Time of Travel down Brachistochrone/Corollary

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Theorem

Let a wire $AB$ be curved into the shape of a brachistochrone.

Let $AB$ be embedded in a constant and uniform gravitational field where Acceleration Due to Gravity is $g$.

Let a bead $P$ be released from anywhere on the wire between $A$ and $B$ to slide down without friction to $B$.


Then the time taken for $P$ to slide to $B$ is:

$T = \pi \sqrt{\dfrac a g}$

where $a$ is the radius of the generating circle of the cycloid which forms $AB$.


Proof

That the curve $AB$ is indeed a cycloid is demonstrated in Brachistochrone is Cycloid.

Let $A$ be located at the origin of a cartesian plane.

Let the bead slide from an intermediate point $\theta_0$.

We have:

$v = \dfrac {\d s} {\d t} = \sqrt {2 g \paren {y - y_0} }$

which leads us, via the same route as for Time of Travel down Brachistochrone, to:


\(\ds T\) \(=\) \(\ds \int_{\theta_0}^\pi \sqrt {\frac {2 a^2 \paren {1 - \cos \theta} } {2 g a \paren {\cos \theta_0 - \cos \theta} } } \rd \theta\)
\(\ds \) \(=\) \(\ds \sqrt {\frac a g} \int_{\theta_0}^\pi \sqrt {\frac {1 - \cos \theta} {\cos \theta_0 - \cos \theta} } \rd \theta\)


Using the Half Angle Formula for Cosine and Half Angle Formula for Sine, this gives:

$\ds T = \sqrt {\frac a g} \int_{\theta_0}^\pi \frac {\map \sin {\theta / 2} } {\sqrt {\map \cos {\theta_0 / 2} - \map \cos {\theta / 2} } } \rd \theta$


Now we make the substitution:

\(\ds u\) \(=\) \(\ds \frac {\map \cos {\theta / 2} } {\map \cos {\theta_0 / 2} }\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d u} {\d \theta}\) \(=\) \(\ds -\frac {\map \sin {\theta / 2} } {2 \map \cos {\theta_0 / 2} }\)


Recalculating the limits:

when $\theta = \theta_0$ we have $u = 1$
when $\theta = \pi$ we have $u = 0$.

So:

\(\ds T\) \(=\) \(\ds -2 \sqrt {\frac a g} \int_1^0 \frac {\d u} {\sqrt {1 - u^2} }\)
\(\ds \) \(=\) \(\ds \intlimits {2 \sqrt {\frac a g} \sin^{-1} u} 0 1\)
\(\ds \) \(=\) \(\ds \pi \sqrt {\frac a g}\)


Thus the time to slide down a brachistochrone from any arbitrary point $\theta_0$ is:

$T = \pi \sqrt {\dfrac a g}$

$\blacksquare$


Sources

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