4
$\begingroup$

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.

$\endgroup$
0

3 Answers 3

2
$\begingroup$

OP can build a consistent Lagrangian formulation in at least 2 ways:

  1. 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 $$\ddot{\theta}~\approx~0. \tag{B}$$

  2. 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.

--

$^1$ The $\approx$ symbol means equality modulo EOMs and constraints.

$\endgroup$
4
  • 1
    $\begingroup$ Makes sense. I can't differentiate with respect to r since it's not a generalized coordinate $\endgroup$
    – Ryan
    Commented Mar 19 at 1:25
  • $\begingroup$ Could you please clarify why you wrote certain expressions using approximations rather than equal signs? $\endgroup$
    – qwerty
    Commented Apr 16 at 23:24
  • $\begingroup$ and I believe there may be a dropped m in your (D) $\endgroup$
    – qwerty
    Commented Apr 16 at 23:52
  • $\begingroup$ Hi @qwerty. Ups, yes. Thank you. I have corrected the answer. $\endgroup$
    – Qmechanic
    Commented Apr 17 at 6:07
1
$\begingroup$

Equations of motion are generated by Lagrangian $L=m\mathbf{\dot{r}}^2/2-V$. Here you want to not have any potential energy, and instead express things in terms of a constraint. So $V=0$

For velocity you have:

$$ \mathbf{\dot{r}}=\frac{d}{dt}\left(r\mathbf{\hat{r}}\right)=\dot{r}\mathbf{\hat{r}}+r\dot{\theta}\boldsymbol{\hat{\theta}} $$

In your case, the radius is constant hence:

$$ L=mr^2\dot{\theta}^2/2 $$

Your only degree of freedom is $\theta$ so the Lagrangian equation is:

$$ \frac{d}{dt}\left(\partial_{\dot{\theta}}L\right)-\partial_{\theta}L=mr^2\ddot{\theta}=0 $$

Which can also be expressed as:

$$ mr^2\dot{\theta}=const,\quad r=const $$

Your error, I recon, was differentiating with respect to $r$, whereas your degrees of freedom here is the angle $\theta$.

There is no place in this answer for centripetal acceleration, since this is taken care of by a constraint. And you get conservation of angular momentum instead of linear momentum. If you remove the constraint and instead bring back the potential energy. If you did you could show that any potential energy that depends solely on radius would have to vary as $V \propto 1/r^2$ for orbit to remain constant

$\endgroup$
1
$\begingroup$

I propose to shift to a more symmetrical setup: a solid object with cilindrical symmetry is spinning around its axis of symmetry.

The solid object is held together by the structural integrity of the material.

The way I understand your setup: normally a Lagrangian for a case in mechanics has a term for kinetic energy and a term for potential energy, but here there is no change of potential energy, so a potential energy term is omitted.



We have that linear mechanics and angular mechanics are to a large extent analogous.

In terms of linear mechanics: if you insert the kinetic energy term in the Euler-Lagrange equation (omitting a potential energy term) then the outcome that you expect (and which of course you get) is that the acceleration is zero.

Angular mechanics has two degrees of freedom; radial distance, and orientation.

I submit that the expression that you ended up with says purely that the time derivative of the orientation coordinate is zero; there is zero angular acceleration.

That is: I submit that the content of the expression

$$ -mr\dot{\theta}^2 = 0 \tag{1} $$

Is actually:

$$ \dot{\theta} = 0 \tag{2} $$


In my opinion the fact that (1) has the same form as the expression for centripetal force does not carry particular significance.

In angular mechanics, if you have a surplus of centripetal force the system will contract, and that contraction causes angular acceleration. (1) expresses that the amount of work done is zero, hence there is zero angular acceleration.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.