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I learned from https://www.math.ubc.ca/~feldman/apps/vlPendulum.pdf that the equation of motion for a pendulum of variable length is $$ l\ddot{\theta} + 2 \dot{l}\dot{\theta} + g \sin \theta = 0 $$ But when I try to reproduce this result using Newtonian mechanics I get an additional term $\ddot{l}\theta$: $$ F=m\ddot{x} = m \frac{d^2}{dt^2}(l\theta) = m (l\ddot{\theta} + 2 \dot{l}\dot{\theta} + \ddot{l}\theta) $$ which together with the tangent force $F_t = - m g \sin \theta$ yields the equation of motion $$ F=F_t \\ l\ddot{\theta} + 2 \dot{l}\dot{\theta} + \ddot{l}\theta + g \sin \theta = 0 $$ What am I doing wrong here? Why does my Newtonian ansatz not generate the same equation of motion as the Lagrangian one?

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  • $\begingroup$ You should try again while marking everything that is a vector as such. $\endgroup$
    – user121330
    Apr 19, 2017 at 21:12
  • $\begingroup$ I think you are making a mistake when taking the derivatives of the $\vec{\mathbf{x}}$ vector. Perhaps this could be useful $\endgroup$
    – Ali
    Apr 19, 2017 at 21:15

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In polar coordinates we have:

$$\frac{d}{dt}\hat{r}=\dot{\theta}\hat{\theta}$$ $$\frac{d}{dt}\hat{\theta}=-\dot{\theta}\hat r$$

So if we are taking the derivative of the position vector or $\vec x=r\hat r$ with respect to time we will get:

$$\dot{\vec x}=\dot r \hat r+r\dot \theta \hat \theta$$ $$\Rightarrow \ddot{\vec{x}} = \ddot{r}\hat r+\dot r \dot \theta \hat \theta + \dot r \dot \theta \hat \theta + r \ddot \theta \hat \theta -r \dot{\theta}^2\hat r = \left(\ddot r-r\dot\theta^2\right)\hat r + \left( r \ddot \theta + 2\dot r \dot \theta\right) \hat \theta$$

Now putting $r=l$, and plugging this into Newton's $\vec F=m\ddot{\vec{ x}}$ you will get the correct result.

$$\begin{array}{rl} mg\cos{\theta} -T& =m\left(\ddot l-l\dot\theta^2\right)\\ -mg\sin{\theta} & =m\left( l \ddot \theta + 2\dot l \dot \theta\right) \end{array}$$

Assuming the length varies slowly in time one can attempt to actually solve these equation using Adiabatic invariants.

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  • $\begingroup$ Sorry, but what is $T$ in your notation? $\endgroup$
    – asmaier
    Apr 20, 2017 at 21:28
  • $\begingroup$ The tension in the string. $\endgroup$
    – Ali
    Apr 21, 2017 at 0:55

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