To find the equation of Motion of a Simple Pendulum using Lagrangian mechanics [closed]

The Equation of the Trajectory of a particle can be obtained by eliminating the variable $$t$$ as we are doing in the equation of Trajectory of a parabolic Projectile. So, my question is if the equation of trajectory of a parabolic projectile is possible, then other simple ones like The Simple Pendulum should also be possible?. So, I defined the degrees of freedom of a simple pendulum using two variables $$t, \theta$$. Here, $$t$$ is the time duration of oscillation and $$\theta$$ is the angle made by the pendulum wrt it's mean equilibrium position. The function $$\theta(t)$$ is assumed to be a function which returns the angle $$\theta$$ wrt it's dependent time variable $$t$$.

So, first I've to define the coordinates and draw a suitable figure for that. So, here's my take over it. Sorry for the bad design.

Note: I want that function in either Classical Mechanics or Newtonian Mechanics Only.

I'm presenting the failed try of the Lagrangian Mechanics Part.

Here, my attack was based on calculating the equation of motion of the pendulum using the lagrangian separately for x-axis and y-axis and then, eliminate $$t$$ f from that.

$$\begin{cases} T_x = (l \dot \cos \theta)^2 \dfrac 12 m \\ T_y = (l \dot \sin \theta)^2 \dfrac 12 m \end{cases}$$

Using them and working separately the equations of motion, we'd get!

$$\text{For x-axis} \\ U = - mgl \cos \theta \\ L(\theta, \dot \theta, t) = T_x - U \\ = \dfrac{l^2 \dot \theta^2 \cos^2 \theta m}{2} + mgl \cos \theta \\ \dfrac{d}{dt} \left ( \dfrac{\partial L}{\partial \dot \theta} \right) - \dfrac{\partial L}{\partial \theta} = 0 \\ \Rightarrow \dfrac{d}{dt} \left ( l^2 \dot \theta \cos \theta m \right) =- mgl \sin \theta\dots(1)$$

$$\text{For y-axis} \\ U = 0\\ L(\theta, \dot \theta, t) = T_y \\ = \dfrac{l^2 \dot \theta^2 \sin^2 \theta m}{2} \\ \dfrac{d}{dt} \left ( \dfrac{\partial L}{\partial \dot \theta} \right)=0 \\ \Rightarrow l^2 \dot \theta \sin^2 \theta m = C\\ \text{Where, C is a constant}$$

Since this is not based on $$t$$ How to get the value of $$\theta$$ as a function, I need help here.

Edit: There is another big problem here, rather. There is no left $$t$$ to eliminate ;(

Edit2: I've defined the motion of a pendulum in polar coordinates because it's easy to tell it's position at an instant at this format $$(l, \theta)$$. I then separated the motion into Rectangular Coordinates, so that I could Eliminate $$t$$, but soon I realized after that, Lagrangians don't work like that. So, I'm updating my try to a better-improved version, (if I get)

closed as off-topic by Aaron Stevens, Kyle Kanos, rob♦Apr 30 at 17:32

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• What do you mean by solving to eliminate $t$? The whole point is to determine these values that are functions of time. – Aaron Stevens Apr 30 at 15:24
• @AaronStevens sorry, I didn't get it. – Abhas Kumar Sinha Apr 30 at 15:26
• @AaronStevens Sorry, updated that – Abhas Kumar Sinha Apr 30 at 15:28
• I'm confused why you're using $x$ and $y$ axis separately when you've defined a polar coordinate system. – Kyle Kanos Apr 30 at 15:31
• Yeah, that's the part that is confusing to me. You are breaking it down into Cartesian axes but leaving the polar coordinates. Either solve it using $(r,\theta)$ or $(x,y)$, don't mix them. – Kyle Kanos Apr 30 at 15:36

The whole point of the Lagrangian approach is to obtain the equations of motion from a single scalar function L.

The Lagrangian for the system is $$L=T-V$$ so $$T=\frac{1}{2}m(\dot x^2+\dot y^2)=\frac{1}{2}m \ell^2\dot{\theta}^2$$ with $$x=\ell\sin\theta, y=\ell(1-\cos\theta)$$. The potential is $$V=mgy= mg\ell(1-\cos\theta)$$ so $$L= \frac{1}{2}m \ell^2\dot{\theta}^2 + mg \ell \cos\theta$$ up to an unimportant constant. $$L$$ does not have components; there is no separate $$L$$ for the $$x$$ or $$y$$ direction.

The equation of motion follows from the usual Euler-Lagrange equation: $$\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\theta}}\right)= \frac{\partial L}{\partial \theta}\, ,$$ is a second order ODE in $$\theta$$ and gives the usual harmonic solution in the limit of small $$\theta$$.

• I don't think T is the tension, looks to have units of KE (and a recent comment indicates it as well) – Kyle Kanos Apr 30 at 15:42
• "and gives usual harmonic solution in the limit of small $\theta$", can you show that please? – Abhas Kumar Sinha Apr 30 at 16:03
• @KyleKanos you’re right. I misread. Now fixed. – ZeroTheHero Apr 30 at 16:42
• @AbhasKumarSinha yes I could but I believe you now have enough info to show that yourself. – ZeroTheHero Apr 30 at 16:43
• You also missed an equals sign, but I fixed that for you – Kyle Kanos Apr 30 at 16:45

In writing Lagrangian, you can not write components like that. You should add Tx and Ty. Then in the Lagrangian the kinetic energy term becomes $$T=(1/2)ml^{2}\dot{\theta}^{2}$$, hence no cosine or sine terms. Then, write down the full Lagrangian which will have only one independent variable, $$\theta$$. Then, write down the Lagrange Equation of motion. Lagrangian is a scalar, not a vector or tensor, you can not find its components!