Time of Travel down Brachistochrone

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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 at $A$ to slide down without friction to $B$.


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

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

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


Corollary

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:

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


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.


We have the equations of the cycloid:

\(\ds x\) \(=\) \(\ds a \paren {\theta - \sin \theta}\)
\(\ds y\) \(=\) \(\ds a \paren {1 - \cos \theta}\)


Differentiating with respect to theta:

\(\ds \frac {\d x} {\d \theta}\) \(=\) \(\ds \paren {1 - \cos \theta}\)
\(\ds \frac {\d y} {\d \theta}\) \(=\) \(\ds a \sin \theta\)


Let $s$ be the distance along $AB$ from the origin.

Then:

\(\ds \paren {\frac {\d s} {\d \theta} }^2\) \(=\) \(\ds \paren {\frac {\d x} {\d \theta} }^2 + \paren {\frac {\d y} {\d \theta} }^2\)
\(\ds \) \(=\) \(\ds a^2 \paren {1 - \cos \theta}^2 + a^2 \sin^2 \theta\)
\(\ds \) \(=\) \(\ds a^2 \paren {1 - 2 \cos \theta + \cos^2 \theta + \sin^2 \theta}\)
\(\ds \) \(=\) \(\ds 2 a^2 \paren {1 - \cos \theta}\)
\(\ds \) \(=\) \(\ds 4 a^2 \map {\sin^2} {\frac \theta 2}\)


Thus:

$\dfrac {\d s} {\d \theta} = 2 a \map \sin {\dfrac \theta 2}$


Now:

\(\ds v\) \(=\) \(\ds \dfrac {\d s} {\d t}\)
\(\ds \leadsto \ \ \) \(\ds \int \rd t\) \(=\) \(\ds t\)
\(\ds \) \(=\) \(\ds \int \frac {\d s} v\)


We also have, from the Principle of Conservation of Energy, that:

\(\ds m g y\) \(=\) \(\ds \frac {m v^2} 2\)
\(\ds \leadsto \ \ \) \(\ds v\) \(=\) \(\ds \dfrac {\d s} {\d t}\)
\(\ds \) \(=\) \(\ds \sqrt {2 g y}\)
\(\ds \) \(=\) \(\ds \sqrt {2 g a \paren {1 - \cos \theta} }\)


Thus:

\(\ds \frac {\d t} {\d \theta}\) \(=\) \(\ds \frac {\d t} {\d s} \theta\)
\(\ds \) \(=\) \(\ds \frac {a \sqrt {2 \paren {1 - \cos \theta} } } {\sqrt {2 g a \paren {1 - \cos \theta} } }\)
\(\ds \) \(=\) \(\ds \sqrt {\frac a g}\)


Let $T$ be the time taken to slide down the brachistochrone.

At the top:

$\theta = 0$

and at the bottom:

$\theta = \pi$

Then:

\(\ds T\) \(=\) \(\ds \int \rd t\)
\(\ds \) \(=\) \(\ds \sqrt {\frac a g} \int_0^\pi \rd \theta\)
\(\ds \) \(=\) \(\ds \pi \sqrt {\frac a g}\)

So the time to slide down this brachistochrone from the top to the bottom is:

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

$\blacksquare$


Sources

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