- #1

Father_Ing

- 33

- 3

- 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?