Given this physical setup: 
I don't want to spend much time on deriving the formula for kinetic and potential energy. In my task these magnitudes are even a given:
$T = \dfrac{1}{2}\,m\left(r^2\,\omega^2+l^2\,\dot{\varphi}^2\right)- m\,r\,l\,\omega\,\dot{\varphi}\,\cos(\omega\,t-\varphi)$
$V = mg[r\,\cos(\omega\,t)-l\,\cos\varphi]$
Now, at the same time on the same file the Lagrangian is written as:
$\mathcal{L} = \dfrac{m\,l^2\,\dot{\varphi}^2}{2}-m\,r\,l\,\omega^2\,\cos(\omega\,t-\varphi)+m\,g\,l\,\cos(\varphi)$
Here I was stunting at first since:
$\mathcal{L} = T-V$ but these don't seem to coincide.
Probably this disparity is worth understanding for solving the main task: showing energy conservation or not. Intuitively I'd say Energy is conserved since no energy is seeping through but on the other side the system is time dependent. Here is my try to decide it mathematically:
$$\dfrac{\mathrm{d}}{\mathrm{dt}}E\left(\dot{\varphi}(t),\varphi(t),t\right) = \partial_{\dot{\varphi}}E\,\ddot{\varphi} + \partial_{\varphi}E\,\dot{\varphi}+ \partial_t E \\ \\
\boxed{E = T+V = \dfrac{1}{2}\,m\,\left(r^2\,\omega^2+l^2\,\dot{\varphi}^2\right)- m\,r\,l\,\omega\,\dot{\varphi}\,\cos(\omega\,t-\varphi) + mg[r\,\cos(\omega\,t)-l\,\cos\varphi]} \\ \\
\begin{align}\dfrac{\mathrm{d}}{\mathrm{dt}}E & = m\,l^2\dot{\varphi}\,\ddot{\varphi}-m\,r\,l\,\omega\,\cos(\omega\,t-\varphi)\,\ddot{\varphi}-m\,r\,l\,\omega\,\dot{\varphi}\,\sin(\omega\,t-\varphi)\,\dot{\varphi}+m\,g\,l\,\sin(\varphi)\,\dot{\varphi}\\ &+m\,r\,l\,\omega^2\,\dot{\varphi}\,\sin(\omega\,t-\varphi)-m\,g\,r\,\omega\,\sin(\omega\,t)\end{align}$$
My problem: I don't really see how this should equal $0$ on it's own.
Best Answer
As in the comments, you may omit terms independent of $\varphi$ in the Lagrangian (they typically arise from changing the zero point of potential energy, etc.).
You can also integrate by parts / remove total derivatives. Here, we note that the derivative of $\cos(\omega t - \varphi)$ means we can exchange $\dot{\varphi}$ for $\omega$. (NB this does not mean $\dot{\varphi} = \omega$!)
As the Hamiltonian has explicit time dependence, energy is not conserved.