplitude.

Homework Equations

The Attempt at a Solution

I know how to prove this with two different methods. One- when you state that the kinetic energy of the pendulum is maximum at the bottom of its rotation and finally at the top when its KE is zero the energy is all potential you can set up an equation and solve (1/2)mv_max² = mgl replacing v with omega etc. The second method is just using the fact that if something undergoes a centripetal force mv²/l = mg replacing v with omega and solve etc. How can this help me in this problem? I've went ahead and found the kinetic energy along the path of rotation by taking the derivative of x and y and then squaring and adding, then multiplying by (1/2)m.

dx/dt = a(φ(dot)-φ(dot)cosφ)
dy/dt = -aφ(dot)sinφ

squaring and adding gives (ds/dt)² = a²[(φ(dot)-φ(dot)cosφ)²+(φ(dot)sinφ)²] and so the kinetic energy is (1/2)ma²[(φ(dot)-φ(dot)cosφ)²+(φ(dot)sinφ)²]. I'm now stuck on the potential energy. I can't just say it's mgl/2 because there is tension and using the fact that this result above is the negative of the potential energy is just going to give me the fact that the addition of the two terms is zero (which makes sense because energy is conserved) but doesn't help in solving for ω=g/l

 

[Vitani11]

I've found arclength of the angle it subtends in one revolution which is just 2l. Now I don't know how to deal with the tension and get rid of it...

 

[kuruman]

Use the first approach, energy conservation.

I can't just say it's mgl/2 because there is tension ...

What does tension have to do with potential energy? Potential energy is a function of vertical position only and you already know y as a function of the angle. So ...

 

[Vitani11]

I know. Tension does no work because it is perpendicular to the path of rotation at all points. (Is this correct?) and that was confusing me because I didn't know how to incorporate it into the equation for potential energy and it turns out I don't need to. Anyway, I solved this now aside from a factor of 2. I basically plugged in the kinetic energy explained above (1/2)ma²[(φ(dot)-φ(dot)cosφ)²+(φ(dot)sinφ)²] into the equation T+U = 0 (since energy is conserved, is this also a correct assumption?) and then I know that U = mgy and I used y in that equation to give me (1/2)ma²[(φ(dot)-φ(dot)cosφ)²+(φ(dot)sinφ)²] = -mga (cos φ − 1) and then everything reduces from here and you get that φ(dot) = sqrt(g/a). dφ=sqrt(g/a)dt integrating φ from 0 to 2π and t from 0 to τ(period) gives 2π=sqrt(g/a)τ. Which rearranging and substituting a = l/4 gives that ω = 2sqrt(g\l). Why is the 2 in there? I spoke to someone from my class about this and they told me you can just get rid of it because it's only needed for one oscillation... but I just integrated over 2π with a period T which is already one oscillation? So did I mess up on my arithmetic?

 

[kuruman]

So did I mess up on my arithmetic?

No. You did not interpret your arithmetic correctly. Look at the cycloid parametric xy plot below. Angle φ goes from 0 to 2π and a = 1.

Imagine the pendulum bob starting at the extreme left riding along the cycloid until it reaches the other end. Angle φ has increased from 0 to 2π. How much time has elapsed? Hint: the pendulum still needs to swing back and return to where it started for a full period to have gone by.

Edit: Another way of looking at it (sans plot) is this. Assume that you start with y = 0. A full period is the time needed for y to become zero for the second time after t = 0. The first time is when φ = 2π; the second time is when φ = 4π.

 

[Vitani11]

Awesome, thank you for the help.