Cartesian and polar coordinate in Simple pendulum, Euler-Lagrange

[Father_Ing]

Homework Statement

Find the equation of motion of a simple pendulum.

Relevant Equations

L = T-V

$$L = \frac {mv^2}{2} - mgy$$
It is clear that $\dot{x}=\dot{\theta}L$ and $y=-Lcos \theta$. After substituting these two equations to Lagrange equation, we will get the answer by simply using this equation: $$\frac {d} {dt} \frac {∂L}{∂\dot{\theta}} - \frac {∂L}{∂\theta }= 0$$

But, What if we only substitute $\dot{x}=\dot{\theta}L$ to the Lagrange?
$$L = \frac {mL^2 \dot{\theta}^2}{2} - mgy$$
Then:
$$\frac {d} {dt} \frac {∂L}{∂\dot{\theta}} - \frac {∂L}{∂\theta }= 0$$ $$\frac {d} {dt} \frac {∂L}{∂\dot{y}} - \frac {∂L}{∂y }= 0$$
So, We get : $$mL^2 \ddot{\theta} = 0$$

But, this answer definitely does not make any sense. Is this caused by using two different types of coordinate system at the same time?
If yes, then why is this prohibited?

 

[PeroK]

But, this answer definitely does not make any sense. Is this caused by using two different types of coordinate system at the same time?
If yes, then why is this prohibited?

The pendulum has only one degree of freedom (independent variable), which is either $\theta$ or $x$ or $y$.

What you have described is a different system where $y$ and $\theta$ are separate degrees of freedom. PS Although it may not make much physical sense, as there is no KE associated with movement in the $y$ direction.

 

[Father_Ing]

The pendulum has only one degree of freedom (independent variable), which is either θ or x or y.

But, all three of them depends on time (because if it is not, I don't think we can differentiate it with respect to time).

 

[ergospherical]

If your system has $n$ degrees of freedom, then the Euler-Lagrange equations in the form you quote (known as the "second kind") require that the Lagrangian is expressed using $n$ independent generalised coordinates. Your pendulum has only one degree of freedom, so you must express $L$ in terms of either $\theta$, or $y$, or $x$, or some other coordinate, but not any more than one of them!

If the number of coordinates $\tilde{n}$ exceeds the number of degrees of freedom $n$, then given $\tilde{n}-n$ holonomic constraint equations expressing the dependencies of these coordinates on each other you can:
1. use the constraint equations to eliminate $\tilde{n} - n$ of these coordinates and get an independent set.
2. or use Lagrange multipliers and Lagrange's equations of the "first kind".

 

[PeroK]

But, all three of them depends on time (because if it is not, I don't think we can differentiate it with respect to time).

I can't see the relevance of that. A simple pendulum has one degree of freedom. Full stop.