Why Does Time Remain Constant on Any Chord in Galileo's Circle?

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Why is it that if you have a circle. If you put any rod from the top center of the circle to any other point on the circle (this would be defined as a chord), then why, if the bead starts at the top, is the time required to slide down any chord independent of the particular chord chosen?

I have no idea how to even start this... any tips would be appreciated.
 
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You know how to find the time taken to slide down any given ramp, given length and height of the ramp, right? So write that down. Now put in the relation between length and height if the ramp is a chord on the circle.
 
What a nice problem -- I'd never heard of this one before! I'll just chime into say that you may find the trig identity for sin(theta/2) useful near the end...
 
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
The value of H equals ## 10^{3}## in natural units, According to : https://en.wikipedia.org/wiki/Natural_units, ## t \sim 10^{-21} sec = 10^{21} Hz ##, and since ## \text{GeV} \sim 10^{24} \text{Hz } ##, ## GeV \sim 10^{24} \times 10^{-21} = 10^3 ## in natural units. So is this conversion correct? Also in the above formula, can I convert H to that natural units , since it’s a constant, while keeping k in Hz ?
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