Brachistochrone: Difference between revisions

From wikiluntti
Line 48: Line 48:
\frac{1-y'{^2}}{\sqrt{2gy(1+y'{^2})}}
\frac{1-y'{^2}}{\sqrt{2gy(1+y'{^2})}}
= C
= C
</math> and multiplying this with the denominator, we have
<math>
1-y'{^2}= C (\sqrt{2gy(1+y'{^2})})
</math>
</math>



Revision as of 19:22, 16 February 2021

Introduction

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

Theory

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 is determined from the the law of conservation of energy . giving . Plugging these in, we get , 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 {\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{2y'{^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{2y'{^2}}{\sqrt{2gy} \sqrt{1+y'^{2}}} = \frac{1-y'{^2}}{\sqrt{2gy(1+y'{^2})}} = C } and multiplying this with the denominator, 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 1-y'{^2}= C (\sqrt{2gy(1+y'{^2})}) }

Friction

Rolling Ball

References

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

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