Swinging chandelier being pulled at a constant rate [tex]\alpha[/tex]

A chandelier of mass $m$ is attached to the end of a rope that passes through a fixed pulley on the ceiling. A person pulls on the other end of the rope at a constant rate $\alpha$ so that the chandelier's distance from the pulley is $l(t)=l_0-\alpha t$. The chandelier swings in a fixed plane, let $\theta$ denote the chandelier's angle relative to the vertical. Assume uniform gravitational field $g$ pointing downwards. Find equation of motion for $\theta$ using Newtonian and Lagrangian methods.

My attempt:
In terms of forces there is tension and gravity influencing the mass, so using Newtonian methods I got (I am taking to the origin to be where the pulley is):
$m(\ddot{r} - r \dot{\theta}^2)\vec{e_r}+m(r \ddot{\theta}+2 \dot{r} \dot{\theta}) \vec{e_\theta} = -(l_0 - \alpha t)cos(\theta)\vec{e_r}-mgcos(\theta)\vec{e_r}-mgsin(\theta)\vec{e_\theta}$.
But also since we have a parametrisation for $l$ in terms of $t$ and also by splitting the whole equation into components:
$\vec{e_r} : -m((l_0-\alpha t)(\dot{\theta})^2 = - (l_0-\alpha t)cos(\theta) - mgcos(\theta)$
$\vec{e_\theta}:m((l_0-\alpha t)\ddot{\theta} - 2\alpha \dot{\theta}) = -mg sin\theta$
Now for Lagrangian methods I have:
$T = 1/2m((\alpha)^2 +\dot{\theta}^2(l_0-\alpha t)^2)$
$U = -mg(l_0-\alpha t)cos\theta$
Applying Euler Lagrange gives:
$m((l_0-\alpha t)\ddot{\theta} - 2\alpha \dot{\theta}) = -mg sin\theta$
which is the second equation from Newtonian methods but what about the 1st equation? Also, I am hoping I got all the forces correct as this would change the solution of course. Thanks in advance!

I don’t get what do you mean by “what about the 1st equation?”. Unless I am missing from the question, the question only expects you to find the equation of motion for θ coordinates.

A side note, I did not work out the problem but using “unit analysis”, there is an error in this equation:
$\vec{e_r} : -m(l_0-\alpha t)\dot{\theta}^2 = - (l_0-\alpha t)\cos(\theta) - mg\cos(\theta)$

.....So, my concern is that maybe something is wrong with my initial equation of motion as it seems weird to have two relationships involving $\theta$ only:
$m(\ddot{r} - r \dot{\theta}^2)\vec{e_r}+m(r \ddot{\theta}+2 \dot{r} \dot{\theta}) \vec{e_\theta} = -(l_0 - \alpha t)cos(\theta)\vec{e_r}-mgcos(\theta)\vec{e_r}-mgsin(\theta)\vec{e_\theta}$

Thank you for your reply! By "what about 1st equation" I meant that in the equation from $\vec{e_r}$ I also get a relationship involving $\theta$. But for some reason we call the equation of motion for $\theta$ the 2nd one, the one from $\vec{e_\theta}$ and not the 1st one, since the 2nd one is what we also obtained from Lagrangian. …..

The main reason of having θ in "two relationships" (I suspect that you already know it) because of the usage of polar coordinates as the generalized coordinates to rewrite Newton's 2nd law.
Acceleration in the radial and angular direction are not independent in a polar coordinates.

As I mention in the previous post there is something wrong in your $\vec{e_r}$ equation and hence the equation
$m(\ddot{r} - r \dot{\theta}^2)\vec{e_r}+m(r \ddot{\theta}+2 \dot{r} \dot{\theta}) \vec{e_\theta} = -(l_0 - \alpha t)cos(\theta)\vec{e_r}-mgcos(\theta)\vec{e_r}-mgsin(\theta)\vec{e_\theta}$
is incorrect, but it does not affect the answer you obtain for the equation of motion for θ.


We call the “2^nd equation” as equation of motion for θ is mainly because it is second order differential equation in θ with respect to time.

Thank you for your help, this clarifies it all now! And yes, I have managed to correct my first equation.