Velocity in a Cycloidal Pendulum

[Paradoxx]

Homework Statement

Cycloidal Pendulum, with x= RΘ+RsinΘ and y = -RcosΘ
I need to find the Lagragian.

Homework Equations

L = T - V

The Attempt at a Solution

I just want to know how do I find the velocity so I can find T, which is 1/2 mv². I thought it would be dx/dΘ but it didn't work. Can anyone explain to me how can I find the velocity?

 

[ZetaOfThree]

Remember that velocity is the derivative of position with respect to time and is a vector. What are its x and y components? Could you use these components to find $v^2$?

 

[Paradoxx]

Ok, I realized that I must consider x and y and derivative in relation with time.

v = dx/dt + dy/dt

then v² would be (dx/dt + dy/dt)².

I did the derivatives:

x'= RΘ' + Rcos(Θ)Θ' and y' = Rsin(Θ)Θ',

so x+y = RΘ' + RΘ' (cosΘ + sinΘ)

Also, v² = ( RΘ' + RΘ' (cosΘ + sinΘ))² = (4R²Θ'² + R²Θ'²cos²Θsin²Θ)
But in my answer it is

4R²Θ²cos²(Θ/2)

What am I doing wrong?

 

[ZetaOfThree]

Ok, I realized that I must consider x and y and derivative in relation with time.

v = dx/dt + dy/dt

then v² would be (dx/dt + dy/dt)².

I did the derivatives:

x'= RΘ' + Rcos(Θ)Θ' and y' = Rsin(Θ)Θ',

so x+y = RΘ' + RΘ' (cosΘ + sinΘ)

Also, v² = ( RΘ' + RΘ' (cosΘ + sinΘ))² = (4R²Θ'² + R²Θ'²cos²Θsin²Θ)
But in my answer it is

4R²Θ²cos²(Θ/2)

What am I doing wrong?

$v=\frac{dx}{dt}+\frac{dy}{dt}$ is wrong. Remember that $\vec{v}$ is a vector! What is the x-component of $\vec{v}$? What is the y-component of $\vec{v}$? After you have the components, you can find $v^2$ by squaring the components and adding them, right?

 

[Paradoxx]

Ohhh, you are right! It worked now, thank you very much!