Consider a pendulum whose string is replaced with a spring.

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It is a system which can be described neatly using polar coordinates, where your radial component is $ l = l_0 + \Delta l $ and your angular component is $ \varphi $.

The Lagrangian of this system is:

$$ L = \frac{1}{2} m (\dot l^2 + l^2 \dot \varphi^2) - \frac{1}{2} k (l - l_0)^2 + mgl \cos(\varphi) $$

With this you get for your Euler-Lagrange-Equations the following terms.

$$ \frac{\partial L}{\partial \varphi} = -mgl \sin(\varphi) $$ $$ \frac{d}{dt}\left(\frac{\partial L}{\partial \dot \varphi}\right) = 2m \dot ll \dot \varphi + ml^2 \ddot \varphi $$

$$ \frac{\partial L}{\partial l} = ml \dot \varphi^2 - kl + kl_0 + mg \cos(\varphi) $$ $$ \frac{d}{dt}\left(\frac{\partial L}{\partial \dot l}\right) = m \ddot l$$

which leads you to the following two equations of motion.

$$ 0 = 2 \dot{l} \dot{\varphi} + l \ddot{\varphi} + g\sin(\varphi) $$

$$ 0 = m (\ddot{l} - l \dot{\varphi}^2 - g \cos(\varphi)) +k (l-l_0) $$

Out of curiosity, i played a little bit with the second law and oddly, it worked like a charm. If you substitute the accelerations with the corresponding terms you have in polar coordinates, you get the following two equations, which lead to the same result.

$$ m a_\varphi = m(l \ddot \varphi + 2 \dot l \dot \varphi) = -mg \sin(\varphi) $$

$$ m a_l = m (\ddot{l} - l \dot{\varphi}^2) = - k(l-l_0) + mg\cos(\varphi) $$

The thing is now, that my professor for dynamics always said, that you cannot use the second law in non inertial frames of reference. And since this coordinate system is rotating with the pendulum (is it though? I mean, the basis vectors change over time, so...?) it can't be inertial, or am i wrong? And if it is indeed not inertial, can you always just add the "fictitious" accelerations, that are relevent and get the second law back to work?

I am a little bit confused by this.

  • $\begingroup$ I think you might get more takers if you at least 'bullet pointed' how you arrived at the first two eqs? $\endgroup$
    – Gert
    Jan 5, 2021 at 12:17
  • $\begingroup$ At what point did you decide that the polar coordinates are rotating with the pendulum? If this were the case, wouldn't $\varphi$ be constant? $\endgroup$
    – J. Murray
    Jan 5, 2021 at 14:18
  • $\begingroup$ @J.Murray I kind of get your point. But your basis vectors are changing with time. This ist the reason why you have all those wierd accelerations in polar coorinates. So $ \varphi $ can't be constant. $\endgroup$
    – Lukas G.
    Jan 5, 2021 at 17:11
  • $\begingroup$ @Gert I updated my post. I hope it is a little bit clearer now? $\endgroup$
    – Lukas G.
    Jan 5, 2021 at 17:12
  • $\begingroup$ Yes, should be clearer to those who want to delve in there. $\endgroup$
    – Gert
    Jan 5, 2021 at 17:18

1 Answer 1


Your second law is applied in an inertial reference frame using polar coordinates. Although the unit vectors (radial and tangential) in polar coordinates have constant magnitude (one), they change direction in time.

If you used a non-inertial reference frame attached to (and moving with) $m$, in that frame the radial and tangential coordinates of $m$ are constant (zeros) and you have to include the fictitious forces. You modify the second law in a non-inertial frame to include the fictitious forces.

See a good physics mechanics text, such as Symon, Mechanics.

  • $\begingroup$ Okey, so let me try to get this straight. If i would center my coordinate system on the mass $m$ (no matter what actual coordinates i use), i would have to add the fictitious forces to explain the movement, since it is more or less moving in a halfcircle-ish trajectory around the centerpoint on which i centered my original polar coordinate system above. And since this coordinate system was not accelerated, it was an inertial one and i could use the second law as always, just with the adjusted accelerations that correspond to my coordinate system of choice, i. e. polar coordinates? Correct? $\endgroup$
    – Lukas G.
    Jan 5, 2021 at 19:36
  • $\begingroup$ For a coordinate system fixed at m you have a non-inertial frame of reference. In this coordinate system, you can use Cartesian, polar, or whatever coordinates are easier to visualize, and apply the second law with coordinates appropriate for the coordinate system (x, y, z for Cartesian, r, $\theta$ for polar), modified to include the necessary fictitious forces for motion in this system. In general there are a number of fictitious forces that can appear; consult a physics mechanics text for these forces and the general equation of motion containing these forces in a non-inertial frame. $\endgroup$
    – John Darby
    Jan 5, 2021 at 22:14

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