Brachistochrone: Difference between revisions

From wikiluntti
Line 32: Line 32:
\frac{2y'}{\sqrt{2gy}}
\frac{2y'}{\sqrt{2gy}}
\frac1 {\sqrt{1+y'^{2}}}
\frac1 {\sqrt{1+y'^{2}}}
</math>. Because the previous statement do not contain explicit <math>x</math>, the derivative is zero, thus
</math>.  
<math>
\frac{2y'}{\sqrt{2gy}}
\frac1 {\sqrt{1+y'^{2}}}
= C
</math>
giving
<math>
\frac{2y'}{\sqrt{2gy}}
\frac1 {\sqrt{1+y'^{2}}}
= C
</math>
 
Since <math>f</math> does not depend on <math>x</math>, we may use the simplified E--L formula <math>f-y' \frac{\partial f}{\partial y'}= \text{Constant}</math>.  
Since <math>f</math> does not depend on <math>x</math>, we may use the simplified E--L formula <math>f-y' \frac{\partial f}{\partial y'}= \text{Constant}</math>.  



Revision as of 19:56, 16 February 2021

Introduction

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

Theory

The time from to is where is the Pythagorean distance measure and is determined from the the law of conservation of energy . giving . Plugging these in, we get , where 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 is satisfied.

No Friction

We get . Since does not depend on , we may use the simplified E--L formula .

Because ,

Friction

Rolling Ball

References

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

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