Are these two expressions for $mathrm{d}E/mathrm{d}t$ compatible?

Question

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,
$$K = sum_{n=1}^Nfrac{1}{2}m_ndot{vec{r}}_n^2 = underbrace{sum_{n=1}^Nfrac{m_n}{2}frac{partialvec{r}_n}{partial t}cdotfrac{partialvec{r}_n}{partial t}}_{K_0} + underbrace{dot{q}^ksum_{n=1}^Nm_nfrac{partialvec{r}_n}{partial q^k}cdotfrac{partialvec{r}_n}{partial t}}_{K_1} + underbrace{dot{q}^kdot{q}^ellsum_{n=1}^Nfrac{m_n}{2}frac{partialvec{r}_n}{partial q^k}cdotfrac{partialvec{r}_n}{partial q^ell}}_{K_2};$$
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} mvec{v}^2 + Vright) =  mvec{v}cdotvec{a} + vec{v}cdotvecnabla 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}^kfrac{partial}{partialdot{q}^k}T_2 + dot{q}^kfrac{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.

Qmechanic · Answer

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
$$ h(q,dot{q},t)~=~left(sum_{j=1}^ndot{q}^jfrac{partial }{partial dot{q}^j}-1 right)L(q,dot{q},t), tag{2.53} $$
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
$$frac{dh}{dt}~=~
sum_{j=1}^ndot{q}^jleft(Q_j-frac{partial{cal F}}{partialdot{q}^j} right)-sum_{ell=1}^mlambda^{ell} a_{ell t}- frac{partial L}{partial t} ,$$
where the notation is borrowed from my Phys.SE answer here.

References:

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

Eli · Answer

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]^Tboldsymbol{f}_aqquadqquad (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}}=0qquadqquadqquad (2)\
  &text{where}
  &mathcal{L}'=mathcal{L}+,left(left[frac{partial boldsymbol{R}}{partial boldsymbol{q}}right]^Tboldsymbol{f}_aright),cdot boldsymbol qqquad
  text{and}~ mathcal{L}=T(boldsymbol{dot{q}}~,boldsymbol q~,t)-U(boldsymbol q~,t)
 end{align*}
begin{align*}
   &textbf{Hamiltonian } 
  &mathcal{H}=boldsymbol pcdot dot{boldsymbol{q}}-
  mathcal{L}'left(dot{boldsymbol{q}}~,boldsymbol q~,tright)
  &text{with}
 &boldsymbol p=frac{partialmathcal{L}'}{partialdot{boldsymbol{q}}}=
 frac{partialmathcal{L}}{partialdot{boldsymbol{q}}}
 &Rightarrow
 mathcal{H}&=frac{partialmathcal{L}left(dot{boldsymbol{q}}~,boldsymbol q~,tright)}{partialdot{boldsymbol{q}}}cdot dot{boldsymbol{q}}-
  mathcal{L}left(dot{boldsymbol{q}}~,boldsymbol q~,tright)-
  left(left[frac{partial boldsymbol{R}}{partial boldsymbol{q}}right]^Tboldsymbol{f}_aright),cdot boldsymbol q
  &=underbrace{frac{partial (T-U)}{partialdot{boldsymbol{q}}}cdot dot{boldsymbol{q}}-
   (T-U)}_{E}-
  left(left[frac{partial boldsymbol{R}}{partial boldsymbol{q}}right]^Tboldsymbol{f}_aright),cdot boldsymbol q
 end{align*}
begin{align*}
  &dot{mathcal{H}}= frac{partial {E}}{partialboldsymbol{dot{q}}}cdot boldsymbol{ddot{q}}+
  frac{partial {E}}{partialboldsymbol{{q}}}cdot boldsymbol{dot{q}}+
  frac{partial {E}}{partial t}-
  left(left[frac{partial boldsymbol{R}}{partial boldsymbol{q}}right]^Tboldsymbol{f}_aright),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:
$$boldsymbol R=L,left[ begin {array}{c} sin left( omega,tau+varphi  right) 
 -cos left( omega,tau+varphi  right) 
 0end {array} right] 
$$
external force
$$boldsymbol f_a=-f_d,left[ begin {array}{c} ,cos left( omega,tau+varphi 
 right) ,sin left( omega,tau+
varphi  right)  0end {array} right] 
$$
where $~f_d~$ is the velocity depending force $~f_d=d,(dot{varphi}+omega)$
The kinetic and potential energy
$$T=frac 12,m{L}^{2} left(omega+dotvarphi
 right)^2 
U=frac 12,varphi , left( varphi +2,omega,tau right) Lc-mgLcos
 left( varphi +omega,tau right) 
$$
$Rightarrow$
The Hamiltonian
$$H=frac 12,m{L}^{2}{dotvarphi }^{2}-1/2,m{L}^{2}{omega}^{2}+1/2,{varphi }
^{2}L,c+varphi ,L,c,omega,tau-m,g,Lcos left( varphi +omega,tau
 right) -varphi ,L{it f_d}
$$
for $~omega=0~$ and $~f_d=0~$ the Hamiltonian is equal to the energy $E=T+Ubigg|_{omega=0}$ which is constant.

Are these two expressions for $mathrm{d}E/mathrm{d}t$ compatible?

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