How Is Time Calculated for a Car Sliding Along a Cycloid?

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Homework Help Overview

The discussion revolves around calculating the time it takes for a car to slide along a cycloidal track from a point P0 to the lowest point P. The problem involves deriving an integral expression for this time based on the parameters of the cycloid and gravitational effects.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to derive the time integral using energy considerations and substitutions involving the cycloid's parametric equations. Some participants question the relationship between the variables x and y, considering their dependence on the parameter theta.

Discussion Status

Participants are actively exploring the mathematical relationships and substitutions needed to simplify the integral. There is acknowledgment of confusion regarding the dependencies of the variables, and some guidance has been offered regarding the correct approach to parameterization.

Contextual Notes

There is mention of the integral being tricky and the potential use of substitutions to facilitate the integration process. The original poster expresses difficulty in achieving the desired form of the integral, indicating a need for further clarification on the setup and relationships involved.

Liquidxlax
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Homework Statement



consider a sigle loop of the cycloid with a fixed value of a. A car is released at a point P0 from rest anywhere on the track between the origin and the lowest point P, that is P0 has a parameter 0<theta0 < pi. show that the time for the cart to roll from P0 to P is given by the integral

time( P0]/sub] -> P) = [itex]\sqrt{\frac{a}{g}}\int \sqrt{\frac{1 - cos\vartheta}{cos\vartheta_{0} - cos\vartheta}}d\vartheta[/itex]

integral is from theta naught to pi

and prove the integral equals [itex]\pi\sqrt{\frac{a}{g}}[/itex] the integral may be tricky and you can use theta = pi -2(alpha)

Homework Equations



[itex]\frac{df}{dx}=\frac{d}{dy}\frac{df/dx'}[/itex]

The Attempt at a Solution



1/2 mv2 = mg(y-y1

v = [itex]\sqrt{2g(y-y<sub>1</sub>}[/itex]

dt = ds/v

T = [itex]\int\frac{\sqrt{1+(x')<sup>2</sup>}}{\sqrt{2g(y-y<sub>1</sub>}}[/itex] dy

test the Euler formula to get

y = a(1 - cos(theta)) y' = asin(theta)

x = a (theta - sin(theta)) x' = a - acos(theta)

This is now the part I am having a problem with

Now i substitute my y and x' in

T = [itex]\int\frac{\sqrt{1 + (a( 1 - cos(\vartheta))<sup>2</sup>}}{\sqrt{2ga(cos(\vartheta) - cos(\vartheta<sub>0</sub>)}}(asin(\vartheta)[/itex]

and somehow that equals the top equation. Then i need to integrate it, which i get no where near close to what it wants no matter what the substitution and trig identities

not sure why all the itex isn't working
 
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Do this help?
 

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Spinnor said:
Do this help?

it sure did sometimes I'm such a retard, x and y do not depend on each other they depend on theta...

because i was doing sqrt ( 1 y'^2) dx

thank you very much
 


Liquidxlax said:
it sure did sometimes I'm such a retard, x and y do not depend on each other they depend on theta...

because i was doing sqrt ( 1 y'^2) dx

thank you very much

x and y do depend on each other but the answer had to be in terms of theta so that was a more useful parameter?
 

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