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

[Liquidxlax]

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-y₁}[/itex]

dt = ds/v

T = [itex]\int\frac{\sqrt{1+(x')²}}{\sqrt{2g(y-y₁}}[/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))²}}{\sqrt{2ga(cos(\vartheta) - cos(\vartheta₀)}}(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

 

[Spinnor]

Do this help?

 

[Liquidxlax]

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

 

[Spinnor]

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?

 

[rwisz]

, but i will try to explain the solution in words.

To solve this problem, we need to use the principle of conservation of energy. The car starts at rest at point P0 and will reach the bottom point P with a certain velocity. The potential energy at P0 is zero and at P it is mgh, where m is the mass of the car, g is the acceleration due to gravity, and h is the height of P from the origin. Therefore, we can equate these two energies and solve for the velocity at point P.

Next, we need to use the formula for time, which is given by T = \int\frac{ds}{v}, where ds is the infinitesimal distance traveled and v is the velocity at that point. However, it is easier to use the formula T = \int\frac{dx}{v}. Using the conservation of energy equation, we can substitute the value of v from there and solve for dx. This will give us the relation between dx and dy, which is dx = \frac{v}{y-y1}dy.

Now we can substitute this value of dx in the formula for time. We also substitute the values of y and x' in terms of theta, which are given in the problem. After simplifying, we get the equation for time as given in the homework statement. To prove that this integral is equal to \pi\sqrt{\frac{a}{g}}, we need to use the substitution theta = \pi - 2\alpha and use the trigonometric identity cos(\pi - \alpha) = -cos(\alpha). After simplifying, we get the desired result.