How to calculate generalized force $Q_\phi$ with D'Alembert's principle? The related post was found here Lagrangian formalism application on a particle falling system with air resistance and also Wikipedia's definition on generalized force. Essential
$$\frac{d}{dt}\frac{\partial L}{\partial \dot q_i}-\frac{\partial L }{\partial q_i}=Q_i.$$
The case concerned was to calculate/translate an arbitrary force into the original system of a Lagrangian.
Example:
Setting generalized coordinates to be $(m,r,\phi)$
$$L=\frac{1}{2}m(\dot r^2 +(r+R_E)^2\dot \phi^2 )+\frac{g_0R_E^2}{R_E+r}m$$
the equation of motion with respect to $r$ and $\phi$ were
$$\frac{d}{dt} (m\dot r)= -\frac{g_0R_E^2}{(R_E+r)^2}m +m(r+R_E)\dot \phi^2 $$
$$\frac{d}{dt} (m(r+R_E)^2\dot\phi)= 0 $$
which, for $r$, was identified to be
$$F_r= -\frac{g_0R_E^2}{(R_E+r)^2}m +m(r+R_E)\dot \phi^2$$
It's straight forward to see that, for an external force in $r$ direction $F_r^{ext}$,
$$Q_r=F_r^{ext}$$
However, for $\phi$ direction it was much more complicated.
Through identification
$$v_\phi=(r+R_E)\dot \phi,$$
without external force,
$$\dot v_\phi=-\frac{\dot r v_\phi }{r+R_E}-\frac{\dot m}{m} v_\phi$$
or, equivalently,
$$m\dot v_\phi+\dot m v_\phi=-\frac{\dot r v_\phi }{r+R_E} m$$
one thus identified
$$F_\phi=-\frac{\dot r v_\phi }{r+R_E} m.$$
But if one wanted to write $Q_\phi$, then
$$\frac{d}{dt} (m(r+R_E)^2\dot\phi)=\frac{d}{dt} (m(r+R_E)v_\phi)= Q_\phi. $$
Clearly,
$$Q_\phi^{ext}=F_\phi^{ext}\cdot (r+R_E) \neq F_\phi^{ext}.$$
Why the generalized force in $\phi$ direction seemed so strange? Is there any other way to actually calculated $Q_\phi^\ext$ rather than making a blind identification? How to calculate/translate arbitrary external force into generalized force in Lagrangian such as resistance?
 A: Try to avoid the awkward variation due to the existence of a term of order $dt$ in the Lagrangian, I reformulate the problem using dissipation function and gneralized force.
$$\tag{1}
  G(r, \dot{\phi}) =  \frac{1}{2} \frac{dm}{dt} \left\{(r+R_E) \dot\phi - u\right\}^2
$$
where $ \dot m = \frac{dm}{dt}$ is the mass changed between  $t$ and $t+dt$, negative for loss. The additional parameter $u$ denotes the enjected tangential volicity of the running masses.
The general force
$$
Q_r = \frac{\partial G}{\partial \dot{r}} = 0;
$$
and
$$\tag{2}
Q_\phi =\frac{\partial G}{\partial \dot{\phi}} =\frac{dm}{dt} \left\{(r+R_E) \dot\phi - u\right\}(r+R_E) = \dot{m} (r+R_E)\left\{ v_\phi - u\right\}
$$
Note that the general force $Q_\phi$ is indeed a unit of torque, as it should be.
The Euler-Lagrangian equation for $\phi$ becomes:
$$
      \frac{d}{dt} \left\{ \frac{\partial L}{\partial \dot{\phi}}\right\} - \frac{\partial L}{\partial \phi} = Q_\phi
 $$
Becomes
$$\tag{4}
\frac{d}{dt} (m(r+R_E) v_\phi) =  \dot{m} (r+R_E) \left\{ v_\phi - u\right\}
$$
An equation for tangential velocity $v_\phi = (r+R_E)\dot\phi$
$$\tag{5}
 \frac{d}{dt} \left( (r+R_E) v_\phi  \right) = -\frac{\dot{m}}{m} (r+R_E) u = \left|\frac{\dot{m}}{m} \right| (r+R_E) u.
$$
Since $Q_r$ is zero, the equation in $r$ is the same as before:
$$ \tag{6} 
 \frac{d}{dt} (m\dot r) +\frac{g_0R_E^2}{(R_E+r)^2}m - m\frac{v_\phi^2}{(r+R_E)}  = 0.
$$
And the $v_\phi$ is solve from the integral:
$$\tag{7}
  v_\phi(t)  = v_\phi(0) + \frac{1}{(r+R_E) } \int_0^t dt' \left|\frac{\dot{m}}{m} \right| (r+R_E) u.
$$
