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 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 P_a} to 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 P_b} 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 t = \int_{P_a}^{P_b} \frac 1 v ds } where 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 ds= \sqrt{1+y'{^2}}dx} is the Pythagorean distance measure 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 v} is determined from the the law of conservation of energy 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 \frac12 mv^2 = mgy } . giving 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 = \sqrt{2gy}} . Plugging these in, 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 t = \int_{P_a}^{P_b} \sqrt{\frac{1+y'^{2}}{2gy}}dx = \int_{P_a}^{P_b} f dx} , where 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=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 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}{d x}\frac{\partial f}{\partial y'} = 0} is satisfied.

No Friction

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 \frac{\partial f}{\partial y'} = \frac{\partial }{\partial y'} \sqrt{\frac{1+y'^{2}}{2gy}} = \frac1{\sqrt{2gy}} \frac{\partial }{\partial y'} \sqrt{1+y'^{2}} = \frac{2y'}{\sqrt{2gy}} \frac1 {2\sqrt{1+y'^{2}}} } .

Since 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} does not depend on 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 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} }

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 }

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

${\begin{aligned}{\vec {F}}&={\vec {G}}-{\vec {F}}_{\mu }\\&=mg{\frac {dy}{ds}}-\mu mg{\frac {dx}{ds}}\end{aligned}}$

Clearly, for the left hand side of NII we have $m{\frac {dv}{dt}}=mv{\frac {dv}{ds}}=m{\frac {1}{2}}{\frac {dv^{2}}{ds}}$, and by including the differential part only, we have

${\begin{aligned}{\frac {1}{2}}v^{2}&=g(y-\mu x)\\v&={\sqrt {2g(y-\mu x)}}\end{aligned}}$

and $f$ for the Euler--Lagrange equation is $f={\sqrt {\frac {1+y'^{2}}{2g(y-\mu x)}}}$

Euler--Lagrange

${\frac {\partial f}{\partial y}}-{\frac {d}{dt}}{\frac {\partial f}{\partial y'}}=0$

${\begin{aligned}{\frac {\partial f}{\partial y}}&={\sqrt {\frac {1+y'^{2}}{2g}}}{\frac {\partial }{\partial y}}(y-\mu x)^{-1/2}\\&={\sqrt {\frac {1+y'^{2}}{2g}}}{\frac {-1}{2(y-\mu x)^{3/2}}}\\&=-{\frac {1}{2(y-\mu x)}}{\sqrt {\frac {1+y'^{2}}{2g(y-\mu x)}}}\\\end{aligned}}$

${\begin{aligned}{\frac {\partial f}{\partial y'}}&={\frac {1}{\sqrt {2g(y-\mu x)}}}{\frac {\partial }{\partial y}}{\sqrt {1+y'^{2}}}\\&={\frac {y'}{2{\sqrt {2g(y-\mu x)}}{\sqrt {1+y'^{2}}}}}\\\end{aligned}}$

${\begin{aligned}{\frac {d}{dt}}{\frac {\partial f}{\partial y'}}&={\frac {-\left(2y''{\sqrt {2g(y-\mu x)}}{\frac {y'}{\sqrt {1+y'^{2}}}}+{\sqrt {2}}g(y'-\mu x'){\frac {\sqrt {1+y'^{2}}}{\sqrt {g(y-\mu x)}}}\right)y'+2{\sqrt {2}}y''{\sqrt {y-\mu x}}{\sqrt {1+y'^{2}}}}{\left(2{\sqrt {2g(y-\mu x)}}{\sqrt {1+y'^{2}}}\right)^{2}}}\end{aligned}}$

Plugging these into EL we have

${\begin{aligned}-{\frac {1}{2(y-\mu x)}}{\sqrt {\frac {1+y'^{2}}{2g(y-\mu x)}}}&={\frac {-\left(2y''{\sqrt {2g(y-\mu x)}}{\frac {y'}{\sqrt {1+y'^{2}}}}+{\sqrt {2}}g(y'-\mu x'){\frac {\sqrt {1+y'^{2}}}{\sqrt {g(y-\mu x)}}}\right)y'+2{\sqrt {2}}y''{\sqrt {y-\mu x}}{\sqrt {1+y'^{2}}}}{\left(2{\sqrt {2g(y-\mu x)}}{\sqrt {1+y'^{2}}}\right)^{2}}}\\\\-4g{\frac {(1+y'^{2})^{3/2}}{\sqrt {2g(y-\mu x)}}}&=-\left(2y''{\sqrt {2g(y-\mu x)}}{\frac {y'}{\sqrt {1+y'^{2}}}}+{\sqrt {2}}g(y'-\mu x'){\frac {\sqrt {1+y'^{2}}}{\sqrt {g(y-\mu x)}}}\right)y'+2{\sqrt {2}}y''{\sqrt {y-\mu x}}{\sqrt {1+y'^{2}}}\\\end{aligned}}$

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