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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?

Qmechanic
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ShoutOutAndCalculate
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1 Answers1

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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:

$$\tag{7} v_\phi(t) = v_\phi(0) + \frac{1}{(r+R_E) } \int_0^t dt' \left|\frac{\dot{m}}{m} \right| (r+R_E) u. $$

ytlu
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