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 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-y' \frac{\partial f}{\partial y'}= \text{Constant}} .

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 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 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{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 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 y\left( 1 + y'^{2}\right) = \frac1{2gC^2} = k^2 }

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

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} x &= \frac12 k^2( \theta - \sin\theta) \\ y &= \frac12 k^2( 1 - \cos\theta) \end{align} }

and is the equation of a cycloid.

Rolling Ball: Angular momentum

The rotational energy 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 E_\text{rot} = \frac12 I \omega^2} and by applying non-slipping condition 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 v = \omega r} we get 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 E_\text{rot} = \frac {v^2}{2r^2} I} . The conservation energy states

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} E_\text{kin} &= E_\text{rot} \\ \frac12 mv^2 + \frac{v^2}{2r^2}I &= mgy \\ v^2 &= \frac{mgy}{m + I/(2r^2)} \\ &= \frac{2mgr^2}{2m + I} \end{align} }

Thus, the path shape is same, than previous.

Friction

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

The normal force follows the path, and thus is given by 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 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

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

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