Brachistochrone

Introduction

To find the shape of the curve which the time is shortest possible. . .

Theory

Variational Calculus and Euler--Lagrange Equation

The time from $P_{a}$ to $P_{b}$ is $t=\int _{P_{a}}^{P_{b}}{\frac {1}{v}}ds$ where $ds={\sqrt {1+y'{^{2}}}}dx$ is the Pythagorean distance measure and $v$ is determined from the the law of conservation of energy ${\frac {1}{2}}mv^{2}=mgy$. giving $v={\sqrt {2gy}}$. Plugging these in, we get $t=\int _{P_{a}}^{P_{b}}{\sqrt {\frac {1+y'^{2}}{2gy}}}dx=\int _{P_{a}}^{P_{b}}fdx$, where $f=f(y,y')$ is the function subject to variational consideration.

According to the Euler--Lagrange differential equation the stationary value is to be found, if E-L equation ${\frac {\partial f}{\partial y}}-{\frac {d}{dx}}{\frac {\partial f}{\partial y'}}=0$ is satisfied.

No Friction

We get

${\frac {\partial f}{\partial y'}}={\frac {\partial }{\partial y'}}{\sqrt {\frac {1+y'^{2}}{2gy}}}={\frac {1}{\sqrt {2gy}}}{\frac {\partial }{\partial y'}}{\sqrt {1+y'^{2}}}={\frac {2y'}{\sqrt {2gy}}}{\frac {1}{2{\sqrt {1+y'^{2}}}}}$.

Since $f$ does not depend on $x$, we may use the simplified E--L formula $f-y'{\frac {\partial f}{\partial y'}}={\text{Constant}}$.

Thus, we have

$f-y'{\frac {\partial f}{\partial y'}}={\sqrt {\frac {1+y'{^{2}}}{2gy}}}-{\frac {y'{^{2}}}{{\sqrt {2gy}}{\sqrt {1+y'^{2}}}}}=C$ So we have ${\frac {1+y'{^{2}}}{\sqrt {2gy(1+y'{^{2}})}}}-{\frac {y'{^{2}}}{{\sqrt {2gy}}{\sqrt {1+y'^{2}}}}}={\frac {1}{\sqrt {2gy(1+y'{^{2}})}}}=C$ and multiplying this with the denominator and rearring, we have $y\left(1+y'^{2}\right)={\frac {1}{2gC^{2}}}=k^{2}$

by redefining the constant. The standard solution to this equation is given by

${\begin{aligned}x&={\frac {1}{2}}k^{2}(\theta -\sin \theta )\\y&={\frac {1}{2}}k^{2}(1-\cos \theta )\end{aligned}}$

and is the equation of a cycloid.

Rolling Ball: Angular momentum

The rotational energy is $E_{\text{rot}}={\frac {1}{2}}I\omega ^{2}$ and by applying non-slipping condition $v=\omega r$ we get $E_{\text{rot}}={\frac {v^{2}}{2r^{2}}}I$. The conservation energy states

${\begin{aligned}E_{\text{kin}}&=E_{\text{rot}}\\{\frac {1}{2}}mv^{2}+{\frac {v^{2}}{2r^{2}}}I&=mgy\\v^{2}&={\frac {mgy}{m+I/(2r^{2})}}\\&={\frac {2mgr^{2}}{2m+I}}\end{aligned}}$

Friction

The forces on the path. Actually the sliding particle is infinitemal small.

The normal force follows the path, and thus is given by ${\vec {T}}={\frac {dx}{ds}}{\vec {x}}+{\frac {dy}{ds}}{\vec {y}}$, but The friction depends on the normal force of the path. The normal force is perpendicular to the previous, thus we have

${\vec {N}}=-{\frac {dy}{ds}}{\vec {x}}+{\frac {dx}{ds}}{\vec {y}}$

The conservation of energy does not apply here, but we have Newton's Second Law, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec F = m \frac{d \vec v}{dt}} . We need the components along the curve Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle s} . Thus we have

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \vec F &= \vec G - \vec F_\mu \\ &= mg \frac{dy}{ds} - \mu mg \frac{dx}{ds} \end{align} }

Clearly, for the left hand side of NII we have Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle m\frac{dv}{dt} = mv \frac{dv}{ds} = m\frac12 \frac {d v^2}{ds}} , and by including the differential part only, we have

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \frac12 v^2 &= g( y - \mu x ) \\ v&= \sqrt{2g(y-\mu x)} \end{align} }

and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} for the Euler--Lagrange equation is Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f = \sqrt{ \frac{1+y'^{2}}{2g(y-\mu x)} } }

Euler--Lagrange

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\partial f}{\partial y} - \frac{d}{dt}\frac{\partial f}{\partial y'} }

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \frac{\partial f}{\partial y} &= \sqrt{\frac{1+y'^{2}}{2g}} \frac{\partial}{\partial y} (y-\mu x)^{-1/2} \\ &= \sqrt{\frac{1+y'^{2}}{2g}} \frac{-1} { 2(y-\mu x)^{3/2} } \\ &= - \frac1{2(y-\mu x)} \sqrt{\frac{1+y'^{2}}{2g(y-\mu x)}} \\ \end{align} }

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \frac{\partial f}{\partial y'} &= \frac1{\sqrt{2g(y-\mu x)}} \frac{\partial}{\partial y} \sqrt{1+y'^{2}} \\ &= \frac{y' }{2\sqrt{2g(y-\mu x)} \sqrt{1+y'^{2}} } \\ \end{align} }

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \frac{d}{dt}\frac{\partial f}{\partial y'} &= \frac{(2\sqrt2 \sqrt{(g(yx))()})^2} \end{align} }

References

https://mathworld.wolfram.com/BrachistochroneProblem.html

https://physicscourses.colorado.edu/phys3210/phys3210_sp20/lecture/lec04-lagrangian-mechanics/

http://hades.mech.northwestern.edu/images/e/e6/Legeza-MechofSolids2010.pdf

https://www.tau.ac.il/~flaxer/edu/course/computerappl/exercise/Brachistochrone%20Curve.pdf