Are these two expressions for $\mathrm{d}E/\mathrm{d}t$ compatible? This is purely recreational, but I'm eager to know the answer.
I was playing around with Hamiltonian systems whose Hamiltonian is not equal to their mechanical energy $E$.
If we split the kinetic energy $K$ and the potential energy $V$ in homogeneous polynomials on $\dot{q}^k$, the time-derivative of the generalised coordinates. For instance,
$$K = \sum_{n=1}^N\frac{1}{2}m_n\dot{\vec{r}}_n^2 = \underbrace{\sum_{n=1}^N\frac{m_n}{2}\frac{\partial\vec{r}_n}{\partial t}\cdot\frac{\partial\vec{r}_n}{\partial t}}_{K_0} + \underbrace{\dot{q}^k\sum_{n=1}^Nm_n\frac{\partial\vec{r}_n}{\partial q^k}\cdot\frac{\partial\vec{r}_n}{\partial t}}_{K_1} + \underbrace{\dot{q}^k\dot{q}^\ell\sum_{n=1}^N\frac{m_n}{2}\frac{\partial\vec{r}_n}{\partial q^k}\cdot\frac{\partial\vec{r}_n}{\partial q^\ell}}_{K_2};$$
with $N$ the number of particles, $m_n$ the mass of the $n$th particle and $\vec{r}_n$ its position. We will assume $V$ is a conservative (so, non-generalised) potential, so it does not depend on these $\dot{q}$s and, thus, $V=V_0$.
It's easy to see, from the definition of $H$ as a Legendre transform of $L$, that
$$H = K + V - (2K_0 + K_1),$$
hence, we can write
$$\frac{\mathrm{d}E}{\mathrm{d}t} = \frac{\mathrm{d}}{\mathrm{d}t}(E-H)+\frac{\mathrm{d}H}{\mathrm{d}t} = \frac{\mathrm{d}}{\mathrm{d}t}(2K_0 + K_1)-\frac{\partial L}{\partial t}$$
using Hamilton's equations, and using $L = K-V$ we finally arrive at
$$\frac{\mathrm{d}E}{\mathrm{d}t} = \frac{\mathrm{d}}{\mathrm{d}t}(2K_0 + K_1)-\frac{\partial K}{\partial t}+\frac{\partial V}{\partial t}\tag{1}.$$
We can also compute this total time-derivative of mechanical energy in a more Newtonian framework, and we find
$$\begin{aligned}\frac{\mathrm{d}E}{\mathrm{d}t} & = \frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{1}{2} m\vec{v}^2 + V\right) =  m\vec{v}\cdot\vec{a} + \vec{v}\cdot\vec\nabla V + \frac{\partial V}{\partial t}
\\
& = \vec{v}\cdot(\underbrace{\vec{F}^\text{consrv} + \vec{F}^\text{constr} + \vec{F}^\text{ncon}}_\text{total force $\vec{F}$}) + \vec{v}\cdot(-\vec{F}^\text{consrv}) + \frac{\partial}{\partial t}V = \dot{W}^\text{ncon} + \frac{\partial V}{\partial t}
\end{aligned}\tag{2}$$
assuming only one particle, and defining $\vec{v} = \dot{\vec{r}}$, $\vec{a} = \ddot{\vec{r}}$, because these equations look pretty ugly already without summations and subindices.
The time-dependences of $K$ should come either from external forces that aren't been taking into account or from being in a non-inertial frame of reference, which would give rise to ficticial forces. In any case, comparing Eq. (1) and Eq. (2), I expect the time-derivative of the work done by these forces, $\dot{W}$, to equal
$$\frac{\mathrm{d}}{\mathrm{d}t}(2K_0 + K_1)-\frac{\partial K}{\partial t},$$
but manipulating that into something that makes sense is quite difficult. I began by rewriting it as
$$\ddot{q}^k\frac{\partial}{\partial\dot{q}^k}T_2 + \dot{q}^k\frac{\partial}{\partial q^k}(2T_0 + T_1) + \frac{\partial}{\partial t}(T_0-T_2),$$
but I get a mess that's difficult to simplify.
I'm asking either for hints on how to simplify it or somehow manifest that those derivatives equal $\dot{W}$, or for someone to point out a flaw in my reasoning that makes all of this meaningless.
 A: Perhaps it is helpful to take a step back and review the definitions:

*

*In this answer, we will assume that the Lagrangian $L=T-U$ is the difference between kinetic and (possibly velocity-dependent) potential energy.


*Consider the (Lagrangian) energy function
$$ h(q,\dot{q},t)~=~\left(\sum_{j=1}^n\dot{q}^j\frac{\partial }{\partial \dot{q}^j}-1 \right)L(q,\dot{q},t), \tag{2.53} $$
which should not be confused with the Hamiltonian function $H(q,p,t)$. They are different functions, although their values agree.


*The energy function $h$ is not necessarily the mechanical energy $T+U$.


*Concerning the relationship between Hamiltonian and energy, see also e.g. this Phys.SE posts and links therein.


*The time-derivative of the energy function is in general given by
$$\frac{dh}{dt}~=~
\sum_{j=1}^n\dot{q}^j\left(Q_j-\frac{\partial{\cal F}}{\partial\dot{q}^j} \right)-\sum_{\ell=1}^m\lambda^{\ell} a_{\ell t}- \frac{\partial L}{\partial t} ,$$
where the notation is borrowed from my Phys.SE answer here.
References:

*

*H. Goldstein, Classical Mechanics, 3rd edition; Chapter 2 + 8.

A: For mechanical system you can use this:
Euler Lagrange
\begin{align*}
   &\mathcal{L}  =T-U\\
   &\frac{d}{dt}\left(\frac{\partial \mathcal{L}}{\partial \dot{\boldsymbol{q}}}\right)-
  \frac{\partial \mathcal{L}}{\partial \boldsymbol{q}}   =\left[\frac{\partial \boldsymbol{R}}{\partial \boldsymbol{q}}\right]^T\boldsymbol{f}_a\qquad\qquad (1)
 \end{align*}
where:

*

*$T$ kinetic energy

*$U$ potential  energy

*$\boldsymbol{q}$ generalized coordinates

*$\boldsymbol{R}$ Position vector

*$\boldsymbol{f}_a$ external forces

I put the velocity depending force components, friction forces and the time depending forces to the external forces.
transferring  equation (1) you obtain:
\begin{align*}
   &\frac{d}{dt}\left(\frac{\partial \mathcal{L}'}{\partial \dot{\boldsymbol{q}}}\right)-
  \frac{\partial \mathcal{L}'}{\partial \boldsymbol{q}}=0\qquad\qquad\qquad (2)\\\\
  &\text{where}\\
  &\mathcal{L}'=\mathcal{L}+\,\left(\left[\frac{\partial \boldsymbol{R}}{\partial \boldsymbol{q}}\right]^T\boldsymbol{f}_a\right)\,\cdot \boldsymbol q\qquad
  \text{and}~ \mathcal{L}=T(\boldsymbol{\dot{q}}~,\boldsymbol q~,t)-U(\boldsymbol q~,t)
 \end{align*}
\begin{align*}
   &\textbf{Hamiltonian } \\
  &\mathcal{H}=\boldsymbol p\cdot \dot{\boldsymbol{q}}-
  \mathcal{L}'\left(\dot{\boldsymbol{q}}~,\boldsymbol q~,t\right)\\
  &\text{with}\\
 &\boldsymbol p=\frac{\partial\mathcal{L}'}{\partial\dot{\boldsymbol{q}}}=
 \frac{\partial\mathcal{L}}{\partial\dot{\boldsymbol{q}}}\\
 &\Rightarrow\\
 \mathcal{H}&=\frac{\partial\mathcal{L}\left(\dot{\boldsymbol{q}}~,\boldsymbol q~,t\right)}{\partial\dot{\boldsymbol{q}}}\cdot \dot{\boldsymbol{q}}-
  \mathcal{L}\left(\dot{\boldsymbol{q}}~,\boldsymbol q~,t\right)-
  \left(\left[\frac{\partial \boldsymbol{R}}{\partial \boldsymbol{q}}\right]^T\boldsymbol{f}_a\right)\,\cdot \boldsymbol q\\
  &=\underbrace{\frac{\partial (T-U)}{\partial\dot{\boldsymbol{q}}}\cdot \dot{\boldsymbol{q}}-
   (T-U)}_{E}-
  \left(\left[\frac{\partial \boldsymbol{R}}{\partial \boldsymbol{q}}\right]^T\boldsymbol{f}_a\right)\,\cdot \boldsymbol q\\
 \end{align*}
\begin{align*}
  &\dot{\mathcal{H}}= \frac{\partial {E}}{\partial\boldsymbol{\dot{q}}}\cdot \boldsymbol{\ddot{q}}+
  \frac{\partial {E}}{\partial\boldsymbol{{q}}}\cdot \boldsymbol{\dot{q}}+
  \frac{\partial {E}}{\partial t}-
  \left(\left[\frac{\partial \boldsymbol{R}}{\partial \boldsymbol{q}}\right]^T\boldsymbol{f}_a\right)\,\cdot \boldsymbol{\dot{q}}
 \end{align*}
for conservative system is the energy $~E~$ constant and $~\boldsymbol f_a=0~$ hence  $~\dot H=0$
Example: pendulum with spring and damper

the pendulum is rotating about the z axes with angular velocity $~\omega$
Position Vector:
$$\boldsymbol R=L\,\left[ \begin {array}{c} \sin \left( \omega\,\tau+\varphi  \right) 
\\ -\cos \left( \omega\,\tau+\varphi  \right) 
\\ 0\end {array} \right] 
$$
external force
$$\boldsymbol f_a=-f_d\,\left[ \begin {array}{c} \,\cos \left( \omega\,\tau+\varphi 
 \right) \\\,\sin \left( \omega\,\tau+
\varphi  \right) \\ 0\end {array} \right] 
$$
where $~f_d~$ is the velocity depending force $~f_d=d\,(\dot{\varphi}+\omega)$
The kinetic and potential energy
$$T=\frac 12\,m{L}^{2} \left(\omega+\dot\varphi
 \right)^2 \\
U=\frac 12\,\varphi \, \left( \varphi +2\,\omega\,\tau \right) Lc-mgL\cos
 \left( \varphi +\omega\,\tau \right) 
$$
$\Rightarrow$
The Hamiltonian
$$H=\frac 12\,m{L}^{2}{\dot\varphi }^{2}-1/2\,m{L}^{2}{\omega}^{2}+1/2\,{\varphi }
^{2}L\,c+\varphi \,L\,c\,\omega\,\tau-m\,g\,L\cos \left( \varphi +\omega\,\tau
 \right) -\varphi \,L{\it f_d}
$$
for $~\omega=0~$ and $~f_d=0~$ the Hamiltonian is equal to the energy $E=T+U\bigg|_{\omega=0}$ which is constant.
