Consider swinging a ball around a center via uniform circular motion. The centripetal acceleration is provided by the tension of a rope. Now, is this force a constraint force? If it is, since it is the only force, the applied force must be zero. Then, the following equation must apply:
$$Q_j = \frac{d}{dt}\left(\frac{\partial T}{\partial \dot{q}_j}\right) – \frac{\partial T}{\partial q_j}.\tag{1}$$
The LHS is the generalized force. Now, assuming the tension force is a constraint force, the generalized force would simply equate to zero, since there are no applied forces to consider.
$$\frac{d}{dt}\left(\frac{\partial T}{\partial \dot{q}_j}\right) – \frac{\partial T}{\partial q_j} = 0,\tag{2}$$
If we do apply this equation in polar co-ordinates, the two generalized co-ordinates would be $r$ and $\theta$. The kinetic energy of the ball in polar co-ordinates would be:
$$T = \frac{1}{2}m\left(\dot r^2 + r^2\dot{\theta}^2\right).\tag{3}$$
Applying the Euler-Lagrange equation for the radial direction:
$$\frac{d}{dt}\left(\frac{\partial T}{\partial \dot{r}_j}\right) – \frac{\partial T}{\partial r_j} = \frac{d}{dt}\left(m\dot r\right) – mr\dot{\theta}^2 = 0.\tag{4}$$
Since the length of the rope is constant:
$$- mr\dot{\theta}^2 = 0.\tag{5}$$
Now, this is not what we would have gotten if Newtonian mechanics was used. I'm getting the expression of centripetal force to be zero since I eliminated tension as a force of constraint. But that can't be true since I'm getting the wrong final result. So what's the solution here? I'm an extreme beginner in Physics (high schooler), so perhaps this may be a very rudimentary question. But I hope I can get a valid answer.
Best Answer
OP can build a consistent Lagrangian formulation in at least 2 ways:
If the holonomic constraint$^1$ $$f(r)~:=~r-R~\approx~0\tag{A}$$ has been used to eliminate $r$, then OP's Lagrange equation (4) wrt. $r$ does not make sense. Rather there is only 1 generalized coordinate $\theta$ left and its Lagrange equation $$\ddot{\theta}~\approx~0. \tag{B}$$
Alternatively, introduce a Lagrange multiplier $\lambda$ for the constraint (A): $$ L(r,\theta,\lambda)~=~\frac{1}{2}m\left(\dot r^2 + r^2\dot{\theta}^2\right)+\lambda f(r).\tag{C} $$ Then the Lagrange equation wrt. $r$ becomes $$ m\ddot{r}~\approx~mr\dot{\theta}^2+\lambda.\tag{D} $$ Since the constraint (A) imposes $$ \ddot{r}~\approx 0 \tag{E},$$ the eq. (D) means that the centrifugal force is balanced by the constraint force $\lambda$, i.e the tension of the string, as one would expect.
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$^1$ The $\approx$ symbol means equality modulo EOMs and constraints.